NZMO Round One

Geometry Difficulty 5.1 AIME, harder Prove it New Zealand

Problem:
Let $ABCD$ be a parallelogram, and let $P$ be a point on the side $AB$ . Let the line through $P$ parallel to $BC$ intersect the diagonal $AC$ at point $Q$ . Prove that
$|DAQ|^{2} = |PAQ| \times |BCD|,$
where $|XYZ|$ denotes the area of triangle $XYZ$ .

Solution

Solution:

Since $PQ$ is parallel to $BC$ , there is a dilation (centred at $A$ ) of factor $AB / AP$ that sends triangle $APQ$ to triangle $ABC$ . Thus
$|PAQ| = |ABC| \times \left(\frac{AP}{AB}\right)^{2}.$
Since $AD$ is parallel to $BC$ , triangle $ABC$ has the same height as triangle $BCD$ . Triangles $ABC$ and $BCD$ also have a common base, so $|ABC| = |BCD|$ . Similarly $|DAQ| = |DAP|$ (because $PQ \parallel AD$ and $AD$ is a common base).
$\therefore |DAP| = |DAB| \times \frac{AP}{AB}$
$\qquad = |BCD| \times \frac{AP}{AB}$
$\qquad |DAP|^{2} = |BCD| \times |BCD| \times \left(\frac{AP}{AB}\right)^{2}$
$\qquad = |BCD| \times |PAQ| \qquad \mathrm{(diagonal~splits~parallelograms~in~half)}$
$\qquad = |BCD| \times |PAQ| \qquad \mathrm{(because~}|PAQ| = |ABC|(AP / AB)^{2})$
as required.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.