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Source: Dutch NMO 2006 p2

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#1 h Sep 21, 2019, 8:50 AM • 1 Y

Y by Adventure10

Given is a acute angled triangle $ABC$ . The lengths of the altitudes from $A, B$ and $C$ are successively $h_A, h_B$ and $h_C$ . Inside the triangle is a point $P$ . The distance from $P$ to $BC$ is $1/3 h_A$ and the distance from $P$ to $AC$ is $1/4 h_B$ . Express the distance from $P$ to $AB$ in terms of $h_C$ .

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#2 h Sep 23, 2019, 8:21 PM • 2 Y

Y by Adventure10, Mango247

The area of the triangle $ABC$ is
$A_{\triangle{ABC}}=A_{\triangle{PBC}}+A_{\triangle{PAC}}+A_{\triangle{PAB}}$ .
Let be $h_1,h_2,h_3$ the distances from $P$ to $BC, AC$ , respectively $AB$ .
$\dfrac{A_{\triangle{PBC}}}{A_{\triangle{ABC}}}=\dfrac{h_1}{h_A}=\dfrac{1}{3}$ .
$\dfrac{A_{\triangle{PAC}}}{A_{\triangle{ABC}}}=\dfrac{h_2}{h_B}=\dfrac{1}{4}$ .
$\dfrac{A_{\triangle{PBC}}}{A_{\triangle{ABC}}}+\dfrac{A_{\triangle{PAC}}}{A_{\triangle{ABC}}}+\dfrac{A_{\triangle{PAB}}}{A_{\triangle{ABC}}}=1\Longrightarrow$
$\Longrightarrow \dfrac{h_3}{h_C}=\dfrac{A_{\triangle{PAB}}}{A_{\triangle{ABC}}}=1-\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{5}{12}\Longrightarrow$
$\Longrightarrow h_3=\dfrac{5}{12}\cdot h_C$ .

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