I Trigonometry problem of interest

[Charles Link]

TL;DR

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.