## Problem 8.1.12

We are given an inscribed triangle $ABC$. Let $R$ denote the circumradius; let $h_a$ denote the altitude AD.

a. Show that triangles $ABD$ and $ALC$ are similar, and hence that $h_a=2R=bc$. b. Show that the area of $\Delta ABC$ is $abc/4R$.

### Solution

a. In triangles $ABC$ and $ALC$, $\angle ABC$ is equal to $\angle ALC$ and $\angle ACL = \angle BAC= 90 ^\circ$. Therefore triangles $ABD$ and $ALC$ are similar.We also have

\begin{align*} \frac{c}{h_a} &= \frac{2R}{b}\\ \implies h_a &= \frac{bc}{2R} \end{align*}

b. Area of the triangle $ABC$ is

\begin{align*} \frac{1}{2}ah_a &= \frac{1}{2}a\frac{bc}{2R} = \frac{abc}{4R} \end{align*}

## Problem 8.1.13

The radius of the inscribed circle of a triangle is $4$,and the segments into which one side is divided by the point of contact are $6$ and $8$.Determine the other two sides.

### Solution

The sides of the triangle $14,6+x$ and $8+x$.Hence, $s=(14+6+x+8+x)/2=14+x$.

\begin{align*} \Delta = \sqrt{(14+x)\cdot x \cdot 8 \cdot 6} &= 4(14+x) \\ \implies 14 + x &= 3x \\ \implies x &= 7 \end{align*}

The other two sides of the triangle are $13$ and $21$.

## Problem 8.1.14

Triangles $ABC$ and $DEF$ are inscribed in the same circle.Prove that

\begin{align*} sin A + sin B+ sin C = sin D + sin E + sin F \end{align*}

if and only if the perimeters of the given triangles are equal.

Let $a_i,b_i,c_i$ for $i=1,2$ be the sides of the two triangles and $R$ the circumradius. We have,

### Solution

\begin{align*} sin A + sin B+ sin C &= sin D + sin E+ sin F \\ \iff \frac{a_1}{2R} + \frac{b_1}{2R} + \frac{c_1}{2R} &= \frac{a_2}{2R} + \frac{b_2}{2R} + \frac{c_2}{2R} \\ \iff a_1 + b_1+ c_1 &= a_2 + b_2+ c_2 \end{align*}

## Problem 8.1.15

In the following figure, $CD$ is a half chord perpendicular to the diameter $AB$ of the semicircle with center $O$. A circle with center $P$ is inscribed as shown in Figure 8.13, touching $AB$ at $E$ and arc $BD$ at $F$.Prove that $\Delta AED$ is isoceles.

### Solution

The key observation is that $O,P$ and $F$ are collinear. This is because the circle $P$ is tangent to semicircle at $F$. Let $r$ be the radius of circle $P$ and $R$ be that of the semicircle. Then $OP=R-r$ and from the right triangle $OPE$,we get

\begin{align*} (R-r)^2 = r^2 + (r+OC)^2 \end{align*}

Also, $R^2=OC^2+CD^2$ from the right triangle $OCD$.Combining these two equations we get:

\begin{align*} DC^2 = r^2 + 2rOC + 2rR \end{align*}

But

\begin{align*} AE = R + r + OC \end{align*}

Therefore,

\begin{align*} AE^2 = r^2 + 2r(R+OC) + (R+OC)^2 = CD^2 + AC^2 \end{align*}

Since $AD^2=CD^2+AC^2$, $AE=AD$.

## Problem 8.1.16

Find the length of a side of an equilateral triangle in which the distances from its vertices to an interior point are $5,7,$ and $8$.

