Let $O$ be the center of circle $\Gamma$, and $A$, $B$ be two points on $\Gamma$ so that $O, A$ and $B$ are not collinear. Let $M$ be the midpoint of $AB$. Let $P$ and $Q$ be points on $OA$ and $OB$, respectively, so that $P \neq A$ and $P, M, Q$ are collinear. Let $X$ be the intersection of the line passing through $P$ and parallel to $AB$ and the line passing through $Q$ and parallel to $OM$. Let $Y$ be the intersection of the line passing through $X$ and parallel to $OA$ and the line passing through $B$ and orthogonal to $OX$. Prove that: if $X$ is on $\Gamma$, then $Y$ is also on $\Gamma$. Proposed by usjl
Problem
Source: 2023 Taiwan Mathematics Olympiad
Tags: Taiwan, geometry
Li4
08.02.2023 09:28
Note that $Y$ lies on $\Gamma$ if and only if $$AO - X = OX - B + 90^\circ \iff 3X = 2A + B + 90^\circ. $$Let $P'$ be the reflection point of $P$ with respect to $OM$. Then $$X(O, B; P', Q) = (O, B; P', Q) = M(O, B; P', P) = -1, $$i.e., $XP'$ is one of the angle bisectors of $\angle OXB$. Hence, $$XO + XB = 2XP' = 2AB \implies 3X = 2A + B + 90^\circ, $$as desired.
sn6dh
08.02.2023 13:29
Let $O=(0, 0), A=(-a, 1), B=(a, 1), M=(0, 1)$. $\because P\in OA,\ Q\in OB$, suppose $P=(-ab, b), Q=(ac, c)$. Since $M, P, Q$ are collinear, $\frac{-ab-0}{b-1}=\frac{0-ac}{1-c}$. $\Rightarrow ab(c-1)+ac(b-1)=0$ $\Rightarrow 2bc=b+c$ Since $PX\parallel AB$ and $QX\parallel OM$, $X=(ac, b)$. Suppose $X$ is on $\Gamma$, $OX^2=a^2c^2+b^2=OA^2=a^2+1$. Since $XY\parallel OA$ and $BY\perp OX$, $Y=(ac-da, b+d)=(a-eb, 1+eac)$ for some $d, e$. $\Rightarrow\begin{cases} ac-da=a-eb\\ b+d=1+eac \end{cases}$. $\Rightarrow ab+ac=2a+ea^2c-eb$ $\Rightarrow e=\frac{a(b+c-2)}{a^2c-b}$. $OY^2=a^2-2eab+e^2b^2+1+2eac+e^2a^2c^2=e^2(b^2+a^2c^2)+2ea(c-b)+a^2+1$ $\because e(b^2+a^2c^2)+2a(c-b)=\frac{a(b+c-2)(b^2+a^2c^2)+(a^2c-b)2a(c-b)}{a^2c-b}$ $=\frac a{a^2c-b}\left((b+c-2)(b^2+a^2c^2)+2(a^2c-b)(c-b)\right)$ $=\frac a{a^2c-b}((b+c)(b^2+a^2c^2)-2b^2-2a^2c^2+2a^2c^2-2a^2bc-2bc+2b^2)$ $=\frac a{a^2c-b}(2bc(b^2+a^2c^2)-2a^2bc-2bc)$ $=\frac a{a^2c-b}(2bc(b^2+a^2c^2)-2(a^2+1)bc)$ $=\frac{2abc}{a^2c-b}(b^2+a^2c^2-a^2-1)$ $=\frac{2abc}{a^2c-b}(OX^2-OA^2)=0$ $\therefore OY^2=e\cdot0+a^2+1=OA^2$, $Y$ is on $\Gamma$.
USJL
13.02.2023 02:27
The motivation behind this problem is actually the rectangular hyperbola defined by the locus of $X$ when $P,Q$ are moving on the line. The striking property it has is that the other three intersections with the circle is an equilateral triangle. See Section 2.8 in https://www.overleaf.com/read/bfbzksthmydz for more stories, where I also talk briefly about the extremely useful tool ("formal sum") that is used by Li4 above.
