Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 8.2 Shortlist Find the answer

We know that the orthocenter reflects over the sides of the triangle on the circumcircle. Therefore the minimal distance $OD\plus{}HD$ equals $R$ . Obviously we can achieve this on all sides, so we assume that $D,E,F$ are the intersection points between $A',B',C'$ the reflections of $H$ across $BC,CA,AB$ respectively. All we have to prove is that $AD$ , $BE$ and $CF$ are concurrent.

In order to do that we need the ratios $\dfrac {BD}{DC}$ , $\dfrac {CE}{EA}$ and $\dfrac {AF}{FB}$ , and then we can apply Ceva's theorem.

We know that the triangle $ABC$ is acute, so $\angle BAH \equal{} 90^\circ\minus{} \angle B \equal{} \angle OAC$ , therefore $\angle HAO \equal{} |\angle A \minus{} 2(90^\circ \minus{}\angle B)| \equal{} |\angle B\minus{} \angle C|$ . In particular this means that $\angle OA'H \equal{} |\angle B\minus{}\angle C|$ . Since $\angle BA'A \equal{} \angle C$ and $\angle AA'C \equal{} \angle B$ , we have that $\angle BA'D \equal{} \angle B$ and $\angle DA'C \equal{} \angle C$ .

By the Sine theorem in the triangles $BA'D$ and $DA'C$ , we get
$\dfrac {BD}{DC} \equal{} \dfrac { \sin B }{\sin C }.$

Using the similar relationships for $\dfrac {CE}{EA}$ and $\dfrac {AF}{FB}$ we get that those three fractions multiply up to 1, and thus by Ceva's, the lines $AD, BE$ and $CF$ are concurrent.

Solution

To show that the lines $AD$ , $BE$ , and $CF$ are concurrent, we need to use Ceva's Theorem. According to Ceva's Theorem, for three lines $AD$ , $BE$ , $CF$ to be concurrent at a single point, it is required that:

$\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = 1.$

Let's find these ratios starting with $\frac{BD}{DC}$ :

1. **Finding $\frac{BD}{DC}$ :**

Since $D$ is the reflection of the orthocenter $H$ over side $BC$ , and given that the triangle $ABC$ is acute, we have the known angle equivalences:
$\angle BAH = 90^\circ - \angle B = \angle OAC.$

Therefore, the angle between the altitude $AH$ and the line $AO$ is:
$\angle HAO = |\angle A - 2 \cdot (90^\circ - \angle B)| = |\angle B - \angle C|.$

Now considering the angles at $A'$ (reflection of $H$ ), we know:
$\angle BA'A = \angle C, \quad \angle AA'C = \angle B.$

Consequently, the angles $\angle BA'D = \angle B$ and $\angle DA'C = \angle C$ .

Applying the Law of Sines in $\triangle BA'D$ and $\triangle DA'C$ :
$\frac{BD}{\sin \angle DA'C} = \frac{AD}{\sin \angle BA'D}, \quad \frac{DC}{\sin \angle BA'D} = \frac{AD}{\sin \angle DA'C}.$

Thus,
$\frac{BD}{DC} = \frac{\sin \angle B}{\sin \angle C}.$

2. **Finding $\frac{CE}{EA}$ :** (similar procedure for remaining reflections)

By reflecting the orthocenter $H$ over each of the sides $CA$ and $AB$ , we use similar reasoning and the Sine Rule in respective triangles to obtain:
$\frac{CE}{EA} = \frac{\sin \angle C}{\sin \angle A},$

and

3. **Finding $\frac{AF}{FB}$ :**

$\frac{AF}{FB} = \frac{\sin \angle A}{\sin \angle B}.$

Substituting these ratios into the product for Ceva's Theorem, we have:
$\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = \frac{\sin \angle B}{\sin \angle C} \cdot \frac{\sin \angle C}{\sin \angle A} \cdot \frac{\sin \angle A}{\sin \angle B} = 1.$

Thus, by Ceva's Theorem, the lines $AD$ , $BE$ , and $CF$ must be concurrent. This completes the proof.

The point of concurrency is known as the symmedian point or the Lemoine point of the triangle. Hence, we have shown that:
$\boxed{\text{The lines } AD, BE, \text{ and } CF \text{ are concurrent.}}$

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