Can You Solve These Summer Math Challenges?

[QuantumQuest]

Summer is coming and brings a new basic math challenge! Enjoy! For more advanced problems you can check our other intermediate level math challenge thread!

RULES:
1) In order for a solution to count, a full derivation or proof must be given. Answers with no proof will be ignored.
2) It is fine to use nontrivial results without proof as long as you cite them and as long as it is "common knowledge to all mathematicians". Whether the latter is satisfied will be decided on a case-by-case basis.
3) If you have seen the problem before and remember the solution, you cannot participate in the solution to that problem.
4) You are allowed to use google, wolframalpha or any other resource. However, you are not allowed to search the question directly. So if the question was to solve an integral, you are allowed to obtain numerical answers from software, you are allowed to search for useful integration techniques, but you cannot type in the integral in wolframalpha to see its solution.
5) Mentors, advisors and homework helpers are kindly requested not to post solutions, not even in spoiler tags, for the challenge problems, until 16th of each month. This gives the opportunity to other people including but not limited to students to feel more comfortable in dealing with / solving the challenge problems. In case of an inadvertent posting of a solution the post will be deleted by @fresh_42$1.$ (resolved in post #62) a) Prove $(e+x)^{e-x}>(e-x)^{e+x}$ for $0<x<e.$

$\space$ $\space$ b) Show that for $0 < b < a$ we have
$\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\dfrac{1}{a} < \dfrac{2}{a+b} < \dfrac{\log (a) - \log (b)}{a-b} < \dfrac{1}{\sqrt{ab}} < \dfrac{1}{b}$

$\space$ $\space$ c) Let $f,g\, : \,[a,b]\longrightarrow \mathbb{R}$ be two monotone integrable functions, either both increasing or both decrasing. $\space$ $\space$ $\space$ $\space$ $\space$ Show that

$\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$$\space$ $\space$ $\int_a^bf(x)g(x)\,dx \ge \int_a^bf(x)\,dx \cdot \int_a^bg(x)\,dx$ $\space$ $\space$ (by @fresh_42)

$2.$ (solved by @PeroK ) One card out of a deck has been lost. The other 51 cards are repeatedly shuffled and then thirteen cards are dealt, face up. The cards are 2 spades, 3 clubs, 4 hearts, and 4 diamonds. What is the chance that the missing card is (a) spade, (b) club, (c) heart, (d)diamond $\space$ $\space$ (by @StoneTemplePython)

$3.$ (solved by @lpetrich ) Farmer Joe bought the blue area for $10,000, the green for $20,000 and the yellow for $30,000. Assuming prices are proportional to the areas, what's the price for his entire field?
Figure:

(by @fresh_42)

$4.$ (solved by @Phylosopher ) Using only geometric reasoning, calculate the average value of $f(x) = \sqrt{2x - x^2}$ on $[0,2]$
(by @QuantumQuest)

$5.$ (resolved in post #64) Can one pack 53 bricks, each of dimension $\text{1 x 1 x 4}$ into a $\text{6 x 6 x 6}$ box?
(by @StoneTemplePython)

$6.$ (solved by @fishturtle1 ) Show that $a^{13} - a \equiv 0(\mod 2730)$ for all $a \in \mathbb{Z}$ $\space$ $\space$ (by @QuantumQuest)

$7.$ (solved by @lpetrich ) The general solution to $y^{(4)}(x)+4y(x)=0$ is given by

$\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $\space$ $y(x)=\alpha e^{-x}\cos (x) + \beta e^{-x}\sin (x) + \gamma e^x \sin (x) + \delta e^x \cos (x)$

a) How do we have to choose the initial conditions at $x=0$ in order to get $y(x)=e^{-x}\cos x$ as unique solution?

b) Which function do we get for the initial conditions

$y'(0)=1 \; , \;y''(0)=0\; , \; y'''(0)=0\; , \; y^{(4)}(0)=0\,$ ? $\space$ $\space$ (by @fresh_42)

$8.$ (resolved in post #63) Find the points where the graphs of $f(x) = e^{-x}$ and $g(x) = e^{-x}\cos x$ are tangent to each other. Find another function whose graph is tangent to the graphs of the aforementioned functions in one of the contact points $\space$ $\space$ (by @QuantumQuest)

$9.$ (solved by @lpetrich ) Decode:

a) ZC ULX QFFY L TBXCFSB FMFS XZYVF ZC RLX AZXVDMFSFA TDSF CULY KZKCB BFLSX LPD, LYA LOO PDDA CUFDSFCZVLO IUBXZVZXCX IHC CUZX YHTQFS HI DY CUFZS RLOO LYA RDSSB LQDHC ZC

b) CO PXLIX BPX Y FPRDMCPO MP KPDOM, Y BDRR LIXCAYMCPO PX VXPPB SDFM ZI WCAIO. YOFJIXF JCMU OP VXPPB JCRR ZI CWOPXIL. CM CF BCOI MP DFI OPOMXCACYR XIFDRMF JCMUPDM VXPPB YF RPOW YF EPD KCMI MUIS YOL YF RPOW YF CM CF KPSSPO QOPJRILWI MP YRR SYMUISYMCKCYOF. JUIMUIX MUI RYMMIX CF FYMCFBCIL JCRR ZI LIKCLIL PO Y KYFIZEKYFI ZYFCF. CB EPD UYAI FIIO MUI VXPZRIS ZIBPXI YOL XISISZIX MUI FPRDMCPO, EPD KYOOPM VYXMCKCVYMI CO MUI FPRDMCPO MP MUYM VXPZRIS $\space$ $\space$ (by @fresh_42)

$10.$ (solved by @Mr Davis 97 ) Calculate
$$ \int_{\pi^{-1}}^\pi \dfrac{1}{x}\sin^2 \left( -x-\dfrac{1}{x} \right) \log x\,dx $$ (by @fresh_42)

 

[Wrichik Basu]

@QuantumQuest cannot view the image for question no. 3. Can you please check?

 

[QuantumQuest]

@Wrichik Basu which browser are you using? I can see it from IE 11, Firefox Quantum 60.0.1 (64-bit) (latest) and Google Chrome 67.0.3396.62 (64-bit) (latest) with no problem.

 

[Wrichik Basu]

@Wrichik Basu which browser are you using? I can see it from IE 11, Firefox Quantum 60.0.1 (64-bit) (latest) and Google Chrome 67.0.3396.62 (64-bit) (latest) with no problem.

Chrome mobile. Not the latest one.

 

[QuantumQuest]

@Wrichik Basu do you see the image now?

 

[Wrichik Basu]

@Wrichik Basu do you see the image now?

Yes, I can.

 

[lpetrich]

My solution for Problem 3:

Spoiler

I like to do analytic geometry, so I assign coordinate values to each of the points:
E = {0,0}, F = {a1, 0}, A = {a1, b1}, D = {0, b1}, B = {a2, b1+b2}, C = {0,
b1+b2}, G = {a2, b1}

H is at the intersection of DF and EG, and its location is thus {a1*a2/(a1+a2), a1*b1/(a1+a2)}.

Using the shoelace formula, the areas are
ABG = (1/2)*(a1-a2)*b2
GBCD = a2*b2
EHD = (1/2)*a1*a2*b1/(a1+a2)
ABCD = a1*b1
Total = ABG + GBCD + ABCD

From the statement of the problem, GBCD = $20,000 and ABG = $10,000, and that gives us
a1*b2 = $40,000
a2*b2 = $20,000

Since we want to find the area of ABCD, we express EHD in terms of it:
EHD = (ABCD) * a2/(a1+a2) = (ABCD) * (1/3)

Since EHD = $30,000, ABCD = $90,000, and the total price is ABCD + GBCD + ABG = $120,000.

