I used a bit of linear algebra to solve

Geometry Level 3

Suppose you have a triangle ABC, with $A = (0, 0), B = (1, 3), C = (3, 1)$ . Point P lies on $\overline{AC}$ , such that the area of triangle APB equals the area of triangle BPC. Point P can be represented in the form $\left( \dfrac{a}{b}, \dfrac{c}{d} \right)$ , where $a, b$ are coprime positive integers and $c, d$ are coprime positive integers. Express your answer in the form $abcd$ by concatenating the digits of $a, b, c \text{ and } d$ . For example, if you think P is $\left( \dfrac{1}{2}, \dfrac{3}{4} \right)$ , enter your answer as $1234$ .

The answer is 3212.

3 solutions

Hobart Pao
Dec 17, 2017

There's a really easy way to do the problem actually using symmetry. However, I found a very interesting formula in one of Felix Klein's Geometry books for the area of a triangle, which expresses it as a determinant and wanted to use it here. If the vertices of the triangle are given by $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ , then the area of the triangle is $\dfrac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$ . Let P be $(x_4, y_4)$ . Then we have:

$\dfrac{1}{2} \begin{vmatrix} 1 & 3 & 1 \\ 0 & 0 & 1 \\ x_4 & y_4 & 1 \end{vmatrix} = \dfrac{1}{2} \begin{vmatrix} 1 & 3 & 1 \\ x_4 & y_4 & 1 \\ 3 & 1 & 1 \end{vmatrix}$

$3x_4 - y_4 = -2x_4 - 2y_4 + 8$ $y_4 = -5x_4 + 8$

Since we know P lies on $\overline{AC}$ , well we know the formula of $\overline{AC}$ is $y = \dfrac{1}{3} x$ , so P has to satisfy this as well and we can substitute $y_4 = \dfrac{1}{3} x_4$ into the equation above obtained from the two determinants:

$\dfrac{1}{3} x = -5x_4 + 8$
$x_4 = \dfrac{3}{2}$ $y_4 = \dfrac{1}{3} \cdot \dfrac{3}{2} = \dfrac{1}{2}$

As these fractions are all in simplest form already, the answer is simply $\boxed{3212}$

Interesting formula but it overcomplicates things here. Since P lies on AC both the triangles have the same height and need the same base to give equal areas. Hence P lies on the midpoint of AC which obviously is the point you found. And this method doesn't only work because of the neat numbers you gave...It's an overall solution in case P lies on AC and A, B and C are random points forming the triangle.

Peter van der Linden - 3 years, 5 months ago

I know that now after doing some work, I just forgot to change the comment.