$P$ is inside square $ABCD$ such that $PD=29$ and $PB=23$. Find the area of $\triangle APC$.

Question

A typical geometry problem that will blow your mind:

Let $ABCD$ be a square and $P$ be the point inside it such that $PD=29$ and $PB=23$. Find the area of $\triangle APC$.

Answer 1

Let us work in a coordinate plane such that :

$$A (0,0), B (a,0), C(a,a), D (0,a), P(x,y)$$

(where $a$ is the square's sidelength).

Distance constraints give :

$$\begin{cases}PB^2&=&(x-a)^2+y^2&=&23^2\\PD^2&=&x^2+(y-a)^2&=&29^2\end{cases}\tag{1}$$

(Geometrical interpretation : $P$ is any of the two intersection points $P_1$ and $P_2$ of circles $C_1$ and $C_2$ with resp. centers $B$ and $D$ and radii $23$ and $29$.)

Subtracting equations in (1), we get:

$$2xa-2ya=312. \tag{2}$$

Therefore, the area of triangle $APC$ is:

$$\frac12 \det(\vec{AP},\vec{AC})=\frac12 \begin{vmatrix}x & a\\y & a\end{vmatrix}=\frac12 a(x-y)=78.\tag{3}$$

(using (2)).

Remarks:

1) There are two paradoxes in this result:

• it is independent from sidelength $a$.

• it is valid for any of the intersection points $P_1$ and $P_2$.

2) Sidelength $a$ must be such that $\dfrac{29-23}{\sqrt{2}}\leq a\leq \dfrac{29+23}{\sqrt{2}}$ in order that circles $C_1$ and $C_2$ have at least one intersection point (not necessarily included in the square).

3) Relationship (2) can be interpreted as an equation for the radical axis of the 2 circles, which here is the equation of the line $P_1P_2$.