 

[lpetrich]

My solution for Problem 9a:

Spoiler

I will assume a simple letter-substitution code. I will make the encoded letters lowercase and the decoded letters uppercase, for ease of distinction. Initial:
zc ulx qffy l tbxcfsb fmfs xzyvf zc rlx azxvdmfsfa tdsf culy kzkcb bflsx lpd, lya loo pdda cufdsfczvlo iubxzvzxcx ihc cuzx yhtqfs hi dy cufzs rloo lya rdssb lqdhc zc

The frequency counts of letters are
{"f", 14}, {"c", 13}, {"l", 12}, {"z", 11}, {"s", 10}, {"x", 10}, {"d", 9}, {"y", 7}, {"b", 6}, {"u", 6}, {"a", 5}, {"o", 5}, {"h", 4}, {"v", 4}, {"i", 3}, {"q", 3}, {"r", 3}, {"t", 3}, {"k", 2}, {"m", 2}, {"p", 2}

suggesting that one of the first letters is "E". In English spelling, the only letter that occurs alone is "A", and there is a lone l, so l -> A. This gives us
zc uAx qffy A tbxcfsb fmfs xzyvf zc rAx azxvdmfsfa tdsf cuAy kzkcb bfAsx Apd, Aya Aoo pdda cufdsfczvAo iubxzvzxcx ihc cuzx yhtqfs hi dy cufzs rAoo Aya rdssb Aqdhc zc

Trying f -> E (14) and c -> T (13) gives us
zT uAx qEEy A tbxTEsb EmEs xzyvE zT rAx azxvdmEsEa tdsE TuAy kzkTb bEAsx Apd, Aya Aoo pdda TuEdsETzvAo iubxzvzxTx ihT Tuzx yhtqEs hi dy TuEzs rAoo Aya rdssb AqdhT zT

A common two-letter word in English is IT, so we do z->I (11):
"IT uAx qEEy A tbxTEsb EmEs xIyvE IT rAx aIxvdmEsEa tdsE TuAy kIkTb bEAsx Apd, Aya Aoo pdda TuEdsETIvAo iubxIvIxTx ihT TuIx yhtqEs hi dy TuEIs rAoo Aya rdssb AqdhT IT"

The second and third words look like they are either "HAS BEEN" or "HAD BEEN". Trying u->H (6), q->B (3), y->N (7) gives us
IT HAx BEEN A tbxTEsb EmEs xINvE IT rAx aIxvdmEsEa tdsE THAN kIkTb bEAsx Apd, ANa Aoo pdda THEdsETIvAo iHbxIvIxTx ihT THIx NhtBEs hi dN THEIs rAoo ANa rdssb ABdhT IT

The word before THAN looks like it could be MORE. Trying t->M (3), d->O (9), s->R (10) gives us
IT HAx BEEN A MbxTERb EmER xINvE IT rAx aIxvOmEREa MORE THAN kIkTb bEARx ApO, ANa Aoo pOOa THEORETIvAo iHbxIvIxTx ihT THIx NhMBER hi ON THEIR rAoo ANa rORRb ABOhT IT

The words THEORETICAL and NUMBER are almost spelled out, giving v->C (4), o->L (5), h->U (4):
IT HAx BEEN A MbxTERb EmER xINCE IT rAx aIxCOmEREa MORE THAN kIkTb bEARx ApO, ANa ALL pOOa THEORETICAL iHbxICIxTx iUT THIx NUMBER Ui ON THEIR rALL ANa rORRb ABOUT IT

I recognize EVER SINCE, PUT THIS NUMBER UP ON THEIR WALL: m->V (2), x->S (10), i->P (3), r->W (3):
IT HAS BEEN A MbSTERb EVER SINCE IT WAS aISCOVEREa MORE THAN kIkTb bEARS ApO, ANa ALL pOOa THEORETICAL PHbSICISTS PUT THIS NUMBER UP ON THEIR WALL ANa WORRb ABOUT IT

AND WORRY ABOUT IT: a->D (5), b->Y (6):
IT HAS BEEN A MYSTERY EVER SINCE IT WAS DISCOVERED MORE THAN kIkTY YEARS ApO, AND ALL pOOD THEORETICAL PHYSICISTS PUT THIS NUMBER UP ON THEIR WALL AND WORRY ABOUT IT

FIFTY YEARS AGO: k->F (2), p->G (2):
IT HAS BEEN A MYSTERY EVER SINCE IT WAS DISCOVERED MORE THAN FIFTY YEARS AGO, AND ALL GOOD THEORETICAL PHYSICISTS PUT THIS NUMBER UP ON THEIR WALL AND WORRY ABOUT IT

Complete code:
{"l" -> "A", "f" -> "E", "c" -> "T", "z" -> "I", "u" -> "H", "q" -> "B", "y" -> "N", "t" -> "M", "d" -> "O", "s" -> "R", "v" -> "C", "o" -> "L", "h" -> "U", "m" -> "V", "x" -> "S", "i" -> "P", "r" -> "W", "a" -> "D", "b" -> "Y", "k" -> "F", "p" -> "G"}

 

[lpetrich]

My solution for Problem 9b:

Spoiler

I use lowercase for encoded and uppercase for decoded, as before. The letter frequencies:
{"i", 39}, {"p", 38}, {"m", 34}, {"c", 31}, {"y", 26}, {"o", 25}, {"f", 24}, {"x", 20}, {"r", 19}, {"d", 12}, {"u", 12}, {"l", 10}, {"s", 10}, {"b", 9}, {"k", 9}, {"z", 9}, {"j", 7}, {"v", 7}, {"w", 5}, {"a", 4}, {"e", 4}, {"q", 1}}

The lone letter must be A: l->A (10):
co pxlix bpx A fprdmcpo mp kpdom, A bdrr lixcaAmcpo px vxppb sdfm zi wcaio. Aofjixf jcmu op vxppb jcrr zi cwopxil. cm cf bcoi mp dfi opomxcacAr xifdrmf jcmupdm vxppb Af rpow Af epd kcmi muis Aol Af rpow Af cm cf kpsspo qopjrilwi mp Arr sAmuisAmckcAof. juimuix mui rAmmix cf fAmcfbcil jcrr zi likclil po A kAfizekAfi zAfcf. cb epd uAai fiio mui vxpzris zibpxi Aol xisiszix mui fprdmcpo, epd kAoopm vAxmckcvAmi co mui fprdmcpo mp muAm vxpzris

The repeated Af's suggest "as", with f -> S (24):
co pxlix bpx A Sprdmcpo mp kpdom, A bdrr lixcaAmcpo px vxppb sdSm zi wcaio. AoSjixS jcmu op vxppb jcrr zi cwopxil. cm cS bcoi mp dSi opomxcacAr xiSdrmS jcmupdm vxppb AS rpow AS epd kcmi muis Aol AS rpow AS cm cS kpsspo qopjrilwi mp Arr sAmuisAmckcAoS. juimuix mui rAmmix cS SAmcSbcil jcrr zi likclil po A kASizekASi zAScS. cb epd uAai Siio mui vxpzris zibpxi Aol xisiszix mui Sprdmcpo, epd kAoopm vAxmckcvAmi co mui Sprdmcpo mp muAm vxpzris

A kASizekASi zAScS suggests some singular word that ends in -S, and such a word likely has an I before it. Thus, c->I (31):
Io pxlix bpx A SprdmIpo mp kpdom, A bdrr lixIaAmIpo px vxppb sdSm zi wIaio. AoSjixS jImu op vxppb jIrr zi Iwopxil. I am IS bIoi mp dSi opomxIaIAr xiSdrmS jImupdm vxppb AS rpow AS epd kImi muis Aol AS rpow AS I am IS kpsspo qopjrilwi mp Arr sAmuisAmIkIAoS. juimuix mui rAmmix IS SAmISbIil jIrr zi likIlil po A kASizekASi zASIS. Ib epd uAai Siio mui vxpzris zibpxi Aol xisiszix mui SprdmIpo, epd kAoopm vAxmIkIvAmi Io mui SprdmIpo mp muAm vxpzris

