perpendicular wanted, circumcircle and perpendicular related

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Source: 2016 Saudi Arabia IMO TST , level 4, II p2

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#1 h Jul 27, 2020, 9:47 PM

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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 $CN \perp CP$ .

This post has been edited 1 time. Last edited by parmenides51, Jul 27, 2020, 10:07 PM

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Diamond-jumper76

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#2 h Jan 12, 2025, 9:20 PM

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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.
Proof

Claim 2: $L$ lies on the circle of $BCMK$ .
Proof

Claim 3: $\triangle LMC$ is similar to $\triangle PBC$ .
Proof

Hence, $CRLP$ is cyclic and From the spiral similarity, we deduce $LC = LP$ since $MC = MB$ . However, $LP = LN = LK$ , so
$\angle NCP = 90^\circ.$

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#3 h Jan 12, 2025, 10:39 PM • 2 Y

Y by HotSinglesInYourArea, Calamarul

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.

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