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BdMO Online Forum • View topic - Geometry Marathon : Season 3

Geometry Marathon : Season 3

For discussing Olympiad level Geometry Problems
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Re: Geometry Marathon : Season 3

Post Number:#81  Unread postby joydip » Mon Feb 27, 2017 9:13 pm

Problem 38 :

In $\triangle ABC$ let the angle bisector of $\angle BAC$ meet $BC$ at $A_o$. Define $B_o,C_o$ similarly.Prove that
the circumcircle of $\triangle A_oB_oC_o$ goes though the Feuerbach point of $\triangle ABC$.
The more I learn, the more I realize how much I don't know.

- Albert Einstein
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Re: Geometry Marathon : Season 3

Post Number:#82  Unread postby Thanic Nur Samin » Tue Feb 28, 2017 2:32 pm

Here is an alternate solution to problem 36

We apply cartesian coordinates.

fig.png
fig.png (25.7 KiB) Viewed 63 times


Note that the problem is equivalent to proving $F',D,E$ collinear in the diagram.

Let $U\equiv (a,1), V\equiv (-a,1), T\equiv (at,t), S\equiv (-as,s)$.

Now, let $E \equiv (p,q)$. We get, $\dfrac{q-t}{p-at}=-a$ and $\dfrac{q-s}{p+as}=a$. From these, we get $E \equiv (\dfrac{(a^2+1)(t-s)}{2a},\dfrac{(a^2+1)(t+s)}{2})$.

Again, $D$ lies on the $y$-axis. From $DU\perp OU$, we get that $D\equiv (a^2+1,0)$.

Now for the *most difficult* part, the coordinates of $F$. The slope of $ST$ is $\dfrac{t-s}{a(t+s)}$. So the slope of $OF$ is $\dfrac{a(t+s)}{s-t}$. So for some real $m$, $F\equiv (m(s-t),ma(s+t))$.

Now, $FS\perp OT$, so $\dfrac{ma(s+t)-s}{m(s-t)+as}=-a$. So we get $m=\dfrac{1-a^2}{2a}$. So $F\equiv (\dfrac{(1-a^2)(s-t)}{2a},\dfrac{(1-a^2)(s+t)}{2})$. So $F'\equiv (\dfrac{(1-a^2)(s-t)}{2a},2-\dfrac{(1-a^2)(s+t)}{2})$.

Now, apply the shoelace formula to $F',D,E$. The det is $0$, so they are collinear.

Time taken: 30 minutes while simultaneously taking english notes. :mrgreen:
Hammer with tact.

Because destroying everything mindlessly isn't cool enough.
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