Im IS -- a two-letter word starting with I that is likely to be a subject is IT: m->T (34):
Io pxlix bpx A SprdTIpo Tp kpdoT, A bdrr lixIaATIpo px vxppb sdST zi wIaio. AoSjixS jITu op vxppb jIrr zi Iwopxil. IT IS bIoi Tp dSi opoTxIaIAr xiSdrTS jITupdT vxppb AS rpow AS epd kITi Tuis Aol AS rpow AS IT IS kpsspo qopjrilwi Tp Arr sATuisATIkIAoS. juiTuix Tui rATTix IS SATISbIil jIrr zi likIlil po A kASizekASi zASIS. Ib epd uAai Siio Tui vxpzris zibpxi Aol xisiszix Tui SprdTIpo, epd kAoopT vAxTIkIvATi Io Tui SprdTIpo Tp TuAT vxpzris

There are some Tp's, and this means p->O (38):
Io Oxlix bOx A SOrdTIOo TO kOdoT, A bdrr lixIaATIOo Ox vxOOb sdST zi wIaio. AoSjixS jITu oO vxOOb jIrr zi IwoOxil. IT IS bIoi TO dSi oOoTxIaIAr xiSdrTS jITuOdT vxOOb AS rOow AS eOd kITi Tuis Aol AS rOow AS IT IS kOssOo qoOjrilwi TO Arr sATuisATIkIAoS. juiTuix Tui rATTix IS SATISbIil jIrr zi likIlil Oo A kASizekASi zASIS. Ib eOd uAai Siio Tui vxOzris zibOxi Aol xisiszix Tui SOrdTIOo, eOd kAooOT vAxTIkIvATi Io Tui SOrdTIOo TO TuAT vxOzris

lixIaATIOo likely ends in -ATION: o->N (25):
IN Oxlix bOx A SOrdTION TO kOdNT, A bdrr lixIaATION Ox vxOOb sdST zi wIaiN. ANSjixS jITu NO vxOOb jIrr zi IwNOxil. IT IS bINi TO dSi NONTxIaIAr xiSdrTS jITuOdT vxOOb AS rONw AS eOd kITi Tuis ANl AS rONw AS IT IS kOssON qNOjrilwi TO Arr sATuisATIkIANS. juiTuix Tui rATTix IS SATISbIil jIrr zi likIlil ON A kASizekASi zASIS. Ib eOd uAai SiiN Tui vxOzris zibOxi ANl xisiszix Tui SOrdTION, eOd kANNOT vAxTIkIvATi IN Tui SOrdTION TO TuAT vxOzris

Tui SOrdTION is likely THE SOLUTION: u->H (12), i->E (39), r->L (19), d->U (12) -- E is the most frequent one, as one might expect, though the letters after it are somewhat out of the order of their usual frequencies.
IN OxlEx bOx A SOLUTION TO kOUNT, A bULL lExIaATION Ox vxOOb sUST zE wIaEN. ANSjExS jITH NO vxOOb jILL zE IwNOxEl. IT IS bINE TO USE NONTxIaIAL xESULTS jITHOUT vxOOb AS LONw AS eOU kITE THEs ANl AS LONw AS IT IS kOssON qNOjLElwE TO ALL sATHEsATIkIANS. jHETHEx THE LATTEx IS SATISbIEl jILL zE lEkIlEl ON A kASEzekASE zASIS. Ib eOU HAaE SEEN THE vxOzLEs zEbOxE ANl xEsEszEx THE SOLUTION, eOU kANNOT vAxTIkIvATE IN THE SOLUTION TO THAT vxOzLEs

kANNOT -> CANNOT, ANl AS LONw AS -> AND AS LONG AS: "k" -> "C" (9), l->D (10), w->G (5):
IN OxDEx bOx A SOLUTION TO COUNT, A bULL DExIaATION Ox vxOOb sUST zE GIaEN. ANSjExS jITH NO vxOOb jILL zE IGNOxED. IT IS bINE TO USE NONTxIaIAL xESULTS jITHOUT vxOOb AS LONG AS eOU CITE THEs AND AS LONG AS IT IS COssON qNOjLEDGE TO ALL sATHEsATICIANS. jHETHEx THE LATTEx IS SATISbIED jILL zE DECIDED ON A CASEzeCASE zASIS. Ib eOU HAaE SEEN THE vxOzLEs zEbOxE AND xEsEszEx THE SOLUTION, eOU CANNOT vAxTICIvATE IN THE SOLUTION TO THAT vxOzLEs

COssON qNOjLEDGE TO ALL sATHEsATICIANS -> COMMON KNOWLEDGE TO ALL MATHEMATICIANS: s->M (M), q->K (1), j->W (7):
IN OxDEx bOx A SOLUTION TO COUNT, A bULL DExIaATION Ox vxOOb MUST zE GIaEN. ANSWExS WITH NO vxOOb WILL zE IGNOxED. IT IS bINE TO USE NONTxIaIAL xESULTS WITHOUT vxOOb AS LONG AS eOU CITE THEM AND AS LONG AS IT IS COMMON KNOWLEDGE TO ALL MATHEMATICIANS. WHETHEx THE LATTEx IS SATISbIED WILL zE DECIDED ON A CASEzeCASE zASIS. Ib eOU HAaE SEEN THE vxOzLEM zEbOxE AND xEMEMzEx THE SOLUTION, eOU CANNOT vAxTICIvATE IN THE SOLUTION TO THAT vxOzLEM

WHETHEx -> WHETHER, LATTEx -> LATTER, SATISbIED -> SATISFIED, eOU -> YOU: x->R (20), b->F (9), e->Y (4)
IN ORDER FOR A SOLUTION TO COUNT, A FULL DERIaATION OR vROOF MUST zE GIaEN. ANSWERS WITH NO vROOF WILL zE IGNORED. IT IS FINE TO USE NONTRIaIAL RESULTS WITHOUT vROOF AS LONG AS YOU CITE THEM AND AS LONG AS IT IS COMMON KNOWLEDGE TO ALL MATHEMATICIANS. WHETHER THE LATTER IS SATISFIED WILL zE DECIDED ON A CASEzYCASE zASIS. IF YOU HAaE SEEN THE vROzLEM zEFORE AND REMEMzER THE SOLUTION, YOU CANNOT vARTICIvATE IN THE SOLUTION TO THAT vROzLEM

DERIaATION, GIaEN, vROOF, WILL zE: a->V (4), z->B (9):
IN ORDER FOR A SOLUTION TO COUNT, A FULL DERIVATION OR vROOF MUST BE GIVEN. ANSWERS WITH NO vROOF WILL BE IGNORED. IT IS FINE TO USE NONTRIVIAL RESULTS WITHOUT vROOF AS LONG AS YOU CITE THEM AND AS LONG AS IT IS COMMON KNOWLEDGE TO ALL MATHEMATICIANS. WHETHER THE LATTER IS SATISFIED WILL BE DECIDED ON A CASEBYCASE BASIS. IF YOU HAVE SEEN THE vROBLEM BEFORE AND REMEMBER THE SOLUTION, YOU CANNOT vARTICIvATE IN THE SOLUTION TO THAT vROBLEM

vROOF, vROBLEM: v->P (7)
IN ORDER FOR A SOLUTION TO COUNT, A FULL DERIVATION OR PROOF MUST BE GIVEN. ANSWERS WITH NO PROOF WILL BE IGNORED. IT IS FINE TO USE NONTRIVIAL RESULTS WITHOUT PROOF AS LONG AS YOU CITE THEM AND AS LONG AS IT IS COMMON KNOWLEDGE TO ALL MATHEMATICIANS. WHETHER THE LATTER IS SATISFIED WILL BE DECIDED ON A CASEBYCASE BASIS. IF YOU HAVE SEEN THE PROBLEM BEFORE AND REMEMBER THE SOLUTION, YOU CANNOT PARTICIPATE IN THE SOLUTION TO THAT PROBLEM

