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1986 AIME Problems/Problem 15

Revision as of 13:20, 18 April 2020 by Aopsuser101 (talk | contribs) (→‎Solution 3)

Problem

Let triangle $ABC$ be a right triangle in the xy-plane with a right angle at $C_{}$. Given that the length of the hypotenuse $AB$ is $60$, and that the medians through $A$ and $B$ lie along the lines $y=x+3$ and $y=2x+4$ respectively, find the area of triangle $ABC$.

Solution 1

Translate so the medians are $y = x$, and $y = 2x$, then model the points $A: (a,a)$ and $B: (b,2b)$. $(0,0)$ is the centroid, and is the average of the vertices, so $C: (- a - b, - a - 2b)$
$AB = 60$ so

$3600 = (a - b)^2 + (2b - a)^2$
$3600 = 2a^2 + 5b^2 - 6ab \ \ \ \ (1)$

$AC$ and $BC$ are perpendicular, so the product of their slopes is $-1$, giving

$\left(\frac {2a + 2b}{2a + b}\right)\left(\frac {a + 4b}{a + 2b}\right) = - 1$
$2a^2 + 5b^2 = - \frac {15}{2}ab \ \ \ \ (2)$

Combining $(1)$ and $(2)$, we get $ab = - \frac {800}{3}$

Using the determinant product for area of a triangle (this simplifies nicely, add columns 1 and 2, add rows 2 and 3), the area is $\left|\frac {3}{2}ab\right|$, so we get the answer to be $400$.

Solution 2

The only relevant part about the xy plane here is that the slopes of the medians determine an angle between them that we will use. This solution uses the tangent subtraction identity $\tan(\alpha -\beta)=\frac{\tan\alpha -\tan\beta}{1+\tan\alpha \tan\beta}$. Therefore, the tangent of the acute angle between the medians from A and B will be $\frac{2-1}{1+2 \cdot 1}= \frac{1}{3}$.

Let $AC =x$ and $BC =y$. Let the midpoint of AC be M and the midpoint of BC be N. Let the centroid be G. By exterior angles, $\angle CMB - \angle CAN = \angle AGM$. However we know that since $\angle AGM$ is the acute angle formed by the medians, $\tan \angle AGM=\frac{1}{3}$. We can express the tangents of the other two angles in terms of $x$ and $y$. $\tan \angle CMB= \frac{y}{\frac{x}{2}}=\frac{2y}{x}$ while $\tan \angle CAN= \frac{\frac{y}{2}}{x}=\frac{y}{2x}$. For simplification, let $\frac{y}{x}=r$.

By the tangent subtraction identity, $\frac{2r-\frac{r}{2}}{1+2r \cdot \frac{r}{2}}=\frac{3r}{2(1+r^2)}=\frac{1}{3}$. We get the quadratic $2r^2 -9r+2=0$, which solves to give $r=\frac{9 \pm \sqrt{65}}{4}$. It does not matter which one is picked, because the two roots multiply to 1, so switching from one root to another is like switching the lengths of AC and BC.

We choose the positive root and we can plug $y=(\frac{9 + \sqrt{65}}{4})x$ into of course the PYTHAGOREAN THEOREM $x^2 + y^2 =3600$ and solve for $x^2=200(9-\sqrt{65})$. We want the area which is $\frac{xy}{2}=\frac{x^2(\frac{9 + \sqrt{65}}{4})}{2}=\frac{200(9-\sqrt{65})(\frac{9 + \sqrt{65}}{4})}{2}=\frac{200\cdot\frac{16}{4}}{2}=400$.

