Revisiting yesterday’s challenging geometry problem

Yesterday Problem #18 from the 2008 AMC 10a gave my older son a bit of trouble. Unfortunately we were a little pressed for time so we didn’t get a chance to go through the problem in detail until today.

The problem is easy to state:

A right triangle has perimeter 32 and area 20. What is the length of its hypotenuse?

My son’s original approach was algebraic. It turned out that bringing the ideas from algebra and geometry together in this problem was difficult for him, so I started off today’s talk about the problem by reviewing the algebraic approach.


With the algebraic approach hopefully a little more clear than it was yesterday I moved on to a geometric approach to this problem. That approach involves a connection between area and perimeter of triangles that you learn from studying some properties of a triangle’s inscribed circle.


After talking about the general ideas you learn about area and perimeter from studying the triangle’s inscribed circle, we applied them to the specific triangle in the problem. It turns out that for right triangles the inscribed circle has a few extra properties that are really helpful in solving this problem:


Finally, since the last two pieces of this project had lots of different ideas, we went back to the beginning to make sure that they at least made a little sense:


I really like this problem. The algebraic approach to the solution is instructive. Although the algebra is a bit difficult, the exercise of bringing algebra and geometry together seems useful. The geometric approach provides a nice opportunity to introduce / review some beautiful ideas from geometry. Happy to have had the opportunity to go through this problem today in a non-rushed way.


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