Complete code:
{"y" -> "A", "f" -> "S", "c" -> "I", "m" -> "T", "p" -> "O", "o" -> "N", "u" -> "H", "i" -> "E", "r" -> "L", "d" -> "U", "k" -> "C", "l" -> "D", "w" -> "G", "s" -> "M", "q" -> "K", "j" -> "W", "x" -> "R", "b" -> "F", "e" -> "Y", "a" -> "V", "z" -> "B", "v" -> "P"}

 

[fresh_42]

My solution for Problem 9a:

Spoiler

I will assume a simple letter-substitution code. I will make the encoded letters lowercase and the decoded letters uppercase, for ease of distinction. Initial:
zc ulx qffy l tbxcfsb fmfs xzyvf zc rlx azxvdmfsfa tdsf culy kzkcb bflsx lpd, lya loo pdda cufdsfczvlo iubxzvzxcx ihc cuzx yhtqfs hi dy cufzs rloo lya rdssb lqdhc zc

The frequency counts of letters are
{"f", 14}, {"c", 13}, {"l", 12}, {"z", 11}, {"s", 10}, {"x", 10}, {"d", 9}, {"y", 7}, {"b", 6}, {"u", 6}, {"a", 5}, {"o", 5}, {"h", 4}, {"v", 4}, {"i", 3}, {"q", 3}, {"r", 3}, {"t", 3}, {"k", 2}, {"m", 2}, {"p", 2}

suggesting that one of the first letters is "E". In English spelling, the only letter that occurs alone is "A", and there is a lone l, so l -> A. This gives us
zc uAx qffy A tbxcfsb fmfs xzyvf zc rAx azxvdmfsfa tdsf cuAy kzkcb bfAsx Apd, Aya Aoo pdda cufdsfczvAo iubxzvzxcx ihc cuzx yhtqfs hi dy cufzs rAoo Aya rdssb Aqdhc zc

Trying f -> E (14) and c -> T (13) gives us
zT uAx qEEy A tbxTEsb EmEs xzyvE zT rAx azxvdmEsEa tdsE TuAy kzkTb bEAsx Apd, Aya Aoo pdda TuEdsETzvAo iubxzvzxTx ihT Tuzx yhtqEs hi dy TuEzs rAoo Aya rdssb AqdhT zT

A common two-letter word in English is IT, so we do z->I (11):
"IT uAx qEEy A tbxTEsb EmEs xIyvE IT rAx aIxvdmEsEa tdsE TuAy kIkTb bEAsx Apd, Aya Aoo pdda TuEdsETIvAo iubxIvIxTx ihT TuIx yhtqEs hi dy TuEIs rAoo Aya rdssb AqdhT IT"

The second and third words look like they are either "HAS BEEN" or "HAD BEEN". Trying u->H (6), q->B (3), y->N (7) gives us
IT HAx BEEN A tbxTEsb EmEs xINvE IT rAx aIxvdmEsEa tdsE THAN kIkTb bEAsx Apd, ANa Aoo pdda THEdsETIvAo iHbxIvIxTx ihT THIx NhtBEs hi dN THEIs rAoo ANa rdssb ABdhT IT

The word before THAN looks like it could be MORE. Trying t->M (3), d->O (9), s->R (10) gives us
IT HAx BEEN A MbxTERb EmER xINvE IT rAx aIxvOmEREa MORE THAN kIkTb bEARx ApO, ANa Aoo pOOa THEORETIvAo iHbxIvIxTx ihT THIx NhMBER hi ON THEIR rAoo ANa rORRb ABOhT IT

The words THEORETICAL and NUMBER are almost spelled out, giving v->C (4), o->L (5), h->U (4):
IT HAx BEEN A MbxTERb EmER xINCE IT rAx aIxCOmEREa MORE THAN kIkTb bEARx ApO, ANa ALL pOOa THEORETICAL iHbxICIxTx iUT THIx NUMBER Ui ON THEIR rALL ANa rORRb ABOUT IT

I recognize EVER SINCE, PUT THIS NUMBER UP ON THEIR WALL: m->V (2), x->S (10), i->P (3), r->W (3):
IT HAS BEEN A MbSTERb EVER SINCE IT WAS aISCOVEREa MORE THAN kIkTb bEARS ApO, ANa ALL pOOa THEORETICAL PHbSICISTS PUT THIS NUMBER UP ON THEIR WALL ANa WORRb ABOUT IT

AND WORRY ABOUT IT: a->D (5), b->Y (6):
IT HAS BEEN A MYSTERY EVER SINCE IT WAS DISCOVERED MORE THAN kIkTY YEARS ApO, AND ALL pOOD THEORETICAL PHYSICISTS PUT THIS NUMBER UP ON THEIR WALL AND WORRY ABOUT IT

FIFTY YEARS AGO: k->F (2), p->G (2):
IT HAS BEEN A MYSTERY EVER SINCE IT WAS DISCOVERED MORE THAN FIFTY YEARS AGO, AND ALL GOOD THEORETICAL PHYSICISTS PUT THIS NUMBER UP ON THEIR WALL AND WORRY ABOUT IT

Complete code:
{"l" -> "A", "f" -> "E", "c" -> "T", "z" -> "I", "u" -> "H", "q" -> "B", "y" -> "N", "t" -> "M", "d" -> "O", "s" -> "R", "v" -> "C", "o" -> "L", "h" -> "U", "m" -> "V", "x" -> "S", "i" -> "P", "r" -> "W", "a" -> "D", "b" -> "Y", "k" -> "F", "p" -> "G"}

This is correct. It is a simple Caesar code: $x \longmapsto 5x+7 (26)$. Do you know who said this?

 

[fresh_42]

My solution for Problem 9b:

Spoiler

I use lowercase for encoded and uppercase for decoded, as before. The letter frequencies:
{"i", 39}, {"p", 38}, {"m", 34}, {"c", 31}, {"y", 26}, {"o", 25}, {"f", 24}, {"x", 20}, {"r", 19}, {"d", 12}, {"u", 12}, {"l", 10}, {"s", 10}, {"b", 9}, {"k", 9}, {"z", 9}, {"j", 7}, {"v", 7}, {"w", 5}, {"a", 4}, {"e", 4}, {"q", 1}}

The lone letter must be A: l->A (10):
co pxlix bpx A fprdmcpo mp kpdom, A bdrr lixcaAmcpo px vxppb sdfm zi wcaio. Aofjixf jcmu op vxppb jcrr zi cwopxil. cm cf bcoi mp dfi opomxcacAr xifdrmf jcmupdm vxppb Af rpow Af epd kcmi muis Aol Af rpow Af cm cf kpsspo qopjrilwi mp Arr sAmuisAmckcAof. juimuix mui rAmmix cf fAmcfbcil jcrr zi likclil po A kAfizekAfi zAfcf. cb epd uAai fiio mui vxpzris zibpxi Aol xisiszix mui fprdmcpo, epd kAoopm vAxmckcvAmi co mui fprdmcpo mp muAm vxpzris

The repeated Af's suggest "as", with f -> S (24):
co pxlix bpx A Sprdmcpo mp kpdom, A bdrr lixcaAmcpo px vxppb sdSm zi wcaio. AoSjixS jcmu op vxppb jcrr zi cwopxil. cm cS bcoi mp dSi opomxcacAr xiSdrmS jcmupdm vxppb AS rpow AS epd kcmi muis Aol AS rpow AS cm cS kpsspo qopjrilwi mp Arr sAmuisAmckcAoS. juimuix mui rAmmix cS SAmcSbcil jcrr zi likclil po A kASizekASi zAScS. cb epd uAai Siio mui vxpzris zibpxi Aol xisiszix mui Sprdmcpo, epd kAoopm vAxmckcvAmi co mui Sprdmcpo mp muAm vxpzris

