# Math Challenge - October 2019

• Challenge
• Featured
Mentor
Questions

1.
(solved by @MathematicalPhysicist ) Show that the difference of the square roots of two consecutive natural numbers which are greater than ##k^2##, is less than ##\dfrac{1}{2k}##, ##k \in \mathbb{N} - \{0\}##. (@QuantumQuest )

2. (solved by @tnich ) Let A, B, C and D be four points in space such that at most one of the distances AB, AC, AD, BC, BD and CD is greater than ##1##. Find the maximum value of the sum of the six distances. (@QuantumQuest )

3. (solved by @PeroK ) A random number chooser can choose only one of the nine integers ##1, 2, \dots, 9## each time, with equal probability for each choice. What is the probability, after n times of choosing (##n \gt 1##), the product of the ##n## chosen numbers be evenly divisible by ##10##. (@QuantumQuest )

4. (solved by @tnich ) Let ##A## be an ##n\times n## matrix with real entries such that ##Ax\cdot x=0## for all ##x\in\mathbb{R}^n##. Show that ##A^2x\cdot x\leq 0## for all ##x\in\mathbb{R}^n##. (@Infrared )

5. Let ##A=\sum_{k=0}^\infty a_k\, , \,B=\sum_{k=0}^\infty b_k## be two convergent series one of which absolutely.
Prove: The Cauchy-product ##C=\sum_{k=0}^\infty c_k## with ##c_k=\sum_{j=0}^ka_jb_{k-j}## converges to ##AB##.
Give an example that absolute convergence of at least one factor is necessary.

6. Prove that a ##T_0## topological group (Kolmogorov space) is already ##T_2## (Hausdorff space).
Show that an infinite linear algebraic group with the Zariski topology is always ##T_0## but never ##T_2##.
Why this discrepancy?

7. Let ##\mathcal{O}(n)## be the group of orthogonal real ##n \times n## matrices. For ##f \in L^p = L^p(\mathbb{R}^n)## we set
$$A.f(x) = f(A^{-1}x)$$
Show that ##\mathcal{O}_f=\{\,A.f\,|\,A\in \mathcal{O}(n)\,\}\subseteq L^p(\mathbb{R}^n)## is compact.

8.
a.)
(solved by @Delta2 ) Let ##f\, : \,\mathbb{R}\longrightarrow \mathbb{R}## be a convex function and ##\lambda_1,\ldots ,\lambda_n## positive weights, i.e. ##\sum_{i=1}^n \lambda_i = 1##. Show that $$f\left(\sum_{i=1}^n \lambda_ix_i \right) \leq \sum_{i=1}^n \lambda_i f(x_i)$$

b.) (solved by @Delta2 ) Let ##g\, : \,[0,1]\longrightarrow \mathbb{R}## be an integrable function such that the continuous function ##f\, : \,\mathbb{R}\longrightarrow \mathbb{R}## is convex on the image of ##g\,.## Prove
$$f\left(\dfrac{1}{b-a}\int_a^b g(x)\,dx \right) \leq \dfrac{1}{b-a}\int_a^b f(g(x))\,dx$$

c.) (solved by @StoneTemplePython ) Prove without differentiation that the cylinder with the least surface area among the ones with given volume ##V## is the cylinder whose height equals the diameter of its base.

d.) (solved by @Delta2 ) Prove that for any sequence ##a_n \geq \ldots \geq a_1 > 0## of positive real numbers
$$\dfrac{1}{\dfrac{1}{a_1}}+\dfrac{2}{\dfrac{1}{a_1}+\dfrac{1}{a_2}}+\ldots +\dfrac{n}{\dfrac{1}{a_1}+\ldots+\dfrac{1}{a_n}} < 2 (a_1+\ldots +a_n)$$

9. Let ##p(x)=x^n+a_{n-1}x^{n-1}+\ldots +a_1x+a_0## be a nonlinear polynomial with ##a_n=1## and suppose ##(x-1)^{k+1}\,|\,p(x)## for some positive integer ##k\,.## Prove that
$$\sum_{j=0}^{n-1} |a_j|> 1+ \dfrac{2k^2}{n}$$
Hint: At some stage of the proof you will need Chebyshev polynomials.

