Tag: equal areas

Take a triangle $ABC$. Place a compass point at $A$, draw a circle, and adjust its radius until half the triangle lies inside the circle. What does the answer look like?

For an equilateral triangle of side $1$, the calculation is pleasantly short. The portion inside the circle is a sector with angle $\pi/3$. Since the triangle has area $\sqrt3/4$, we want

$$\frac{\pi r^2}{6}=\frac{\sqrt3}{8}, \qquad\text{hence}\qquad r=\sqrt{\frac{3\sqrt3}{4\pi}}.$$

The circle has not yet reached the opposite side $BC$, so the sector calculation really does describe the part of the triangle inside the circle. A vertex-centered circular arc bisecting an equilateral triangle also appears in Sanjoy Mahajan’s 2008 MIT course materials.

Now change the triangle. The circle might meet $BC$ before it captures half the area. Or it might grow past one of the other vertices. The figure above shows the three possible pictures.

A radius always exists

As the radius increases from $0$, the area of the triangle inside the circle increases continuously from $0$ to the area of the whole triangle. It increases strictly until the farther vertex is reached. Therefore exactly one radius divides the triangle into equal areas.

We can scale the triangle without changing which of the three pictures occurs. To simplify the calculations, label the vertices so that $AC\le AB$, and set $AC=1$. Write $A,B,C$ for the angles, measured in radians. By the sine rule,

$$AB=L=\frac{\sin C}{\sin B}, \qquad T=\operatorname{area}(ABC)=\frac{L\sin A}{2}.$$

The desired area inside the circle is $T/2$.

Case 1: A sector is enough

While the circle remains inside the triangle’s angle at $A$, its area inside the triangle is $Ar^2/2$. The candidate radius is thus

$$\boxed{r_0=\sqrt{\frac{T}{A}}=\sqrt{\frac{L\sin A}{2A}}.}$$

We still need to check whether the circle reaches $BC$. If both base angles are acute, the distance from $A$ to $BC$ is $h=\sin C$, so this is the answer precisely when $r_0\le h$. The test can be written entirely in angles:

$$\boxed{\cot B+\cot C\le 2A.}$$

If $C$ is right or obtuse, the nearest point of the segment $BC$ to $A$ is $C$. In that situation the sector answer works when $r_0\le AC=1$.

Case 2: The circle crosses $BC$ twice

Suppose both base angles are acute, $h<r<1$, and the circle cuts $BC$ at two points. Start with a sector and subtract the circular segment lying beyond $BC$:

$$S(r)=\frac{Ar^2}{2} -r^2\arccos\!\left(\frac{h}{r}\right) +h\sqrt{r^2-h^2}, \qquad h=\sin C.$$

The equal-area radius is the unique solution of $S(r)=T/2$ in $(h,1)$. We can determine before solving whether it lies there. At $r=1$,

$$S(1)=\frac{C-B}{2}+\sin C\cos C.$$

If the sector candidate already exceeds $h$, and $S(1)\ge T/2$, we are in this case. Equality gives the boundary case $r=1$.

For example, a triangle with $A=150^\circ$ and $B=C=15^\circ$ has an equal-area radius of approximately $0.335593$ when $AC=1$. Its circle crosses $BC$ twice.

Case 3: The circle passes a vertex

If half the area has still not been captured at $r=1$, the circle passes the nearer vertex $C$. The remaining part outside the circle sits near $B$, and $1<r<L$.

Let $D$ be the circle’s intersection with $BC$, and set $\phi=\angle BAD$. The sine rule in triangle $ABD$ gives

$$\phi=\arcsin\!\left(\frac{\sin C}{r}\right)-B.$$

The outside area equals the area of triangle $ABD$ minus a sector centered at $A$:

$$U(r)=\frac{Lr\sin\phi-r^2\phi}{2}.$$

Set $U(r)=T/2$, or equivalently $Lr\sin\phi-r^2\phi=T$. This determines the unique radius between $1$ and $L$. As an example, $A=80^\circ, B=20^\circ, C=80^\circ$ gives $r\approx1.014031$ when $AC=1$.

The equilateral triangle leads to a square-root formula. Other triangles lead to a circular segment, and some require us to measure the outside area instead. The same question changes character as the angles change.

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Another Equal-Area Problem

Here the challenge was to use a circle to divide the area of a triangle exactly in half. What happens if we reverse the roles and ask a circle to divide the area of another circle in half? Surprisingly, we can even require the center of the cutting circle to lie outside the original circle.

Continue exploring: Can a Circle Centered Outside Another Circle Cut Its Area Exactly in Half?