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Difference between revisions of "1964 IMO Problems/Problem 3"

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== Solution ==
 
== Solution ==
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{{solution}}
 
 
Let the tangent to the in circle parallel to BC cut AB,AC at D & E respectively. Similarly let the tangent to the same parallel to AB cut AC,BC at F & G respectively and the tangent to the same parallel to AC cuts BC,AB at H,M respectively. Let the incircle touch the sides BC,CA,AB at P,Q,R respectively and let the points of contact of MH,FG,DE with the in circle be X,Y,Z respectively. Then perimeter of BHM = BH+HX+XM+MB=BH+HP+MR+BM=BP+BQ=2(s-b) and similar results follow!
 
Let the tangent to the in circle parallel to BC cut AB,AC at D & E respectively. Similarly let the tangent to the same parallel to AB cut AC,BC at F & G respectively and the tangent to the same parallel to AC cuts BC,AB at H,M respectively. Let the incircle touch the sides BC,CA,AB at P,Q,R respectively and let the points of contact of MH,FG,DE with the in circle be X,Y,Z respectively. Then perimeter of BHM = BH+HX+XM+MB=BH+HP+MR+BM=BP+BQ=2(s-b) and similar results follow!
 
  Each of the triangles BHM,CGF,ADE are similar to ABC.  
 
  Each of the triangles BHM,CGF,ADE are similar to ABC.  
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[BHM]/[BCA] = (perimeter of BHM/perimeter of BCA)^2 ={(2s-2b)/(a+b+c)}^2 ={(c+a-b)/(c+a+b)}^2
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<math>[BHM]/[BCA] = \text{(perimeter of BHM/perimeter of BCA)}^2 ={(2s-2b)/(a+b+c)}^2 ={(c+a-b)/(c+a+b)}^2</math>
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  Thus, [BHM]+[CGF]+[ADE]=[ABC]/(a+b+c)^2 { ( b+c-a)^2 + (c+a-b)^2 + (a+b-c)^2}
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  Thus, <math>[BHM]+[CGF]+[ADE]=[ABC]/(a+b+c)^2 { ( b+c-a)^2 + (c+a-b)^2 + (a+b-c)^2}</math>
  
  

Revision as of 11:47, 29 January 2021

Problem

A circle is inscribed in a triangle $ABC$ with sides $a,b,c$. Tangents to the circle parallel to the sides of the triangle are contructed. Each of these tangents cuts off a triangle from $\triangle ABC$. In each of these triangles, a circle is inscribed. Find the sum of the areas of all four inscribed circles (in terms of $a,b,c$).

Solution

Let the tangent to the in circle parallel to BC cut AB,AC at D & E respectively. Similarly let the tangent to the same parallel to AB cut AC,BC at F & G respectively and the tangent to the same parallel to AC cuts BC,AB at H,M respectively. Let the incircle touch the sides BC,CA,AB at P,Q,R respectively and let the points of contact of MH,FG,DE with the in circle be X,Y,Z respectively. Then perimeter of BHM = BH+HX+XM+MB=BH+HP+MR+BM=BP+BQ=2(s-b) and similar results follow! 

Each of the triangles BHM,CGF,ADE are similar to ABC.

$[BHM]/[BCA] = \text{(perimeter of BHM/perimeter of BCA)}^2 ={(2s-2b)/(a+b+c)}^2 ={(c+a-b)/(c+a+b)}^2$ 

Thus, $[BHM]+[CGF]+[ADE]=[ABC]/(a+b+c)^2 { ( b+c-a)^2 + (c+a-b)^2 + (a+b-c)^2}$


Denote $[ABC]$ the area of $\triangle ABC$ and $(ABC)$ the perimeter of $\triangle ABC$.

Then $\frac{[BHM]}{[ABC]} = \left(\frac{(BHM)}{(ABC)}\right)^{2} =\left(\frac{c+a-b}{c+a+b}\right)^{2}$.

So $[BHM]=\left(\frac{c+a-b}{c+a+b}\right)^{2}\cdot [ABC]$.

We know, $r_{1}$ is the radius of the incircle of $\triangle BHM$: $r_{1}= 2 \cdot \frac{[BHM]}{(BHM)}$.

Area of the incircle of $\triangle BHM$ \[= \pi \cdot 4 \cdot (\frac{[BHM]}{(BHM)})^{2}= 4\pi \frac{(c+a-b)^{2}}{(c+a+b)^{4}} \cdot ([ABC])^{2}\] Area of the incircle of $\triangle ABC$:$4\pi (\frac{[ABC]}{a+b+c})^{2}$.

Sum of the area of the 4 incircles: \[4 \pi ([ABC])^{2}\left[\frac{(c+a-b)^{2}+(b+c-a)^{2}+(c+a-b)^{2}}{(a+b+c)^{4}}\right]+4\pi (\frac{[ABC]}{a+b+c})^{2}\] \[=4 \pi \frac{([ABC])^{2}}{(a+b+c)^{2}}\left[\frac{(c+a-b)^{2}+(b+c-a)^{2}+(c+a-b)^{2}+(a+b+c)^{2}}{(a+b+c)^{2}}\right]\] \[=16 \pi \frac{([ABC])^{2}(a^{2}+b^{2}+c^{2})}{(a+b+c)^{4}}\]

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