Competition problem — Geometry — Maths Olympiad Prep

Library / /733 of 860

Geometry Difficulty 5.5 AIME, harder Find the answer

Let $ABC$ be a triangle with $AB=5, BC=4$ and $AC=3$ . Let $\mathcal{P}$ and $\mathcal{Q}$ be squares inside $ABC$ with disjoint interiors such that they both have one side lying on $AB$ . Also, the two squares each have an edge lying on a common line perpendicular to $AB$ , and $\mathcal{P}$ has one vertex on $AC$ and $\mathcal{Q}$ has one vertex on $BC$ . Determine the minimum value of the sum of the areas of the two squares.

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Let the side lengths of $\mathcal{P}$ and $\mathcal{Q}$ be $a$ and $b$ , respectively. Label two of the vertices of $\mathcal{P}$ as $D$ and $E$ so that $D$ lies on $AB$ and $E$ lies on $AC$ , and so that $DE$ is perpendicular to $AB$ . The triangle $ADE$ is similar to $ACB$ . So $AD=\frac{3}{4}a$ . Using similar arguments, we find that $\frac{3a}{4}+a+b+\frac{4b}{3}=AB=5$ so $\frac{a}{4}+\frac{b}{3}=\frac{5}{7}$ Using Cauchy-Schwarz inequality, we get $\left(a^{2}+b^{2}\right)\left(\frac{1}{4^{2}}+\frac{1}{3^{2}}\right) \geq\left(\frac{a}{4}+\frac{b}{3}\right)^{2}=\frac{25}{49}$ It follows that $a^{2}+b^{2} \geq \frac{144}{49}$ Equality occurs at $a=\frac{36}{35}$ and $b=\frac{48}{35}$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.