## A Triangle and Two Circles

A circle of radius R is circumscribed around an isosceles triangle of height h and base b. The triangle is circumscribed around a unit circle (radius = 1).
1. Find b in terms of h.
2. Find R in terms of h. If h=4, find R.
3. Find h such that the base of the triangle is the diameter of the large circle (b=2R).
4. Find h in terms of R. If R=4, find h.
An isosceles triangle has two sides equal in length, with the third side called the "base"; for this problem, the height is measured perpendicular to the base. To be circumscribed means that each circle touches the triangle in three places.

Source: Original.

Solutions were received from Al Zimmerman, Philippe Fondanaiche, John Hewson, Larry Baum, Henry Bottomley, and Denis Borris . Larry Baum's solutions are below:
```To simplify typing, let b=2c and let m be the side length of the of the
non-base sides of ABC.

Then:
1) area(ABC) = bh/2 = ch
2) Let C1 be the inscribed unit circle with center pt P.  Draw lines from P
to the three vertices.  This gives three triangles APB, BPC, APC with
respective areas: b*1/2 (= c), m*1/2, m*1/2 (since each has altitude 1).  So:
area(ABC) = c + m = c + sqrt(c^2 + h^2)

Thus: ch = c + sqrt(c^2 + h^2) => c(h-1) = sqrt(c^2 + h^2) =>
c^2(h-1)^2 = c^2 + h^2 => c^2 = h/(h-2) =>

A) b = 2sqrt(h/(h-2))    (b in terms of h)

Let Q be the center of circle CR that inscribes the triangle.  Then the
altitude of ABC passes thru Q.  Let x = length of QK, where K is the point
where the altitude bisects the base.  Then

h = R+x
R^2 = c^2 + x^2 = c^2 + (h-R)^2 = c^2 + h^2 - 2hR + R^2 =>
2hR = h/(h-2) + h^2 => 2R = 1/(h-2) + h = (h-1)^2/(h-2) =>

B) R = (h-1)^2/2(h-2)
When h=4, R=9/4

If b=2R, then ABC is a right triangle and its altitude = R.
Thus (from A) 2R = 2sqrt(R/(R-2)) => R^2 = R/R-2 => R^2-2R-1 = 0 =>

C) h = R = 1+sqrt(2)

By B, 2hR - 4R = h^2 - 2h + 1 => h^2 - (2+2R)h + (1+4R) = 0 =>
h = [(2+2R) +/- sqrt((2+2R)^2 - 4(1+4R))]/2 =>

D) h = 1+R +/- sqrt(R^2-2R)
When R=4, h = 5+/-sqrt(8) = 7.828427125 or 2.171572875
```

Mail to Ken