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Mock Geometry AIME 2011 Problems/Problem 14

Problem

The point $(10,26)$ is a focus of a non-degenerate ellipse tangent to the positive $x$ and $y$ axes. the locus of the center of the ellipse lies along graph of, $ax-by+c=0,$ where $a,b,c$ are positive integers with no common factor other than $1$. Find $a+b+c.$

Solution

We utilize (the proof of) a highly useful lemma.

LEMMA: Let $\mathcal{E}$ be an ellipse with foci $A$ and $B$, and let $P$ be a point on $\mathcal{E}$. Furthermore, let $\ell$ be the line tangent to $\mathcal{E}$ at $P$. Then the acute angles formed by $PA$ and $PB$ with respect to $\ell$ are equal. (This is analogous to the theorem stating that the angle of incidence is equal to the angle of reflection in optics.)

Proof. Let $a$ be the length of the semimajor axis of the ellipse. Denote by $B'$ the reflection of $B$ across $\ell$. Remark that $AP+PB'=AP+PB=2a$, which is fixed. Furthermore, for any other point $Q$ on $\ell$, the distance $AQ+QB'=AQ+QB$ is intuitively larger than $2a$ (since a "bigger" ellipse is needed to pass through $Q$). Hence $P$ is the point for which $AP+PB'$ is minimized, and so $A$, $P$, $B'$ are collinear. Thus the two acute vertical angles formed by $AB'$ and $\ell$ are congruent - tracing back the reflection gives our desired. (For more detail, see http://www.maa.org/sites/default/files/0746834207514.di020724.02p0009e.pdf .)

This reflecting business motivates what to do next. Let $X(10,-26)$ and $Y(-10,26)$ be the reflections of $A(10,26)$ across the $x$ and $y$ axes respectively, and let $B$ be the location of the other focus of our ellipse. Note that said axes are tangent to the ellipse, so we are basically replicating the proof above. Now recall that by the proof $XB$ and $YB$ both equal $2a$, so they are equal in length to each other. Thus the locus of points $B$ is the perpendicular bisector of segment $XY$. To find the equation of this perpendicular bisector, remark that the midpoint of $\overline{XY}$ is the origin $(0,0)$ of the plane and that the slope of $XY$ is $-\tfrac{13}5$, so the equation is $y=\tfrac5{13}x$.

Finally, we compute the locus of the centers. We could go through a slight coordinate bash (and it isn't that hard to do), but here we shall use Euclidean geometry. Consider the homothety $\mathcal{H}$ centered at $A$ with scale factor $\tfrac12$. For an arbitrary point $B_0$ on this line, said homothety will take $B_0$ to the midpoint of line segment $\overline{AB_0}$ - which is just the center of the ellipse with foci $A$ and $B_0$! Hence the desired locus of centers is the line $\ell_2$ that results when the entire line $y=\tfrac5{13}x$ is transformed under $\mathcal{H}$. By basic homothety rules, the line $\ell_2$ is parallel to $y=\tfrac5{13}x$. Furthermore, this line must pass through the midpoint of the segment connecting the origin to $(10,26)$, or the point $(5,13)$. Thus, the equation of the desired locus is \[y-13=\tfrac5{13}(x-5)\implies 5x-13y+144=0.\] Our requested answer is $5+13+144=\boxed{162}$.

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