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Difference between revisions of "1992 OIM Problems/Problem 6"
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<math>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</math>
 
<math>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</math>
 +
 +
<math>2H=(a^2+b^2+c^2)(sin(A)+sin(B)+sin(C))+(2ab.sin(C)+2ac.sin(B)+2bc.sin(A))-4T</math>
 +
 +
Using the extended law of sines:
 +
 +
<math>sin(A)=\frac{a}{2R}</math>; <math>sin(B)=\frac{b}{2R}</math>; <math>sin(C)=\frac{c}{2R}</math> where <math>R</math> is the circumradius of the triangle
 +
 +
Substitute into <math>2H</math>:
 +
 +
<math>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</math>
 +
 +
<math>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</math>
 +
 +
<math>2H= \frac{(a^2+b^2+c^2)(a+b+c)}{2R} +\frac{3abc}{R}-4T</math>
 +
 +
<math>H= \frac{(a^2+b^2+c^2)(a+b+c)}{4R} +\frac{3abc}{2R}-2T</math>
 +
 +
Also from the extended law of sines, <math>T=\frac{abc}{4R}</math> therefore, <math>\frac{3abc}{2R}=6T</math>
 +
 +
<math>H= \frac{(a^2+b^2+c^2)(a+b+c)}{4R} +4T</math>
 +
 +
Applying AM-GM:
 +
 +
<math>\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}</math>
 +
 +
Therefore,
 +
 +
<math>\frac{(a^2+b^2+c^2)(a+b+c)}{9} \ge abc</math>
 +
 +
<math>(a^2+b^2+c^2)(a+b+c) \ge 9abc</math>
 +
 +
<math>\frac{(a^2+b^2+c^2)(a+b+c)}{4R} \ge 9\frac{abc}{4R}</math>
 +
 +
<math>\frac{(a^2+b^2+c^2)(a+b+c)}{4R} \ge 9T</math>
 +
 +
<math>\frac{(a^2+b^2+c^2)(a+b+c)}{4R}+4T \ge 9T+4T</math>
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 +
Therefore,
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<math>H \ge 13T</math>, which proves: <math>area(H) \ge 13.area(T)</math>
 +
 +
Equality holds with equilateral triangle:
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 +
<math>T=\frac{a^2\sqrt{3}}{4}</math>
 +
 +
<math>R=\frac{a}{2sin(A)}=\frac{a}{2\frac{\sqrt{3}}{2}}=\frac{a}{\sqrt{3}}</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.
 
* 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}}
 
{{alternate solutions}}
  
 
== See also ==
 
== See also ==
 +
[[OIM Problems and Solutions]]
 +
 
https://www.oma.org.ar/enunciados/ibe7.htm
 
https://www.oma.org.ar/enunciados/ibe7.htm

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:

\[area(H) \ge 13.area(T)\]

Ibe7 2.gif

~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.

See also

OIM Problems and Solutions

https://www.oma.org.ar/enunciados/ibe7.htm

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