### Solution

We have

\begin{align*} cos\theta = \frac{5^2+8^2-7^2}{2\cdot5\cdot8} = \frac{1}{2} \end{align*}

We also have

\begin{align*} cos (60^\circ + \theta) = \frac{1}{2}\cdot\frac{1}{2} - \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{3}}{2} &= \frac{5^2+8^2-s^2}{2\cdot5\cdot8}\\ \implies s^2 &= 129 \end{align*}

## Problem 8.4.5

Let $A_0,A_1,A_2,A_3,A_4$ divide a unit circle into five equal parts.Prove that the chords $A_0A_1$, $A_0A_2$ satisfy

\begin{align*} (A_0A_1 \cdot A_0A_2)^2 = 5 \end{align*}

### Solution

If $A_0=1$,then $A_1=e^{i2\pi/5}$ and $A_2=e^{-i2\pi/5}$.We have

\begin{align*} (A_0A_1 \cdot A_0A_2)^2 \\ = (e^{i2\pi/5}-1)( e^{-i2\pi/5}-1)(e^{-i2\pi/5}-e^{i2\pi/5})(e^{i2\pi/5}-e^{-i2\pi/5}) \\ = 4(1-cos(2\pi/5))(1-cos(4\pi/5)) \\ = 8(1-cos^2(\pi/5))(1+cos(\pi/5)) \end{align*}

When $5\theta=\pi$, we have

\begin{align*} cos3\theta &= -cos2\theta \\ \implies 4cos^3\theta - 3cos\theta &= -2cos^2\theta+1 \\ \implies 4cos^3\theta +2cos^2\theta- 3cos\theta -1 &=0 \end{align*}

\begin{align*} cos4\theta &= -cos\theta \\ \implies 8cos^4\theta - 8cos^2\theta + 1 &= -cos\theta \\ \implies 8cos^2\theta(cos^2\theta -1) &= -(1+cos\theta) \\ \implies 8cos^2\theta(cos\theta -1) &= -1 \\ \implies 8cos^3\theta - 8cos^2\theta +1 &= 0 \end{align*}

From the above two equations, we have

\begin{align*} 4cos^2\theta-2cos\theta -1 = 0 \end{align*}

Therefore, $cos\theta = \frac{\sqrt{5}+1}{4}$.

Therefore,

\begin{align*} (A_0A_1 \cdot A_0A_2)^2 &= 8(1-cos^2(\pi/5))(1+cos(\pi/5)) \\ &= 8(\frac{10-2\sqrt{5}}{16})(\frac{\sqrt{5}+5}{4}) = 5 \end{align*}

## Problem 8.4.6

Given a point on the circumference of a unit circle and the vertices $A_1,A_2,\dots,A_n$ of an inscribed regular polygon of $n$ sides,prove that $PA_1^4+PA_2^4+\dots+PA_n^4$ is a constant (i.e., independent of the position of $P$ on the circumference).

### Solution

The vertices of the regular polygon can be represented by $e^{i2\pi k/n}$ where $k=0,1,2,\dots,n-1$. Let $z$ be any point on the circumference of the circle. We have

\begin{align*} PA_1^4+PA_2^4+\dots+PA_n^4 \\ = \sum_{k=0}^{n-1} \left((z-e^\frac{i2\pi k}{n})(\overline{z}-e^{-\frac{i2\pi k}{n}}\right)^2 \\ = \sum_{k=0}^{n-1} \left(2-z e^{-\frac{i2\pi k}{n}} -\overline{z}e^\frac{i2\pi k}{n}\right)^2 \\ = \sum_{k=0}^{n-1} 4 + z^2e^{-\frac{i4\pi k}{n}} + \overline{z}^2e^\frac{i4\pi k}{n} -4z e^{-\frac{i2\pi k}{n}} -4\overline{z}e^\frac{i2\pi k}{n} + 2\\ = 6n \end{align*}

## Problem 8.4.7

Let $G$ denote the centroid of triangle $ABC$.Prove that

\begin{align*} 3(GA^2+GB^2+GC^2)=AB^2+BC^2+CA^2 \end{align*}

### Solution

WLOG, we can assume that the centroid of the triangle with vertices at $A(z_1)$,$B(z_2)$ and $C(z_3)$ is located at $0$. We then have $z_1+z_2+z+3=0$.

