Category Archives: Geometry

October 13, 2024 · 4:59 am

Tricky Area Ratio Question

A circle of radius $r$ , is inscribed within an isosceles triangle $ABC$ . $CA=CB=5r$ . Given that $\angle{ACB}$ is acute, find the ratio of the area of the circle to that of triangle $ABC$ .

Mathematics Specialist 3AB Question15 page 55

I came across this question while searching for an area of sectors and segments question.

Here’s a diagram

We know $AC, AB and BC$ are tangents to the circle. Because the triangle is isosceles, the distance from $A$ to the circle is the same as the distance from $B$ to the circle.

$CP$ is perpendicular to $AB$ (because the triangle is isosceles).

Because it is proportional, i.e. $r$ and $5r$ , we can let $r=1$

Let $h=CP$

(1) $\begin{equation*}h=\sqrt{25-a^2}\end{equation*}$

but $h$ also equals

(2) $\begin{equation*}h=1+\sqrt{(5-a)^2+1}\end{equation*}$

Set equation $1$ equal to equation $2$

$\begin{equation*}\sqrt{25-a^2}={1+\sqrt{(5-a)^2+1}\end{equation}$

Square both sides

$\begin{equation*}25-a^2=1+(5-a)^2+1+2\sqrt{(5-a)^2+1}\end{equation}$

Expand and simplify

$\begin{equation*}25-a^2=2+25-10a+a^2+2\sqrt{(5-a)^2+1}\end{equation}$

$\begin{equation*}0=2-10a+2a^2+2\sqrt{(5-a)^2+1}\end{equation}$

Divide by $2$

$\begin{equation*}0=1-5a+a^2+\sqrt{(5-a)^2+1}\end{equation}$

$\begin{equation*}-\sqrt{(5-a)^2+1}=a^2-5a+1\end{equation}$

Square both sides

$\begin{equation*}(5-a)^2+1=a^4-5a^3+a^2-5a^3+25a^2-5a+a^2-5a+1\end{equation}$

$\begin{equation*}25-10a+a^2+1=a^4-10a^3+27a^2-10a+1\end{equation}$

$\begin{equation*}0=a^4-10a^3+26a^2-25 \end{equation}$

I solved this using a graphics calculator

$a=-0.8434, a=1.3068, a=4.5367, a=5$

We can reject $a=-0.8434$ , $a=4.5367$ , and $a=5$ . If $a=5$ , there isn’t a triangle, and if $a=4.5367$ $\angle{ACB}$ is not acute.

Hence the area of triangle $ABC=a\times h$

$\begin{equation*}A_{\Delta}=1.3068\times\sqrt{25-1.3068^2}\end{equation}$

(3) $\begin{equation*}A_{\Delta}=6.3069\end{equation*}$

(4) $\begin{equation*}A_{\circ}=\pi\end{equation*}$

Hence, equation $4$ divided by equation $3$ is

$\begin{equation*}\frac{\pi}{6.3069}\approx 0.5\end{equation}$

Perhaps I approached this question in the wrong way. Is there an easier process?

Intersecting Circles

Two circles of radius $L$ and $2L$ intersect as shown. What is the area of the shaded region?

From Professor Povey’s Perplexing Problems

My plan is to find the sum of the area of the two segments (see below).

Construct triangles

The diagonal of the square (the pink line above) has length

$=\sqrt{(2L)^2+(2L)^2}=\sqrt{8L^2}=2\sqrt{2}L$

From the pink triangle in the above diagram, I am going to find the angles using the cosine rule.

$cos\theta=\frac{L^2+(2\sqrt{2}L)^2-(2L)^2}{2(L)(2\sqrt{2}L)}$

$cos\theta=\frac{5L^2}{4\sqrt{2}L^2}$

$cos\theta=\frac{5}{4\sqrt{2}}$

$\theta=0.487$

$cos\alpha=\frac{(2L)^2+(2\sqrt{2}L)^2-L^2}{(2\sqrt{2}L)(2L)}$

$cos\alpha=\frac{11L^2}{8\sqrt{2}L^2}$

$cos\alpha=\frac{11}{8\sqrt{2}}$

$\alpha=0.236$

The green quadrilateral is a kite, which means the diagonals are perpendicular.

This means the segment angles are $2\theta$ and $2\alpha$ (because the triangles are isosceles and the diagonal is perpendicular to the base of the triangles).

Area of green segment

$A=\frac{1}{2}r^2(\theta-sin\theta)$

$A=\frac{1}{2}L^2(0.9734-sin(0.9734))=0.0733L^2$

Area of yellow segment

$A=\frac{1}{2}4L^2(0.4721-sin(0.4721))=0.0035$

Total Area = $0.0733L^2+0.0035L^2=0.108L^2$

Geometry Problem

This problem is from Geometry Snacks by Ed Southall and Vincent Pantaloni – it’s a great book.

Two squares are constructed such that three vertices are collinear as shown. Find the value of the marked angle.

I started by marking in the right angles. And I added the diagonal of the larger square (pink line).

Because there are right angles at $O$ and $P$ , we know there is a circle, which has the diagonal of the square as its diameter (see second image below).