AIMO 2015 Q10

Geometry Level 5

X X is a point inside an equilateral triangle A B C ABC . Y Y is the foot of the perpendicular from X X to A C AC , Z Z is the foot of the perpendicular from X X to A B AB , and W W is the foot of the perpendicular from X X to B C BC .

The ratio of the distances of X X from the three sides of the triangle (respectively, A C AC , A B AB and B C BC ) is 1 : 2 : 4 1 : 2 : 4 .

If the area of A Z X Y AZXY is 13 cm 2 13 \text{ cm}^2 , find the area of A B C ABC .

Investigation

If X Y : X Z : X W = a : b : c XY:XZ:XW = a:b:c , find the ratio of the areas of A Z X Y AZXY and A B C ABC . Write your answers with proof in the solutions.


The answer is 98.

Sharky Kesa by Sharky Kesa , 19, United Kingdom

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1 solution

Sum of two opposite angles in each of the three below quadrilaterals =90 + 90=180. ∴ AZXY, BWXZ, and CYXW are cyclic. They are also similar, corresponding angles equal. Δ X Y Z : − a p p l y i n g C o s L a w , Z Y 2 = K 2 + ( 2 K ) 2 − 2 ∗ K ∗ 2 K ∗ C o s 60 = 7 K 2 . a p p l y i n g S i n L a w , S i n α K = S i n 120 Z Y . ⟹ α = S i n − 1 3 2 ∗ 7 = 19.1066053 5 o B u t ∠ A X Y = α , s u b t e n d e d b y c h o r d X Y = K . ∴ A Y = K T a n α . \text{Sum of two opposite angles in each of the three below quadrilaterals =90 + 90=180.}\\ \therefore ~\text{AZXY, BWXZ, and CYXW are cyclic. They are also similar, corresponding angles equal.}\\ \Delta XYZ :- ~~applying ~Cos ~Law, ~ZY^2=K^2+(2K)^2 - 2*K*2K*Cos60=7K^2.\\ ~~~~~~~~~applying ~Sin~Law, ~ \dfrac {Sin\alpha} K=\dfrac{Sin120} {ZY}. ~\implies~\color{#3D99F6}{\alpha=Sin^{ - 1}\dfrac{\sqrt3}{2*\sqrt7}}=19.10660535^o\\ But ~\angle ~ AXY=\alpha, ~subtended ~ by ~ chord ~ XY=K. ~~ \therefore~AY =\dfrac {K}{Tan\alpha}. \\ ∴ A r e a o f Δ A X Y = 1 2 ∗ X Y ∗ K T a n α = 1 2 ∗ K 2 T a n α . S i m i l a r l y A r e a o f Δ A Z X = 1 2 ∗ ( 2 K ) 2 T a n ( 60 − α ) . ∴ A r e a A Z X Y = 1 2 { K 2 T a n α + 4 K 2 T a n ( 60 − α ) } = 1 2 { 1 T a n α + 4 T a n ( 60 − α ) } K 2 = 3.75277675 ∗ K 2 = 13 \therefore ~ Area ~of ~ \Delta AXY=\frac 1 2 *XY*\dfrac {K}{Tan\alpha}=\frac 1 2*\dfrac {K^2}{Tan\alpha}. \\ Similarly~Area ~of ~ \Delta AZX=\frac 1 2*\dfrac {(2K)^2}{Tan(60 - \alpha)}. \\ \therefore ~Area_{AZXY}=\frac 1 2\{\dfrac {K^2}{Tan\alpha} + \dfrac {4K^2}{Tan(60 - \alpha)}\}= \frac 1 2\{\dfrac 1{Tan\alpha} + \dfrac 4 {Tan(60 - \alpha)}\}K^2=3 .75277675*K^2=13 \\ ∴ K 2 = 3.464101615 T h e A l t i t u t e = ( 1 + 2 + 4 ) K = 7 K . ∴ A r e a A B C = A l t i t u t e 2 3 K 2 = 49 ∗ 3.464101615 3 = 98 \therefore ~\color{#EC7300}{K^2}=3.464101615 \\ The Altitute=(1+2+4)K=7K. ~~\\ \therefore Area_{ABC}=\dfrac{ Altitute^2}{\sqrt3}K^2=49*\dfrac{3.464101615 }{\sqrt3} = \Large ~~~~\color{#D61F06}{98} \\~~~~\\

A r e a A B C = ( a + b + c ) 2 3 ∗ G i v e n A r e a L . W h e r e L = 1 2 { a 2 T a n α + b 2 T a n ( 60 − α ) } If X is a point in an equilateral triangle, sum of three perpendiculars from X to the sides = a l t i t u d e o f t h e t r i a n g l e . \color{#3D99F6}{Area_{ABC}=\dfrac{ (a+b+c)^2}{\sqrt3}*\dfrac{Given Area} L. ~~\\ Where ~L =\frac 1 2\{\dfrac {a^2}{Tan\alpha} + \dfrac {b^2}{Tan(60 - \alpha)}\}\\ \text{If X is a point in an equilateral triangle, sum of three perpendiculars from X to the sides} \\ =altitude ~of~ the~ triangle.}

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