Solution 3

We first seek to find the angle between the lines $y = x + 3$ and $y = 2x + 4$. [asy] import graph; size(150); Label f; f.p=fontsize(6); xaxis(-8,8,Ticks(f, 2.0)); yaxis(-8,8,Ticks(f, 2.0)); real f(real x) { return (x + 3); } real g( real x){ return (2x + 4); } draw(graph(f,-8,5),red+linewidth(1)); draw(graph(g,-6,2),blue+linewidth(1)); [/asy] Let the acute angle the red line makes with the $x-$ axis be $\alpha$ and the acute angle the blue line makes with the $x-$ axis be $\beta$. Then, we know that $\tan \alpha = 1$ and $\tan \beta = 2$. Note that the acute angle between the red and blue lines is clearly $\beta - \alpha$. Therefore, we have that: \[\tan (\beta - \alpha) = \frac{2 - 1}{1 + 2} = \frac{1}{3}\]It follows that $\cos (\beta - \alpha) = \frac{1}{\sqrt{10}}$ and $\sin (\beta - \alpha) = \frac{3}{\sqrt{10}}$. From now on, refer to $\theta = \beta - \alpha$ [asy] import graph; size(4cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -5.96, xmax = 24.22, ymin = -10.06, ymax = 14.62; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); draw(arc((2,2.6666666666666665),0.6,290.55604521958344,326.3099324740202)--(2,2.6666666666666665)--cycle, linewidth(1)); draw((0,8)--(0,0), linewidth(1) + wrwrwr); draw((0,0)--(6,0), linewidth(1) + wrwrwr); draw((6,0)--(0,8), linewidth(1) + wrwrwr); draw((0,8)--(3,0), linewidth(1) + wrwrwr); draw((0,4)--(6,0), linewidth(1) + wrwrwr); dot((0,0),dotstyle); label("$A$", (0.08,0.2), NE * labelscalefactor); dot((6,0),dotstyle); label("$B$", (6.08,0.2), NE * labelscalefactor); dot((0,8),dotstyle); label("$C$", (0.08,8.2), NE * labelscalefactor); dot((3,0),linewidth(4pt) + dotstyle); label("$D$", (3.08,0.16), NE * labelscalefactor); dot((0,4),linewidth(4pt) + dotstyle); label("$E$", (0.08,4.16), NE * labelscalefactor); label("theta ", (2.14,2.3), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy] Suppose $ABC$ is our desired triangle. Let $E$ be the midpoint of $CA$ and $D$ be the midpoint of $AB$ such that $CE = EA = m$ and $AD = DB = n$. Let $G$ be the centroid of the triangle (in other words, the intersection of $EB$ and $CD$). It follows that if $CG = 2x$, $GD = x$ and if $GB = 2y$, $GE = y$. By the Law of Cosines on $\triangle GEB$, we get that: \[GE^2 + GC^2 - 2GE \cdot GC \cdot \cos \theta = CE^2 \Longleftrightarrow\]\[(2x)^2 + y^2 - (2x)(y)(2)\cos \theta = m^2\]By the Law of Cosines on $\triangle GDB$, we get that: \[DG^2 + GB^2 -2 DG \cdot GB \cdot \cos \theta = DB^2 \Longleftrightarrow\]\[(2y)^2+x^2-(2y)(x)(2)\cos \theta = n^2\]Adding yields that: \[5x^2+5y^2 - 8xy\cos \theta = m^2 + n^2\]However, note that $(2m)^2 + (2n)^2 = 60^2 \Longleftrightarrow m^2+n^2 = 30^2$. Therefore, \[5x^2+5y^2 - 8xy\cos \theta = 30^2\]We also know that by the Law of Cosines on $\triangle CGB$, \[CG^2 + GB^2 - 2CG \cdot GB \cdot \cos (180 - \theta) = CB^2 \Longleftrightarrow\]\[(2x)^2 + (2y)^2 + 2(2x)(2y) \cos \theta = 60^2 \Longleftrightarrow\]\[x^2+y^2+2xy \cos \theta = 30^2 \Longleftrightarrow\]\[5x^2+5y^2 + 10xy \cos \theta = 30^2 \cdot 5\]Subtracting this from the $5x^2+5y^2 - 8xy\cos \theta = 30^2$ we got earlier yields that: \[xy \cos \theta = 200\]but recall that $\cos \theta = \frac{3}{\sqrt{10}}$ to get that: \[xy = \frac{200 \sqrt{10}}{3}\]Plugging this into the $x^2 + y^2 + 2xy \cos \theta = 30^2$, we get that: \[x^2 + y^2 = 500\]Aha! How convenient. Recall that $(2x)^2 + y^2 - (2x)(y)(2)\cos \theta = m^2$ and $(2y)^2+x^2-(2y)(x)(2)\cos \theta = n^2$. Then, we clearly have that: \[m^2n^2 = ((2x)^2 + y^2 - (2x)(y)(2)\cos \theta)((2y)^2+x^2-(2y)(x)(2)\cos \theta = n^2) \Longleftrightarrow\]\[m^2n^2 = 17x^2y^2 - 4xy \cos \theta (5x^2 + 5y^2) + 16x^2y^2 \cos ^2 \theta + 4x^4 + 4y^4 \Longleftrightarrow\]\[m^2n^2 = 17 \cdot \frac{200^2 \cdot 10}{9} - 800(5 \cdot 500) + 16 \cdot 200^2 + 4 \left( (x^2+y^2)^2-2x^2y^2\right) \Longleftrightarrow\]\[\frac{m^2n^2}{100^2} = \frac{680}{9} - 8 \cdot 5 \cdot 5 + 16 \cdot 2^2 + 4 \left( 5^2 - 2 \cdot \frac{2^2 \cdot 10}{9}\right) \Longleftrightarrow\]\[\frac{m^2n^2}{100^2} = 4\]But note that the area of our triangle is $2m \cdot 2n \cdot \frac {1}{2} = 2mn$. As $mn = 200$, we get a final answer of $\boxed{400}$.

See also

1986 AIME (Problems • Answer Key • Resources)
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