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2013 Mock AIME I Problems/Problem 12

Revision as of 21:37, 8 June 2022 by Sugar rush (talk | contribs)

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 $0<t<24$.

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]

See also

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