# Problem 10 of the 2011 Euclid Math Contest (remix)

I wrote the Euclid contest this year two days ago, on tuesday. There were 10 problems, and the tenth problem was a nice geometry problem. Three subproblems with a nice neat triangle on the right, with the subproblems getting progressively harder and harder. As I proceeded with the problem, I couldn’t help but imagine it as a Project Euler problem — instead of finding one such integer triangle, it would ask you to find the sum of the perimeters of all such integer triangles with perimeter less than 10^12 or some large number.

### A modified problem

In the above diagram, $\angle CAK = 2 \angle KAB$ and $\angle ABC = 2 \angle KAB$. Let $a = CB$, $b = AC$, $c = AB$, $d = AK$, and $x = KB$. Write an algorithm to find triangles satisfying these conditions where a, b, c are all integers.

### Similar triangles

It is difficult to try to find integer triangles with such strange requirements as these. It seems that the line $AK$ is completely unnecessary, but if we take it out, there doesn’t seem to be any way to relate the angle ratios to integer side lengths.

We can prove that $\triangle CAK$ is similar to $\triangle ABC$. Being an exterior angle, $CKA = 3 \theta$, and also $\angle CAB = 3 \theta$. Both of the triangles have an angle of measure $2 \theta$ and another angle of measure $3 \theta$, thus they are similar.

From the similar triangles ratio

$\frac{b}{d} = \frac{a}{c}$

We can write d in terms of the three sides of the triangle:

$d = \frac{bc}{a}$

Similarly, the side $CK$ can be written as $a-x$. Then we have the ratio

$\frac{a}{b} = \frac{b}{a-x}$

Solving for x allows us to express it in terms of the three sides of the triangle, again:

$x = \frac{a^2 - b^2}{a}$

### Constructing another similar triangle

Our goal here is to relate the lengths a, b, c with a simple equation, which then the problem turns into a number theory problem. Since we can write the lengths d and x in terms of a, b, and c, we can also relate any of a, b, c, d, x in an equation.

Again, there doesn’t seem to be a way to relate all of the variables together, in a way that any solution implies the original angle ratio required — unless we make a construction.

Here we extend AB to F, so that $KB = BF$ and $\triangle KBF$ is an isosceles triangle.

Again, since the exterior angle here is $2 \theta$, both $\angle BKF = \angle BFK = \theta$. Also with this construction, $\triangle AKF \sim \triangle KBF$, and is also isosceles, hence $d = AK = KF$.

With this construction, we can write the ratio

$\frac{x}{d} = \frac{d}{c+x}$

Perfect! Cross multiplying and then substituting in the original sides of the triangles gives

$(\frac{a^2-b^2}{a})(c+\frac{a^2-b^2}{a}) = (\frac{bc}{a})^2$

Simplifying this gives

$(a^2 - b^2) (a^2 - b^2 + ac) = b^2 c^2$

### Number theory magic

Now that we have an equation relating the three side lengths — it’s easy to verify that any three integers satisfying the triangle inequality gives a triangle with the required conditions — we can use number theory to try to generate integers that fit the equation.

If we expand the equation, we get

$a^4+a^3 c-2 a^2 b^2-a b^2 c+b^4-b^2 c^2 = 0$

It makes sense to solve for c, which can be done using just the quadratic formula:

$c = \frac{a^3 - ab^2 \pm \sqrt{(ab^2-a^3)^2 + 4b^2(b^4 + a^4 - 2a^2 b^2)}}{2b^2}$

We need the discriminant D — the expression inside the square root — to be a perfect square, where we are allowed to have integer values for a and b. If we can get D to be a perfect square, then c will turn out to be a rational number. Then multiplying all three variables by a constant gives integer values for all three.

So we defined D:

$D = (ab^2-a^3)^2 + 4b^2(b^4 + a^4 - 2a^2 b^2)$

Expanding this gives

$D = a^6 - 7a^2 b^4 + 2 a^4 b^2 + 4b^6$

Fortunately, this expression has an interesting factorization:

$D = (a^2+4b^2) (a+b)^2 (a-b)^2$

Or we can also write

$\sqrt{D} = (a+b) (a-b) \sqrt{a^2 + 4b^2}$

We’ve simplified this problem to finding values where $a^2 + 4b^2$ is a perfect square, that is:

$a^2 + (2b)^2 = k^2$

This is just finding Pythagorean triples where one of the two sides are even! For instance, in the triple (3,4,5), we have a=3 and b=2. However, substituting a=3, b=2 into the quadratic formula gives c=5. This is almost a solution, only that the sides have to satisfy the triangle inequality (two sides have to add up to more than the third side).

The next Pythagorean triple (5,12,13) gives a=5 and b=6. Substituting this in gives c=11/9, which does satisfy the triangle inequality. Multiplying everything by 9 gives a=45, b=54, c=11 as the smallest working combination.

With this method, it is possible to quickly find arbitrarily many such triples, using Pythagorean triples as a starting point (which can be generated quickly with known methods).

## One thought on “Problem 10 of the 2011 Euclid Math Contest (remix)”

1. Emidio Iacobucci says:

Isn’t a the largest side since it is opposite 3theta and b is opposite 2theta?