Triangle Post

by math_explorer, May 24, 2011, 10:13 AM

(Obsolete, now that BigSams made this wonderful thread + PDF.)

For probably all triangles $\triangle ABC$:
common centers:
$O$: circumcenter
$H$: orthocenter
$G$: centroid
$I$: incenter
$I_A, I_B, I_C$: excenters

side lengths: $a, b, c$
substitution: $a = y + z$, $b = z + x$, $c = x + y$
triangle inequality turns into the fact that $x, y, z$ are positive (zero if degenerate but w/e)

common variables:
$\triangle$: area (sometimes $A$ or $S$)
$s$: semiperimeter, or half of perimeter, $\frac{a + b + c}{2} = x + y + z$ (sometimes $p$)
$r$: inradius, radius of incircle
$R$: circumradius, radius of circumcircle (also twice radius of 9-point circle)
$r_A, r_B, r_C$: exradii, radii of excircles

Law of Sines: $a/\sin A = b/\sin B = c/\sin C = 2R$
Area: $\triangle = ab\sin C/2 = abc/4R$

Heron's: $\triangle = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{xyz(x+y+z)}$
inradius-semiperimeter (by splitting into $IAB$, $IBC$, $ICA$) $\triangle = rs$
therefore, $r = \sqrt{\frac{xyz}{s}}$

corollary: $abc = 4Rrs$

exradius-semiperimeter: analogously $\triangle = r_Ax = r_By = r_Cz$
therefore $r_A = \sqrt{\frac{yzs}{x}}$ a.s.o.

Euler's formula: $d^2 = R(R - 2r)$, $d$ is distance between circumcenter and incenter
Euler's inequality (implied by above): $R \geq 2r$, equality iff $\triangle ABC$ is equilateral

Law of Cosines: $c^2 = a^2 + b^2 - 2ab\cos C$
$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$
bashing yields $\sum \cos C = 1 + \frac{r}{R}$

distances to orthocenter: $AH = 2R \cos A$ a.s.o.

Erdos-Mordell: twice the sum of distances from $P$ to each side $\leq$ the sum of distances from $P$ to each vertex; equality iff $\triangle ABC$ is equilateral and $P$ is its center

Hadwiger-Finsler: $\sum_{\text{cyc}} a^2 \geq \sum_{\text{cyc}} (a-b)^2 + 4\sqrt{3}\triangle$
Weitzenbock (obv. weaker): $\sum_{\text{cyc}} a^2 \geq 4\sqrt{3}\triangle$
(equality in both cases iff $\triangle ABC$ is equilateral)

Half-angle trig (simply use $AI$ as hypotenuse, either side as adjacent side)
opposite: $r$, adjacent: $x$, hypotenuse: $\sqrt{r^2 + x^2} = \sqrt{\frac{xyz + x^2s}{s}} = \sqrt{\frac{x(x+y)(x+z)}{s}}$

corollary: $\prod_{\text{cyc}} \sin A/2 = \frac{r^3}{\sqrt{xyz((x+y)(y+z)(z+x))^2/s^3}}$

$= \frac{r^3}{\sqrt{\frac{xyz}{s}}\times\frac{abc}{s}}$
$= \frac{r^3}{r\times\frac{4R\triangle}{s}}$
$= \frac{r^3}{r\times 4Rr} = \frac{4r}{R}$
This post has been edited 9 times. Last edited by math_explorer, Nov 5, 2016, 10:56 PM

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What's the difference between Hadwiger-Finsler and Weitzenbock?

by chaotic_iak, May 24, 2011, 11:31 AM

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darn another fail
Fixed.

by math_explorer, May 24, 2011, 12:04 PM

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