2 parallelograms on consecutive sides of ABCD

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2 parallelograms on consecutive sides of ABCD

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Source: IGO Elementary 2016 3

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#1 h Jul 22, 2018, 11:51 AM • 1 Y

Y by Adventure10

Suppose that $ABCD$ is a convex quadrilateral with no parallel sides. Make a parallelogram on each two consecutive sides. Show that among these $4$ new points, there is only one point inside the quadrilateral $ABCD$ .

by Morteza Saghafian

This post has been edited 1 time. Last edited by parmenides51, Jul 22, 2018, 11:52 AM
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TheCoolDinosuar

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TheCoolDinosuar

#2 h Oct 14, 2022, 8:17 PM • 1 Y

Y by Mango247

Let the reflections of $A$ across $BD$ be $A'$ , and similar for $B',C',D'$ so that we get our parallelograms. Finally, let $P=AC\cap BD$ .
We first show that there cannot be more than one point outside $ABCD$ , on the assumption that there is at least $1$ point outside. Assume WLOG that $B'$ is outside $ABCD$ . Then $\angle B'AC<\angle DAC$ and $\angle B'CA<\angle DCA$ , so
$\begin{align*} \angle BCA&<\angle DAC\\ \angle BAC&<\angle DCA. \end{align*}$ Clearly $D$ cannot satisfy its respective angle inequalities. Assume for contradiction that $A$ satisfies its angle inequalities. That is,
$\begin{align*} \angle ABD&<\angle BDC\\ \angle ADB&<\angle DBC. \end{align*}$ Then $\angle PAB<\angle PCD$ and $\angle PBD<\angle PDC$ . However, this would contradict the fact that $\angle APB=\angle DPC$ . So we cannot have more than $1$ of $A',B',C',D'$ inside $ABCD$ .

Now we show that at least of one $A',B',C',D'$ is inside $ABCD$ . WLOG we can say that this implies
$\begin{align} \angle BAC&<\angle ACD\\ \angle CAD&<\angle BCA\\ \angle ABD&<\angle BDC\\ \angle CBD&<\angle BDA. \end{align}$ However, taking (1) and (3) gives the same contradiction again. Therefore, exactly one of $A',B',C',D'$ lies inside $ABCD$ .

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