1986 AIME Problem 15

Question

Let triangle

ABC

be a right triangle in the

xy

-plane with a right angle at

C.

Given that the hypotenuse

AB

has length

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.

Answer

The intersection of the two median lines is the centroid

Parametrize

A

and

B

from the centroid along direction vectors

(1,1)

and

(1,2)

The median lines meet at the centroid

G=(-1,2).

For real

u

and

v,

write

\begin{aligned} A&=G+(u,u),\\ B&=G+(v,2v). \end{aligned}

Since

A+B+C=3G,

C=G-(u,u)-(v,2v).

The condition

AC\perp BC

gives

\begin{aligned} &(2u+v,2u+2v)\\ &\qquad\mathbin{\cdot}(u+2v,u+4v)=0, \end{aligned}

or

4u^2+15uv+10v^2=0.

Meanwhile

AB=60

gives

\begin{aligned} &(u-v)^2+(u-2v)^2\\ &\qquad=2u^2-6uv+5v^2\\ &\qquad=3600. \end{aligned}

Subtracting half of the first equation from the second yields

-\frac{27}{2}uv=3600,

so

uv=-\frac{800}{3}.

Using the two perpendicular legs from

C,

the area is half the absolute determinant:

\begin{aligned} [ABC] &=\frac12\left| \det\begin{pmatrix}2u+v&2u+2v\\u+2v&u+4v\end{pmatrix} \right|\\ &=\frac12|3uv| =400. \end{aligned}

Problem 15 in other years