1982 USAMO Problems/Problem 3: Difference between revisions
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Proof. <cmath>\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)}</cmath> | Proof. <cmath>\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)}</cmath> | ||
<cmath>= 1 - \frac{2}{1 + \cos 2x + \tan A \sin 2x} = 1 - \frac{2}{1 + \sec A \cos (A - 2x)</cmath> is increasing on the desired interval, because <math>\cos (A - 2x)</math> is increasing on <math>0 < x < \frac{A}{2}.</math> | <cmath>= 1 - \frac{2}{1 + \cos 2x + \tan A \sin 2x} = 1 - \frac{2}{1 + \sec A \cos (A - 2x)}</cmath> is increasing on the desired interval, because <math>\cos (A - 2x)</math> is increasing on <math>0 < x < \frac{A}{2}.</math> | ||
Let <math>x_1, y_1, z_1</math> and <math>x_2, y_2, z_2</math> be half of the angles of triangles <math>A_1 BC</math> and <math>A_2 BC</math> in that order, respectively. Then it is immediate that <math>30^\circ > y_1 > y_2</math>, <math>30^\circ > z_1 > z_2</math>, and <math>x_1 + y_1 + z_1 = x_2 + y_2 + z_2 = 90^\circ</math>. Hence, by Lemma it follows that | Let <math>x_1, y_1, z_1</math> and <math>x_2, y_2, z_2</math> be half of the angles of triangles <math>A_1 BC</math> and <math>A_2 BC</math> in that order, respectively. Then it is immediate that <math>30^\circ > y_1 > y_2</math>, <math>30^\circ > z_1 > z_2</math>, and <math>x_1 + y_1 + z_1 = x_2 + y_2 + z_2 = 90^\circ</math>. Hence, by Lemma it follows that | ||
Latest revision as of 22:29, 18 May 2015
Problem
If a point 
is in the interior of an equilateral triangle 
and point 
is in the interior of 
, prove that
,
where the isoperimetric quotient of a figure 
is defined by
![$I.Q.(F) = \frac{\text{Area (F)}}{\text{[Perimeter (F)]}^2}$](http://latex.artofproblemsolving.com/5/9/6/59605bef5fa499b868bb9d25da48807658139b0a.png)
Solution
First, an arbitrary triangle has isoperimetric quotient (using the notation ![$[ABC]$](http://latex.artofproblemsolving.com/d/3/3/d33cc80fa8f093e155c5be46d2e5d9da3d7e1ef5.png)
for area and 
):
![\[\frac{[ABC]}{4s^2} = \frac{[ABC]^3}{4s^2 [ABC]^2} = \frac{r^3 s^3}{4s^2 \cdot s(s-a)(s-b)(s-c)} = \frac{r^3}{4(s-a)(s-b)(s-c)}\]](http://latex.artofproblemsolving.com/5/4/4/5446232498508437b7a797bd9e7e49d1e0d6eaca.png)
![\[= \frac{1}{4} \cdot \frac{r}{s-a} \cdot \frac{r}{s-b} \cdot \frac{r}{s-c} = \frac{1}{4} \tan A/2 \tan B/2 \tan C/2.\]](http://latex.artofproblemsolving.com/9/8/9/989660aefdfcde9e881ab39eeaacb5798951a179.png)
Lemma. 
is increasing on 
, where 
.
Proof. ![\[\tan x \tan (A - x) = \tan x \cdot \frac{\tan A - \tan x}{1 + \tan A \tan x} = 1 - \frac{1}{\cos^2 x (1 + \tan A \tan x)}\]](http://latex.artofproblemsolving.com/0/2/2/0227a0403c25294417508b570735a601402ba27c.png)
![\[= 1 - \frac{2}{1 + \cos 2x + \tan A \sin 2x} = 1 - \frac{2}{1 + \sec A \cos (A - 2x)}\]](http://latex.artofproblemsolving.com/5/2/4/5247911c108dc1ab453b8def453fe271d56a61ed.png)
is increasing on the desired interval, because 
is increasing on 
Let 
and 
be half of the angles of triangles 
and 
in that order, respectively. Then it is immediate that 
, 
, and 
. Hence, by Lemma it follows that
![\[\tan x_1 \tan y_1 \tan z_1 = \tan (90^\circ - y_1 - z_1) \tan y_1 \tan z_1 > \tan (90^\circ - y_1 - z_2) \tan y_1 \tan z_2\]](http://latex.artofproblemsolving.com/0/f/d/0fd9cb051af690cb5e8423c1c8f27b1680628105.png)
![\[> \tan (90^\circ - y_2 - z_2) \tan y_2 \tan z_2 = \tan x_2 \tan y_2 \tan z_2.\]](http://latex.artofproblemsolving.com/a/0/7/a079f11e251c3e377f1ba0542d33856815e9fcf1.png)
Multiplying this inequality by 
gives that ![$I.Q[A_1 BC] > I.Q[A_2 BC]$](http://latex.artofproblemsolving.com/d/2/9/d29336270fb31f616ba75fa9409d76d2593d75e8.png)
, as desired.
See Also
| 1982 USAMO (Problems • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America. 
