Tuesday, August 8, 2006

Back On Track. Topic: Algebra/Geometry. Level: AMC/AIME.

Problem: Two isosceles triangles with side lengths $ x $, $ x $, $ a $ and $ x $, $ x $, $ b $ (with $ a \neq b $) have equal areas. Find $ x $ in terms of $ a $ and $ b $.

Solution: Drop the altitude to the base in each of the triangles and use the Pythagorean Theorem to find the heights to be

$ \sqrt{x^2-\frac{a^2}{4}} $ and $ \sqrt{x^2-\frac{b^2}{4}} $,

respectively. From here, we find the areas and equate them to get

$ \frac{a}{2}\sqrt{x^2-\frac{a^2}{4}} = \frac{b}{2} \sqrt{x^2-\frac{b^2}{4}} $.

Squaring and multiplying through by $ 4 $, we have

$ a^2\left(x^2-\frac{a^2}{4}\right) = b^2\left(x^2-\frac{b^2}{4}\right) $,

which we can solve for $ x $ to obtain

$ x = \frac{\sqrt{a^2+b^2}}{2} $.



Comment: Not a tough problem; just getting back on my feet after taking a long break from problem solving.


Practice Problem: Given two figures in a plane, show that there exists a line that bisects both of them.


  1. welcome back! good luck this year, you kept me motivated last year (my senior year, no fun math for me in college, bah.) with your daily postings :)

  2. Consider the set of lines, for a given figure F, that bisect that figure. We claim that they intersect at a common point P_{F}. Proof by obviosity! :P

    Simply find that point for both figures, and the line connecting them bisects both.