A kASizekASi zAScS suggests some singular word that ends in -S, and such a word likely has an I before it. Thus, c->I (31):
Io pxlix bpx A SprdmIpo mp kpdom, A bdrr lixIaAmIpo px vxppb sdSm zi wIaio. AoSjixS jImu op vxppb jIrr zi Iwopxil. I am IS bIoi mp dSi opomxIaIAr xiSdrmS jImupdm vxppb AS rpow AS epd kImi muis Aol AS rpow AS I am IS kpsspo qopjrilwi mp Arr sAmuisAmIkIAoS. juimuix mui rAmmix IS SAmISbIil jIrr zi likIlil po A kASizekASi zASIS. Ib epd uAai Siio mui vxpzris zibpxi Aol xisiszix mui SprdmIpo, epd kAoopm vAxmIkIvAmi Io mui SprdmIpo mp muAm vxpzris

Im IS -- a two-letter word starting with I that is likely to be a subject is IT: m->T (34):
Io pxlix bpx A SprdTIpo Tp kpdoT, A bdrr lixIaATIpo px vxppb sdST zi wIaio. AoSjixS jITu op vxppb jIrr zi Iwopxil. IT IS bIoi Tp dSi opoTxIaIAr xiSdrTS jITupdT vxppb AS rpow AS epd kITi Tuis Aol AS rpow AS IT IS kpsspo qopjrilwi Tp Arr sATuisATIkIAoS. juiTuix Tui rATTix IS SATISbIil jIrr zi likIlil po A kASizekASi zASIS. Ib epd uAai Siio Tui vxpzris zibpxi Aol xisiszix Tui SprdTIpo, epd kAoopT vAxTIkIvATi Io Tui SprdTIpo Tp TuAT vxpzris

There are some Tp's, and this means p->O (38):
Io Oxlix bOx A SOrdTIOo TO kOdoT, A bdrr lixIaATIOo Ox vxOOb sdST zi wIaio. AoSjixS jITu oO vxOOb jIrr zi IwoOxil. IT IS bIoi TO dSi oOoTxIaIAr xiSdrTS jITuOdT vxOOb AS rOow AS eOd kITi Tuis Aol AS rOow AS IT IS kOssOo qoOjrilwi TO Arr sATuisATIkIAoS. juiTuix Tui rATTix IS SATISbIil jIrr zi likIlil Oo A kASizekASi zASIS. Ib eOd uAai Siio Tui vxOzris zibOxi Aol xisiszix Tui SOrdTIOo, eOd kAooOT vAxTIkIvATi Io Tui SOrdTIOo TO TuAT vxOzris

lixIaATIOo likely ends in -ATION: o->N (25):
IN Oxlix bOx A SOrdTION TO kOdNT, A bdrr lixIaATION Ox vxOOb sdST zi wIaiN. ANSjixS jITu NO vxOOb jIrr zi IwNOxil. IT IS bINi TO dSi NONTxIaIAr xiSdrTS jITuOdT vxOOb AS rONw AS eOd kITi Tuis ANl AS rONw AS IT IS kOssON qNOjrilwi TO Arr sATuisATIkIANS. juiTuix Tui rATTix IS SATISbIil jIrr zi likIlil ON A kASizekASi zASIS. Ib eOd uAai SiiN Tui vxOzris zibOxi ANl xisiszix Tui SOrdTION, eOd kANNOT vAxTIkIvATi IN Tui SOrdTION TO TuAT vxOzris

Tui SOrdTION is likely THE SOLUTION: u->H (12), i->E (39), r->L (19), d->U (12) -- E is the most frequent one, as one might expect, though the letters after it are somewhat out of the order of their usual frequencies.
IN OxlEx bOx A SOLUTION TO kOUNT, A bULL lExIaATION Ox vxOOb sUST zE wIaEN. ANSjExS jITH NO vxOOb jILL zE IwNOxEl. IT IS bINE TO USE NONTxIaIAL xESULTS jITHOUT vxOOb AS LONw AS eOU kITE THEs ANl AS LONw AS IT IS kOssON qNOjLElwE TO ALL sATHEsATIkIANS. jHETHEx THE LATTEx IS SATISbIEl jILL zE lEkIlEl ON A kASEzekASE zASIS. Ib eOU HAaE SEEN THE vxOzLEs zEbOxE ANl xEsEszEx THE SOLUTION, eOU kANNOT vAxTIkIvATE IN THE SOLUTION TO THAT vxOzLEs

kANNOT -> CANNOT, ANl AS LONw AS -> AND AS LONG AS: "k" -> "C" (9), l->D (10), w->G (5):
IN OxDEx bOx A SOLUTION TO COUNT, A bULL DExIaATION Ox vxOOb sUST zE GIaEN. ANSjExS jITH NO vxOOb jILL zE IGNOxED. IT IS bINE TO USE NONTxIaIAL xESULTS jITHOUT vxOOb AS LONG AS eOU CITE THEs AND AS LONG AS IT IS COssON qNOjLEDGE TO ALL sATHEsATICIANS. jHETHEx THE LATTEx IS SATISbIED jILL zE DECIDED ON A CASEzeCASE zASIS. Ib eOU HAaE SEEN THE vxOzLEs zEbOxE AND xEsEszEx THE SOLUTION, eOU CANNOT vAxTICIvATE IN THE SOLUTION TO THAT vxOzLEs

COssON qNOjLEDGE TO ALL sATHEsATICIANS -> COMMON KNOWLEDGE TO ALL MATHEMATICIANS: s->M (M), q->K (1), j->W (7):
IN OxDEx bOx A SOLUTION TO COUNT, A bULL DExIaATION Ox vxOOb MUST zE GIaEN. ANSWExS WITH NO vxOOb WILL zE IGNOxED. IT IS bINE TO USE NONTxIaIAL xESULTS WITHOUT vxOOb AS LONG AS eOU CITE THEM AND AS LONG AS IT IS COMMON KNOWLEDGE TO ALL MATHEMATICIANS. WHETHEx THE LATTEx IS SATISbIED WILL zE DECIDED ON A CASEzeCASE zASIS. Ib eOU HAaE SEEN THE vxOzLEM zEbOxE AND xEMEMzEx THE SOLUTION, eOU CANNOT vAxTICIvATE IN THE SOLUTION TO THAT vxOzLEM

WHETHEx -> WHETHER, LATTEx -> LATTER, SATISbIED -> SATISFIED, eOU -> YOU: x->R (20), b->F (9), e->Y (4)
IN ORDER FOR A SOLUTION TO COUNT, A FULL DERIaATION OR vROOF MUST zE GIaEN. ANSWERS WITH NO vROOF WILL zE IGNORED. IT IS FINE TO USE NONTRIaIAL RESULTS WITHOUT vROOF AS LONG AS YOU CITE THEM AND AS LONG AS IT IS COMMON KNOWLEDGE TO ALL MATHEMATICIANS. WHETHER THE LATTER IS SATISFIED WILL zE DECIDED ON A CASEzYCASE zASIS. IF YOU HAaE SEEN THE vROzLEM zEFORE AND REMEMzER THE SOLUTION, YOU CANNOT vARTICIvATE IN THE SOLUTION TO THAT vROzLEM

DERIaATION, GIaEN, vROOF, WILL zE: a->V (4), z->B (9):
IN ORDER FOR A SOLUTION TO COUNT, A FULL DERIVATION OR vROOF MUST BE GIVEN. ANSWERS WITH NO vROOF WILL BE IGNORED. IT IS FINE TO USE NONTRIVIAL RESULTS WITHOUT vROOF AS LONG AS YOU CITE THEM AND AS LONG AS IT IS COMMON KNOWLEDGE TO ALL MATHEMATICIANS. WHETHER THE LATTER IS SATISFIED WILL BE DECIDED ON A CASEBYCASE BASIS. IF YOU HAVE SEEN THE vROBLEM BEFORE AND REMEMBER THE SOLUTION, YOU CANNOT vARTICIvATE IN THE SOLUTION TO THAT vROBLEM

vROOF, vROBLEM: v->P (7)
IN ORDER FOR A SOLUTION TO COUNT, A FULL DERIVATION OR PROOF MUST BE GIVEN. ANSWERS WITH NO PROOF WILL BE IGNORED. IT IS FINE TO USE NONTRIVIAL RESULTS WITHOUT PROOF AS LONG AS YOU CITE THEM AND AS LONG AS IT IS COMMON KNOWLEDGE TO ALL MATHEMATICIANS. WHETHER THE LATTER IS SATISFIED WILL BE DECIDED ON A CASEBYCASE BASIS. IF YOU HAVE SEEN THE PROBLEM BEFORE AND REMEMBER THE SOLUTION, YOU CANNOT PARTICIPATE IN THE SOLUTION TO THAT PROBLEM

Complete code:
{"y" -> "A", "f" -> "S", "c" -> "I", "m" -> "T", "p" -> "O", "o" -> "N", "u" -> "H", "i" -> "E", "r" -> "L", "d" -> "U", "k" -> "C", "l" -> "D", "w" -> "G", "s" -> "M", "q" -> "K", "j" -> "W", "x" -> "R", "b" -> "F", "e" -> "Y", "a" -> "V", "z" -> "B", "v" -> "P"}

Yes, correct, too. Here I used a random, but fixed mapping of letters. The frequency analysis is thus the way to decode it.

