# 2013 Mock AIME I Problems/Problem 12

## Problem

In acute triangle $ABC$, the orthocenter $H$ lies on the line connecting the midpoint of segment $AB$ to the midpoint of segment $BC$. If $AC=24$, and the altitude from $B$ has length $14$, find $AB\cdot BC$.

## Solution (easy coordinate bash)

Toss on the coordinate plane with $A=(0,0)$, $B=(t,14)$, and $C=(24,0)$, where $t$ is a real number and $0.

Then, the line connecting the midpoints of $AB$ and $BC$ runs from $\left(\frac{t}{2}, 7\right)$ to $\left(\frac{24+t}{2}, 7\right)$, or more simply the line $y=7$.

The orthocenter of $\triangle ABC$ will be at the intersection of the altitudes from $A$ and $B$.

The slope of the altitude from $A$ is the negative reciprocal of the slope of $\overline{BC}$. The slope of $\overline{BC}$ is $-\frac{14}{24-t}$, and its negative reciprocal is $\frac{24-t}{14}$. Since the altitude from $A$ passes through the origin, its equation is $y=\frac{24-t}{14}x$.

The altitude from $B$ is the vertical line running through $B=(t, 14)$ which has equation $x=t$.

Thus the lines $x=t$ and $y=\frac{24-t}{14}x$ meet on the line $y=7$. Substituting the first equation into the second, $\frac{t(24-t)}{14}=7$.

Multiplying both sides by $14$, we have $t(24-t)=98$.

This rearranges to the quadratic $t^{2}-24t+98=0$, and completing the square by adding $46$ to each side gives us $(t-12)^{2}=46$. Thus $t=12\pm\sqrt{46}$.

The cases where $t=12-\sqrt{46}$ and $t=12+\sqrt{46}$ are similar; they merely correspond to two triangles that can each be obtained by reflecting the other across the perpendicular bisector of $\overline{AC}$, so we consider the case where $t=12-\sqrt{46}$.

So $A=(0,0),B=(12-\sqrt{46},14),C=(24,0)$.

Thus $$AB\cdot BC=\sqrt{14^{2}+\left(12-\sqrt{46}\right)^{2}}\sqrt{14^{2}+\left(12+\sqrt{46}\right)^{2}}=\sqrt{\left(386-24\sqrt{46}\right)\left(386+24\sqrt{46}\right)}=\sqrt{386^{2}-46\cdot 24^{2}}=\boxed{350}.$$

The cases where $t=12-\sqrt{46}$ and $t=12+\sqrt{46}$ are shown below, labeled $\triangle ABC$ and $\triangle AB^{\prime}C$, respectively, where the dotted line is a midline in both triangles. As you can see, the orthocenter falls perfectly on that line for both triangles, and the value of $AB\cdot BC$ is the same for both triangles.

$[asy] real t = 12 - sqrt(46); pair A = (0, 0); pair B = (t, 14); pair C = (24, 0); draw(A--B--C--cycle); draw((A+B)/2--(B+C)/2, dashed); draw(B--(t, 0), blue); draw(A--(8.572, 11.55), red); draw(C--(2.927, 7.854), green); label("A", A, W); label("B", B, N); label("C", C, E); real t = 12 + sqrt(46); pair A = (0, 0); pair B = (t, 14); pair C = (24, 0); draw(A--B--C--cycle); draw((A+B)/2--(B+C)/2, dashed); draw(B--(t, 0), blue); draw(A--(21.073, 7.854), red); draw(C--(15.428, 11.55), green); label("A", A, W); label("B^{\prime}", B, N); label("C", C, E); [/asy]$

## Solution 2 (no coordinates)

$[asy] import geometry; point A = origin; point B = (12-sqrt(46),14); point C = (24,0); point M = midpoint(A--B); point N = midpoint(B--C); triangle t = triangle(A,B,C); point H = orthocentercenter(t); point D, E, F; line a = altitude(t.BC); line b = altitude(t.AC); line c = altitude(t.AB); // Triangle ABC and Segment MN draw(t); draw(M--N); // Altitudes pair[] e = intersectionpoints(b,A--C); E = e[0]; draw(B--E); pair[] d = intersectionpoints(a,B--C); D = d[0]; draw(A--D); pair[] f = intersectionpoints(c,A--B); F = f[0]; draw(C--F); // Labeling Points dot(A); label("A",A,SW); dot(B); label("B",B,NW); dot(C); label("C",C,SE); dot(M); label("M",M,WSW); dot(N); label("N",N,ENE); dot(H); label("H",H,SSE); dot(D); label("D",D,NE); dot(E); label("E",E,S); dot(F); label("F",F,NW); // Right angle marks markscalefactor = 0.15; draw(rightanglemark(B,H,M)); draw(rightanglemark(B,D,C)); [/asy]$

Let $M$ be the midpoint of $\overline{AB}$ and $N$ be the midpoint of $\overline{BC}$. Further let $D$ be the foot of the altitude from $A$, $E$ from $B$, and $F$ from $C$, as in the diagram.

Because $\overline{MN}$ is a midpoint connector of $\triangle ABC$ amd $H$ is on $\overline{MN}$ and $\overline{BE}$, we know that $H$ is the midpoint of altitude $BE$. Thus, because, from the problem, $BE=14$, $BH=HE=7$. Now we see that $\overline{MH}$ is a midpoint connector of $\triangle BAE$, so $MH=\tfrac12 AE$.

Now, let $\measuredangle CHE = \theta$. We know that $\measuredangle BHF = \measuredangle CHE = \theta$, because they are vertical angles. Because $\triangle BHF$ is right (by the definition of an altitude), we know that $\measuredangle MBH = 90^{\circ}-\theta$. $\triangle BHM$ is also right, so $\measuredangle BMH = 90^{\circ} -(90^{\circ}-\theta)=\theta$.

From $\triangle HEC$, we know that $\tan\theta = \tfrac{CE}7$. From $\triangle BHM$, we know that $\tan\theta = \tfrac7{AE/2} = \tfrac{14}{AE}$. Equating these two expressions for $\tan\theta$, we see that $AE \cdot CE = 7 \cdot 14 = 98$. From the problem, we know that $AE+CE=AC=24$.

Now, we can proceed as in Solution 1 by using the quadratic formula to solve for $AE$ and the Pythagorean Theorem to find $AB$ and $BC$. We do this to obtain our answer $AB \cdot BC = \boxed{350}$.