3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – ArsenBerk Jan 13 '18 at 9:29
  • $\begingroup$ How can I prove that there is no other triangle? $\endgroup$ – G. Amber Jan 13 '18 at 11:06
1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Thanks for devoting time to my problem! $\endgroup$ – G. Amber Jan 14 '18 at 9:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.