 

[fresh_42]

My solution for Problem 3:

Spoiler

I like to do analytic geometry, so I assign coordinate values to each of the points:
E = {0,0}, F = {a1, 0}, A = {a1, b1}, D = {0, b1}, B = {a2, b1+b2}, C = {0,
b1+b2}, G = {a2, b1}

H is at the intersection of DF and EG, and its location is thus {a1*a2/(a1+a2), a1*b1/(a1+a2)}.

Using the shoelace formula, the areas are
ABG = (1/2)*(a1-a2)*b2
GBCD = a2*b2
EHD = (1/2)*a1*a2*b1/(a1+a2)
ABCD = a1*b1
Total = ABG + GBCD + ABCD

From the statement of the problem, GBCD = $20,000 and ABG = $10,000, and that gives us
a1*b2 = $40,000
a2*b2 = $20,000

Since we want to find the area of ABCD, we express EHD in terms of it:
EHD = (ABCD) * a2/(a1+a2) = (ABCD) * (1/3)

Since EHD = $30,000, ABCD = $90,000, and the total price is ABCD + GBCD + ABG = $120,000.

We already have $\,60,000$ for the known area. This means according to your calculation, that the pink area equals the sum of the others. Although the image isn't correctly scaled, I think this can't be true. One difficulty I saw, is the naming of your distances. While the $b_i$ add up, $a_1$ includes $a_2$. That's a bit disturbing and might have caused the error. I couldn't follow your formula for the height of $EHD$, which is the key to the problem.

Btw.: Could you use LaTeX code rather than to write all formulas, esp. the quotients linear?

 

[Mr Davis 97]

Solution to problem 10:

Spoiler

Let $u = \log x$. Then $\displaystyle \int_{\pi^{-1}}^\pi \dfrac{1}{x}\sin^2 \left( -x-\dfrac{1}{x} \right) \log x\,dx = \int_{- \log \pi}^{\log \pi} u\sin^2 (e^u+e^{-u}) \,du$. The integrand can be seen to be an odd function. Since the limits of integration are symmetric about $0$, the value of the integral is $0$.

 

[fresh_42]

Solution to problem 10:

Spoiler

Let $u = \log x$. Then $\displaystyle \int_{\pi^{-1}}^\pi \dfrac{1}{x}\sin^2 \left( -x-\dfrac{1}{x} \right) \log x\,dx = \int_{- \log \pi}^{\log \pi} u\sin^2 (e^u+e^{-u}) \,du$. The integrand can be seen to be an odd function. Since the limits of integration are symmetric about $0$, the value of the integral is $0$.

Yes, that's correct. You could have left the minus sign in the sine function, because making it a plus isn't necessary and must also be shown. The idea was to demonstrate, that there can also be a multiplicative symmetry in integrals, not only an additive. This can be seen by the substitution $u=x^{-1}$ and proving $\int = -\int \,.$ Taking the logarithm reduces it to the additive case.

 

[lpetrich]

LaTexified version of my solution of Problem 3:

Spoiler

I like to do analytic geometry, so I assign coordinate values to each of the points:
E = {0,0}, F = {a1, 0}, A = {a1, b1}, D = {0, b1}, B = {a2, b1+b2}, C = {0,
b1+b2}, G = {a2, b1}

H is at the intersection of DF and EG, and its location is thus {$\frac{a1 \cdot a2}{a1+a2}$, $\frac{a1 \cdot b1}{a1+a2}$}.

Using the shoelace formula, the areas are
ABG = $\frac12 (a1-a2) \cdot b2$
GBCD = $a2 \cdot b2$
EHD = $\frac12 \frac{a1 \cdot a2 \cdot b1}{a1+a2}$
ABCD = $a1 \cdot b1$
Total = ABG + GBCD + ABCD

From the statement of the problem, GBCD = $20,000 and ABG = $10,000, and that gives us
$a1 \cdot b2$ = $40,000
$a2 \cdot b2$ = $20,000

Since we want to find the area of ABCD, we express EHD in terms of it:
EHD = (ABCD) * $\frac{a2}{a1+a2}$ = (ABCD) * $\frac13$

Since EHD = $30,000, ABCD = $90,000, and the total price is ABCD + GBCD + ABG = $120,000.

 

[fresh_42]

LaTexified version of my solution of Problem 3:

Spoiler

I like to do analytic geometry, so I assign coordinate values to each of the points:
E = {0,0}, F = {a1, 0}, A = {a1, b1}, D = {0, b1}, B = {a2, b1+b2}, C = {0,
b1+b2}, G = {a2, b1}

H is at the intersection of DF and EG, and its location is thus {$\frac{a1 \cdot a2}{a1+a2}$, $\frac{a1 \cdot b1}{a1+a2}$}.

Using the shoelace formula, the areas are
ABG = $\frac12 (a1-a2) \cdot b2$
GBCD = $a2 \cdot b2$
EHD = $\frac12 \frac{a1 \cdot a2 \cdot b1}{a1+a2}$
ABCD = $a1 \cdot b1$
Total = ABG + GBCD + ABCD

From the statement of the problem, GBCD = $20,000 and ABG = $10,000, and that gives us
$a1 \cdot b2$ = $40,000
$a2 \cdot b2$ = $20,000

Since we want to find the area of ABCD, we express EHD in terms of it:
EHD = (ABCD) * $\frac{a2}{a1+a2}$ = (ABCD) * $\frac13$

Since EHD = $30,000, ABCD = $90,000, and the total price is ABCD + GBCD + ABG = $120,000.

You have to prove the coordinates for H, resp. the height of the triangle, and it is asked for the prize of the area of ABCEF which is not $120,000.

 

[lpetrich]

The missing parts of Problem 3, and correction of an error in my original solution:

Spoiler

H is at the intersection of EG and DF. So its location is
(H) = (1-t1)*(E) + (t1)*(G) = (1-t2)*(D) + (t2)*(F)
where t1 and t2 are parameters for points along those two lines.
$$ (1-t1)\{0,0\} + (t1)\{a2,b1\} = (1-t2)\{0,b1\} + (t2)\{a1,0\} $$
giving
$$ t1 \cdot a2 = t2 \cdot a1 \\ t1 \cdot b1 = (1 - t2) \cdot b1 $$
This gives us t2 = 1 - t1 and
$$ \frac{t1}{1-t1} = \frac{a1}{a2} $$
Solving for t1 gives us
$$ t1 = \frac{a1}{a1 + a2} ,\ t2 = \frac{a2}{a1 + a2} $$
and the coordinates of point H:
$$ (H) = \frac{a1}{a1+a2} \{ a2, b1 \} $$
Applying a triangle area formula with (base) = b1 and (height) = first coordinate of above, I find
$$ \text{(area of EHD)} = \frac12 \frac{a1 \cdot a2 \cdot b1}{a1 + a2} $$
The total area (ABCEF) = (ABG) + (BCDG) + (ADEF) with
$$ \text{(ABG)} = \frac12 (a1 - a2) \cdot b2 ,\, \text{(BCDG)} = a2 \cdot b2 ,\, \text{(ADEF)} = a1 \cdot b1 $$
From the problem statement, the area ratio (BCDG)/(ABG) = ($20,000)/($10,000) = 2, and (EHD)/(ABG) = 3, giving
$$ a2 \cdot b2 = 2 \cdot \frac12 (a1 - a2) \cdot b2 ,\, \frac12 \frac{a1 \cdot a2 \cdot b1}{a1 + a2} = 3 \cdot \frac12 (a1 - a2) \cdot b2 $$
Solving the first one gives a1 = 2*a2, and solving the second one gives b1 = (9/2)*b2. Thus, area (ADEF) = a1*b1 = 9*(a2*b2).