10. (solved by @aheight ) Consider the triangle ##A=(0,0)\, , \,B=(2\sqrt{3},0)\, , \,C=(3-\sqrt{3}\, , \,-3+3\sqrt{3})##. Now choose on each side a point, ##M_a,M_b,M_c##, such that the new triangle built by those points is of minimal perimeter.
What is the area of the ##\triangle (M_a,M_b,M_c)\,?##

High Schoolers only.

11.
(solved by @etotheipi ) Show that ##\lim_{x\to 0}\dfrac{\tan(\sin x)}{\sin(\tan x)} = 1##. (@QuantumQuest )

12. (solved by @etotheipi ) Calculate the masses of Sun, Earth and Jupiter. You may assume circular orbits. We further calculate with the following data:

• the gravitational constant ##G=6.67 \cdot 10^{-11}\,\dfrac{m^3}{kg\cdot s^2}##
• the Kepler constant for our solar system ##C= \dfrac{T^2}{R^3}=0.29 \cdot 10^{-18}\,\dfrac{s^2}{m^3}##
• the acceleration by gravity on earth ##\gamma = 9.81\,\dfrac{m}{s^2}##
• Io's orbital radius ##R_I=4.22 \cdot 10^{8}\,m##
• Io's orbital period ##T_I=1.77 \,d##

13. (solved by @etotheipi ) A car drives at ##72## km/h directly past a resting observer when the driver presses its horn. By what interval does the pitch of the horn change as the car passes the observer? (Speed of sound ##s= 340## m/s.)

14. (solved by @archaic and @lekh2003 ) Consider the sphere ##\mathbb{S}^2=\{\,(x,y,z)\,|\,x^2+y^2+z^2=r^2\,\}## and a point ##P\in \mathbb{S}^2##.
Determine the set of all center points of all chords starting in ##P##.

15. (solved by @YoungPhysicist ) At the monthly meeting of former mathematics students, six members choose a real number ##a##, which has to be guessed by a seventh mathematician who had left the room before. He gets the following information after he returned:

(1) ##a## is rational.
(2) ##a## is an integer divisible by ##14##.
(3) ##a## is real and its square equals ##13##.
(4) ##a## is an integer divisible by ##7##.
(5) ##a## is real and the inequality ##0<a^3+a<8,000## holds.
(6) ##a## is even.

He is told, that all pairs ##(1,2),(3,4),(5,6)## always consist of a true and a false statement. What is ##a##?

Last edited:
Anti Hydrogen, Not anonymous and YoungPhysicist

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tnich
Homework Helper
##5+\sqrt 3##
Arrange the four points to form a quadrilateral with each side of length 1 and the distance between one pair of non-adjacent points equal to 1. This causes the distance between the other pair to be ##\sqrt 3##. Since none of the other sides can be lengthened, this gives the maximum sum of the six distances.

scottdave and QuantumQuest
Delta2
Homework Helper
Gold Member
Question 8a)
for ##n=2## the proof is the definition of the convex function
Suppose it holds for ##n## (induction hypothesis)
We gonna prove it for ##n+1##
we define $$x_{12}=\frac{l_1x_1+l_2x_2}{l_1+l_2}$$ and then we have
##f(l_1x_1+l_2x_2+...+l_{n+1}x_{n+1})=f((l_1+l_2)x_{12}+...+l_{n+1}x_{n+1})##

Now we use the induction hypothesis on the n-uple ##(x_{12},x_3,...,x_{n+1})## so we will have
$$f((l_1+l_2)x_{12}+l_3x_3..+l_{n+1}x_{n+1})\leq (l_1+l_2)f(x_{12})+\sum_3^{n+1}l_if(x_i)$$

Now all that is left to prove is that $$(l_1+l_2)f(x_{12})\leq l_1f(x_1)+l_2f(x_2) (1)$$
We apply again the definition of the convex function
$$f(x_{12})=f(\frac{l_1}{l_1+l_2}x_1+\frac{l_2}{l_1+l_2}x_2)\leq \frac{l_1}{l_1+l_2}f(x_1)+\frac{l_2}{l_1+l_2}f(x_2)$$
from which it follows that (1) holds

MathematicalPhysicist
Gold Member
Question 1:
$$|\sqrt{n+1}-\sqrt{n}| = |\frac{1}{\sqrt{n+1}+\sqrt{n}}|\le 1/ (2\sqrt{k^2})$$
the last inequality follows from the fact that both $n$ and $n$ are greater than ##k^2##.

