Bashing

by math_explorer, Dec 9, 2011, 9:49 AM

Quote:

Also, in triangle $ABC$ , $H$ is the orthocenter and $M$ is the midpoint of $BC$ . The foot of the altitude from $H$ to $AM$ is $P$ . Prove that $AM \times PM = BM^2$

I don't know why I failed to solve this for so long. I was reluctant to bring any trig into the picture. Not unjustified, considering I had to rederive $\cos (\pi/2 - \theta) = \sin \theta$ from the angle-difference formula. Sigh.

Monstrous trig bash setup!
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First we construct $P'$ on line $AM$ , on the same side of $M$ as $A$ , such that $AM \times P'M = BM^2$ . We now have to prove that $P'$ is the foot of the altitude from $H$ to $AM$ , or that $P'H \perp AM$ .

Draw circles $\odot U$ passing through $A, P', B$ and $\odot V$ passing through $A, P', C$ . The power of $M$ w.r.t. both circles is $AM \times P'M = BM^2 = CM^2$ so $BC$ is tangent to both circles, that is $UB \perp BC$ and $VC \perp BC$ .

Now, $AP'$ is the connection of the two circles' intersections, so (by symmetry) is perpendicular to $UV$ . Therefore we can equivalently prove that $P'H \parallel UV$ . Also by symmetry, the midpoint of $AP'$ is on $UV$ . Thus, if we let $J$ be the midpoint of $AH$ , it's also equivalent to proving that $J$ is on $UV$ .

Let the midpoint of $AB$ be $N$ ; then since $U$ is equidistant from $A$ and $B$ we have $UN \perp AB$ , and $UB = BN \sec \angle UBN = \frac{c}{2\sin B}$ (using the typical bash notation for $\triangle ABC$ ). Similarly $VC = \frac{b}{2\sin C}$ .

Let the foot of the altitude from $A$ be $D$ . Then $BD = c \cos B$ and $CD = b \cos C$ .

Now, let $J'$ be the intersection of $UV$ with $AD$ . By playing with similarities based on the three parallel segments $UB, J'D, VC$ (let $UV$ and $BC$ intersect and find similar triangles; treat the case where they're parallel specially or just insist that it can be taken as a limit case) we have

\[ \begin{align*}
J'D &= \frac{UB\times CD + VC\times BD}{BC}\\
&= \frac{\frac{bc\cos C}{2\sin B} + \frac{bc\cos B}{2\sin C}}{a}\\
&= \frac{bc(\sin C\cos C + \sin B\cos B)}{2a\sin B \sin C}\\
&= \frac{2R^2(\sin C\cos C + \sin B\cos B)}{a}\\
&= \frac{R(\sin C\cos C + \sin B\cos B)}{\sin A}\\
&= \frac{R(\sin C\cos C + \sin B\cos B)}{\sin (B+C)}\\
&= \frac{R(\sin C\cos C + \sin B\cos B)}{\sin B\cos C + \cos C\sin B}\\
&= R\left(\frac{(\sin C\cos C(\sin^2 B + \cos^2 B) + \sin B\cos B(\sin^2 C + \cos^2 C))}{\sin B\cos C + \cos C\sin B}\right)\\
&= R\left(\frac{(\sin B\cos C + \cos C\sin B)(\sin B\sin C + \cos B\cos C)}{\sin B\cos C + \cos C\sin B}\right)\\
&= R\cos(B-C)
\end{align*} \]

(I can't imagine how to come up with that magical $\sin^2 B + \cos^2 B$ if you're not secretly doing the other side first to see what you want to prove)

Next, $AD = c \sin B = 2R \sin B \sin C$ and $HD = BD \tan \angle HBC = c \cos B \cot C = 2R \cos B \cos C$ , so $JD = \frac{AD+HD}{2} = R(\sin B \sin C + \cos B \cos C) = R\cos (B-C)$ .

Thus $J = J'$ , so $U, J, V$ are collinear, which means that $HP' \parallel UV \perp AM$ , so that $P'$ is the foot of the altitude from $H$ to $AM$ .

geo solution

geometry

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Darn trig bash and indirect proofs (let something has a property, pove that that thing is one we want to prove to have that property).

Magical trig Pythagorean identity