\begin{align*} AB^2+BC^2+CA^2 = \\ (z_1-z_2)(\overline z_1-\overline z_2)+ (z_2-z_3)(\overline z_2-\overline z_3)+ (z_3-z_1)(\overline z_3-\overline z_1) \\ = z_1\overline z_1 + z_2\overline z_2-z_2\overline z_1 - z_1\overline z_2 + z_2\overline z_2 + z_3\overline z_3-z_2\overline z_3 - z_3\overline z_2 + \\ z_3\overline z_3 + z_1\overline z_1-z_1\overline z_3 - z_3\overline z_1 \\ = 2(z_1\overline z_1 + z_2\overline z_2 +z_3\overline z_3) \\ - \overline z_1(z_2 + z_3) - \overline z_2(z_1 + z_3) - \overline z_3(z_1 + z_2)\\ = 2(z_1\overline z_1 + z_2\overline z_2 +z_3\overline z_3)- \overline z_1(-z_1) - \overline z_2(-z_2) - \overline z_3(-z_3)\\ = 3(z_1\overline z_1 + z_2\overline z_2 +z_3\overline z_3) = 3(GA^2+ GB^2+ GC^2) \end{align*}

## Problem 8.4.8

Let $ABCDEF$ be a hexagon in a circle of radius $r$. Show that if $AB=CD=EF=r$, then the midpoints of $BC,DE$ and $FA$ are the vertices of an equilateral triangle.

### Solution

If $O$ is the center of the circle circumscribing the hexagon, $OAB$, $OCD$ and $OEF$ are equilateral triangles. If $A,C$, and $E$ are represented by $z_1,z_2$ and $z_3$, the coordinates of $B,D$ and $F$ are $z_1e^{i\pi/3}$, $z_2e^{i\pi/3}$ and $z_3e^{i\pi/3}$.The midpoints of $BC$, $DE$ and $FA$ are $(z_1e^{i\pi/3} + z_2)/2$, $(z_2e^{i\pi/3} + z_3)/2$ and $(z_3e^{i\pi/3} + z_1)/2$.

Whenever $c_1,c_2$ and $c_3$ are complex numbers that are the vertices of an equilateral triangle, we have

\begin{align*} c_1^2+c_2^2+c_3^2 = c_1 c_2 + c_2 c_3 + c_3 c_1 \end{align*}

We have,

\begin{align*} \left(\frac{z_1e^{i\pi/3} + z_2}{2}\right)^2 + \left(\frac{z_2e^{i\pi/3} + z_3}{2}\right)^2 + \left(\frac{z_3e^{i\pi/3} + z_1}{2}\right)^2 \\ = \frac{z_1^2e^{i2\pi/3} + z_2^2 + 2z_1 z_2 e^{i\pi/3}}{4} + \frac{z_2^2e^{i2\pi/3} + z_3^2 + 2z_1 z_3 e^{i\pi/3}}{4} \\ + \frac{z_3^2e^{i2\pi/3} + z_1^2 + 2z_1 z_3 e^{i\pi/3}}{4} \\ = (z_1^2(e^{i2\pi/3}+1) + z_2^2(e^{i2\pi/3}+1) + z_3^2(e^{i2\pi/3}+1) + 2z_1 z_2 e^{i\pi/3} \\ + 2z_2 z_3 e^{i\pi/3} +2z_3 z_1 e^{i\pi/3})/4\\ = (z_1^2e^{i\pi/3} + z_2^2e^{i\pi/3} + z_3^2e^{i\pi/3} + \\ 2z_1 z_2 e^{i\pi/3}+ 2z_2 z_3 e^{i\pi/3} +2z_3 z_1 e^{i\pi/3})/4 \end{align*}

