I Trigonometry problem of interest

[Charles Link]

TL;DR Summary

The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer of exactly 7, but the arithmetic is so complex that the integer result is rather surprising.

Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the integer result is rather surprising. Might anyone see some method that simplifies things, or is this perhaps simply a case of mathematical coincidence?

 

[jedishrfu]

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.

 

[Charles Link]

Writing law of cosines twice $11^2=2r^2(1-\cos(\theta))$ and $2^2=2r^2(1-\cos(120-\theta))$ and trigonometric expansion on $\cos(120-\theta)$, one can eliminate the $\cos(\theta)$, and solve for $r$, but $\sin(\theta)$ shows up in the trigonometric expansion, so that the final algebraic result is a quadratic in $r^2$, and the arithmetic is complex. One gets two solutions for $r$: $r=7$, and an extraneous $r$ just slightly less than 6.

Edit: To provide a little more detail $r^2=(250 \pm \sqrt{1936})/6$.

I anticipate there might be a simpler solution to this that gets the result that $r=7$, but it is not readily apparent to me.

 

[kuruman]

TL;DR Summary: The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer of exactly 7, but the arithmetic is so complex that the integer result is rather surprising.

So what exactly are you given and what are you asked to find?

 

[Charles Link]

So what exactly are you given and what are you asked to find?

The problem is to find x. In the lower semi-circle they had one half of a regular hexagon with all 3 sides of length x, so that says immediately that x=r. In the upper semi-circle they had 3 sides of length 2, x , and 11. The problem is to find x. The correct answer is 7, but I'm looking for a simpler way to do it than the cumbersome way I presented with more detail in post 3.

 

[phinds]

I guess it's too much trouble for you to post a graphic of what you are talking about.

 

[Charles Link]

I guess it's too much trouble for you to post a graphic of what you are talking about.

I am all thumbs with computer graphics. Hopefully the explanation is clear enough to follow. :)
The original problem was on-line on Facebook, but normally the share feature isn't allowed on PF.

 

[FactChecker]

I am all thumbs with computer graphics. Hopefully the explanation is clear enough to follow. :)
The original problem was on-line on Facebook, but normally the share feature isn't allowed on PF.

Can you take a screenshot and copy it here?

 

[kuruman]

Is this it?

 

[Charles Link]

In the original problem, x is in between the 2 and the 11, but mathematically it makes no difference.

Thank you @kuruman :)

and what I wonder, is did the authors of the problem find a mathematical coincidence, or might there be a very simple way to show x=r=7? (In the other semi-circle they had one half of a regular hexagon with all 3 sides =x.) Given that x=r, find x.

So far, especially after working through the trigonometry, algebra, and arithmetic, I see nothing that should indicate the answer would be an integer. :) I even tried eliminating $r^2$ and solving for $\cos{\theta}$, but that came with some very complicated arithmetic. I'm attempting to do this without resorting to the use of a calculator to evaluate coefficients, etc. I did get the correct answer, r=7, by solving for $r^2$, but I'm looking for any possible much simpler solution.

 

[jedishrfu]

I guess it's too much trouble for you to post a graphic of what you are talking about.

Let us not forget the marvelous, dapper Snowmen that Charles has posted in the past.

---

When I tested the central angles using 7 for the x and the radius,

it worked out, so there may be a subtle linkage.

 

[jedishrfu]

So, given the picture, if x is a chord and x is the radius, we have an equilateral triangle, reducing the problem to a quadrilateral with three inscribed corners on the circle, where two of the sides are both radii as x. The angle between the two x's is 120 degrees.

and so then we can divide the quadrilateral into two triangles, where for triangle 2 we have an its central angle relationship: $sin(2*\theta_2)=x$ and $sin(2*\theta_{11})=(11/2)x=$

and separately we have $2*\theta_2 + 2*\theta_{11} = 120$ which reduces to $\theta_2 + \theta_{11} = 60$

Does that make sense?

