Balancing with Geometry

Problem

An isosceles triangle is to be cut from a square plate of uniform density $\rho$, and width $a$ (as shown below). What is the height of the removed isosceles triangle such that the resulting shape balances at the tip of the triangle?

Solution

Partition the shape into two zones: the rectangle, and the triangles. The pivot point is the tip of the removed isosceles triangle. The two triangles forms an isosceles triangle equal in area to the removed segment. Due to its symmetry, this problem is equivalent to inverting the triangle regions. Therefore, to balance the torques on both sides, we calculate:

${\tau}_{rectangle} = {\tau}_{triangle}$.

To do so, we must know where the center-of-mass of the rectangle, and the center-of-mass of the isosceles triangle are located. Since the material is of uniform density, the torques are proportional to the areas of each shape. Let $a$ be the width of the square, $h$ be the height of the triangle, and $a - h$ be the height of the rectangle. The center-of-mass of an isosceles triangle is located at $\frac{1}{3}h$ from its base. Relative to the pivot, the center-of-mass for the isosceles triangle is $\frac{2}{3}h$. It is visually obvious (perhaps intuitive is a better word) that the center-of-mass of the rectangle is located at $\frac{1}{2}(a - h)$.

Consequently,

${\tau}_{rectangle} = {\tau}_{triangle}$

$\rho \left({\frac {1}{2}}ah\right)\left({\frac {2}{3}}h\right)=\rho \left(a(a-h)\right)\left({\frac {1}{2}}(a-h)\right)$.

This torque balancing equation reduces to

$\frac{1}{3}h^2 = \frac{1}{2}(a - h)^2$

$0=3a^{2}-6ah+2h^{2}$.

The solutions for the height of the triangle is $h = a\left(\frac{3 \pm \sqrt{3}}{2}\right)$. However, the plus sign solution makes no sense because it would imply that the pivot is located outside the plate. Therefore, the height of the isosceles triangle is

$h=a\left({\frac {3-{\sqrt {3}}}{2}}\right)$.