Subfield: Geometry


Question: Given a triangle $A B C$, let $H$ and $O$ be its orthocentre and circumcentre, respectively. Let $K$ be the midpoint of the line segment $A H$. Let further $\ell$ be a line through $O$, and let $P$ and $Q$ be the orthogonal projections of $B$ and $C$ onto $\ell$, respectively. Prove that $K P+K Q \geq B C$.

Solution:
Fix the origin at $O$ and the real axis along $\ell$. A lower case letter denotes the complex coordinate of the corresponding point in the configuration. For convenience, let $|a|=|b|=|c|=1$.



Clearly, $k=a+\frac{1}{2}(b+c), p=a+\frac{1}{2}\left(b+\frac{1}{b}\right)$ and $q=a+\frac{1}{2}\left(c+\frac{1}{c}\right)$.



Then $|k-p|=\left|a+\frac{1}{2}\left(c-\frac{1}{b}\right)\right|=\frac{1}{2}|2 a b+b c-1|$, since $|b|=1$.



Similarly, $|k-q|=\frac{1}{2}|2 a c+b c-1|$, so, since $|a|=1$,



$$

\begin{gathered}

|k-p|+|k-q|=\frac{1}{2}|2 a b+b c-1|+\frac{1}{2}|2 a c+b c-1| \\

\geq \frac{1}{2}|2 a(b-c)|=|b-c|,

\end{gathered}

$$



as required.
Let $M$ be the midpoint of $B C$, and let $R$ be the projection of $M$ onto $\ell$. In other words, $R$ is the midpoint of $P Q$. Since $\angle B P O=\angle B M O=$ $90^{\circ}$, the points $B, P, O$, and $M$ are concyclic, so $\angle(O M, O B)=\angle(P M, P B)=\angle(P M, M R)$, so the right triangles $M R P$ and $O M B$ are similar and have different orientation. Similarly, the triangles $M R Q$ and $O M C$ are similar and have different orientation, hence so are the triangles $O B C$ and $M P Q$.



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Recall that $\overrightarrow{A H}=2 \overrightarrow{O M}$, so $\overrightarrow{O M}=\overrightarrow{A K}$. Hence $A O M K$ is a parallelogram, so $M K=O A=O B=$ $O C$.



Consider the rotation through $\angle(\overrightarrow{O C}, \overrightarrow{O B})$ about $M$. It maps $P$ to $Q$; let it map $K$ to some point $L$. Then $M K=M L=O B=O C$ and $\angle L M K=\angle B O C$, so the triangles $O B C$ and $M K L$ are congruent. Hence $B C=K L \leq K Q+L Q=$ $K Q+K P$, as required.
Let $\alpha=\angle(P B, B C)=\angle(Q C, B C)$. Since $P$ lies on the circle of diameter $O B$, $\angle(O P, O M)=\alpha$. Since also $Q$ lies on the circle of diameter $O C$, it immediately follows that $M P=M Q=R \sin \alpha$ by sine theorem in triangles $\triangle O P M$ and $\triangle O Q M$.



Because $P Q$ is the projection of $B C$ on line $\ell$, it follows that $P Q=B C \sin \alpha$. Just like in the first solution, $K M=A O=R$ (the circumradius of triangle $\triangle A B C$ ).



Now apply Ptolemy's inequality for the quadrilateral $K P M Q: K P \cdot M Q+K Q \cdot M P \geq P Q \cdot K M$, and now substitute the relations from above, leading to



$$

R \sin \alpha(K P+K Q) \geq R \sin \alpha \cdot B C

$$



which is precisely the conclusion whenever $\sin \alpha \neq 0$. The case when $\sin \alpha=0$ can be treated either directly, or via a limit argument.
Denote by $R$ and $O$ the circumradius and the circumcentre of triangle $A B C$, respectively. As in Solution 1, we see that $M K=R$.



Assume now that $\ell$ is fixed, while $A$ moves along the fixed circle $(A B C)$. Then $K$ will move along a cricle centred at $M$ with radius $R$. We must show that for each point $K$ on this circle we have $B C \leq K P+K Q$. In doing so, we prove that the afore-mentioned circle contains an ellipse with foci at $Q$ and $P$ with distance $B C$.



Let $S$ be the foot of the perpendicular from $M$ to $P Q$, it is easy to verify that $S$ is the center of the ellipse. We shall then consider it as the origin. Let $u=\frac{B C}{2}$ and $t=\frac{P Q}{2}$; notice that $u$ is the major semi-axis of the ellipse and $\sqrt{u^{2}-t^{2}}$ is the minor one. Assume $X(x, y)$ is a point on this ellipse. We now need to prove $M X \leq R$.



Since $X$ is on the ellipse, we can write $(x, y)=$ $\left(u \cos \theta, \sqrt{u^{2}-t^{2}} \sin \theta\right)$, for some $\theta \in(0,2 \pi)$. Since $M X^{2}=x^{2}+(y+M S)^{2}$, we can expand and obtain



$M X^{2}=u^{2}+M S^{2}-t^{2} \cdot \sin ^{2} \theta+2 M S \cdot \sqrt{u^{2}-t^{2}} \cdot \sin \theta$.



Add and subtract $M S^{2}\left(u^{2}-t^{2}\right) / t^{2}$ in order to obtain a square on the right hand side: $M X^{2}=u^{2}+$ $M S^{2}+\frac{M S^{2}\left(u^{2}-t^{2}\right)}{t^{2}}-\left(t \sin \theta-\frac{M S \sqrt{u^{2}-t^{2}}}{t}\right)^{2}$. It now suffices to show that $u^{2}+M S^{2}+\frac{M S^{2}\left(u^{2}-t^{2}\right)}{t^{2}}=$ $R^{2}$, since then it would immediately follow that $M X^{2} \leq R^{2}$.



Applying Pythagorean theorem in triangles $O B M$ and $O S M$, we obtain $R^{2}=u^{2}+O M^{2}$ and $O M^{2}=M S^{2}+O S^{2}$, so it remains to prove that $O S^{2}=\frac{M S^{2}\left(u^{2}-t^{2}\right)}{t^{2}}$. Let $\alpha=\angle(O P, B M)$, then $O S / M S=\tan \alpha$ and $t / u=\cos \alpha$, so $O S^{2}=$ $M S^{2} \tan ^{2} \alpha=M S^{2}\left(\frac{1-\cos ^{2} \alpha}{\cos ^{2} \alpha}\right)=M S^{2} \cdot \frac{u^{2}-t^{2}}{t^{2}}$, which is the desired result.