Area (ABG) = (1/2)*(a2*b2), thus (ADEF) = 18 * (ABG), and is thus $180,000. Adding the prices of ABG and BCDG, $10,000 and $20,000, I find $210,000.

 

[fresh_42]

The missing parts of Problem 3, and correction of an error in my original solution:

Spoiler

H is at the intersection of EG and DF. So its location is
(H) = (1-t1)*(E) + (t1)*(G) = (1-t2)*(D) + (t2)*(F)
where t1 and t2 are parameters for points along those two lines.
$$ (1-t1)\{0,0\} + (t1)\{a2,b1\} = (1-t2)\{0,b1\} + (t2)\{a1,0\} $$
giving
$$ t1 \cdot a2 = t2 \cdot a1 \\ t1 \cdot b1 = (1 - t2) \cdot b1 $$
This gives us t2 = 1 - t1 and
$$ \frac{t1}{1-t1} = \frac{a1}{a2} $$
Solving for t1 gives us
$$ t1 = \frac{a1}{a1 + a2} ,\ t2 = \frac{a2}{a1 + a2} $$
and the coordinates of point H:
$$ (H) = \frac{a1}{a1+a2} \{ a2, b1 \} $$
Applying a triangle area formula with (base) = b1 and (height) = first coordinate of above, I find
$$ \text{(area of EHD)} = \frac12 \frac{a1 \cdot a2 \cdot b1}{a1 + a2} $$
The total area (ABCEF) = (ABG) + (BCDG) + (ADEF) with
$$ \text{(ABG)} = \frac12 (a1 - a2) \cdot b2 ,\, \text{(BCDG)} = a2 \cdot b2 ,\, \text{(ADEF)} = a1 \cdot b1 $$
From the problem statement, the area ratio (BCDG)/(ABG) = ($20,000)/($10,000) = 2, and (EHD)/(ABG) = 3, giving
$$ a2 \cdot b2 = 2 \cdot \frac12 (a1 - a2) \cdot b2 ,\, \frac12 \frac{a1 \cdot a2 \cdot b1}{a1 + a2} = 3 \cdot \frac12 (a1 - a2) \cdot b2 $$
Solving the first one gives a1 = 2*a2, and solving the second one gives b1 = (9/2)*b2. Thus, area (ADEF) = a1*b1 = 9*(a2*b2).

Area (ABG) = (1/2)*(a2*b2), thus (ADEF) = 18 * (ABG), and is thus $180,000. Adding the prices of ABG and BCDG, $10,000 and $20,000, I find $210,000.

That's better, as one almost actually can read what you've done, and correct. Just a little request for future posts: Please write your variables $a2=a_2$ which is a_2. Otherwise your readers might confuse indices with coefficients. It looks more like code than mathematics.

 

[lpetrich]

Solution for Problem 7:

Spoiler

I will express the general solution of $y^{(4)}(x) + 4y(x) = 0$ as
$$ y(x) = c_1 e^x \cos x + c_2 e^x \sin x + c_3 e^{-x} \cos x + c_4 e^{-x} \sin x $$
I calculated the Wronskian of the four terms, and I found it to be 32. So the solutions are linearly independent, and the values of derivatives 0, 1, 2, and 3 at any x value thus suffice to determine the c's.

(a) To find the initial conditions, take the first three derivatives of $y(x) = e^{-x} \cos x$ and find their values at 0.
$$ y(x) = e^{-x} \cos x \\ y'(x) = -e^{-x} (\cos x + \sin x) \\ y''(x) = 2e^{-x} \sin x \\ y'''(x) = 2e^{-x} (\cos x - \sin x) $$
giving us
$$ y(0) = 1 ,\, y'(0) = -1 \,\ y''(0) = 0, y'''(0) = 2 $$

(b) We take four derivatives of the general solution, and then find their value at x = 0.
$$ y(x) = c_1 e^x \cos x + c_2 e^x \sin x + c_3 e^{-x} \cos x + c_4 e^{-x} \sin x \\ y'(x) = c_1 e^x (\cos x - \sin x) + c_2 e^x (\cos x + \sin x) - \\ c_3 e^{-x} (\cos x + \sin x) + c_4 e^{-x} (\cos x - \sin x) \\ y''(x) = -2 c_1 e^x \sin x + 2 c_2 e^x \cos x + 2 c_3 e^{-x} \sin x - 2 c_4 e^{-x} \cos x \\ y'''(x) = -2 c_1 e^x (\cos x + \sin x) + 2 c_2 e^x (\cos x - \sin x) + \\ 2 c_3 e^{-x} (\cos x - \sin x) + 2 c_4 e^{-x} (\cos x + \sin x) \\ y^{(4)}(x) = -4 y(x) $$
At x = 0,
$$ y(0) = c_1 + c_3 ,\, y'(0) = c_1 + c_2 - c_3 + c_4 ,\, y''(0) = 2 (c_2 - c_4) ,\, \\ y'''(0) = 2(-c_1 + c_2 + c_3 + c_4) ,\, y^{(4)}(0) = -4(c1 + c3) $$
Setting all but the first of them equal to
$$ y'(0) = 1 ,\, y''(0) = 0 ,\, y'''(0) = 0 ,\, y^{(4)}(0) = 0 $$
gives the solution
$$ c_1 = \frac14 ,\, c_2 = \frac14 ,\, c_3 = - \frac14 ,\, c_4 = \frac14 $$
and
$$ y(x) = \frac14 \left( e^x \cos x + e^x \sin x - e^{-x} \cos x + e^{-x} \sin x \right) $$

 

[fishturtle1]

For 6)

Spoiler

Proof: We make use of Euler's totient function and Euler's theorem through out the proof. Observe, odds are closed under multiplication and evens are closed under multiplication so $a^{13} \equiv a (\operatorname{mod} 2)$. Also, $a^{13} \equiv (a^2)^6\cdot a \equiv a (\operatorname{mod} 3)$. So $a^{13} \equiv a (\operatorname{mod} 3)$. Also, $a^{13} \equiv (a^4)^3\cdot a (\operatorname{mod} 5)$. So $a^{13} \equiv a (\operatorname{mod} 5)$. Also, $a^{13} \equiv (a^6)^2\cdot a \equiv a (\operatorname{mod} 7)$. So $a^{13} \equiv a (\operatorname{mod} 7)$. Finally, $a^{13} \equiv a^{12}\cdot a (\operatorname{mod} 13)$. So $a^{13} \equiv a (\operatorname{mod} 13$.

By Chinese Remainder theorem, we can conclude $a^{13} \equiv a (\operatorname{mod} 2\cdot 3\cdot 5\cdot 7\cdot 13)$ that is, $a^{13} \equiv a (\operatorname{mod} 2730)$.

[\SPOILER]

 

[QuantumQuest]

For 6)

Well done fishturtle1!

