Trigonometry problem of interest

[Charles Link]

Very good. That reduces to $R^2=(a^2+ab+b^2)/3$, and is in agreement with my post 24, but far simpler. Thank you @kuruman :)

Edit: Even though it looks simpler, I find it easier to work with what I have in post 24 for finding solutions, calculating the $4x^2-a^2$, and seeing if it gives a square upon dividing by 3, (i.e. factoring out 3 to give 3^2).

 

[renormalize]

I've been trying to find others, but so far only 2,7, and 11 along with 1, 13, and 22 plus multiples thereof. The combinations that work seem to be very limited. Perhaps someone else can find one or two more. Hopefully I got the arithmetic right on the other one that I found. :)

Edit: I did find one more: 11,19, and 26 with $x=r=19$. Hopefully I got the arithmetic right again. I will double-check that result. :) and I checked the angles on this last one=they do add up to 120 degrees.

@Charles Link, given that $a\geq1,b\geq1$, you can solve your formula $(a+2b)^2=3(4x^2-a^2)$ to get $x=\sqrt{a^{2}+ab+b^{2}}/\sqrt{3}\,$. I thought it might be fun to write a couple of lines of Mathematica code to step through the integers $1\leq a\leq100,a\leq b\leq100$ and list all the solutions for $x$ in which $x$ is both an integer and also a prime. (This eliminates all the results that are simple integer multiples of others.) I also drop all the $a=b=x$ solutions. Here are the results for your amusement:

Let me know if you want to see any larger integer solutions.

 

[Charles Link]

@renormalize Thank you very much. :) You did verify the 3 combinations that I had previously found, plus the one in the Olympiad. You also got a whole bunch more that would have been very hard to get by hand. Glad you also found this interesting. :)

 

[jedishrfu]

I took your suggestion, but no luck so far. I had more luck yesterday coming up with the video of post 18 with a search. I'll need to keep trying. :)

Yeah me too. I was hoping it was a somewhat obscure set of triples that were used in other areas of math but sadly, nothing popped up.

Calling All Snowmen

It's time for you to step forward and develop an Insights article about what you've discovered. You could title the triples as the Snowman sequence because it snowed so many people trying to find a clean solution as shown in the video.

 

[Herman Trivilino]

View attachment 366563Is this it?

Does the unlabeled chord pass through the circle's center?

 

[fresh_42]

My main dyslexic difficulty was distinguishing 9x6 from 8x7.

Funny, that's mine, too, together with 3x14 and 4x13. I still "calculate" them, whereas others are stored.

 

[kuruman]

Does the unlabeled chord pass through the circle's center?

Yes, it's a diameter.

 

[Charles Link]

@kuruman You essentially have the same solution for $x=R$ in post 30 that @renormalize has, but you should recognize that the difference of cubes factors.

This form is even simpler than what the video of post 18 has where you get $(a+2b)^2=3(4x^2-a^2)$ with $a$ appearing on both sides of the equation as I have in post 24.

This final simple form that @renormalize has in post 32 looks to be the reason why there are as many integer solutions for this thing as there are. Certainly the clumsy quadratic equation in $R^2$ that I used in my very first solutions in this thread looked like it would be about a one in ten thousand chance or less of getting an integer solution for $x$. When the answer for the $a=2$ and $b=11$ combination in the Olympiad problem was exactly 7, it almost looked like some kind of magic.

This solution of $x=\sqrt{\frac{a^2+ab+b^2}{3}}$ is indeed a rather simple form.

Thank you for everyone who contributed. :)

 

[kuruman]

Here is the derivation mentioned in post #30.

Refer to the figure on the right. It shows half the hexagon as a quadrilateral inscribed in a circle of radius $R$. The full hexagon can be obtained by reflecting the quadrilateral about the diameter or by appending to it itself rotated by $180^{\circ}$.

I have labeled the angles subtended by the chords, by Greek letters related to their label in Latin, e.g. angle $\alpha$ for chord $a$ etc. I state without proof the following.

1. The angle subtended by a chord is invariant if the point that subtends it on the circumference is moved along the circumference. For example, $\alpha = \angle DCA=\angle DBA.$

2. The angle subtended by a chord at the circumference is half the angle subtended by the same chord at the center of the circle. For example, the angle subtended by a diameter is $180^{\circ}$ at the center and $90^{\circ}$ at any point on the circumference.

