IMO 1996 SL G3

Origin: GBR | Category: Geometry

Problem

Let $\triangle ABC$ be an acute-angled triangle with $BC > CA$. Let $O$ be the circumcenter, $H$ its orthocenter, and $F$ the foot of its altitude $CH$. Let the perpendicular to $OF$ at $F$ meet the side $CA$ at $P$. Prove that

$\angle FHP = \angle BAC.$

Possible second part: What happens if $|BC| \leq |CA|$ (the triangle still being acute-angled)?

Solution

It is easy to see that $P$ lies on the segment $AC$. Let $E$ be the foot of the altitude $BH$ and $Y, Z$ the midpoints of $AC, AB$ respectively. Draw the perpendicular $HR$ to $FP$ ($R \in FP$). Since $Y$ is the circumcenter of $\triangle FCA$, we have

$\angle FYA = 180^{\circ} - 2\angle A.$

Also, $OFPY$ is cyclic; hence

$\angle OPF = \angle OYF = 2\angle A - 90^{\circ}.$

Next, $\triangle OZF$ and $\triangle HRF$ are similar, so

$\frac{OZ}{OF} = \frac{HR}{HF}.$

This leads to

$HR \cdot OF = HF \cdot OZ = 2 HF \cdot HC = \frac{1}{2} HE \cdot HB = HE \cdot OY \Rightarrow \frac{HR}{HE} = \frac{OY}{OF}.$

Moreover,

$\angle EHR = \angle FOY;$

hence the triangles $EHR$ and $FOY$ are similar. Consequently,

$\angle HPC = \angle HRE = \angle OYF = 2\angle A - 90^{\circ},$

and finally,

$\angle FHP = \angle HPC + \angle HCP = \angle A.$

(Figure: A, B, C, Y, Z, O, F, H, E, P, R)

Second solution. As before, $\angle HFY = 90^{\circ} - \angle A$, so it suffices to show that $HP \perp FY$. The points $O, F, P, Y$ lie on a circle, say $\Omega_1$ with center at the midpoint $Q$ of $OP$. Furthermore, the points $F, Y$ lie on the nine-point circle $\Omega$ of $\triangle ABC$ with center at the midpoint $N$ of $OH$. The segment $FY$ is the common chord of $\Omega_1$ and $\Omega$, from which we deduce that $NQ \perp FY$. However, $NQ \parallel HP$, and the result follows.

Third solution. Let $H'$ be the point symmetric to $H$ with respect to $AB$. Then $H'$ lies on the circumcircle of $\triangle ABC$. Let the line $FP$ meet the circumcircle at $U, V$ and meet $H'B$ at $P'$. Since $OF \perp UV$, $F$ is the midpoint of $UV$. By the butterfly theorem, $F$ is also the midpoint of $PP'$. Therefore

$\triangle H'FP' \cong \triangle FHP;$

hence

$\angle FHP = \angle FH'B = \angle A.$

Remark. It is possible to solve the problem using trigonometry. For example,

$\frac{FZ}{ZO} = \frac{FK}{KP} = \frac{\sin(A-B)}{\cos C},$

where $K$ is on $CF$ with $PK \perp CF$. Then

$\frac{CF}{KP} = \frac{\sin(A-B)}{\cos C} + \tan A,$

from which one obtains formulas for $KP$ and $KH$. Finally, we can calculate

$\tan \angle FHP = \frac{KP}{KH} = \cdots = \tan A.$

Second remark. Here is what happens when $BC \leq CA$. If $\angle A > 45^{\circ}$, then $\angle FHP = \angle A$. If $\angle A = 45^{\circ}$, the point $P$ escapes to infinity. If $\angle A < 45^{\circ}$, the point $P$ appears on the extension of $AC$ beyond $C$, and

$\angle FHP = 180^{\circ} - \angle A.$