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China Hong Kong Mathematical Olympiad

Geometry Difficulty 6.1 National Olympiad Prove it Hong Kong

Given △ABC\triangle ABC with CA>BC>ABCA > BC > AB, let OO and HH be the circumcentre and orthocentre of △ABC\triangle ABC respectively. Denote by DD and EE the midpoints of arcs AB^\widehat{AB} and AC^\widehat{AC} of the circumcircle of △ABC\triangle ABC not containing the opposite vertices. Let D′D' be the reflection of DD in side ABAB and E′E' be the reflection of EE in side ACAC. Prove that O,H,D′,E′O, H, D', E' are concyclic if and only if A,D′,E′A, D', E' are collinear.

Solution

We claim that both statements are equivalent to ∠A=60∘\angle A = 60^\circ.

Firstly,
A,D′,E′A, D', E' are collinear
⇔∠BAD′+∠CAE′=∠BAC \Leftrightarrow \angle BAD' + \angle CAE' = \angle BAC
⇔∠BAD+∠CAE=∠BAC \Leftrightarrow \angle BAD + \angle CAE = \angle BAC
⇔C2+B2=A \Leftrightarrow \frac{C}{2} + \frac{B}{2} = A
⇔90∘−A2=A \Leftrightarrow 90^\circ - \frac{A}{2} = A
⇔A=60∘. \Leftrightarrow A = 60^\circ.

Secondly, it is well-known that the reflection of HH in ABAB lies on (ABC)(ABC). Therefore, A,B,H,D′A, B, H, D' are concyclic. Also, note that D,D′,OD, D', O are collinear since all of them lie on the perpendicular bisector of ABAB. It follows that
∠DD′H=∠DD′B+∠BD′H=(90∘−∠ABD′)+∠BAH=(90∘−∠ABD)+∠BAH=(90∘−C2)+90∘−B=A+C2. \begin{align*} \angle DD'H &= \angle DD'B + \angle BD'H \\ &= (90^\circ - \angle ABD') + \angle BAH \\ &= (90^\circ - \angle ABD) + \angle BAH \\ &= \left(90^\circ - \frac{C}{2}\right) + 90^\circ - B \\ &= A + \frac{C}{2}. \end{align*}

Similarly, C,A,H,E′C, A, H, E' are concyclic, and we have
∠OE′H=∠CE′H−∠CE′E=(180∘−∠CAH)−(90∘−∠ACE)=(90∘+C)−(90∘−B2)=C+B2. \begin{align*} \angle OE'H &= \angle CE'H - \angle CE'E \\ &= (180^\circ - \angle CAH) - (90^\circ - \angle ACE) \\ &= (90^\circ + C) - \left(90^\circ - \frac{B}{2}\right) \\ &= C + \frac{B}{2}. \end{align*}
Now,
O,H,D′,E′O, H, D', E' are concyclic
∠DD′H=∠OE′HA+C2=C+B2A=90∘−A2A=60∘. \begin{align*} \angle DD'H &= \angle OE'H \\ A + \frac{C}{2} &= C + \frac{B}{2} \\ A &= 90^\circ - \frac{A}{2} \\ A &= 60^\circ. \end{align*}

Therefore, the two statements are equivalent.

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