# Random Math Problems (2)

Okay, these problems are not really random, they’re from the same section of a textbook:

This book is called Geometry Revisited (by Coxeter and Greitzer). It’s often recommended as a olympiad level geometry textbook. I would recommend this book too (although I haven’t gone through the entire book yet).

Section 1.1 of this book proves a theorem referred to as the Extended Law of Sines:

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$

where $R$ is the circumradius of triangle $ABC$.

The extended Law of Sines is often given in a truncated form, without mention to the circumradius.

After the proof the book gives four exercises, which I shall rephrase and solve here.

### Problem 1

Show that in any triangle $\triangle ABC$,

$a = b \cos C + c \cos B$

From this, deduce the ‘addition formula’ for sines:

$\sin (B+C) = \sin B \cos C + \sin C \cos B$

#### Solution

From the Law of Cosines:

$\begin{array}{rcl} c^2 &=& a^2 + b^2 - 2ab \cos C \\ b^2 &=& a^2 + c^2 - 2ac \cos B \end{array}$

Adding the two:

$\begin{array}{rcl} c^2 + b^2 &=& 2a^2 + c^2 + b^2 - 2ab \cos C - 2ac \cos B \\ 2a^2 &=& 2ab \cos C + 2ac \cos B \\ a &=& b \cos C + c \cos B \end{array}$

as desired.

We now prove the addition formula. From the law of sines:

$\begin{array}{rcl} \frac{a}{\sin A} &=& 2R \\ a &=& 2R \sin A \end{array}$

Similarly:

$\begin{array}{rcl} b &=& 2R \sin B \\ c &=& 2R \sin C \end{array}$

Substituting into our previous equation:

$\begin{array}{rcl} a &=& b \cos C + c \cos B \\ 2R \sin A &=& 2R \sin B \cos C + 2R \sin C \cos B \\ \sin A &=& \sin B \cos C + \sin C \cos B \end{array}$

Since $\triangle ABC$ is a triangle, $A+B+C = 180$. Thus, $A = 180 - (B+C)$.

Note that $\sin \theta = \sin (180 - \theta)$. Thus,

$\sin (B+C) = \sin A = \sin B \cos C + \sin C \cos B$

as desired.

### Problem 2

Show that in any triangle $\triangle ABC$,

$a (\sin B - \sin C) + b (\sin C - \sin A) + c(\sin A -\sin B) = 0$

#### Solution

Using the law of sines again:

$\begin{array}{rcl} \frac{a}{\sin A} &=& \frac{b}{\sin B} \\ a \sin B &=& b \sin A \\ a \sin B - b \sin A &=& 0 \end{array}$

Similarly,

$\begin{array}{rcl} b \sin C - c \sin B &=& 0 \\ c \sin A - a \sin C &=& 0 \end{array}$

Adding the three equations:

$\begin{array}{rcl} a \sin B - b \sin A + b \sin C - c \sin B + c \sin A - a \sin C &=& 0 \\ a (\sin B - \sin C) + b(\sin C - \sin A) + c(\sin A - \sin B) &=& 0 \end{array}$

as desired.

### Problem 3

Show that in any triangle $\triangle ABC$,

$[ABC] = \frac{abc}{4R}$

(although the book uses $(ABC)$, I’ve accustomed to writing $[ABC]$ on my blog to denote area of $\triangle ABC$)

#### Solution

Let h be the height of $\triangle ABC$ (from A perpendicular to BC), so that $[ABC] = \frac{1}{2} ah$:

Using trigonometry,

$h = b \sin C$

so we can write the area as

$[ABC] = \frac{1}{2} ab \sin C$

From the extended Law of Sines,

$\begin{array}{rcl} \frac{c}{\sin C} &=& 2R \\ \sin C &=& \frac{c}{2R} \end{array}$

Now we can rewrite the area again:

$\begin{array}{rcl} [ABC] &=& \frac{1}{2}ab \frac{c}{2R} \\ &=& \frac{abc}{4R} \end{array}$

as desired.

### Problem 4

For a triangle $\triangle ABC$, let p be the radius of a circle going through A and tangent to BC at B. Let q be the radius of a circle going through A and tangent to BC at C.

Show that:

$pq = R^2$

#### Solution

Let M be the center of the circle with radius p, and N be the center of the circle with radius q. Also K and L are two points on circles M and N respectively:

Using the law of sines on circle N and M, we have:

$\begin{array}{rcl} \frac{b}{\sin L} &=& 2q \\ \frac{c}{\sin K} &=& 2p \end{array}$

Because BC is tangent to the two circles, $\angle L = \angle C$ and $\angle K = \angle B$.

Thus,

$\begin{array}{rcl} \frac{b}{\sin C} = 2q & \Rightarrow & b = 2q \sin C \\ \frac{c}{\sin B} = 2p & \Rightarrow & c = 2p \sin B \end{array}$

Now applying the Law of Sines again in $\triangle ABC$:

$\begin{array}{rcl} \frac{b}{\sin B} = 2R & \Rightarrow & \sin B = \frac{b}{2R} \\ \frac{c}{\sin C} = 2R & \Rightarrow & \sin C = \frac{c}{2R} \end{array}$

Substituting these values back into the previous equation:

$\begin{array}{rcl} b = 2q \frac{c}{2R} = \frac{qc}{R} & \Rightarrow & q = \frac{bR}{c} \\ c = 2p \frac{b}{2R} = \frac{pb}{R} & \Rightarrow & p = \frac{cR}{b}\end{array}$

By multiplication,

$pq = \frac{bR}{c} \frac{cR}{b} = R^2$

as desired.