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} $.

QED.

--------------------

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.

2 comments:

  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 :)

    ReplyDelete
  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.

    ReplyDelete