 

[fresh_42]

For 6)

Spoiler

Proof: We make use of Euler's totient function and Euler's theorem through out the proof. Observe, odds are closed under multiplication and evens are closed under multiplication so $a^{13} \equiv a (\operatorname{mod} 2)$. Also, $a^{13} \equiv (a^2)^6\cdot a \equiv a (\operatorname{mod} 3)$. So $a^{13} \equiv a (\operatorname{mod} 3)$. Also, $a^{13} \equiv (a^4)^3\cdot a (\operatorname{mod} 5)$. So $a^{13} \equiv a (\operatorname{mod} 5)$. Also, $a^{13} \equiv (a^6)^2\cdot a \equiv a (\operatorname{mod} 7)$. So $a^{13} \equiv a (\operatorname{mod} 7)$. Finally, $a^{13} \equiv a^{12}\cdot a (\operatorname{mod} 13)$. So $a^{13} \equiv a (\operatorname{mod} 13$.

By Chinese Remainder theorem, we can conclude $a^{13} \equiv a (\operatorname{mod} 2\cdot 3\cdot 5\cdot 7\cdot 13)$ that is, $a^{13} \equiv a (\operatorname{mod} 2730)$.

[\SPOILER]

I think you meant Fermat's little theorem, not Euler's.

 

[PeroK]

For 6)

More generally, if $n$ is a product of distinct primes, $p_i$, and $l = lcm \{p_i - 1 \}$ then

$\forall x: \ x^{l+1} = x (mod \ n)$

I found a further generalisation of this a few years ago - a maximal extension of Fermat's little theorem - but I don't remember it exactly, except that it involved the Euler totient function.

 

[fishturtle1]

Well done fishturtle1!

thanks!

I think you meant Fermat's little theorem, not Euler's.

thanks for pointing that out, I guess since the moduli are all prime I did mean FLT to be more specific

 

[fresh_42]

thanks for pointing that out, I guess since the moduli are all prime I did mean FLT to be more specific

No criticism, I only thought, as you didn't explicitly mention the theorem, that Fermat's little theorem is faster to find for our readers than to search for Euler's theorem. Euler has left so many theorems ...

 

[fishturtle1]

No criticism, I only thought, as you didn't explicitly mention the theorem, that Fermat's little theorem is faster to find for our readers than to search for Euler's theorem. Euler has left so many theorems ...

I sincerely mean this, I always appreciate your comments and had not thought of it from that perspective, that writing Euler's theorem is ambiguous and that writing FLT could save time for the reader. I definitely think the proof I wrote could be improved by being more specific.

 

[QuantumQuest]

I definitely think the proof I wrote could be improved by being more specific.

The steps you have in the proof are correct and lead to the correct result. But @fresh_42's comment is really worthwhile because the point in a proof is besides being correct to be also easily followed. Fermat's little theorem is a special case of Euler's theorem $a^{\phi (n)} \equiv 1(\mod n)$ where $\phi (n)$ is the Euler totient function, so you're essentially right in saying that you use these two things but as you apply FLT in your proof it is necessary for your readers in order to easily follow along, to state that explicitly.

 

[opus]

What level of math is typically sufficient for these basic challenges? I am always interested but they still look pretty heavy!

 

[fresh_42]

What level of math is typically sufficient for these basic challenges? I am always interested but they still look pretty heavy!

It is the same as it is so often: once you have the correct entry point, the proofs themselves aren't hard. I only solved mine, but I think it's similar with all of them. Do you have one particular in mind? Maybe we can give some hints.

 

[opus]

It is the same as it is so often: once you have the correct entry point, the proofs themselves aren't hard. I only solved mine, but I think it's similar with all of them. Do you have one particular in mind? Maybe we can give some hints.

I don't even think a hint would be beneficial at this point for me. I'm taking Calculus I in the fall, so I don't have any of those tools in my belt yet. What's a common requisite for being able to solve these? Calculus? Discrete math?

 

[fresh_42]

I don't even think a hint would be beneficial at this point for me. I'm taking Calculus I in the fall, so I don't have any of those tools in my belt yet. What's a common requisite for being able to solve these? Calculus? Discrete math?

The examples are quite a variety of very different problems, i.e. not subject to a single area.

1. calculus
2. combinatorics
3. geometry (area formulas) with linear algebra (equations of straight lines)
4. geometry (of a circle)
5. combinatorics
   
6. discrete mathematics, resp. abstract algebra
7. differentiations and solving a system of linear equations
8. calculus
9. simply a riddle
10. integration, i.e. calculus

Farmer Joe's field wasn't very difficult, the key was basically to find the height of the yellow area, and the key was to use a coordinate system, find the equations of the lines and solve it. The coding was a bit like a crossword puzzle in the weekend edition of a newspaper, and for #7 one basically only needed to know what the differentials of the functions $e^x\; , \;\sin(x)\; , \;\cos(x)$ are.

I choose the first problem, because I found that those inequalities are useful to know. And, yes, it's again about differentiation, integration and the power series expansion of $\log \,$. The first part can be solved by differentiation. The second part is a bit tricky, and the third is actually the easiest of them, although it might not appear as such, but one doesn't even have to know a lot about integration. They all can be solved within a few lines, but I admit one has to find an idea.

The basic difficulty for us is, that there are so many different people around. We already introduced this two weeks ban for members, we expected them to find the answers, in order to give the younger ones a chance. So if we added easier questions, then there are definitely some people who first had to promise us not to spoil those.

Why don't you try number 5? This one looks fun. I would start by an example with smaller numbers to get an idea about the difficulty, but I would expect, that it has something to do with symmetries. It reminds me a bit of my very first exercise at university:

"Can there be a walk through Königsberg which crosses all seven bridges exactly once, and if so, can there also be a round trip?"

 

[Phylosopher]

4.4. Using only geometric reasoning, calculate the average value of f(x)=√2x−x2f(x) = \sqrt{2x - x^2} on [0,2] [0,2]
(by @QuantumQuest)

I am not sure really what the question is requiring ^^".

4) I assume the question just doesn't want me to use calculus/integration. If so then:
the equation is just half a circle:
$$f(x)=\sqrt{2x-x^2}=\sqrt{2x-x^2+1-1}=\sqrt{1-(x^2-2x+1)^2}=\sqrt{1-(x-1)^2} \rightarrow y^2+(x-1)^2=1$$

Thus the answer is half the half area of the circle. $$Average=\frac{\pi}{4}$$

Note: y is restricted to be positive.

 

[StoneTemplePython]

I don't even think a hint would be beneficial at this point for me. I'm taking Calculus I in the fall, so I don't have any of those tools in my belt yet. What's a common requisite for being able to solve these? Calculus? Discrete math?

I think @fresh_42's response is on point. A couple things I would add.

1.) To the extent you are in the US, I am sympathetic. I don't think I would have been able to solve much (or any?) of these in high school even after I took calc 1. But so what -- if there are problems you can at least visualize, then try to struggle to the understand the associated solutions and you'll learn something interesting in the process. For example, I'm not sure I would have solved the cypher problem but it certainly is easy to visualize what is being discussed and follow the solution.

2.) I think we have a slight bias toward geometric problems in here as geometry in various forms is quite important in math and physics. It is also has a certain visual flavor. (Also reference part 1.)

 

[QuantumQuest]

I am not sure really what the question is requiring ^^".

4) I assume the question just doesn't want me to use calculus/integration

The point in this problem is to translate "using only geometric reasoning" correctly. You interpreted it correctly and your solution is right. For completeness, one thing that I want to add is that what we must find is $\frac{1}{2}\int_{0}^{2}\sqrt{2x - x^2} dx$. Then, using the way you used in your solution, we finally find that the graph of $y = f(x)$ between $x = 0$ and $x = 2$ is the upper half of the circle you found. So, the aforementioned integral is the area $\frac{\pi}{2}$ of that semicircle, so the average value is $\frac{\pi}{4}$.

 

[opus]

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Thanks guys. I'll try my hand at that problem fresh_42.
And yes I'm a U.S student- a horrendous public education, but that's no excuse for me. My lack of mathematical skills is my own doing as I didn't care in the least for it when I was in high school. Trying to make up for that now!
I think these problems are really great and helpful and I'm going to spend some time on trying to understand them so I can have meaningful participation down the road.