 

[Lnewqban]

The problem is to find x. In the lower semi-circle they had one half of a regular hexagon with all 3 sides of length x, so that says immediately that x=r. In the upper semi-circle they had 3 sides of length 2, x , and 11. The problem is to find x. The correct answer is 7, but I'm looking for a simpler way to do it than the cumbersome way I presented with more detail in post 3.

Graphically, x = 7.00

 

[Charles Link]

Does that make sense?

You got the right idea, and I was able to solve it. However, I am rather puzzled why the final answer, when the problem is so arithmetically complex in the way that I solved it, has such a simple integer result. Oftentimes that implies there may be some kind of hidden symmetry involved, but in this case, so far I don't see any.

@Lnewqban very good,(post 13), but the largest circle with R=11 doesn't have its center on the other center. Perhaps I'm overlooking something.

 

[Lnewqban]

@Lnewqban very good,(post 13), but the largest circle with R=11 doesn't have its center on the other center. Perhaps I'm overlooking something.

I located the center of the R = 11.0 largest blue circle at the point I knew with certainty, which was the interception of the of the R = 2.0 smallest blue circle and the R = 7.0 magenta circumcircle connecting all the vertices of our polygon.

 

[Lnewqban]

Using another point as center of the R = 11.0 largest blue circle:

 

[Charles Link]

@Lnewqban Very good graphics. Thank you. :)

I'm still looking for a simpler solution to this one. Perhaps the combination of 2,7, and 11 on the semi-circle of radius $r=7$ are simply a mathematical coincidence. I've looked for things that could be obvious like Pythagorean relations, but the geometry here doesn't seem to present right triangles that would be the reason behind the integer $x=r=7$ result.

Meanwhile using the answer that $x=r=7$ doesn't offer anything obvious upon computing $\theta$ or $120-\theta$. There doesn't seem to be anything special about those two angles.

Edit: In exploring a path like this one, it is possible we will come up empty, but I would not be too surprised if someone comes up with a reason why this one has a simple integer result.

 

[Charles Link]

See

He has kind of a simple solution. Still no hidden symmetry, but he does have a good solution. His arithmetic is kind of simple, because he makes some good algebraic choices.

 

[Lnewqban]

@Lnewqban Very good graphics. Thank you. :)

I'm still looking for a simpler solution to this one. Perhaps the combination of 2,7, and 11 on the semi-circle of radius $r=7$ are simply a mathematical coincidence. I've looked for things that could be obvious like Pythagorean relations, but the geometry here doesn't seem to present right triangles that would be the reason behind the integer $x=r=7$ result.

You are welcome, Charles.

Please, take a second look, right triangles are hidden in your diagram:
See the Thales' theorem in the following link:

https://www.mathsisfun.com/geometry/circle-theorems.html

​

 

[jedishrfu]

That's a great video description of the steps needed to solve the problem, and it eliminates the complex, difficult-to-reduce equations we encountered.

In some ways, it reminds me of the admonishment I received as a young undergrad trying to reduce a matrix to a simpler form, but ended up with complex fractions. The professor reminded us to subtract rows to achieve the magic 1 in a column, allowing us to reduce the matrix further.

In another case, the professor reminded us that integrating over y before x could produce a helpful answer, and he gave a quiz with that exact scenario.

 

[Lnewqban]

 

[Charles Link]

@Lnewqban I didn't know the name of it, but Thales' theorem was one of a couple of things I was trying to apply to simplify things. It turns out though, the video has some very simple formulas to see the results. It is likely IMO that the authors of this problem may have begun with one side of 2, and tried a bunch of integers for the second side using the simple formulas of the video, and then found that 11 works with the final side and radius being a 7.

I am pleased with the solution that the video has. It makes things simple enough that we don't need mathematical coincidence to explain where the combination of 2, 7, 11 and came from. :)

Edit: He gets on the video, (with the first side = 2), that $(11+1)^2=3(x^2-1)$. It is sort of easy to see from this how they got the combination of 7 and 11, but I don't see any others that work besides the simple case of x=r=2 with the third side =2. This case of 2, 7 and 11 might just be a rare combination...and then I see x=26 gives 44 for the 3rd side. :) That would imply that the combination of 1,13, and 22 would also work, with x=r=13.

