Geometry Intersections Problem

  • Thread starter rizardon
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In summary, the author is trying to solve a homework problem from a 2009 Math paper for the Japanese Government Scholarship Qualifying Exam. They use the result from (i) to find OP and then express it in terms of a, b, and c.
  • #1
rizardon
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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.
 
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  • #2
Consider △ABP and △CBP,
∵they have the same base BP, and Area of △ABP : Area of △CBP = 2 : 3,
∴Height of △ABP : Height of △CBP = 2 : 3
As △APQ and △CPQ also have the same base PQ,
AQ : QC = 2 : 3

Then, Area of △APQ : Area of △CPQ = 2 : 3,
Consider △ABP and △APQ,
Area of △ABP : Area of △APQ = 1 : 1
∵They have the same height,
∴Base of △ABP : Base of △APQ = BP : PQ = 1 : 1
∴[itex]\vec{BP}[/itex] = [itex]\frac{1}{2}[/itex][itex]\vec{BQ}[/itex]
 
  • #3
Thanks a lot. I get the first part now. But I still don't get the second part. We only know the two triangles share the same height so how can we deduce that their areas are in the same proportion without knowing the relationship between the heights.

Sorry, but I really suck in geometry.
 
  • #4
Let Area of △X be AX, Base of △X be bX, Height of △X be hX
AX = (0.5)(bX)(hX)
Similarly for △Y, AY = (0.5)(bY)(hY)

If AX = AY and hX = hY
bX = bY
 
  • #5
Thanks, I get it now.
 

Related to Geometry Intersections Problem

1. What is a geometry intersections problem?

A geometry intersections problem involves finding the points or lines where two or more geometric shapes intersect. This can be done by solving equations or using geometric constructions.

2. What are the common types of geometric shapes in intersection problems?

The most common types of geometric shapes used in intersection problems are points, lines, circles, and polygons. These shapes can be intersected with each other to find intersection points or lines.

3. How do you solve a geometry intersections problem?

To solve a geometry intersections problem, you can use algebraic methods such as solving equations, or geometric methods such as constructing figures and using properties of intersecting lines and shapes. It is important to carefully analyze the problem and choose the appropriate method.

4. What are some real-life applications of geometry intersections?

Geometry intersections have many real-life applications, including in architecture, engineering, and computer graphics. They are used to determine the angles and positions of intersecting lines in buildings and structures, as well as in creating 3D models and animations.

5. Can geometry intersections be used in non-Euclidean geometries?

Yes, geometry intersections can be applied in non-Euclidean geometries such as spherical geometry and hyperbolic geometry. The concepts of points, lines, and intersections are still applicable, but the properties and rules may differ from Euclidean geometry.

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