## 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.

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

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