QuantumQuest and Delta2
Delta2
Homework Helper
Gold Member
Question 8b)
We gonna use the definition of the definite integral as a Riemann sum
$$f(\frac{1}{b-a}\int_a^b g(x)dx)=f(\frac{1}{b-a}\lim_{n\to +\infty}\sum_{i=1}^n g(x_i)\frac{b-a}{n})=\lim_{n\to +\infty}f(\sum_{i=1}^n g(x_i)\frac{1}{n})$$
The last equality holds because f is continuous so we can take the limit outside.

Now we apply 8a) for ##\lambda_i=\frac{1}{n}## and ##x_i=g(x_i)## so we ll have
$$f(\frac{1}{b-a}\int_a^b g(x)dx)\leq \lim_{n\to +\infty}\sum_{i=1}^n f(g(x_i))\frac{1}{n}$$

and it is easy to see that $$\frac{1}{b-a}\int_a^b f(g(x))dx=\frac{1}{b-a}\lim_{n\to +\infty}\sum_{i=1}^nf(g(x_i))\frac{b-a}{n}=\lim_{n\to +\infty}\sum_{i=1}^n f(g(x_i))\frac{1}{n}$$

Mentor
Question 8b)
The two (8.a and 8.b) are called Jensen's inequalities. They are extremely useful, esp. in functional analysis, but elsewhere, too. They can shorten complicated estimations significantly. E.g. the power and exponential functions are convex, and the negative of a concave function, too.

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Delta2
Delta2
Homework Helper
Gold Member
Question 8d)
This looks hard with all the nasty fractions out there but it turns out to be easy, unless I am overlooking something and my solution is maybe wrong.
for the k-th term of the sum we have
$$\frac{k}{\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_k}}\leq\frac{k}{\frac{1}{a_k}+\frac{1}{a_k}+...+\frac{1}{a_k}}=\frac{k}{\frac{k}{a_k}}=a_k$$
since it is $$a_k\geq a_{k-1}\geq…\geq a_1$$ therefore $$\frac{1}{a_k}\leq\frac{1}{a_{k-1}}\leq…\leq\frac{1}{a_1}$$
hence the given sum of all these fractions is S $$S\leq\sum_{k=1}^n a_k<2\sum_{k=1}^n a_k$$

Mentor
Question 8d)
This looks hard with all the nasty fractions out there but it turns out to be easy, unless I am overlooking something and my solution is maybe wrong.
No, you didn't overlook something. I had a formal induction in mind, but the idea is the same.

Delta2
MathematicalPhysicist
Gold Member
The two (8.a and 8.b) are called Jensen's inequalities. They are extremely useful, esp. in functional analysis, but elsewhere, too. They can shorten complicated estimations significantly. E.g. the power and exponential functions are convex, and the negative of a concave function, too.
Jarzinsky's inequality in non-equilibrium statistics uses Jensen's IIRC.
https://en.wikipedia.org/wiki/Jarzynski_equality

Delta2
DEvens
Gold Member
Maybe question 3 is really high school? You made them do integrals last month.

To get a factor of 10 there must be at least one 5, and at least one even number.

So the chance of getting a 10 is 1 minus the chance not to get a 10.

Divide the nine numbers into ##5##, ##E## (for even nums), and ##O## (for other nums, odd not 5).

To get no 10 it means there must be some combination of either ##5^a O^{n-a}## or ##E^a O^{n-a}##. But we must be careful not to double count the all ##O## case. There are four elements in ##E## and four in ##O##. So each time there is a ##E^a## we get a factor of ##4^a##. And each time there is an ##O^{n-a}## we get a factor of ##4^{n-a}##. Plus, for each combo there is an "n choose a" to get the order of elements. So we need to total up all the chances to get all ##5## and ##O##, and the chances to get all ##E## and ##O##, subtract off the double-count of all ##O##, then take 1 minus this fraction.

