Let $ABC$ be a triangle inscribed in the circle $(O)$ and $P$ is a point inside the triangle $ABC$. Let $D$ be a point on $(O)$ such that $AD \perp AP$. The line $CD$ cuts the perpendicular bisector of $BC$ at $M$. The line $AD$ cuts the line passing through $B$ and is perpendicular to $BP$ at $Q$. Let $N$ be the reflection of $Q$ through $M$. Prove that $CN \perp CP$.
Problem
Source: 2016 Saudi Arabia IMO TST , level 4, II p2
Tags: geometry, circle, perpendicular, circumcircle
Diamond-jumper76
13.01.2025 00:20
Let $L$ be the midpoint of $PN$, and let $K$ be the foot of $P$ on $MN$. Lastly, define $R$ to be the intersection of $ML$ and $BP$. Claim 1: $BCMK$ is cyclic.
\[
\angle BSQ = \angle BAQ = \angle BCM
\]Note that $KP$ is an angle bisector of $\angle BKC$.
Claim 2: $L$ lies on the circle of $BCMK$.
\[
\angle(ML, MC) = \angle (QP, MC) = \angle (QP, MQ) + \angle (MQ, MC) = \angle (PB, SB) + \angle (SB, SC)
\]\[
= \angle (SB, SC) - \angle (SB, PB) = \angle (RB, RC).
\]
Claim 3: $\triangle LMC$ is similar to $\triangle PBC$.
We already have a common angle; we just need to prove the ratios:
\[
\frac{PB}{LM} = 2 \cdot \frac{PB}{PQ} = 2 \sin(\angle BQP) = 2 \sin(\angle BKP) = 2 \sin\left(\frac{1}{2} \angle BKC\right) = 2 \sin\left(\frac{1}{2} \angle BMC\right).
\]\[
= 2 \cdot \left(\frac{\frac{BC}{2}}{AC}\right).
\]
Hence, $CRLP$ is cyclic and From the spiral similarity, we deduce $LC = LP$ since $MC = MB$. However, $LP = LN = LK$, so \[ \angle NCP = 90^\circ. \]
Double07
13.01.2025 01:39
Really really nice with complex bashing: Take $(O)$ the unit circle and consider $E$ and $F$ on $(O)$ such that $BE\perp PB$ and $CF\perp PC$ $AD\perp AP\iff \dfrac{a-p}{\overline{a}-\overline{p}}=-\dfrac{a-d}{\overline{a}-\overline{d}}=ad\implies d=\dfrac{a-p}{1-a\overline{p}}$. Similarly, $e=\dfrac{b-p}{1-b\overline{p}}$ and $f=\dfrac{c-p}{1-c\overline{p}}$. Suppose the bisector of segment $[BC]$ intersects $(O)$ at $X$ and $Y$, which are the midpoints of arcs $\widehat{BC}$, so $x+y=0$ and $xy=-bc$. $M=CD\cap XY\implies m=\dfrac{cd(x+y)-xy(c+d)}{cd-xy}=\dfrac{bc(c+d)}{cd+bc}=\dfrac{bc+bd}{b+d}=\dfrac{bc-abc\overline{p}+ab-bp}{a+b-p-ab\overline{p}}$ $Q=AD\cap BE\implies q=\dfrac{ad(b+e)-be(a+d)}{ad-be}=\dfrac{\dfrac{a^2-ap}{1-a\overline{p}}\cdot\dfrac{2b-p-b^2\overline{p}}{1-b\overline{p}}-\dfrac{b^2-bp}{1-b\overline{p}}\cdot\dfrac{2a-p-a^2\overline{p}}{1-a\overline{p}}}{\dfrac{a^2-ap}{1-a\overline{p}}-\dfrac{b^2-bp}{1-b\overline{p}}}=...=$ $=\dfrac{(a-b)(2ab-ap-bp+p^2-abp\overline{p})}{(a-b)(a+b-p-ab\overline{p})}=\dfrac{2ab-ap-bp+p^2-abp\overline{p}}{a+b-p-ab\overline{p}}$ $M$ is the midpoint of $[QN]\implies n=2m-q=\dfrac{2bc-2abc\overline{p}-bp+ap-p^2+abp\overline{p}}{a+b-p-ab\overline{p}}$. Now we're just left to prove that $N$ is on chord $CF$, which is equivallent to proving that $n=c+f-cf\overline{n}$. $c+f-cf\overline{n}=\dfrac{2c-p-c^2\overline{p}}{1-c\overline{p}}-\dfrac{c(c-p)}{1-c\overline{p}}\cdot\dfrac{\dfrac{2}{bc}-\dfrac{2p}{abc}-\dfrac{\overline{p}}{b}+\dfrac{\overline{p}}{a}-\overline{p^2}+\dfrac{p\overline{p}}{ab}}{\dfrac{a+b-ab\overline{p}-p}{ab}}=\dfrac{2c-p-c^2\overline{p}}{1-c\overline{p}}+\dfrac{p-c}{1-c\overline{p}}\cdot\dfrac{2a-2p-ac\overline{p}+bc\overline{p}-abc\overline{p^2}+cp\overline{p}}{a+b-p-ab\overline{p}}=$ $=\dfrac{(2c-p-c^2\overline{p})(a+b-p-ab\overline{p})+(p-c)(2a-2p-ac\overline{p}+bc\overline{p}-abc\overline{p^2}+cp\overline{p})}{(1-c\overline{p})(a+b-p-ab\overline{p})}=$ $=\dfrac{2bc-2abc\overline{p}-bp+abp\overline{p}-2bc^2\overline{p}+2abc^2\overline{p^2}+ap-p^2-acp\overline{p}+bcp\overline{p}-abcp\overline{p^2}+cp^2\overline{p}}{(1-c\overline{p})(a+b-p-ab\overline{p})}=$ $=\dfrac{(1-c\overline{p})(2bc-2abc\overline{p}-bp+abp\overline{p}+ap-p^2)}{(1-c\overline{p})(a+b-p-ab\overline{p})}=\dfrac{2bc-2abc\overline{p}-bp+abp\overline{p}+ap-p^2}{a+b-p-ab\overline{p}}=n$, so we're done.