Hakurei_Reimu
14.02.2023 10:16
Assume $OA=OB=r$, $\vec{OA}\cdot\vec{OB}=\delta$, $OP=x\vec{OA}$. By Menelaus theorem we have $\vec{OQ}=\dfrac x{2x-1}\vec{OB}$. Besides, $\vec{OM}=\dfrac12(\vec{OA}+\vec{OB})$. Now we can easily get $\vec{OX}=\dfrac12\left((x-\dfrac x{2x-1})\vec{OA}+(x+\dfrac x{2x-1})\vec{OB}\right)$. Since $XY$ is parallel to $OA$, we can assume $\vec{OY}=\vec{OX}-y\vec{OA}$. From $BY$ perpendicular to $OX$ we can get $$\left(\vec{OY}-\vec{OB}\right)\cdot\vec{OX}=\left(\vec{OX}-y\vec{OA}-\vec{OB}\right)\cdot\vec{OX}=0.$$Here we can calculate these inner products:\begin{align*}\vec{OX}\cdot\vec{OX}&=\dfrac12\left(\left(x^2+\left(\dfrac x{2x-1}\right)^2\right)r^2+\left(x^2-\left(\dfrac x{2x-1}\right)^2\right)\delta\right)\\ \vec{OB}\cdot\vec{OX}&=\dfrac12\left(\left(x+\dfrac x{2x-1}\right)r^2+\left(x-\dfrac x{2x-1}\right)\delta\right)\\ \vec{OA}\cdot\vec{OX}&=\dfrac12\left(\left(x-\dfrac x{2x-1}\right)r^2+\left(x+\dfrac x{2x-1}\right)\delta\right)\end{align*}so we get this:$$y=\dfrac{\left(x^2-x-\frac x{2x-1}+\left(\frac x{2x-1}\right)^2\right)r^2+\left(x^2-x+\frac x{2x-1}-\left(\frac x{2x-1}\right)^2\right)\delta}{\left(x-\frac x{2x-1}\right)r^2+\left(x+\frac x{2x-1}\right)\delta}$$Now $X$ is on $\Gamma$ iff $\|\vec{OX}\|=r$, i.e.$$\left(x^2+\left(\dfrac x{2x-1}\right)^2\right)r^2+\left(x^2-\left(\dfrac x{2x-1}\right)^2\right)\delta=2r^2.$$Simplify it will get $$\dfrac{(x-1)(2(r^2+\delta)x^3-3xr^2+r^2)}{2x-1}=0$$When $x=1\Rightarrow y=0$, which means $Y=X$. Therefore $Y$ is on $\Gamma$. When $x=\dfrac12$, $X=\infty_{AB}$, which doesn't lie on $\Gamma$. Consider $x\neq 1$ or $\dfrac12$, now the above equation will be$$2(r^2+\delta)x^3=3xr^2-r^2\ \ \ \ \ \ \ ...(1)$$On the other side, we also need to find $\|\vec{OY}\|^2$, which is expected to be $r^2$. \begin{align*}\|\vec{OY}\|^2&=\vec{OY}\cdot\vec{OY}\\&=\vec{OX}\cdot\vec{OX}-2y\vec{OA}\cdot\vec{OX}+y^2\vec{OA}\cdot\vec{OA}\\&=r^2-2y\vec{OA}\cdot\vec{OX}+y^2r^2\end{align*}Claim : $2\vec{OA}\cdot\vec{OX}=yr^2$ proof : Expand it will get:$$\left(x-\dfrac x{2x-1}\right)r^2+\left(x+\dfrac x{2x-1}\right)\delta\stackrel{?}{=}\dfrac{\left(x^2-x-\frac x{2x-1}+\left(\frac x{2x-1}\right)^2\right)r^4+\left(x^2-x+\frac x{2x-1}-\left(\frac x{2x-1}\right)^2\right)r^2\delta}{\left(x-\frac x{2x-1}\right)r^2+\left(x+\frac x{2x-1}\right)\delta}$$$$\Rightarrow\left(x^2-\dfrac{2x^2}{2x-1}+\left(\dfrac x{2x-1}\right)^2\right)r^4+2\left(x^2-\left(\dfrac x{2x-1}\right)^2\right)r^2\delta+\dfrac{4x^4}{(2x-1)^2}\delta^2\stackrel{?}{=}\left(x^2-x-\frac x{2x-1}+\left(\frac x{2x-1}\right)^2\right)r^4+\left(x^2-x+\frac x{2x-1}-\left(\frac x{2x-1}\right)^2\right)r^2\delta$$$$\Rightarrow 4x^4\delta\stackrel{?}{=} -2x(2x^3-3x+1)r^2$$$$\Rightarrow 2x^3(r^2+\delta)\stackrel{?}{=}3xr^2-r^2$$This equation is exatly $(1)$, so the proof is complete. Now using this claim we have:$$\|\vec{OY}\|^2=r^2\ \ \ \ \ \ \ \blacksquare$$
hukilau17
15.02.2023 23:09
Identify $\Gamma$ with the unit circle, and assume $X$ lies on $\Gamma$, so that
$$|a|=|b|=|x|=1$$$$o = 0$$$$m = \frac{a+b}2$$Now $P$ lies on line $OA$ and also on the line through $X$ parallel to $AB$, so
$$p = \frac{a(-a)\left(x+\frac{ab}x\right) - x\left(\frac{ab}x\right)(a-a)}{a(-a) - x\left(\frac{ab}x\right)} = \frac{a(ab+x^2)}{x(a+b)}$$Similarly, $Q$ lies on line $OB$ and also one the line through $X$ parallel to $OM$, so
