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.

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