1982 USAMO Problems/Problem 3

Problem

If a point $A_1$ is in the interior of an equilateral triangle $ABC$ and point $A_2$ is in the interior of $\triangle{A_1BC}$ , prove that

$I.Q. (A_1BC) > I.Q.(A_2BC)$ ,

where the isoperimetric quotient of a figure $F$ is defined by

$I.Q.(F) = \frac{\text{Area (F)}}{\text{[Perimeter (F)]}^2}$

Solution

First, an arbitrary triangle $ABC$ has isoperimetric quotient (using the notation $[ABC]$ for area and $s = \frac{a + b + c}{2}$ ): $\frac{[ABC]}{4s^2} = \frac{[ABC]^3}{4s^2 [ABC]^2} = \frac{r^3 s^3}{4s^2 \cdot s(s-a)(s-b)(s-c)} = \frac{r^3}{4(s-a)(s-b)(s-c)}$ $= \frac{1}{4} \cdot \frac{r}{s-a} \cdot \frac{r}{s-b} \cdot \frac{r}{s-c} = \frac{1}{4} \tan A/2 \tan B/2 \tan C/2.$

Lemma. $\tan x \tan (A - x)$ is increasing on $0 < x < \frac{A}{2}$ , where $0 < A < 90^\circ$ .

Proof. $\tan x \tan (A - x) = \tan x \cdot \frac{\tan A - \tan x}{1 + \tan A \tan x} = 1 - \frac{1}{\cos^2 x (1 + \tan A \tan x)}$ $= 1 - \frac{2}{1 + \cos 2x + \tan A \sin 2x} = 1 - \frac{2}{1 + \sec A \cos (A - 2x)}$ is increasing on the desired interval, because $\cos (A - 2x)$ is increasing on $0 < x < \frac{A}{2}.$

Let $x_1, y_1, z_1$ and $x_2, y_2, z_2$ be half of the angles of triangles $A_1 BC$ and $A_2 BC$ in that order, respectively. Then it is immediate that $30^\circ > y_1 > y_2$ , $30^\circ > z_1 > z_2$ , and $x_1 + y_1 + z_1 = x_2 + y_2 + z_2 = 90^\circ$ . Hence, by Lemma it follows that $\tan x_1 \tan y_1 \tan z_1 = \tan (90^\circ - y_1 - z_1) \tan y_1 \tan z_1 > \tan (90^\circ - y_1 - z_2) \tan y_1 \tan z_2$ $> \tan (90^\circ - y_2 - z_2) \tan y_2 \tan z_2 = \tan x_2 \tan y_2 \tan z_2.$ Multiplying this inequality by $\frac{1}{4}$ gives that $I.Q[A_1 BC] > I.Q[A_2 BC]$ , as desired.

1982 USAMO (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5
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1982 USAMO Problems/Problem 3

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