## Saturday, January 27, 2007

### Vector Magic. Topic: Geometry. Level: AIME.

Problem: (1997 Putnam - A1) A rectangle, $HOMF$, has sides $HO = 11$ and $OM = 5$. A triangle $ABC$ has $H$ as the intersection of its altitudes, $O$ the center of its circumscribed circle, $M$ the midpoint of $BC$, and $F$ the foot of the altitude from $A$. What is the length of $BC$?

Solution: Since everyone loves algebra and hates geometry, let's use the standard algebrization of a triangle problem: vectors. Set $O$ to be the origin, $H$ to $(-11, 0)$, from which we automatically obtain $M = (0, -5)$ and $F = (-11, -5)$. Let $a, b, c$ be the vectors from $O$ to points $A, B, C$, respectively.

We know $|a| = |b| = |c|$ and $a+b+c = H$ from here, so we'll use these results. Also, $b+c = 2M$ by the definition of a midpoint. Using these three equations, we can solve for $a, b, c$.

$a+b+c = (-11, 0)$ and $b+c = (0, -10)$ implies that $a = (-11, 10)$.

Also, the $x$-coordinates of $b$ and $c$ are the same value with opposite signs. So if $|a| = |b|$, where $a = (-11, 10)$ and $b = (-x, -5)$, we must have

$11^2+10^2 = x^2+5^2 \Rightarrow x^2 = 196 \Rightarrow x = 14$.

So $BC = |b-c| = |(-28,0)| = 28$. QED.

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Comment: There are various other synthetic ways to approach this problem, but using vectors seems to make life really easy as we know lots of relationships about the circumcenter, orthocenter, and midpoints. See if you can convince me that there is a simpler way to do it without vectors.

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Practice Problem: "Prove" that the vector representation of the orthocenter $H$ has to be symmetric in $a, b, c$ (i.e. explain why you know $H = a(b+c)$ is wrong).