Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.2 AIME, harder Find the answer Italy

Problem:

Let $ABC$ be an isosceles triangle with base $BC = 10$ and $AB = AC$ . On its two slanted sides, construct externally two other isosceles triangles $DAB$ and $EAC$ , both similar to $ABC$ , with $DA = DB$ and $EA = EC$ . Given that $DE = 45$ , find the length of $AB$ .

Pick one

Solution

Solution:

The answer is (A). Since triangles $ABC$ , $DAB$ , $EAC$ are similar, their respective base angles are congruent; this means that $\widehat{DAB} = \widehat{ABC} = \widehat{ACB} = \widehat{CAE}$ . Now we can compute the measure of angle $\widehat{DAE} = \widehat{DAB} + \widehat{BAC} + \widehat{CAE} = \widehat{ABC} + \widehat{BAC} + \widehat{ACB} = 180^\circ$ , that is, $D, A, E$ are collinear.

Triangles $DAB$ and $EAC$ are similar, and their bases $AB$ , $AC$ have the same length, so they are also congruent and $DA = EA$ ; this, combined with the collinearity of $D, A, E$ , leads to $DA = \frac{DE}{2}$ .

From the similarity of $DAB$ and $ABC$ we know that their sides are in proportion, so $\frac{DA}{AB} = \frac{AB}{BC}$ , that is $AB^2 = DA \cdot BC$ and therefore
$AB = \sqrt{DA \cdot BC} = \sqrt{\frac{DE \cdot BC}{2}} = \sqrt{\frac{45 \cdot 10}{2}} = 15$

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