Chance to get a 10 is therefore:

## 1 - \frac{1}{9^n} \left\{ \left[ \sum_{a=0}^{n} \left(\frac{n}{a} \right) (4^n + 4^{n-a}) \right] -4^n\right\} ##

Note that by ##\left(\frac{n}{a} \right)## I mean "n choose a" since I couldn't figure out the correct tags for "choose." Let's try it for n=2. We should get 4 for ## E 5## plus 4 for ## 5 E##.

##1 - \frac{1}{81} ( 16 + 16 + 2 * (16 + 4) + 16 + 1 - 16) = 1 - \frac{73}{81} = \frac{8}{81}##

Seems to match.

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BWV
Get a different answer to 3

With the same criteria the odds to not get a 5 on n draws is (8/9)^n
the prob to get all odd numbers on n draws is (5/9)^n

so the odds of not getting a product divisible by 10 is
P(no10)=(8/9)^n + (1-(8/9)^n)*(5/9)^(n-1),
i.e. the probability of either a) not getting a 5 or b) getting a 5 and only odd numbers on the remaining draws, the complement then is the probability of getting a multiple of 10, 1-P(no10)

for n=2 it is 0.093 which is different than the above
for n=10 it is 0.689

PeroK
PeroK
Homework Helper
Gold Member

To get a multiple of 10 we need at least one 5 and at least one even.

The probability of not getting a multiple of 10 is:

##(5/9)^n## (all odd)
##(8/9)^n## (no 5's)

But, this double counts:

##(4/9)^n## (all odd but no 5)

Putting these together and taking the complement gives:

##\frac{9^n + 4^n - 5^n - 8^n}{9^n}##

[\spoiler]

Delta2, QuantumQuest and StoneTemplePython
DEvens
Gold Member
I think you have at least one problem. You don't include a mix of even and odd with no 5's.

DEvens
Gold Member
Get a different answer to 3
I think you also don't include a mix of even and odd with no 5's. And you are multi-counting odd and 5.

PeroK
Homework Helper
Gold Member
I think you have at least one problem. You don't include a mix of even and odd with no 5's.
I don't know how you come to that conclusion.

If you do it by hand, then the answer for ##n =3## is 156/729. Which is what my formula gives. Mine is correct for ##n =2## as well.

Can you check your formula for ##n=2,3##?

I agree with @BWV for ##n = 10##.

PS my formula works for ##n=1## as well!

BWV
I think you also don't include a mix of even and odd with no 5's. And you are multi-counting odd and 5.
Dont think so
(8/9)^n is getting no 5's in n draws, even and odd mix doesn't matter in the case of no 5. The other way to not get a power of 10 is to get at least one 5 and then all odd which is (1-(8/9)^n)*(5/9)^(n-1) no double counting that I see. The problem with using combinations is they assume no repetition of the same number, which is counter to the problem as stated.

PeroK
PeroK
Homework Helper
Gold Member
I think you also don't include a mix of even and odd with no 5's. And you are multi-counting odd and 5.
As far as I can see, your formula gives the same as mine for ##n = 2,3##.

But, clearly, mine is a lot simpler.

QuantumQuest
Gold Member
Maybe question 3 is really high school? You made them do integrals last month.
Well, as far as I know, high school curricula vary from country to country. For my country, when I was given this problem - some 33+ years before, it was above high school and I think - at least to the extend that I'm aware of, it is above high school now too, for a number of countries.

tnich
Homework Helper
If ##X^TAX = 0## for all ##X##, then ##A## is anti-symmetric. That is ##a_{ij} = -a_{ji} \forall \, i, j##. To see this , suppose that ##X## has one non-zero element ##x_i##. Then ##X^TAX= x_i^2a_{ii}=0## implies that ##a_{ii}=0##. If ##X## has two non-zero elements ##x_i## and ##x_j,## then ##X^TAX= x_ix_j(a_{ij}+a_{ji})=0##, and ##a_{ij}=-a_{ji}##. Since by hypothesis ##X^TAX = 0## for all ##X##, ##a_{ij}=-a_{ji}## for all ##i,j##.
Now we have ##-A^T=A##, and therefore ##A^2 = -A^TA##. Then ##X^TA^2X=-X^TA^TAX=-(AX)^T(AX)\leq0## for all ##X \in \mathbb R^n##.

