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:
where is the circumradius of triangle
.
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 ,
From this, deduce the ‘addition formula’ for sines:
Solution
From the Law of Cosines:
Adding the two:
as desired.
We now prove the addition formula. From the law of sines:
Similarly:
Substituting into our previous equation:
Since is a triangle,
. Thus,
.
Note that . Thus,
as desired.
Problem 2
Show that in any triangle ,
Solution
Using the law of sines again:
Similarly,
Adding the three equations:
as desired.
Problem 3
Show that in any triangle ,
(although the book uses , I’ve accustomed to writing
on my blog to denote area of
)
Solution
Let h be the height of (from A perpendicular to BC), so that
:
Using trigonometry,
so we can write the area as
From the extended Law of Sines,
Now we can rewrite the area again:
as desired.
Problem 4
For a triangle , 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:
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:
Because BC is tangent to the two circles, and
.
Thus,
Now applying the Law of Sines again in :
Substituting these values back into the previous equation:
By multiplication,
as desired.