Probably any of these problems would be too small to include in a blog post on their own, so it might be more efficient (?) to put several of them together into one post.

The problems are not mine, and the solutions are not mine. They are just interesting problems I’ve come across when studying for math olympiads.

I hope to publish more collections of these random math problems in the future. That’s why I’m appending (1) to this post.

### Problem 1

(From the 2010 Fermat contest)

Spheres can be stacked together into a tetrahedron (triangular pyramid). Here’s a visual (not created by me, from Wikipedia):

Each of the spheres in the tetrahedron has a number written on it. The top sphere is 1, and the number of any sphere is* the sum of the numbers on the spheres in the layer above*.

An *internal sphere* is defined as a sphere that is not on the outside, or in other words, touching exactly three spheres in the layer above.

In a tetrahedron with 13 layers, what is the sum of the numbers on all of the internal spheres?

#### Solution

Let’s draw out a few layers of the tetrahedron.

Look for patterns.

There’s a few things that can be noticed-

- The numbers at the three corners of a layer are 1
- Any layer is highly symmetrical: in the last diagram the sequence (1 4 6 4 1) appears in the bottom row, as well as the left and right diagonals. We can rotate the tetrahedron and the numbers would still be the same.
- The numbers on the outside form Pascal’s triangle. The sequence (1 3 3 1) is the 4th row of Pascal’s triangle; (1 4 6 4 1) is the 5th, and so on.

The most important thing to look for is the fact that Pascal’s triangle appears in the tetrahedron.

Each row of Pascal’s triangle is one longer than the previous row. To calculate a row, overlap the previous row with the new row, like this:

Here the number in a purple box is the number to the top-right of it plus the number to the top-left of it:

Simple.

Notice that each number in the top gets added into exactly two boxes at the bottom.

Therefore the sum of the numbers in the bottom is twice the sum of the numbers in the top. , and .

This applies to any two rows. The sum of any row is twice the sum of the previous row.

Since the sum of the first row is 1, the sum of the second is 2, the sum of the third 4, and so on. We can even express this as a formula. The sum of the *n*th row is:

A similar idea applies to the layers of the tetrahedron.

In Pascal’s triangle, each number gets added to two places in the next row.

The tetrahedron is like Pascal’s triangle, in three dimensions. Each sphere in the tetrahedron sits on top of three spheres in the layer under it. So each number gets added to three different numbers in the next row.

Thus the sum of the numbers in any layer of the tetrahedron is 3 times the sum of numbers in the previous layer.

A formula for the sum of the *n*th layer of the tetrahedron:

What we really want is a formula for the middle numbers, or internal spheres. The internal spheres are the spheres that are not on the outer layer.

We know the formula for the sum of the whole layer, and we know the formula for any of the three ‘sides’ of a layer.

The internal numbers would be the sum of the entire layer minus three times the sum of a side. To cover up for the three 1’s we’ve counted more than once, we add 3 to the result.

The formula for the sum of the internal numbers on the *n*th layer:

The problem asks for the sum of the first to the thirteenth layer. Well, the first three layers don’t have any internal spheres so we start from the forth:

We can use the sum of geometric series formula, or more easily calculate it by brute force. The answer would be **772626**.

### Problem 2

(From the 2008 Canadian Math Olympiad)

In quadrilateral , splits the quadrilateral into two parts of equal area. also splits the quadrilateral into two parts of equal area. is the longest side of the quadrilateral.

Prove that bisects .

#### Solution

I’m going to use the equal sign to mean two figures with equal areas, not two figures that are congruent.

The triangles and are both half of , so their areas are equal. Then .

can be split into , and similarly can be split into .

We know that . By subtraction, .

The triangles and share a common base, . Because their areas are equal, the perpendicular from to is equal to the perpendicular from to . Therefore .

Now we draw a line through , parallel to and . We extend and to meet this line (and we draw a couple of additional lines):

Because , .

Here .

We can write as . Since , .

It follows that , and .

Similarly, . Then , and .

Because and , the result follows that .

**Problem 3**

(From the 2010 USAMO)

This problem actually appears in both the junior and senior USAMO taken about a week ago from time of writing. It’s problem 4 in the senior and problem 6 in the junior.

is a right triangle. bisects , and bisects .

Prove that the lines cannot all have integer lengths.

#### Solution

I was really surprised at how short and simple the solution can be for this problem.

Because , and because of the angle bisectors.

Then .

First we apply the pythagorean theorem:

We then apply the Law of Cosines:

Combining the two, and substituting for :

This is absurd, if *AB*, *AC*, *IB*, and *IC* are all integers and nonzero.