We also have,

\begin{align*} \left(\frac{z_1e^{i\pi/3} + z_2}{2}\right) \left(\frac{z_2e^{i\pi/3} + z_3}{2}\right) + \left(\frac{z_2e^{i\pi/3} + z_3}{2}\right) \left(\frac{z_3e^{i\pi/3} + z_1}{2}\right) \\ +\left(\frac{z_3e^{i\pi/3} + z_1}{2}\right) \left(\frac{z_1e^{i\pi/3} + z_2}{2}\right) \\ = (z_1^2e^{i\pi/3} + z_2^2e^{i\pi/3} + z_3^2e^{i\pi/3} \\ + z_1 z_2(1 + e^{i\pi/3}+e^{i2\pi/3}) + z_2 z_3(1 + e^{i\pi/3}+ e^{i2\pi/3}) \\ + z_3 z_1(1 + e^{i\pi/3}+ e^{i2\pi/3})/4 \\ = (z_1^2e^{i\pi/3} + z_2^2e^{i\pi/3} + z_3^2e^{i\pi/3} \\ + 2z_1 z_2 e^{i\pi/3}+ 2z_2 z_3 e^{i\pi/3} +2z_3 z_1 e^{i\pi/3})/4 \end{align*}

Therefore, the midpoints of $BC$, $DE$ and $FA$ form an equilateral triangle.

## Problem 8.4.9

If $z_1,z_2,z_3$ are such that $\vert z_1 \vert=\vert z_2 \vert= \vert z_3 \vert=1$ and $z_1+z_2+z_3=0$,show that $z_1,z_2,z_3$ are the vertices of an equilateral triangle inscribed in a unit circle.

### Solution

The lengths of the three medians of the triangle are given by

\begin{align*} \left\lvert \frac{z_1+z_2-2z_3}{2} \right\rvert &= \left\lvert \frac{-z_3-2z_3}{2} \right\rvert = \frac{3}{2}\\ \left\lvert \frac{z_1+z_3-2z_2}{2} \right\rvert &= \left\lvert \frac{-z_2-2z_2}{2} \right\rvert = \frac{3}{2}\\ \left\lvert \frac{z_2+z_3-2z_1}{2} \right\rvert &= \left\lvert \frac{-z_1-2z_1}{2} \right\rvert = \frac{3}{2} \end{align*}

As the lengths of the three medians are equal, the triangle whose vertices are $z_1,z_2$ and $z_3$ is an equilateral triangle.

## Problem 8.4.10

Show that $z_1,z_2,z_3$ form an equilateral triangle if and only if

\begin{align*} z_1^2+z_2^2+z_3^2 = z_1z_2 + z_2z_3+z_3z_1 \end{align*}

### Solution

Rotating one side by $60^{\circ}$ we get the second side in an equilateral triangle, so we have

\begin{align*} z_3-z_1 = (z_2-z_1)e^{i\pi/3} \\ \iff \frac{z_3-z_1}{z_2-z_1} - \frac{1}{2} = i\frac{\sqrt{3}}{2} \\ \iff \frac{(2z_3-z_1-z_2)^2}{4(z_2-z_1)^2} = \frac{-3}{4} \\ \iff 4z_3^2 + z_1^2 + z_2^2 +2z_1z_2-4z_3z_1-4z_3z_2 \\ = -3z_2^2-3z_1^2+6z_2z_1 \\ \iff 4z_3^2 + 4z_1^2 + 4z_2^2 = 4z_1z_2 + 4z_2z_3 + 4z_3z_1 \\ \iff z_1^2+z_2^2+z_3^2 = z_1z_2 + z_2z_3+z_3z_1 \end{align*}

## Problem 8.4.11

The three points in the complex plane which correspond to the roots of the equation

\begin{align*} z^3 -3pz^2 + 3qz -r = 0 \end{align*}

are the vertices of a triangle.

a. Prove that the centroid of the triangle is the point corresponding to $p$.

b. Prove that $ABC$ is an equilateral triangle if and only if $p^2=q$.

### Solution

If $z_1,z_2,z_3$ are the roots of the above equation, we have

\begin{align*} z_1+z_2+z_3 &= 3p\\ z_1z_2+z_2z_3+z_3z_1 &= 3q\\ \end{align*}

a. If $z_1,z_2,z_3$ are the vertices of a triangle,we have the centroid at $(z_1+z_2+z_3)/3 = p$.

b. If $z_1,z_2,z_3$ are the vertices of an equilateral triangle, we have

\begin{align*} z_1^2+z_2^2+z_3^2 &= z_1z_2 + z_2z_3+z_3z_1 \\ \implies 9p^2-6q &= 3q \\ \implies p^2 &= q \end{align*}