Difference between revisions of "1992 OIM Problems/Problem 6"

Latest revision as of 08:41, 23 December 2023

Problem

From the triangle $T$ with vertices $A$ , $B$ and $C$ , the hexagon $H$ with vertices $A_1$ , $A_2$ , $B_1$ , $B_2$ , $C_1$ , $C_2$ is constructed as shown in the figure.

Show that:

$area(H) \ge 13.area(T)$

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

For this problem were going to denote $H$ and $T$ as the area of hexagon $H$ and triangle $T$ respectively.

We know that the area of an isosceles triangle is given by $\frac{a^2sin(\theta)}{2}$ where $a$ is the length of the sides that are equal and $\theta$ is the angle between them.

We notice that there are 6 isosceles triangles in the hexagon with three of them sharing the area of the inside triangle in common Therefore we can write an equation for the area of $H$ by adding all of those triangles and then subtracting 2 times the area of the triangle:

$H=\frac{a^2sin(A)}{2}+\frac{b^2sin(B)}{2}+\frac{c^2sin(C)}{2}+\frac{(a+b)^2sin(C)}{2}+\frac{(a+c)^2sin(B)}{2}+\frac{(b+c)^2sin(A)}{2}-2T$

$2H=(a^2+b^2+c^2)(sin(A)+sin(B)+sin(C))+(2ab.sin(C)+2ac.sin(B)+2bc.sin(A))-4T$

Using the extended law of sines:

$sin(A)=\frac{a}{2R}$ ; $sin(B)=\frac{b}{2R}$ ; $sin(C)=\frac{c}{2R}$ where $R$ is the circumradius of the triangle

Substitute into $2H$ :

$2H=(a^2+b^2+c^2)\left( \frac{a}{2R}+\frac{b}{2R}+\frac{c}{2R} \right)+\left(2ab\frac{c}{2R}+2ac\frac{b}{2R}+2bc\frac{a}{2R}\right)-4T$

$2H=\left( \frac{(a^2+b^2+c^2)(a+b+c)}{2R} \right)+\left(2ab\frac{c}{2R}+2ac\frac{b}{2R}+2bc\frac{a}{2R}\right)-4T$

$2H= \frac{(a^2+b^2+c^2)(a+b+c)}{2R} +\frac{3abc}{R}-4T$

$H= \frac{(a^2+b^2+c^2)(a+b+c)}{4R} +\frac{3abc}{2R}-2T$

Also from the extended law of sines, $T=\frac{abc}{4R}$ therefore, $\frac{3abc}{2R}=6T$

$H= \frac{(a^2+b^2+c^2)(a+b+c)}{4R} +4T$

Applying AM-GM:

$\frac{(a^2+b^2+c^2)(a+b+c)}{9}=\frac{a^3+b^3+c^3+ac^2+ab^2+ba^2+bc^2+ca^2+cb^2}{9} \ge \sqrt[9]{a^9b^9c^9}$

Therefore,

$\frac{(a^2+b^2+c^2)(a+b+c)}{9} \ge abc$

$(a^2+b^2+c^2)(a+b+c) \ge 9abc$

$\frac{(a^2+b^2+c^2)(a+b+c)}{4R} \ge 9\frac{abc}{4R}$

$\frac{(a^2+b^2+c^2)(a+b+c)}{4R} \ge 9T$

$\frac{(a^2+b^2+c^2)(a+b+c)}{4R}+4T \ge 9T+4T$

Therefore,

$H \ge 13T$ , which proves: $area(H) \ge 13.area(T)$

Equality holds with equilateral triangle:

$T=\frac{a^2\sqrt{3}}{4}$

$R=\frac{a}{2sin(A)}=\frac{a}{2\frac{\sqrt{3}}{2}}=\frac{a}{\sqrt{3}}$

$H= \frac{(a^2+b^2+c^2)(a+b+c)}{4R} +4T=\frac{(3a^2)(3a)}{4\frac{a}{\sqrt{3}}} +4T=\frac{9a\sqrt{3}}{4}+4T=9T+4T=13T$

Note. I actually competed at this event in Venezuela when I was in High School representing Puerto Rico. I think I got 1 or 2 points out of 10 on this one. I don't remember what I did.

~Tomas Diaz. orders@tomasdiaz.com

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

@@ Line 65: / Line 65: @@
 <math>R=\frac{a}{2sin(A)}=\frac{a}{2\frac{\sqrt{3}}{2}}=\frac{a}{\sqrt{3}}</math>
-<math>H= \frac{(3a^2)(3a)}{4\frac{a}{\sqrt{3}}} +4T=\frac{9a\sqrt{3}}{4}+4T=9T+4T=13T</math>
+<math>H= \frac{(a^2+b^2+c^2)(a+b+c)}{4R} +4T=\frac{(3a^2)(3a)}{4\frac{a}{\sqrt{3}}} +4T=\frac{9a\sqrt{3}}{4}+4T=9T+4T=13T</math>
 * Note.  I actually competed at this event in Venezuela when I was in High School representing Puerto Rico.  I think I got 1 or 2 points out of 10 on this one.  I don't remember what I did.
+~Tomas Diaz. orders@tomasdiaz.com
 {{alternate solutions}}
 == See also ==
+[[OIM Problems and Solutions]]
 https://www.oma.org.ar/enunciados/ibe7.htm

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Difference between revisions of "1992 OIM Problems/Problem 6"

Latest revision as of 08:41, 23 December 2023

Problem

Solution

See also