- #1

rizardon

- 20

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## Homework Statement

Let A, B, C be three points on a plane and O be the origin point on this plane. Put a = OA and b = OB, and c = OC, (a,b and c are vectors). P is a point inside the triangle ABC. Suppose that the ratio of the areas of the triangles PAB, PBC and PCA is 2:3:5

(i) The straight line BP intersects the side AC at the point Q.

Find AQ:QC

(ii) Express OP in terms of a, b and c.

The problem is from the 2009 Math paper (B) for the Japanese Government Scholarship Qualifying Exam. They have a solution and here's the link.

http://www.studyjapan.go.jp/pdf/questions/09/ga-answers.pdf

They are in Japanese by the way.

## Homework Equations

## The Attempt at a Solution

For (i) The answer is 2:3, but I don't get it at all. What is the basis on saying that the line BQ drawn by extending BP will divide the areas of the triangle ABQ and BQC in the same ratio as that of ABP and APC? And why is it necessary that if ABQ and BQC are in the same ratio then AQ and QC are in the same ratio as well?

Using the result from (i) I have the following:

OP = OB + BP

From the problem OB = b

BP = kBQ

BQ = BA + AQ

BA = a-b

AQ = (2/5)AC

AC = c-a

So BQ = (a-b)+(2/5)(c-a) = (3/5)a - b + (2/5)C = (1/5)(3a-5b+2c)

The problem is I don't know what proportion of BQ, BP is. How can I determine the value of k?

Thanks.