Midpoints of altitudes

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Midpoints of altitudes

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Source: RMO 2003, Grade 9, Problem 3

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#1 h Oct 23, 2008, 9:22 AM • 2 Y

Y by Adventure10, Mango247

Prove that the midpoints of the altitudes of a triangle are collinear if and only if the triangle is right.

Dorin Popovici

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joh

#2 h Oct 26, 2008, 7:22 AM • 2 Y

Y by Adventure10, Mango247

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#3 h Oct 26, 2008, 9:01 AM • 3 Y

Y by galav, Adventure10, Mango247

The middles $H_1,H_2,H_3$ of the altitudes of the triangle are on the sides of the median triangle.It isn't possible that 2midpoints to be the same.The only posibility to be on the same line is that two of them to form a segment composed by the midpoints of two sides and the third one to be in the interior of this segment.This can hapen only if the triangle has a $90$ degree angle.

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#4 h Mar 11, 2024, 4:31 PM

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Here's the complex numbers solution from the OTIS walkthrough.

Let $(ABC)$ be the unit circle, as always, and note that the midpoints of the altitudes are $x = \frac 12\left(a+\frac 12\left(a+b+c-\frac{bc}a\right)\right)$ and cyclic permutations. So it suffices to show that $0 = \begin{vmatrix} 3a+b+c-\frac{bc}a & \frac 3a+\frac 1b+\frac 1c - \frac a{bc} & 1 \\ 3b+a+c-\frac{ac}b & \frac 3b+\frac 1a+\frac 1c-\frac b{ac} & 1 \\ 3c+a+b-\frac{ab}c & \frac 3c+\frac 1a+\frac 1b-\frac c{ab} & 1 \end{vmatrix} = \begin{vmatrix} 2a-\frac{bc}a & \frac 2a - \frac a{bc} & 1 \\ 2b-\frac{ac}b & \frac 2b - \frac b{ac} & 1 \\ 2c-\frac{ab}c & \frac 2c - \frac c{ab} & 1 \end{vmatrix}$ if and only if $0 \in \{a+b, b+c, c+a\}$ . It turns out that this factors as $\sum_{\mathrm{cyc}} \left(2a-\frac{bc}a\right)\left(\frac 2b-\frac b{ac}\right) - \left(\frac 2a-\frac a{bc}\right)\left(2b-\frac{ac}b\right) = -\frac{(a-b)(b-c)(c-a)(a+b)(b+c)(c+a)}{a^2b^2c^2}$ as needed.

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