# Chaitanya's Random Pages

## December 23, 2015

### Areas of sections of a triangle from distances to its sides

Filed under: mathematics — ckrao @ 12:35 pm

If a point $P$ is in the interior of triangle $ABC$ distance $x, y$ and $z$ from the sides, what is the ratio of the area of quadrilateral $BXPZ$ to that of $ABC$?

One way of determining this is to draw parallels to the sides of the triangles through $P$. Let $X_1$ and $X_2$ be where these parallels meet side $BC$ as shown below.

Let the sides of the triangles have lengths $a, b, c$ with corresponding altitudes $h_a, h_b, h_c$.

Then as $\triangle PX_1 X_2$ and $\triangle ACB$ are similar,

\begin{aligned}|PX_1X| &= |PX_1X_2| \frac{X_1X}{X_1X_2}\\ &= |PX_1X_2| \frac{b\cos C}{a}\\ &= |PX_1X_2| \frac{b (a^2 + b^2 - c^2)}{2a^2b} \quad \text{ (cosine rule)}\\ &= \left(\frac{x}{h_a} \right)^2|ABC|\frac{(a^2 + b^2 - c^2)}{2a^2}\\ &= |ABC|\left(\frac{ax}{ax+by+cz}\right)^2\frac{(a^2 + b^2 - c^2)}{2a^2}\\&= \frac{|ABC|x^2(a^2 + b^2 - c^2)}{2(ax+by+cz)^2},\quad\quad (1) \end{aligned}

where the second last line follows from twice the area of |ABC| being $ah_a = ax + by + cz$.

Similarly,

$\displaystyle |PY_1Z| = \frac{|ABC|z^2(b^2 + c^2 - a^2)}{2(ax+by+cz)^2}.\quad \quad (2)$

Finally,

\begin{aligned}|X_1Y_1B| &= \left(\frac{h_b-y}{h_b}\right)^2|ABC|\\ &= \left(1-\frac{by}{bh_b}\right)^2 |ABC|\\ &= \left(1-\frac{by}{2|ABC|}\right)^2|ABC|\\ &= \left(1-\frac{by}{ax+by+cz}\right)^2|ABC|\\ &= \left(\frac{ax +cz}{ax+by+cz}\right)^2|ABC|. \quad\quad(3)\end{aligned}

Combining (1), (2) and (3), we obtain our desired answer as

\begin{aligned} \frac{|BXPZ|}{|ABC|} &= \frac{|X_1Y_1B|-|PX_1X|-|PY_1Z|}{|ABC|}\\&= \left(\frac{ax +cz}{ax+by+cz}\right)^2-\frac{x^2(a^2 + b^2 - c^2)}{2(ax+by+cz)^2}-\frac{z^2(b^2 + c^2 - a^2)}{2(ax+by+cz)^2}\\&=\frac{(ax+cz)^2 - x^2(a^2 +b^2-c^2)/2 - z^2(b^2+c^2-a^2)/2}{(ax+by+cz)^2}\\ &= \frac{2axcz + x^2(a^2 - b^2 + c^2) + z^2(c^2 +a^2-b^2)}{(ax+by+cz)^2}\\&= \frac{4axcz + (x^2 + z^2)(a^2 - b^2 + c^2)}{2(ax+by+cz)^2}.\quad\quad(4)\end{aligned}

Similar formulas can be found for quadrilaterals $XPYC$ and $YPZA$ by permuting variables. Note that if $P$ is outside the triangle or if the triangle is obtuse-angled, care must be taken in the signs of the areas (the quadrilaterals may not be convex) and variables $x, y, z$.

Note that (4) may also be written as

$\displaystyle \frac{|BXPZ|}{|ABC|} = \frac{ac(2xz + (x^2 + z^2)\cos B)}{(ax+by+cz)^2}.\quad\quad(5)$

### Special cases

1) If $\triangle ABC$ is equilateral, $a=b=c$ and from (4) we obtain

\begin{aligned} \frac{|BXPZ|}{|ABC|} &= \frac{4axcz + (x^2 + z^2)(a^2 - b^2 + c^2)}{2(ax+by+cz)^2}\\ &= \frac{4a^2xz + (x^2 + z^2)(a^2)}{2a^2(x+y+z)^2}\\ &= \frac{4xz + x^2 + z^2}{2(x+y+z)^2}.\quad\quad(6)\end{aligned}

2) If $P$ is at the incentre of $\triangle ABC$, then $x = y = z = r$ (the inradius) and from (4) we have

\begin{aligned} \frac{|BXPZ|}{|ABC|} &= \frac{4axcz + (x^2 + z^2)(a^2 - b^2 + c^2)}{2(ax+by+cz)^2}\\ &= \frac{4a^2xz + (x^2 + z^2)(a^2)}{2a^2(x+y+z)^2}\\ &= \frac{4xz + x^2 + z^2}{2(x+y+z)^2}.\quad\quad(7)\end{aligned}

3) If $\triangle P$ is right-angled at $B$, then quadrilateral $BXPZ$ is a rectangle with area $xz$ and $\triangle ABC$ has area $ac/2$ and from (5),

\begin{aligned} \frac{|BXPZ|}{|ABC|} &= \frac{2acxz )}{(ax+by+cz)^2}\\ &= \frac{2acxz )}{(ac)^2}\\ &= \frac{2xz}{ac}.\quad \quad (8)\end{aligned}

as expected.

4) If $a=c$ and $x=z$ (symmetric isosceles triangle case) then from (4),

\begin{aligned} \frac{|BXPZ|}{|ABC|} &= \frac{4axcz + (x^2 + z^2)(a^2 - b^2 + c^2)}{2(ax+by+cz)^2}\\ &= \frac{4a^2x^2 + 2x^2(2a^2-b^2)}{2(2ax+by)^2}\\ &= \frac{x^2(4a^2 -b^2)}{(2ax+by)^2}.\quad\quad(9)\end{aligned}