# A Math Olympiad question regarding Geometry

A little bit of a backstory (you may skip this if you want): My high school math teacher knows that I love math, but he also knows that I usually drift off during my classes, perhaps because it's too easy, perhaps because it's too slow, I don't know why exactly, but I'm sure many frequenters of this site know how it feels. When I drift off I do think about math, but about different things, for example, in my most recent math class in the previous week (we were discussing the basics of integrals, which I already did for fun a few months before) I proved the quadratic equation. My teacher saw I was bored and asked my why I never participated in any Olympiads, and quite frankly I never knew about them (it is now too late for me, this was the final year I was eligible). He gave me the paper for the second round of the Dutch Olympiad and the final problem immediately catched my eye; the paper said that 0% of the participants solved this problem. So, me being me, I disregarded all problems and immediately focused on this one. It took me 20 minutes at most which surprised me, but I can't find the solutions anywhere, so I want to ask if my answer is correct on this site. I'm not quite aware if this is against the rules, I'm new here.

The original problem:

A flag in the form of an equilateral triangle is connected to the tops of 2 vertical poles. One of the pole has a length of 4 and the other pole has a length of 3. You also know that the third vertex touches the ground perfectly. Calculate the length of a side. Calculators aren't allowed. The following picture was appended:

My solution:

For some reason I immediately knew how to solve the problem. A picture says a 1000 words:

Which leaves us with $$x^2 = 1 + (\sqrt{x^2-16} + \sqrt{x^2-9})^2$$

$$x^2 = 1 + x^2-16 + x^2-9 + 2\sqrt{(x^2-16)(x^2-9)}$$

$$-x^2 + 24 = 2\sqrt{(x^2-16)(x^2-9)}$$

Then just square, fast forward a couple of extremely messy steps (I could include but I think it is clear) and we get $x = \sqrt{17 \dfrac{1}{3}}$

Questions:

• Is my answer correct? I know I might have made a mistake in the algebra, but is the main reasoning correct?

• Why would this problem be considered as hard? I mean if (almost) nobody actually got it, there must be a reason why? This was a year that the Netherlands got just 1 bronze and 3 honourable mentions, so it wasn't the brighest generation, but I'm still confused.

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I guess he meant $17 + 1/3$, which it is quite common to write as $17 \frac13$ (otherwise, he would have written $\frac{17}{3}$. $\sqrt{17\frac13}$ is just a bit more than $4$, which looks reasonable. (And the reasononing is reasonable, too IMHO). (Hmm, the comment I was referring to just disappeared...) – Elmar Zander Feb 14 '13 at 19:41
I get the same answer. Maybe no one else answered it because it was the last problem listed and they just didn't get to it. :) – Blue Feb 14 '13 at 19:42
@ElmarZander If you're referring to my short-lived comment, I realized that seconds after posting, and it is long gone. I almost forgot about "mixed fractions" since they are so useless... – rschwieb Feb 14 '13 at 19:42
@rschwieb Sorry, but I can't edit my comment any more. As to the mixed fractions: I fully agree. I've also never seen them outside school books. – Elmar Zander Feb 14 '13 at 19:55
I'm sure it's not an IMO question.at least excepted one. – user59671 Feb 14 '13 at 20:10

It's certainly a valid way to solve the problem. There might be a cute way of reasoning that cuts out all the computation, but there is always a little bit of luck involved when finding something like that.

Why would it be considered hard? Lots of highschoolers (in the US at least) have trouble even setting up word problems like this, and even if they can, they might be defeated in trying to solve the resulting equation. Maybe it is "olympic" just to do the computation.

Anyhow, I understand your disappointment with this problem. Maybe someone will see a key that unravels the problem without a lot of writing!