$$q = \frac{b(-b)\left(x-\frac{ab}x\right) - x\left(-\frac{ab}x\right)(b-b)}{b(-b)-x\left(-\frac{ab}x\right)} = \frac{b(ab-x^2)}{x(a-b)}$$Now $P, M, Q$ are collinear. We have the vectors
$$p-m = \frac{a(ab+x^2)}{x(a+b)} - \frac{a+b}2 = \frac{2a^2b+2ax^2-a^2x-2abx-b^2x}{2x(a+b)}$$$$q-m = \frac{b(ab-x^2)}{x(a-b)} - \frac{a+b}2 = \frac{2ab^2-2bx^2-a^2x+b^2x}{2x(a-b)}$$Thus
$$\frac{p-m}{q-m} \in \mathbb{R}$$and so
$$\frac{(a-b)(2a^2b+2ax^2-a^2x-2abx-b^2x)}{(a+b)(2ab^2-2bx^2-a^2x+b^2x)} = \frac{(a-b)(2ab^2+2bx^2-a^2x-2abx-b^2x)}{(a+b)(2a^2b-2ax^2-a^2x+b^2x)}$$$$(2a^2b-2ax^2-a^2x+b^2x)(2a^2b+2ax^2-a^2x-2abx-b^2x) = (2ab^2-2bx^2-a^2x+b^2x)(2ab^2+2bx^2-a^2x-2abx-b^2x)$$$$\left[2ax^2 + x(a^2-b^2) - 2a^2b\right]\left[2ax^2 - x(a+b)^2 + 2a^2b\right] = \left[2bx^2 + x(a^2-b^2)-2ab^2\right]\left[2bx^2 - x(a+b)^2 + 2ab^2\right]$$$$4a^2x^4 - 4abx^3(a+b) - x^2(a-b)(a+b)^3 + 4a^3bx(a+b) - 4a^4b^2 = 4b^2x^4 - 4b^2x^3(a+b) - x^2(a-b)(a+b)^3 + 4a^2b^2x(a+b) - 4a^2b^4$$$$4x^4(a+b)(a-b) - 4bx^3(a+b)(a-b) + 4a^2bx(a+b)(a-b) - 4a^2b^2(a+b)(a-b) = 0$$$$(a+b)(a-b)(b-x)(a^2b+x^3) = 0$$Now $(a+b)(a-b) \neq 0$ because $A, B, O$ are not collinear, and $b-x\neq 0$ or else $P = A$. So
$$a^2b + x^3 = 0$$Now the second intersection of the line through $X$ parallel to line $OA$ with $\Gamma$ has coordinate
$$\frac{a(-a)}x = -\frac{a^2}x$$And the second intersection of the line through $B$ perpendicular to line $OX$ with $\Gamma$ has coordinate
$$-\frac{x(-x)}b = \frac{x^2}b$$Since $b = -\frac{x^3}{a^2}$, these two coordinates are the same. So the point $Y$ given by this coordinate lies on both lines and on $\Gamma$, as desired. $\blacksquare$
PRMOisTheHardestExam
04.05.2023 07:38
Li4 wrote: Note that $Y$ lies on $\Gamma$ if and only if $$AO - X = OX - B + 90^\circ \iff 3X = 2A + B + 90^\circ. $$Let $P'$ be the reflection point of $P$ with respect to $OM$. Then $$X(O, B; P', Q) = (O, B; P', Q) = M(O, B; P', P) = -1, $$i.e., $XP'$ is one of the angle bisectors of $\angle OXB$. Hence, $$XO + XB = 2XP' = 2AB \implies 3X = 2A + B + 90^\circ, $$as desired. what is this method someone can tell pls angles written by letters? or what?
USJL
05.05.2023 06:49
PRMOisTheHardestExam wrote: Li4 wrote: Note that $Y$ lies on $\Gamma$ if and only if $$AO - X = OX - B + 90^\circ \iff 3X = 2A + B + 90^\circ. $$Let $P'$ be the reflection point of $P$ with respect to $OM$. Then $$X(O, B; P', Q) = (O, B; P', Q) = M(O, B; P', P) = -1, $$i.e., $XP'$ is one of the angle bisectors of $\angle OXB$. Hence, $$XO + XB = 2XP' = 2AB \implies 3X = 2A + B + 90^\circ, $$as desired. what is this method someone can tell pls angles written by letters? or what? That is "formal sum" that is coined by Lii4 (and it's getting more and more popular in Taiwan now). Check out my handout above (Section 2.8) to see a brief introduction for that. Lii4 also has a handout that explains it in much more depth, but it's currently in Traditional Chinese :3
axolotlx7
21.04.2024 10:59
When $X$ is on $\Gamma$ (WLoG as shown), we claim that $Y$ is the reflection of $B$ across $OX$. It suffices to check that $\angle OXY = \angle OXB$. Note that if $PX$ and $OB$ intersect at $N$, then \[ (Q,N;O,B)\stackrel{P}{=} (M, \infty; A, B)=-1 \]and since $\angle NXQ = 90^\circ$, we get that $NX$ is the angle bisector of $\angle OXB$. Now \[ \angle OXY = \angle AOX = 2\angle ABX = 2 \angle NXB = \angle OXB, \]finishing the problem.