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$ABC$ is a right triangle with $m\angle C=90^{\circ}$. $AC=10\text{ units}^2$ and $BC=24\text{ units}^2$. Point $P$ is located inside $ABC$ such that the distance from $P$ to $AB$ is twice the distance from $P$ to $AC$, and the distance from $P$ to $AC$ is twice the distance from $P$ to $BC$. In $\text{units}^2$, what is the distance from $P$ to $AB$?

 Feb 4, 2021
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Let $AC\to P = x$. Call $P\perp AC=K$. Call $P\perp AB = K_1$. Call $P\perp BC=K_2$. We know that $AK=AK_1=10-2x$. We know that $BK_1=BK_2=24-x$. Therefore, $AK_1+BK_1=26\implies 10-2x+24-x=26\implies 34-3x=26\implies x=\frac{8}{3}$. And we want $2x=\boxed{\frac{16}{3}}$. 

 Feb 4, 2021

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