Infrared and Delta2
tnich
Homework Helper
Let ##D>0## be the diameter and height for the optimal configuration and let ##X## be an increment to the diameter. We can assume that the altered diameter ##X+D## is positive. The volume for the optimal configuration is ##\frac 1 4 \pi D^3## and the surface area is ##\frac 3 2 \pi D^2##. With the increment ##X## added to the diameter and the volume remaining the same, the height becomes ##\frac {D^3} {(D+X)^2}##. The surface area is then $$\pi \frac {D^3} {(D+X)^2}(D+X) + \frac 1 2 (D+X)^2,$$ and the difference between this surface area and the optimal surface area is $$\pi \frac {D^3} {(D+X)^2}(D+X) + \frac 1 2 (D+X)^2-\frac 3 2\pi D^3 = (3D+X)\frac{X^2}{2(D+X)}>0$$.
The final inequality follows from ##3D+X > D+X >0##.

Mentor
Let ##D>0## be the diameter and height for the optimal configuration and let ##X## be an increment to the diameter. We can assume that the altered diameter ##X+D## is positive. The volume for the optimal configuration is ##\frac 1 4 \pi D^3## and the surface area is ##\frac 3 2 \pi D^2##. With the increment ##X## added to the diameter and the volume remaining the same, the height becomes ##\frac {D^3} {(D+X)^2}##. The surface area is then $$\pi \frac {D^3} {(D+X)^2}(D+X) + \frac 1 2 (D+X)^2,$$ and the difference between this surface area and the optimal surface area is $$\pi \frac {D^3} {(D+X)^2}(D+X) + \frac 1 2 (D+X)^2-\frac 3 2\pi D^3 = (3D+X)\frac{X^2}{2(D+X)}>0$$.
The final inequality follows from ##3D+X > D+X >0##.
Why does the height change if the diameter does? Under which assumptions? And you cannot start with "the optimal configuration", since this should be the end of the proof, not its first sentence. I guess, i.e. you didn't tell me, that you want to show that any other configuration leads to an increased surface?!? And why can't diameter and height change independently?

Anyway, you used differentiation disguised as increment: vary the variable a little bit and see what the function does. That's differentiation.

tnich
Homework Helper
Why does the height change if the diameter does? Under which assumptions? And you cannot start with "the optimal configuration", since this should be the end of the proof, not its first sentence. I guess, i.e. you didn't tell me, that you want to show that any other configuration leads to an increased surface?!? And why can't diameter and height change independently?

Anyway, you used differentiation disguised as increment: vary the variable a little bit and see what the function does. That's differentiation.
If you posed the question, then I guess you had another proof in mind, but to answer you questions:
1) To maintain a fixed volume (as specified in the problem), if the diameter changes, then the height must also change.
2) The optimal configuration is given in problem statement. It is enough to prove that any other configuration results in a larger surface area. I do see your point, though. It would have been better not to start by calling it "optimal".
3) I used an increment, but I did not find the limit of the surface area as the increment approaches zero. I didn't even restrict the increment to small values, so I did not use differentiation.

Still, if this was not the answer you were looking for, I will bow to your judgement.

Mentor
so I did not use differentiation
You used ##f(D+X)-f(D)=o(X)## which is the Weierstraß equation of a differential which is zero, as it is for the optimum. Hence it's differentiation.

The proof without this method can be given in a single line.

