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, 2022

Geometry Difficulty 4.8 AIME Prove it Bulgaria

In an acute triangle $ABC$ , point $M$ is the midpoint of $AB$ and $AH$ is the altitude. Let $CP$ be the perpendicular to the line $MH$ . If $AB = 21$ , $BH = 7$ and $BP = CP$ find the length of $AC$ .

Solution

Answer: $AC = \sqrt{473}$ .

Let $PE \perp BC$ and denote $\angle ABC = \beta$ . Since $HM$ is a median in the right triangle we have $\angle MHA = 90^\circ - \beta$ , $\angle PHC = \beta$ and $\angle PCH = 90^\circ - \beta$ . Therefore $\triangle ABH \sim \triangle PHE \sim \triangle CPH$ and thus
$\frac{7}{21} = \frac{BH}{AB} = \frac{EH}{HP} = \frac{PH}{CH}. \quad (1)$
If $EH = x$ then $PH = 3x$ , $CH = 9x$ , $CE = 9x - x = 8x$ and using that $BP = CP$ we obtain $BE = 8x$ and $BH = 7x$ , i.e. $x = 1$ . Finally: $AC = \sqrt{AB^2 - BH^2 + CH^2} = \sqrt{21^2 - 7^2 + 9^2} = \sqrt{473}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.