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Another possibility is that the judges had the wrong answer. – rschwieb Feb 14 '13 at 19:50
"Maybe it is 'olympic' just to do the computation." One of the USAMO problems from 2011 certainly felt like this. – Joe Z. Feb 14 '13 at 19:53
@JoeZeng I sure hope there's a pretty answer. If there isn't, then this is another infuriating instance of mathematics being viewed as something you train people to do (as if it were a diving competition, or something...) If they want people to be captured by math's beauty, the problems should really have a kernel of insight in them... – rschwieb Feb 14 '13 at 19:57
I suppose you could set up something contrived with complex numbers. – Joe Z. Feb 14 '13 at 20:09
Something like $\displaystyle Re(r e^{i\alpha}) = Re(r e^{i\(\alpha + \frac{2\pi}{3})}) + 1 = Re(r e^{i\(\alpha + \frac{4\pi}{3})}) + 3$, solve for $r$. But that just makes it even uglier. – Joe Z. Feb 14 '13 at 20:12

Or vectors. If you think a little bit, you'll realize that the height of the center is just $1/3$ of the sum of the heights of the corners, so the center is at the height $7/3$ from the ground. The next thing you may realize is that the sum of the squares of the distances from a line through the center to the vertices also does not depend on the rotation (can you see why?). Thus, this sum is $\frac{49+4+25}9=\frac{26}3$. To relate it to the side length $a$ is a piece of cake (just consider the line that is an altitude of the triangle and get $2(a/2)^2=a^2/2$). Thus $\frac{a^2}{2}=\frac{26}3$ and $a=\sqrt{\frac{52}3}$.

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One possible approach to this problem is to use trigonometry and polar coordinates. Consider the center of the triangle to be the origin, and each point on the triangle being a fixed radius $r$ and some bearing $\theta$ away from it. These bearings can be expressed as $\alpha$, $\alpha+\frac{2\pi}{3}$, and $\alpha+\frac{4\pi}{3}$ respectively, and the $y$-coordinates of these points can be represented as $r \sin \alpha$, $r \sin (\alpha+\frac{2\pi}{3})$, and $r \sin (\alpha+\frac{4\pi}{3})$ respectively.

Given your description, we have the following problem first: Find the value of $\alpha$ (or some representation of it using a trigonometric function) such that $4 (\sin \alpha - \sin (\alpha+\frac{2\pi}{3})) = \sin (\alpha+\frac{4\pi}{3}) - \sin (\alpha+\frac{2\pi}{3})$. Then use the angle-sum identities to piece this out, find the value of $r$ so that $\sin \alpha - \sin (\alpha+\frac{2\pi}{3}) = 1$, and your answer is $r\sqrt{3}$.

This solution still involves a lot of writing, though, to actually compute what $\alpha$ and $r$ are.

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Here's a relatively simple way with just some basic trigonometry. Based on this diagram:

We have $$\sin\alpha=\frac3s$$ $$s\sin\beta=4$$ However, notice that $\beta=120^{\circ}-\alpha$. Applying our subtraction rule for $\sin$, we have $$s\left(\sin120^{\circ}\cos\alpha-\cos120^{\circ}\sin\alpha\right)=4\\ \implies s\left( \frac{\sqrt3}{2}\cos\alpha+\frac12\sin\alpha \right)=4$$ Now notice that $\cos\alpha=\sqrt{1-\sin^2\alpha}$. Combining this with the first equation, we have $$s\left( \frac{\sqrt3}{2}\cos\alpha+\frac12\sin\alpha \right)=4\\ \implies s\left( \frac{\sqrt3}{2}\sqrt{1-\frac{9}{s^2}}+\frac{3}{2s} \right)=4\\ \implies 3+\sqrt{3s^2-27}=8\\ \implies s=\sqrt{\frac{52}3}$$ I don't know if this is the "pretty answer" they intended, but it is fairly simple and direct deriviation.

In fact, on playing further with this, there an elegant generalisation. If you need to fit an equilateral triangle into a trapezoid with heights $a$ and $b$ (the original had heights 3 and 4), the formula for the side length is $$s=\frac{2}{\sqrt3}\sqrt{\frac{a^3+b^3}{a+b}}=\frac{1}{\sin{60^{\circ}}}\sqrt{\frac{a^3+b^3}{a+b}}$$

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