Problem 10 is a statement about the orthic Triangle. A proof that the orthic triangle has minimum perimeter can be found here: Orthic Triangle. Also, this one: UofG Proof

Referring to the diagram below, the angles ##\alpha,\beta,\gamma## are the angles of ##\triangle ABC ## and ##a=AC,b=CB,c=AB##. Using Law of Cosines we obtain:
\begin{align*} \alpha=\frac{5\pi}{12} \\ \beta=\frac{\pi}{3} \\ \gamma=\frac{\pi}{4} \end{align*}
The side lengths of the orthic triangle ##\triangle LMN## are given by the expressions:
\begin{align*} a\cos(\gamma)=\sqrt{6(7-4\sqrt{3})} \\ b\cos(\beta)=3\sqrt{2-\sqrt{3}} \\ c\cos(\alpha)=\sqrt{6-3\sqrt{3}} \end{align*}
Then using Heron's Formula, the area is ##3\sqrt{3}-9/2##. Or can bash it out without these formulas or even compute it numerically by minimizing a function of three variables representing the points M, L, and N (see below for code) .

For fun, here's the Mathematica code to solve it as a minimizing problem:

Mathematica code:
aPoint = {0, 0};
bPoint = {2*Sqrt[3], 0};
cPoint = {3 - Sqrt[3], -3 + 3*Sqrt[3]};
theTriangle = Triangle[{aPoint, bPoint, cPoint}];
abf[t_] := 3*(Sqrt[3] - 1)*t;
thex[t_] := (3 - Sqrt[3])*t;
acf[t_] := 2*Sqrt[3]*t;
bcfx[t_] := (3 - Sqrt[3]) + Abs[(3 - Sqrt[3]) - 2*Sqrt[3]]*t;
bcfy[t_] := ((3*(1 - Sqrt[3]))/(2*Sqrt[3] - 3 + Sqrt[3]))*(bcfx[t] - 3 + Sqrt[3]) + 3*(Sqrt[3] - 1);
rgLength[myT_, bct_] := EuclideanDistance[{thex[myT], abf[myT]}, {bcfx[bct], bcfy[bct]}];
gbLength[bct_, myP_] := EuclideanDistance[{bcfx[bct], bcfy[bct]}, {acf[myP], 0}];
rbLength[myT_, myP_] := EuclideanDistance[{thex[myT], abf[myT]}, {acf[myP], 0}];
semiPerim[x_, y_, z_] := (rgLength[x, y] + gbLength[y, z] + rbLength[x, z])/2;
theArea[x_, y_, z_] := Sqrt[semiPerim[x, y, z]*(semiPerim[x, y, z] - rgLength[x, y])*(semiPerim[x, y, z] - gbLength[y, z])*(semiPerim[x, y, z] - rbLength[x, z])];
minResults = FindMinimum[rgLength[x, y] + gbLength[y, z] + rbLength[x, z], {{x, 0.5125}, {y, 0.5125}, {z, 0.5125}}]
sp = semiPerim[x, y, z] /. minResults[[2]]
rg1 = rgLength[x, y] /. minResults[[2]]
rg2 = gbLength[y, z] /. minResults[[2]]
rg3 = rbLength[x, z] /. minResults[[2]]
theArea[x, y, z] /. minResults[[2]]
2.12132
0.896575
1.79315
1.55291
0.696152

The last line is the area of the orthic triangle and agrees with the results above.

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fresh_42
Mentor
Problem 10 is a statement about the orthic Triangle.
Yes, well done. The orthic triangle is also called Fagnano triangle and the question is known as Fagnano problem. I have found some interesting remarks on Wikipedia:
Fagnano's original proof used calculus methods and an intermediate result given by his father Giulio Carlo de'Toschi di Fagnano. Later, however, several geometric proofs were discovered as well, amongst others by Hermann Schwarz and Lipót Fejér. These proofs use the geometrical properties of reflections to determine some minimal path representing the perimeter.

A solution from physics is found by imagining putting a rubber band that follows Hooke's Law around the three sides of a triangular frame ##{\displaystyle ABC}##, such that it could slide around smoothly. Then the rubber band would end up in a position that minimizes its elastic energy, and therefore minimize its total length. This position gives the minimal perimeter triangle. The tension inside the rubber band is the same everywhere in the rubber band, so in its resting position, we have, by Lami's theorem, ##{\displaystyle \angle bcA=\angle acB,\angle caB=\angle baC,\angle abC=\angle cbA}##

aheight