 

[jedishrfu]

Historically, these are likely known triplets for this kind of problem similar to the pythagorean triplets.

 

[Charles Link]

Historically, these are likely known triplets for this kind of problem similar to the pythagorean triplets.

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.

Perhaps I should post a little detail on what I am solving. Following the video of post 18, I'm looking for integer solutions of $(a+2b)^2=3(4x^2-a^2)$, for integers $a,b,$ and $x$. I start with an $a$, and then look for an $x$ that gives the expression in parentheses the result of 3 times a perfect square. From that I then compute $b$.

It should be noted that $x=a=b$ always works. This is the trivial solution of the regular hexagon.

and note, with the combinations I found ( 2 of them), I could make up my own version of the Olympiad problem. e.g. they could start with 11 and 26 and find the third side x with 19 as the answer. LOL

See https://www.facebook.com/MATEMATICA...tif_t=page_post_liker_invite_follow&ref=notif

It may be of interest that the above is where I first saw this Polish Olympiad problem.

 

[Charles Link]

and I found a 4th one that works: 13,31, and 35 (correction =46 instead of 35) with $x=r=31$. Like @jedishrfu said above,( post 23), it is likely these have been found previously and are well known, but it is fun to do my own exploring. :)

I also thought I had another with $a=13$, and $x=7$, but I found that $b=-2$ for that case.

Perhaps this is a very well known problem and someone can post a list of the known solutions. :)

 

[jedishrfu]

Once you tire of these triplets, you can move onto the Euler brick and then the perfect Euler brick where no one has found a solution.

That should keep you real busy with Pythagorean triplets and diophantine equations before the onset of winter snow and the Snowmen calling out your name.

 

[Charles Link]

@jedishrfu I am pleased with what I got so far on this one. It was my instincts that the integer combinations that would work for this semi-circle problem were somewhat rare, but it was really kind of neat to find 3 others besides the one that the Polish Olympiad problem had.

It would still be neat if someone could post a link of solutions of this type. If this is not already a carefully researched problem, perhaps there are even college students who would find it of interest to research it and see how many solutions they can find. I've been doing the arithmetic by hand, but it helps in doing the math by hand to know things like 28^2=784, etc. without needing to multiply them.

 

[jedishrfu]

That's an interesting point about knowing numbers. In the movie, The Man Who Knew Infinity, on the life of Ramanujan, there's a scene where Ramanujan is challenged to compute a value, and he does, and the professor who challenged Ramanujan accepts the counter and was capable of rapid calculations too. It's a gift for some people.

In schools throughout the US, we're taught the times ten tables. Other countries teach times twelve or even times twenty. Knowing these tables can give you a huge boost in solving math problems faster than your peers.

I hated the memorization part and as a fourth grader searched the library, that's searched very loosely. I found the thin Trachtenberg Number System book and was hooked. Here was a means to banish memorization. After the euphoria wore off and I realized that my teacher would never go for it and it in fact required keeping a tally inside your head and that was harder to do than memorizing the tables.

My main dyslexic difficulty was distinguishing 9x6 from 8x7. It wasn't until a few grades later that I learned the nines rule. It helped me unequivocally realize that 9x6 was 54 since 5+4=9 and I still go through this train of thought whenever 9x6 comes up.

If you search using the solution values it might identify a name and many additional values. There may even be some other use for them.

 

[Charles Link]

If you search using the solution values it might identify a name and many additional values.

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. :)

 

[kuruman]

Using the law of cosines and reasonably simple algebra, I got the following expression for the radius-squared in terms of the given sides $$R^2=\frac{a^3-b^3}{3(a-b)}.$$I will post my derivaton in a few hours because it's very late where I am.

 

[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.