Altitudes of a convex pentagon.

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Altitudes of a convex pentagon.

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Source: 2017 Thailand TST Day 1 Problem 1

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#1 h Jul 26, 2017, 5:22 PM • 2 Y

Y by Mathmick51, Adventure10

In a convex pentagon, five altitudes are drawn from each vertex to its opposite side. Prove that if four of them meet at a single point, then the fifth altitude must pass through that point.

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#2 h Jul 27, 2017, 9:50 PM • 3 Y

Y by FlakeLCR, Mathmick51, Adventure10

Nice.

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#3 h Nov 21, 2021, 2:01 PM • 1 Y

Y by centslordm

Let the pentagon be $ABCDE$ with the feet of the altitudes $A',\ldots,E'$ such that $\overline{BB'},\ldots,\overline{EE'}$ concur at $P$ , so we wish to show that $P \in \overline{AA'}$ .
Clearly, $BB'DD',CC'EE',EE'BB'$ are cyclic, so
$CP\cdot CP'=EP\cdot E'P=BP\cdot B'P=DP\cdot D'P,$ hence $CC'DD'$ is cyclic too. Further, since $AA'DD'$ and $AC'PD'$ are cyclic, we have
$\angle D'AA'=\angle D'DA'=\angle D'DC=\angle D'C'C=\angle D'C'P.$ Letting $\overline{C'P} \cap \overline{AA'}=Q$ , this means that $AC'QD'$ is cyclic, but since $AC'PD'$ is also cyclic we require $P=Q$ ( $P,Q \neq C$ both lie on $\overline{C'P}$ ), so $P \in \overline{AA'}$ as desired.

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#4 h May 6, 2022, 2:56 PM

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Let the five vertices of the pentagon be represented by the complex numbers $a$ , $b$ , $c$ , $d$ , and $e$ , respectively, and let the point of concurrency be the origin. Then, the given condition rewrites as follows:
$\begin{align*} (a - 0) (\overline{c} - \overline{d}) + (\overline{a} - \overline{0}) (c - d) = a (\overline{c} - \overline{d}) + \overline{a} (c - d) = 0 \\ (b - 0) (\overline{d} - \overline{e}) + (\overline{b} - \overline{0}) (d - e) = b (\overline{d} - \overline{e}) + \overline{b} (d - e) = 0 \\ (c - 0) (\overline{e} - \overline{a}) + (\overline{c} - \overline{0}) (e - a) = c (\overline{e} - \overline{a}) + \overline{c} (e - a) = 0 \\ (d - 0) (\overline{a} - \overline{b}) + (\overline{d} - \overline{0}) (a - b) = d (\overline{a} - \overline{b}) + \overline{d} (a - b) = 0 \end{align*}$ Summing the four equations yields a mass cancellation of most of the terms and leaves us with $-b\overline{e} - \overline{b}e + c\overline{e} + \overline{c} e = 0$ , which after negating translates to $e (\overline{b} - \overline{c}) + \overline{e} (b - c) = 0$ , the requested result. $\blacksquare$

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