ISL 2005 G5

by Wolstenholme, Aug 1, 2014, 9:15 PM

Let $\triangle ABC$ be an acute-angled triangle with $AB \not= AC$ . Let $H$ be the orthocenter of triangle $ABC$ , and let $M$ be the midpoint of the side $BC$ . Let D be a point on the side $AB$ and $E$ a point on the side $AC$ such that $AE = AD$ and the points $D, H, E$ are on the same line. Prove that the line $HM$ is perpendicular to the common chord of the circumscribed circles of triangle $\triangle ABC$ and triangle $\triangle ADE$ .

Proof:

This problem can be done with complex numbers. WLOG let $\omega$ , the circumcircle of $\triangle{ABC}$ , be the unit circle. Let the centers of $\omega$ and the circumcircle of $\triangle{ADE}$ be denoted by $O$ and $O_1$ respectively. Let $X$ be the second intersection of $\omega$ and the circumcircle of $\triangle{ADE}$ . Let $A, B, C, D, E, H, M, X, O_1$ have complex coordinates $a^2, b^2, c^2, d, e, h, m, x, o_1$ respectively.

First let us find $d$ . Since $DH$ is perpendicular to the $A$ -angle bisector of $\triangle{ABC}$ we have that $\frac{d - h}{\overline{d} - \overline{h}} = -\frac{a^2+ bc}{\overline{a}^2 + \overline{bc}} = -a^2bc$ . Using the fact that $h = a^2 + b^2 + c^2$ this becomes $d = \frac{(b^2 + bc + c^2)(a^2 + bc)}{bc} - a^{2}bc\overline{d}$ . Moreover since $D \in AB$ we have that $d = a^2 + b^2 - a^{2}b^{2}\overline{d}$ .

Solving these two equations for $\overline{d}$ we find that $\overline{d} = \frac{bc^3 + a^{2}b^{2} + b^{2}c^{2} + c^{2}a^{2}}{a^{2}b^{2}c(c - b)}$ . This implies that $d = \frac{b(a^{2}b + c^3 + a^{2}c + b^{2}c)}{c(b - c)}$ . Similarly we have that $e = \frac{ c(a^{2}b + b^3 + a^{2}c + bc^2)}{b(c - b)}$ .

Now I shall show that $o_1 = -\frac{bc(2a^2 + b^2 + c^2)}{(b - c)^2}$ . Since this expression is symmetric in $b$ and $c$ it suffices to show that $\left|a^2 + \frac{bc(2a^2 + b^2 + c^2)}{(b - c)^2}\right| = \left|d + \frac{bc(2a^2 + b^2 + c^2)}{(b - c)^2}\right|$ .

Plugging in our expression for $d$ , we need $\left|a^2 + \frac{bc(2a^2 + b^2 + c^2)}{(b - c)^2}\right| = \left|\frac{b(a^{2}b + c^3 + a^{2}c + b^{2}c)}{c(b - c)} + \frac{bc(2a^2 + b^2 + c^2)}{(b - c)^2}\right|$ . Factoring out $|b| = 1$ from the RHS, multiplying both sides by $|c(b - c)^2|$ , and then factoring $|c| = 1$ from the LHS it suffices to show that $|a^2(b - c)^2 + bc(2a^2 + b^2 + c^2)| = |(b - c)(a^{2}b + c^3 + a^{2}c + b^{2}c) + c^2(2a^2 + b^2 + c^2)|$ . Upon expanding both sides, we find that they are equal and so the desired result holds.

Now, since the chord between $\omega$ and the circumcircle of $\triangle{ADE}$ is the two circles' radical axis, it is perpendicular to $OO_1$ . Therefore to solve the problem it suffices to show that $OO_1 \parallel MH$ . So it suffices to show that $\frac{o_1}{\overline{o_1}} = \frac{h - m}{\overline{h} - \overline{m}} = \frac{a^2 + b^2 + c^2 - \frac{b^2 + c^2}{2}}{\overline{a}^2 + \overline{b}^2 + \overline{c}^2 - \frac{\overline{b}^2 + \overline{c}^2}{2}} = \frac{(2a^2 + b^2 + c^2)a^{2}b^{2}c^{2}}{2b^{2}c^{2} + a^{2}b^{2} + a^{2}c^{2}}$ .

Now we know $o_1 = -\frac{bc(2a^2 + b^2 + c^2)}{(b - c)^2}$ and we easily find that $\overline{o_1} = -\frac{2b^{2}c^{2} + a^{2}b^{2} + a^{2}c^{2}}{a^{2}bc(b - c)^2}$ so $\frac{o_1}{\overline{o_1}} = \frac{(2a^2 + b^2 + c^2)a^{2}b^{2}c^{2}}{2b^{2}c^{2} + a^{2}b^{2} + a^{2}c^{2}}$ as desired and so we are done.

Now I want to discuss the motivation for this solution. First of all, $M$ and $H$ are very nice in complex which is the first clue to use complex numbers. Moreover since we immediately see that line $DE$ is perpendicular to the $A$ -angle bisector in $\triangle{ABC}$ , whose complex equation is not bad, we see that $d$ and $e$ can be found by solving a simple system of linear equations. Now since we are dealing with a radical axis - one of whose circles is $\omega$ - we see a nice way to get $O$ involved. The only tough thing about this solution is how to find $o_1$ . In my solution I seemingly pulled its value from a magic hat and showed that it worked. However, since I knew the problem statement was true, I found the value before I proved that it worked. We knew that $OO_1 \parallel MH$ and moreover that $O_1$ lied on the $A$ -angle bisector of $\triangle{ABC}$ so, given that the problem statement was correct, we could find $o_1$ with a system of linear equations! This insight led to the solution above.

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lol 30 minutes ago i posted this in my blog too

by bcp123, Aug 1, 2014, 9:17 PM

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by nukelauncher, Aug 13, 2020, 6:40 AM
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by fukano_2, Jun 14, 2020, 6:24 AM
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by briantix, Mar 18, 2016, 9:57 PM
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by mishka1980, Sep 12, 2015, 10:33 PM
This is late, but where is the ARML results post?
by donot, Aug 31, 2015, 11:07 PM
"I am Sam"
"Sam I am"
by mathwizard888, Aug 12, 2015, 9:13 PM
HW $\textcolor{white}{}$
by Eugenis, Apr 20, 2015, 10:10 PM
Uh-oh ARML practice is Thursday... I should start the homework.
by nosaj, Apr 20, 2015, 12:34 AM
Yes I am Sam, and Chebyshev polynomials aren't trivial, although they do make some problems trivial
by Wolstenholme, Apr 15, 2015, 10:00 PM
How are Chebyshev Polynomials trivial?
by nosaj, Apr 13, 2015, 4:10 AM
Are you Sam?
by Eugenis, Apr 4, 2015, 2:05 AM
@Brian: yes, yes I did #whoneedsalgskillz?

@gauss1181; hey!
by Wolstenholme, Mar 1, 2015, 11:25 PM
hello!!!
by gauss1181, Nov 27, 2014, 12:19 AM
Hi Wolstenholme did you actually use calc on that tstst problem
by briantix, Aug 2, 2014, 12:25 AM

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ISL 2005 G5

by Wolstenholme, Aug 1, 2014, 9:15 PM

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