Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.3 AIME Find the answer Italy

Problem:

Let $ABC$ be a triangle and let $A'$ be the reflection of $A$ with respect to $BC$ ; let $DAA'$ be similar to $ABC$ and let $D'$ be the reflection of $D$ with respect to $AA'$ . Knowing that the product of the areas of the quadrilaterals $ABA'C$ and $ADA'D'$ is $16$ , can we say that $AA'$ ...

Pick one

Solution

Solution:

The answer is (D). Let $H$ be the foot of the altitude from $A$ to $BC$ , let $K$ be the foot of the altitude from $D$ to $AA'$ ; the product of the areas of the indicated quadrilaterals is $(AH \cdot BC)(DK \cdot AA') = 16$ . We have the following equalities: $AA' = 2AH$ by construction and $AA' \cdot AH = BC \cdot DK$ by similarity between $ABC$ and $DAA'$ . Substituting the second into the product we get $AH \cdot AA' \cdot AH \cdot AA' = 16$ , that is (using the first equality) $AA'^4 = 4 \cdot 16 = 64$ , from which $AA' = 2\sqrt{2}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.