Find the ratio of areas of triangles - [UKRMO 2009 Grade 10]

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Find the ratio of areas of triangles - [UKRMO 2009 Grade 10]

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#1 h Jan 23, 2011, 2:19 AM • 2 Y

Y by Adventure10, Mango247

Let $ABCD$ be a parallelogram with $\angle BAC = 45^\circ,$ and $AC > BD .$ Let $w_1$ and $w_2$ be two circles with diameters $AC$ and $DC,$ respectively. The circle $w_1$ intersects $AB$ at $E$ and the circle $w_2$ intersects $AC$ at $O$ and $C$ , and $AD$ at $F.$ Find the ratio of areas of triangles $AOE$ and $COF$ if $AO = a,$ and $FO = b .$

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#2 h Jan 23, 2011, 10:20 AM • 2 Y

Y by Adventure10, Mango247

Join CE. angle CEA = 90.
Let CO = x.
=> AE = (a+x)/sqrt(2).
=> area of triangle AEO is (a)*((a+x)/(sqrt(2))*sin45/2 = (a)*(a+x)/4.
Let angle CAD = T.
In triangle ACF,
CF = (a+x)sinT.
In cyclic quad DOCF,
angle CFO = angle CDO = 90 - angle DCO = 90 - angle OAB = 90 - 45 = 45.
Hence area of triangle COF is (b)*(a+x)sinT*sin45/2
=> ar(COF)=(b)*(a+x)sinT/(sqrt(2)*2).
Now, in triangle AOF, by sine rule,
b/sinT=a/sinOFD=a/sinOCD=a/sinOAB=a/sin45=a*sqrt(2).
=>sinT=b/(a*sqrt(2)).
Hence the ratio of areas of AOE and COF turns out to be (a*a)/(b*b).

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#3 h Jan 24, 2011, 12:35 PM • 1 Y

Y by Adventure10

$\angle OAE=45=\angle ODC=\angle OFC$ . Since $AECF$ is cyclic, $E,O,F$ are collinear, so $\angle AOE=\angle FOC$ . Thus $\triangle COF\sim\triangle EOA$ . It follows that $\frac{[AOE]}{[COF]}=\frac{a^2}{b^2}$ .

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