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Let $AA_1$ and $BB_1$ be the bisectors of angles in triangle $ABC$. The bisectors intercept at the point $I$. How do I find all the triangles for which the perpendicular line from $I$ to $AB$ halves the line segment $A_{1}B_1$?

Recently I discovered the following:

If $AA_1$ and $BB_1$ are bisectors of sharp angles in a right triangle $ABC$ intersect at the point $I$, then the perpendicular line from $I$ to $AB$ halves the line segment $A_1B_1$.

How do I prove the converse? Are there any other triangles that satisfy the aforementioned condition?

G. Amber
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  • There is a trivial example such as $\Delta ABC$ being isosceles triangle with $|AC| = |BC|$ (This includes $\Delta ABC$ being equilateral triangle of course). But trivial examples do not count unless they are the only examples satisfying the conditions. So I'm trying to find a more general solution. – ArsenBerk Jan 13 '18 at 09:29
  • How can I prove that there is no other triangle? – G. Amber Jan 13 '18 at 11:06

1 Answers1

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The answer is as follows : $$\text{$\triangle{ABC}$ is either a right triangle with $\angle C=90^\circ$ or an isosceles triangle with $|\overline{CA}|=|\overline{CB}|$}$$

We may suppose that $A(-1,0),B(1,0),C(c,d)$ where $c\ge 0$ and $d\gt 0$.

The equation of the line $AC,BC$ is $dx-(c+1)y+d=0,dx-(c-1)y-d=0$ respectively.

For a point $(x,y)$ on the line $AA_1$, we have $$\small\frac{|dx-(c+1)y+d|}{\sqrt{d^2+(c+1)^2}}=|y|\implies \frac{dx-(c+1)y+d}{\sqrt{d^2+(c+1)^2}}=\pm y\ \rightarrow\ y=\frac{dx+d}{c+1+t}$$ which is the equation of $AA_1$ where $t=\sqrt{d^2+(c+1)^2}$ since the slope of $AA_1$ is positive.

For a point $(x,y)$ on the line $BB_1$, we have $$\small\frac{|dx-(c-1)y-d|}{\sqrt{d^2+(c-1)^2}}=|y|\implies \frac{dx-(c-1)y-d}{\sqrt{d^2+(c-1)^2}}=\pm y\ \rightarrow\ y=\frac{dx-d}{c-1-s}$$ which is the equation of $BB_1$ where $s=\sqrt{d^2+(c-1)^2}$ since the slope of $BB_1$ is negative.

It follows that $$ I\left(\frac{2c+t-s}{2+t+s},\frac{2d(c-1)}{(c-1-s)(2+t+s)}\right),\quad A_1\left(\frac{2c+t}{t+2},\frac{2d}{t+2}\right),\quad B_1\left(\frac{2c-s}{s+2},\frac{2d}{s+2}\right)$$

Therefore, we have $$\begin{align}&\text{the perpendicular line from $I$ to $AB$ halves the line segment $A_{1}B_1$}\\\\&\iff \frac 12\left(\frac{2c+t}{t+2}+\frac{2c-s}{s+2}\right)=\frac{2c+t-s}{2+t+s}\\\\&\iff \frac{2c+t}{t+2}+\frac{2c-s}{s+2}=\frac{4c+2t-2s}{2+t+s}\\\\&\iff (2c+t)(s+2)(2+t+s)+(2c-s)(t+2)(2+t+s)=(4c+2t-2s)(t+2)(s+2)\\\\&\iff (c^2+d^2-1)(2c-s+t)=0\\\\&\iff c^2+d^2=1\quad\text{or}\quad c=0\end{align}$$ where $$2c-s+t=0\iff (2c+t)^2=s^2\iff 4c\left(c+1+\sqrt{d^2+(c+1)^2}\right)=0\iff c=0$$ from which the conclusion written at the top follows.

mathlove
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