area chasing, AX =// 2AB,BY =//3BC, CZ =// 4CA (Argentina 1996 OMA L1 p4)

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area chasing, AX =// 2AB,BY =//3BC, CZ =// 4CA (Argentina 1996 OMA L1 p4)

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#1 h May 8, 2020, 1:34 PM • 1 Y

Y by jhu08

Let $ABC$ be an area triangle $7$ . The triangle $XYZ$ is constructed as follows: side $AB$ is extended such that $AX = 2AB$ , side $BC$ is extended such that $BY = 3BC$ , and side $CA$ is extended such that $CZ = 4CA$ . Find the area of the triangle $XYZ$ .

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#2 h May 8, 2020, 3:38 PM • 1 Y

Y by jhu08

Basic Area Ratio Finding:
This solution assumes X is on ray AB, Y is on ray BC, Z is on ray CA.
[BCX]=1*[ABC], [CAY]=2*[ABC], [ABZ]=3*[ABC].
[CXY]=2*[BCX]=2*[ABC], [AYZ]=3*[CAY]=6*[ABC], [BXZ]=1*[ABZ]=3*[ABC].

[XYZ]=[ABC]+[BCX]+[CAY]+[ABZ]+[CXY]+[AYZ]+[BXZ]= [ABC]*(1+1+2+3+2+6+3)=7*18=126.

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#3 h Sep 11, 2021, 3:01 AM • 1 Y

Y by jhu08

Solution

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