Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.7 AIME, harder Prove it Argentina

Let $ABC$ be an acute-angled and scalene triangle. Consider the altitudes $BE$ and $CD$ , that intersect in $H$ . The bisector of the angle $B\hat{A}C$ intersects the altitudes $BE$ and $CD$ in $P$ and $Q$ respectively. Let $T$ be the orthocenter of the triangle $HPQ$ . Prove that the triangles $TDA$ and $TEA$ have the same area.

Solution

As $T$ is the orthocenter of the triangle $HPQ$ , we have that $QT$ is perpendicular to $BE$ ; then, since $BE$ is perpendicular to $AC$ , it follows that $QT$ is parallel to $AC$ . Similarly, $TP$ is parallel to $AB$ . Assume $QT$ intersects $AB$ at the point $F$ and $TP$ intersects $AC$ at a point $G$ .

Since $AP$ is the bisector of the angle $B̂AC$ , the triangles $AFQ$ and $AGP$ are isosceles and similar; then:
$\frac{AQ}{AF} = \frac{AP}{AG} \qquad (5)$

On the other hand, as $ÂBP = ÂCQ$ , the triangles $ABP$ and $ACQ$ are also similar, so,
$\frac{AQ}{AC} = \frac{AP}{AB} \qquad (6)$

From (5) and (6), we obtain that $\frac{AF}{AG} = \frac{AC}{AB}$ . Since the triangles $CAD$ and $BAE$ are similar, then $\frac{AC}{AB} = \frac{AD}{AE}$ . Therefore,
$\frac{AF}{AG} = \frac{AD}{AE}.$

Finally, recalling that $AFTG$ is a parallelogram, we conclude that:
$\text{area}(TDA) = \frac{AD}{AF} \cdot \text{area}(TFA) = \frac{AE}{AG} \cdot \text{area}(TGA) = \text{area}(TAE).$

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