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Difference between revisions of "1986 AIME Problems/Problem 15"

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We get the quadratic <math>2r^2 -9r+2=0</math>, which solves to give <math>r=\frac{9 \pm \sqrt{65}}{4}</math>. 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 get the quadratic <math>2r^2 -9r+2=0</math>, which solves to give <math>r=\frac{9 \pm \sqrt{65}}{4}</math>. 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 <math>y=(\frac{9 + \sqrt{65}}{4})x</math> into of course the PYTHAGOREAN THEOREM <math>x^2 + y^2 =3600</math> and solve for <math>x^2=200(9-\sqrt{65})</math>. We want the area which is <math>\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\frac{16}{4}}{2}=400</math>.
+
We choose the positive root and we can plug <math>y=(\frac{9 + \sqrt{65}}{4})x</math> into of course the PYTHAGOREAN THEOREM <math>x^2 + y^2 =3600</math> and solve for <math>x^2=200(9-\sqrt{65})</math>. We want the area which is <math>\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</math>.
  
 
== See also ==
 
== See also ==

Revision as of 23:01, 26 December 2019

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 $\frac{xy}{2}=\frac{x^2(\frac{9 + \sqrt{65}}{4})}{2}=\frac{200(9-\sqrt{65})(\frac{9 + \sqrt{65}}{4})}{2}=\frac{200\cdot\frac{16}{4}}{2}=400$.

See also

1986 AIME (Problems • Answer Key • Resources)
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