3. The length of a chord is given by the the diameter times the sine of the subtended angle. For example,$~DA=a=2R\sin{\alpha}.$

Now for the derivation. We write the diagonal $DB$ in two ways using the law of cosines for triangle $(DAB)$ and the Pythagorean theorem for right triangle $(DBC).$ Note that the angle between vectors $\mathbf a$ and $\mathbf x$ is $\theta = 180^{\circ}-(90^{\circ}+\beta)=(90^{\circ}-\beta)$. Therefore, $\mathbf a\cdot\mathbf x=ax\sin\beta.$ Thus, $$d^2=a^2+x^2+2ax\sin\beta=4R^2-b^2.$$ We now set $x=R$, $\sin\beta=\dfrac{b}{2R}~$ and rearrange to get, $$ a^2+ab+b^2=3R^2.$$ Using the identity $a^3-b^3=(a-b)(a^2+ab+b^3)$ and rearranging, yields $$R^2=\frac{a^3-b^3}{3(a-b)}.$$While thinking about this, I stumbled upon something that might be of interest. I will post it on a separate thread on this forum as a solved problem.

 

[Charles Link]

@kuruman Very good. :) Excellent derivation. Between you and @renormalize of post 32, we have a simpler expression for $R=x$ than the video of post 18 and my post 24 has.

Even though @renormalize got additional solutions with the computer from what I got by hand, I am pleased that my hand calculations were good enough, that I found all of the ones that I attempted. I only got to $a=15$ by hand, and the calculations were getting more difficult as $a$ got larger.

@jedishrfu My computer search didn't give the best results, but what we derived as a group surpasses what we might have found on the computer. Our posts on Physics Forums did appear a couple of times in my computer search, but I think the search engines would do well to steer people to Physics Forums more often than they do these days. :)

 

[daverusin]

As others have noted, if the four sides of that quadrilateral have lengths a, x, b, and the diameter 2x, then 3x^2 = a^2+ab+b^2. As to the question of how the heck they came up with the 11, 2, and 7: we can scale the problem so that the diameter x is 1, and then seek *rational* values of a and b for which a^2+ab+b^2=3. This is a classic situation: we can find all the rational points on a conic curve with a parameterization. In this case, one parameterization would give (a,b) = (-m^2+2m+2, 2m^2+2m-1)/(m^2+m+1) ; taking all rational values for m gives all rational solutions (a,b). The particular one in the Polish Olympiad corresponds to m=2. The other solutions in the table posted by @renormalize correspond to m= 5/2, 2, 3/2, 7/4, 8/5, 4/3, 10/7, and 5/4 respectively; not sure what happened to (23, 71, 49) which comes from m=5/3. (Replacing m by 1/m swaps a and b; requiring a and b to be positive requires 0.366<m<2.732. So you can effectively generate all solutions of interest by running through all rationals m = i/j with j < i < (1+sqrt(3))*j and i coprime to j .)

 

[renormalize]

not sure what happened to (23, 71, 49) which comes from m=5/3.

Good catch. To eliminate solution triples that are integer multiples of other triples I must verify that either $b$ or $x$ (or both) are prime. (My code in post #32 only tested for $x$ being prime.) Here is the corrected table of solutions:

The table now includes the previously missed triple $(23,71,49)$.

 

[Gavran]

One more approach based on
https://en.wikipedia.org/wiki/Cyclic_quadrilateral and the cosine theorem.

$\angle FGB=120^\circ\implies\angle FEB=60^\circ$ (Supplementary angles)
$\cos(\angle FEB)=(c^2-7^2-7^2+11^2)/(2\cdot(11\cdot c+7\cdot7))\implies c=13$ (Angle formulas)
$\cos(\angle EFG)=(c^2-11^2-7^2+7^2)/(2\cdot(7\cdot c+11\cdot7))=1/7$ (Angle formulas)
$d^2=c^2+7^2-2\cdot c\cdot7\cdot\cos(\angle EFG)\implies d=8\sqrt{3}$ (The cosine theorem)

In the same way it can be calculated that:
$\angle ADG=60^\circ$
$b=13$
$\cos(\angle DGF)=11/14$
$a=5\sqrt{3}$.

$c\cdot b=7\cdot x+a\cdot d\implies x=7$ (Ptolemy's theorem)

 

[Charles Link]

This is a classic situation: we can find all the rational points on a conic curve with a parameterization. In this case, one parameterization would give (a,b) = (-m^2+2m+2, 2m^2+2m-1)/(m^2+m+1) ; taking all rational values for m gives all rational solutions (a,b).

Could you elaborate on this a little? (post 41). My mathematics with calculus, etc. is reasonably advanced, but this looks like it is more advanced than anything I have previously seen.

 

[Charles Link]

I found the following video that might explain some of post 41. It seems to be a specialized topic, and not one that is found in the curriculum for most physicists.

