(China TST 2011) Let $\Gamma$ be the circumcircle of a triangle $ABC$. Assume $AA', BB',CC'$ are diameters of $\Gamma$. Let $P$ be a point inside $\triangle ABC$ and let $D, E, F$ be the feet from $P$ to $BC, CA, AB$. Let $X$ be the reflection of $A'$ across $D$; define $Y$ and $Z$
similarly. Prove that $\triangle XYZ\sim\triangle ABC$.
I found this problem and tried to use complex numbers to solve it, but every time it failed. Can someone please tell me what is wrong with my computations?
First, I defined $\Gamma$ to be the unit circle. Then, its easy to notice that $a' = – a$ , $b' = -b$ and $c' = -c$ ( Since zero is the midpoint, and thus $\frac{1}{2} (a + a') = 0 \implies a' = -a$). As $D$ is the midpoint of $A'$ and $X$, we have $d = \frac12(x + a') = \frac12(x – a) \implies x = 2d + a$. Similarly, $y = 2e + b$ and $z = 2f + c$. Now the only thing left is to express $d$, $e$ and $f$ in therms of $p$. As $A$, $B$ and $C$ lie on the unit circle, we can use the formula: "given $T$ and $S$ on the unit circle and a point $H$, then the foot from $H$ to $TS$ is:
$$\frac12( t + s + h – ts\bar{h})$$
where $\bar{h}$ is the conjugate of $h$"
And thus
\begin{align*}
x &= 2d + a = a + b + c + p – bc\bar{p},\\
y &= a + b + c + p – ca\bar{p}\\
z &= a + b + c + p – ab\bar{p}
\end{align*}
For the triangles to similar, we must have $(c – a)/(b – a) = (z – x)/(y – x)$, but
$$
\frac{z – x}{y – x} = \frac{\bar{p}b( c – a)}{\bar{p}c(b – a)} = \frac{b(c – a)}{c (b – a)}
$$
This means: $(c – a)/(b – a) = (b(c – a))/(c(b – a)) \implies b = c$, which is a lie.
Best Answer
Note that $$\frac{z-x}{y-x}=\frac{ab-bc}{ac-bc}=\frac{1/c-1/a}{1/b-1/a}=\frac{\bar{c}-\bar{a}}{\bar{b}-\bar{a}}$$
This means $\triangle XYZ$ is inversely similar to $\triangle ABC$. That is if $A,B,C$ are oriented counter-clockwise, $X,Y,Z$ are oriented clockwise.
Since the triangle formed by $\bar{a},\bar{b},\bar{c}$ is reflection of that formed by $a,b,c$ in real axis, it has its orientation flipped.