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Pre-RMO 2014/10

In a triangle \(ABC\), \(X\) and \(Y\) are points on the segments \(AB\) and \(AC\), respectively, such that \(AX : XB = 1 : 2\) and \(AY : YC = 2 : 1\). If area of triangle \(AXY\) is \(10\) then what is the area of triangle \(ABC\)?


This note is part of the set Pre-RMO 2014

Note by Pranshu Gaba
2 years, 7 months ago

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Answer is 45... Gaurav Singh · 1 year, 9 months ago

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Let side \(AB = c\), \(AC = b\), and \(\angle BAC = \theta\). Then \(AX = \frac{c}{3}\) and \(AY = \frac{2b}{3}\).

Area of triangle \(ABC = \frac{1}{2} bc \sin\theta\) and area of triangle \(AXY\) = \(\frac{1}{2} \times \frac{2b}{3} \times \frac{c}{3} \sin \theta = 10\)

\[\text{Area }\triangle AXY = \dfrac{2}{9}( \text{Area }\triangle ABC)\]

\[\text{Area }\triangle ABC = \dfrac{9}{2} ( \text{Area }\triangle AXY)\]

\[\text{Area }\triangle ABC = \dfrac{9}{2} \times 10 = \boxed{45}\] Pranshu Gaba · 2 years, 7 months ago

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Answer is 45. Ar Agarwal · 2 years, 7 months ago

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90?? Harsh Khatri · 2 years, 7 months ago

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