Euler's Inequality

One of the oldest inequalities about triangles is that relating the radii of the circumcircle and incircle. It was proved by Euler and is contained in the following theorems. Proofs are given in Geometry Revisited by Coxeter and Greitzer. It is published by the Mathematical Association of America and should be on the bookshelf of everyone interested in geometry.

Theorem 1 (Euler 1765)   Let O and I be the circumcenter and incenter, respectively, of a triangle with circumradius R and inradius r; let d be the distance OI. Then

\begin{displaymath}d^2 = R^2-2Rr\end{displaymath}

Theorem 2   In a triangle with circumradius R and inradius r, R $\ge$ 2r.

Here are seven other interesting and useful facts about triangles. Let $s$ denote the semiperimeter of triangle $ABC$, $\alpha ,\beta, \gamma$ the angles, $a,b,c$ the opposite sides, and $K$ the area.
  1. $ K = \frac {1}{2}ab\sin\gamma =\frac {1}{2}ac\sin\beta =\frac {1}{2}bc\sin\alpha.$
  2. $ K = \sqrt{s(s-a)(s-b)(s-c)}.$ (Heron's formula)
  3. $ K = rs$
  4. $ 2R = \frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{c}{\sin\gamma}$ (Law of Sines).
  5. $ K = \frac{abc}{4R}$.
  6. $ 1+ \cos\alpha = {\frac{(a+b+c)(-a+b+c)}{2bc}}$ $ 1- \cos\alpha = {\frac{(a-b+c)(a+b-c)}{2bc}}$.
  7. $ \sin\frac{\alpha}{2}=\sqrt{\frac{(s-b)(s-c)}{bc}}$ $ \cos\frac{\alpha}{2}=\sqrt{\frac{s(s-a)}{bc}}$ $ \tan\frac{\alpha}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}={\frac{r}{s-a}}$.
Formulas similar to those in (6) and (7) can also be written for the angles $\beta$ and $\gamma$. To see (1), drop an altitude from $C$ to $c$ forming a right triangle. The area is one-half the product of the base $c$ and the altitude. But the altitude equals $a\sin\beta$. To see (2), again drop an altitude, $h$, forming two right triangles with bases $x$ and $c-x$. Use the Pythagorean Theorem twice and eliminate the altitude to solve for $x = \frac{a^2-b^2 +c^2}{2c}$ (Note $x = a\cos\beta$). Now, substitute $x$ back into $ h^2 = a^2-x^2$. Use $A^2-B^2=(A-B)(A+B)$ and $ A^2+2AB+B^2 = (A+B)^2$ to expand. Then multiply by $4c^2$ giving $(b+c-a)(a+b-c)(a+c-b)(a+b+c)$. For more details see pages 337-338 of Geometry, Second Edition by Harold Jacobs. For a proof using trigonometry see Cyclic quadrangles; Brahmagupta's formula on pages 56-59 of Geometry Revisited by Coxeter and Greitzer. Heron's formula is then seen to be a corollary to Brahmagupta's formula. To see (3), divide the triangle into three triangles with segments from the incenter to the vertices. To see (4), circumscribe the triangle and draw a diameter from one of the vertices. Draw a chord from the other endpoint of the diameter to a second vertex of the triangle. Note that the angle at the third vertex is equal to the angle formed by the diameter and the chord, or supplementary to it, if the third angle is not acute. Therefore, the two angles have equal sines. To see (5), use (1) and (4). To see (6), solve the Law of Cosines for $\cos\alpha$ and add $1$ or subtract from $1$. To see (7), use the half-angle formulas $\sin^2\frac{\alpha}{2}=\frac{1-\cos\alpha}{2}$, $\cos^2\frac{\alpha}{2}=\frac{1+\cos\alpha}{2}$, and (6). For the final part of (7) use the first two parts of (7) and formulas (2) and (3).