 

[Charles Link]

I think I figured out now what @daverusin did in post 41. One begins with $x^2+xy+y^2-3=0$, and to find the rational solutions, you first pick out one rational solution. In this case $(-1,-1)$ makes a good point to use. You then have a line of slope $m$ though that point, and that line is found by taking $m=\frac{y--1}{x--1}$ and solving for $y$. You then plug into $x^2+xy+y^2 -3 =0$ and solve the quadratic equation for $x$. It will have two roots where it intersects what I believe is a rotated ellipse: Both must be rational if one of them is rational, and one will be $x=-1$ with $y=-1$, The other is $x=(-m^2+2m+2)/(m^2+m+1)$ with $y=(2m^2+2m-1)/(m^2+m+1)$. The numerators become the $a$ and $b$, and the denominator is the scaling factor $x$ where we scaled down to the case of $x=1$, and to scale up again, that factor $x$ is the denominator. We thereby have the $a,b,$ and $x$ that we are looking for by putting in various rational values for $m$.

additional comment: The ellipse above is rotated 45 degrees and has its major axis on the line $y=-x$ with a half length of $\sqrt{6}$ and minor axis on the line $y=x$ with half width of $\sqrt{2}$.

and thank you @daverusin for a very excellent post 41. :)

 

[Charles Link]

and a follow-on to the above: The case of $m=1$ gives the point $(1,1)$ for $(x,y)$. It may be worth mentioning that I tried using the point $(1,1)$ as a starting point on the ellipse for the known rational solution, but it seemed to give poor results. The resulting $m's$ that connected this point to the other points were too hard to find.

It seems like the approach of post 41 is a lucky one for this problem because it did have a very convenient as well as simple rational solution in $(-1,-1)$ that is at a location on the ellipse where the sought after positive $x$ and $y$ are conveniently located basically across the ellipse from the point $(-1,-1)$. Without this conveniently located point, I think applying the method of parameterization could have been a difficult one. With this point, it made for some rather simple results.

Edit: and note it is only necessary to check for $m>1$, because that will give you $b>a$. The $x$ and $y$ are symmetrical, and checking for $m<1$ would simply duplicate solutions.

 

[Charles Link]

Here is an aside item, and in a way a little simpler. The problem of finding integer solutions for $x^2=(a^2+ab+b^2)/3$ reminds me of something I tried a few years back. We know that $x^2+2x+1$ is always a perfect square, but could the factor that appears from the difference of two cubes, ($x^3-1^3$), $x^2+x+1$ be a perfect square? I did find a number of rational solutions with a little trial and error, and perhaps I can find my old calculations which I'll look for later.

Edit: and I couldn't find my original calculations, but I was able to duplicate the results: Besides the somewhat trivial $x=-1$, $x=3/5, 5/3$ and $8/7$ all work if my arithmetic is correct.

Edit 2: I now (a few days later) was successful at applying the method of posts 41 and 46 to $x^2+x+1=y^2$, using as a starting point $(-1,1)$ and finding positive rational solutions for $x$ and $y$ using the line with slope $m$ that passes through this point, and finding the other point where it intersects what is a hyperbola rotated 90 degrees. We just need to use $0<m<1$, because the asymptote has slope of $1$. Just like with the ellipse where I was getting solutions that work with every $m$ I tried in post 61, I'm getting the same kind of success with this hyperbola.

 

[hutchphd]

I think you mean a 6-sided polygon inscribed in a circle of radius 7.

Folks solve it by determining the central angles and determine if they add up to 360 degrees.

One way to play with the problem is to use GeoGebra to visualize it.

There is also a lovely theorem by Ptolemy that deals with cyclic quadrilaterals. I think that will work......not an intuitive result! Will write it up if time/energy permits.

https://en.wikipedia.org/wiki/Ptolemy's_theorem

 

[Charles Link]

@hutchphd a super result. :) With a couple algebraic steps one gets $3x^2=11^2+2^2+(2)(11)$. I'll let you write it out, but if you don't get to it in a day or so, I could post the 3 equations and their rather straightforward solution.

 

[hutchphd]

@hutchphd a super result. :) With a couple algebraic steps one gets $3x^2=11^2+2^2+(2)(11)$. I'll let you write it out, but if you don't get to it in a day or so, I could post the 3 equations and their rather straightforward solution.

Please feel free. I have had an exhausting week and seem disinclined to do anything (feeling each of my 73 yrs I guess.....ah, well. I wish LaTeX was not such a chore for me)

 

[Charles Link]

@hutchphd Here is the solution:

$(ac)(bd)=2x^2+(11)(2)$ from Ptolemy
$(ac)^2=4x^2-2^2$ from Pythagoras and Thales
$(bd)^2=4x^2-11^2$ from Pythagoras and Thales

This gives $(ac)^2(bd)^2=(4x^2-2^2)(4x^2-11^2)=(2x^2+(11)(2))^2$.
We get $16x^4-4x^2(2^2+11^2)+11^2 *2^2=4x^4+8(11)x^2+11^2*2^2$
Thereby $12x^4=4(2^2+11^2+(2)(11) )x^2$.

