# Locus of a point on triangle side

Let $P$ be a point on line segment $XY$. On the perpendicular line to $XY$ through $P$ lies point $Z$. Prove that for any position of $Z$ there is a unique point $Q$ on line segment $XZ$ for which $\angle ZYQ = \angle ZXY$. The perpendicular line on $XZ$ through $Q$ then always passes through the same (fixed) point $R$ on line segment $XY$, independent of the position of $Z$.

From geogebra it seems that $$\angle ZSQ = \angle ZYQ =\angle ZXY$$ and $$\angle YZP = \angle YQS$$ $S$ is the intersection of $ZP$ and $QR$. Holding $Z$ and shifting $Q$ on $XZ$ does not affect $\angle ZSQ$.

• where is point $S$? – Vasya Jul 5 '18 at 15:29
• Re-edited question – jeremiah Jul 5 '18 at 18:06
• Can you add a figure? I tried visualizing this but came up with something where $R$ very much depends on $Z$, so I guess I misunderstood some of your description and a picture should help clarify this. – MvG Jul 5 '18 at 19:37
• @MvG: pls refer to John Glenn’s link below – jeremiah Jul 5 '18 at 21:00

WLOG, suppose $XY$ is on the $x$-axis. Let us define points $X,Y$ as $(-a,0)$ and $(a,0)$, respectively. Then, $Z$ should be defined as point $(b,z)$. Additionally, point $Q=\left(q,-\frac{z}{a+b}\left(q+a\right)\right)$. For $a,b,z,q \in \Bbb R$.

We have the following equations for lines $XZ, QY$ and $ZY$

$$XZ: y=\frac{z}{a+b}\left(x+a\right)\qquad m_1=\frac{z}{a+b}\\ ZY:y=-\frac{z}{a-b}\left(x-a\right)\qquad m_2=-\frac{z}{a-b}\\ QY:y=-\frac{z(a+q)}{(a+b)(a-q)}\left(x-a\right))\qquad m_3=-\frac{z(a+q)}{(a+b)(a-q)}$$

Since we should have $\angle ZYQ=\angle ZXY$, we want the angle between lines $QY$ and $ZY$ be equal to $\angle ZXY$ Thus, we have: $$\frac{m_2-m_3}{1+m_2m_3}=m_1$$ Substituting the slopes from the equations above, we get that $m_2$ should be: $$m_2\to-\frac{2 a z (a+b)}{(a+b)^2 (a-q)-z^2 (a+q)}$$ However, we already know that $m_2=-\frac z{a-b}$, therefore: $$-\frac{2 a z (a+b)}{(a+b)^2 (a-q)-z^2 (a+q)}=-\frac{z}{a-b}\\ q\to -\frac{a \left((a+b) (a-3 b)+z^2\right)}{(a+b)^2+z^2}\tag{1}$$

This gives us $q$ such that $\angle ZXY=\angle ZYQ$.

Now the line perpendicular to $XZ$ through $Q$ is defined by the line: $$y=-\frac{a+b}{z}\left(x-q\right)+\frac{z}{a+b}\left(q+a\right)$$ Which has a zero of (i.e. $R=(x,0)$): $$x\to \frac{z^2 (a+q)}{(a+b)^2}+q\tag{2}$$ Since we already know $q$, substituting $(1)$ in $(2)$, and we get: $$\bbox[20px,border:1px black solid]{x\to-\frac{a (a-3 b)}{a+b}\implies \therefore R \text{ is definitely invariant to } Z}$$

You can check this implementation.

• What a beautiful answer - incredible. – jeremiah Jul 5 '18 at 18:56
• @jeremiah Hardly. A more intuitive, geometric solution definitely exists. – John Glenn Jul 5 '18 at 19:01
• Perhaps straightforward or direct would be more suitable? I liked it, however. But if you have a hunch for a more intuitive solution I´d appreciate any hint or direction. Thnx – jeremiah Jul 5 '18 at 19:13

$\triangle XYZ \sim \triangle YQZ$, therefore $QZ/YZ = YZ/XZ$.

$PQRZ$ is cyclic, therefore $$PX \cdot RX = QX \cdot XZ = (QZ-XZ) \cdot XZ = YZ^2 - XZ^2 = \\ (PY^2 + PZ^2) - (PX^2 + PZ^2) = PY^2 - PX^2.$$ Note that the argument doesn't go through if both $\angle ZYQ$ and $\angle ZXY$ are taken in the clockwise direction, because the two triangles won't be similar. There is an implicit condition in the problem.

• Point $Q$ should be on line segment $XZ$, i.e. in between those points. Same applies to point $R$ which should be on line segment $XY$. The formulation n the question is unclear, so I re-edited it. – jeremiah Jul 6 '18 at 5:46
• That condition excludes the possibility of both angles having the same orientation, but then the required point $Q$ doesn't exist for all $P$ (if it does, then, for that $P$, it exists for all $Z$). The condition is too restrictive though. The proof holds for $Q$ inside the segment $XZ$ as well as for $P$ outside of the segment $XY$ with the possible modification that, in terms of unsigned segments, we may have $QX=XZ−QZ$ instead of $QX=QZ−XZ$. The essential condition is the different orientation of the angles. – Maxim Jul 6 '18 at 12:16
• I believe that the orientation of the angles does not matter. – jeremiah Jul 6 '18 at 12:24
• Please see here: i.imgur.com/GWUtUbL.png. – Maxim Jul 6 '18 at 13:07

# Solution

It's easy to have that the circumcircle of $\triangle QXY$ touches $ZY$ at $Y$,and $Z,P,Q,R$ are cyclic.

Hence, $$ZY^2=ZX \cdot ZQ=ZX \cdot (ZX+XQ)=ZX^2+ZX \cdot XQ=ZX^2+PX \cdot XR.$$

This shows that $$XR=\frac{ZY^2-ZX^2}{PX}=\frac{PY^2-PX^2}{PX}=\frac{PY^2}{PX}-PX.$$

Notice that $PX,PY$ are fixed length. Then the length of $XR$ is a constant. As a result, $R$ is a fixed point independent of $Z.$

• I'm curious, how do you prove that $ZY$ is tangent to the circle without using $ZY^2=ZX \cdot ZQ$? – Maxim Jul 6 '18 at 2:09
• Do you know the chord tangent angle theorem? If so, you can consider applying the "coincidence method". – mengdie1982 Jul 6 '18 at 2:44
• I believe a polite answer would look like this. Let $\angle QXY = \pi - \angle ZXY = \pi - \phi$. By the properties of inscribed angles, the arc $QXY$ is $2\phi$, therefore the angle between $QY$ and the tangent at $Y$ is $\phi$. – Maxim Jul 6 '18 at 10:55
• I´m happy that all of you took on this problem and provided good insights from different perspectives. Thank you for your time and effort. – jeremiah Jul 6 '18 at 12:06