Finally $3 x^2=2^2+11^2+(2)(11)$, which agrees with our previous results from @kuruman of post 30 and @renormalize of post 32.

Doing the arithmetic, we get of course that $x=7$. :) (I almost forgot to include that part. It was more important in some ways to get agreement with posts 30 and 32. LOL )

 

[Charles Link]

Looking back to the other posts, I see @Gavran also used Ptolemy's theorem previously in post 43, but the post 52 method gets the answer of $x=7$ as well as the arithmetically simple result of posts 30 and 32 without the extra calculations of post 43.

 

[Gavran]

Looking back to the other posts, I see @Gavran also used Ptolemy's theorem previously in post 43, but the post 52 method gets the answer of $x=7$ as well as the arithmetically simple result of posts 30 and 32 without the extra calculations of post 43.

You are right.

There is one more approach which is based on the figure from the post #39 and it is simpler than the one I have provided in the post #43.

$\cos(\angle ADC)=a/(2R)$ from The definition of cosine in the context of a right-angled triangle
$\angle ADC+\angle ABC=180^\circ$ from The inscribed angle theorem
$a^2+(2R)^2-2\cdot a\cdot2R\cdot\cos(\angle ADC)=x^2+b^2-2\cdot x\cdot b\cdot\cos(\angle ABC)$ from The cosine theorem

Above three equations imply
$\cos(\angle ADC)=(a^2-x^2-b^2+(2R)^2)/(2\cdot(a\cdot2R+x\cdot b))=a/(2R)\implies x=7$.

 

[Charles Link]

@Gavran I presume in your solution that you plugged in $R=x$.

You then get $a^2-b^2+3x^2 =2a^2+ab$
so that $3x^2=a^2+b^2+ab$.

Very good. :) You have a simple solution also, and these last two algebraic steps that I included show that your method here is also in complete agreement with posts 30 and 32.

You might also include that when you used the cosine theorem that both are equal to $c^2$.

 

[Charles Link]

Looking back over things, I do think @daverusin in post 41 has something of considerable merit. With his method, which I figured out in posts 45 and 46, he was able to spot a solution that @renormalize missed in his post 32. The algebra of finding where the line with slope of $m$ that passes through the point $(-1,-1)$, (which lies on the rotated ellipse), and intersects the rotated ellipse at the other points isn't terribly difficult, and I think it would be a good exercise for some of those with good math skills to try, to see how you then get the $a,b,$ and $x$, basically in their entirety, from his results. So far I have been the only one to give his post 41 a "like", so I think many have probably overlooked his post or didn't try to work out the details to see what he did.

Edit: The rotated ellipse is $x^2+xy+y^2=3$. It may not even be well-known to many, but those who have good math skills should know you can do the transformation $x'=x \cos{\theta}+y \sin{\theta}$ and $y'=-x \sin{\theta}+y \cos{\theta}$, and know that with $\theta=\pi /4$ that the $xy$ term becomes $(x')^2/2-(y')^2/2$, with the $x^2+y^2=(x')^2+(y')^2$ staying the same with the transformation. Thereby an ellipse with $(x')^2/2+(y')^2/6=1$ is the form it takes in the coordinate system that has the axes rotated 45 degrees, which is now in the standard form for an ellipse.

 

[hutchphd]

You are right.

Yes indeed. Through laziness I did not really look at the attempt.......bad form on my part (apologies!)

 

[Charles Link]

@hutchphd It gets harder as we get older. I am 70 now. We sort of have an excuse. Too many of the younger ones though I see are relying too much on google and want instantaneous answers, and don't want to take the time to wrestle with math problems the way our generation did. :)

 

[hutchphd]

Although I slighted @Gavran by not mentioning his attempt, I don't think anyone had called out Ptolemy's Theorem by name. How often do you get to cite Ptolemy directly?? Who wouldn't want to add that in.... lotsa fun....and pedagogically an impressive theorem IMHO.

 

[Charles Link]

Although I slighted @Gavran by not mentioning his attempt, I don't think anyone had called out Ptolemy's Theorem by name. How often do you get to cite Ptolemy directly??

In post 43 @Gavran cited it on his very last line in parentheses. It is so easy to overlook something like that though, because this thread has a lot of posts, and it can also be difficult to go back to the previous page of posts. The solution we have though, written out in post 52, is a more direct application of Ptolemy's theorem, where the answer of $x=7$ appears in short order.