Triangle Area Ratios and Point Positioning

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Homework Help Overview

The problem involves point P located within triangle ABC, with given area ratios of sub-triangles PAB, PBC, and PCA. Participants are tasked with finding the ratio AQ : QC and demonstrating that BP = 0.5 BQ.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between area ratios and segment ratios, with some attempting to relate AQ : QC to areas of triangles QAB and QBC. Questions arise regarding the definition and role of point Q, as well as the geometric properties of the triangles involved.

Discussion Status

Some participants have provided hints about using properties of triangles with shared altitudes and bases, while others express uncertainty about their understanding of the geometry involved. There is ongoing exploration of assumptions and clarifications regarding the setup of the problem.

Contextual Notes

There is some confusion regarding the terminology used, particularly the term 'altitude' and its application in this context. Additionally, the role of point Q and its relationship to point P is under discussion, with varying interpretations among participants.

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Homework Statement


P is a point on triangle ABC such that area PAB : area PBC : area PCA = 2 : 3 : 5
a. Find AQ : QC
b. Show that BP = 0.5 BQ

Homework Equations





The Attempt at a Solution


a. i found that AQ : QC = area QAB : area QBC, but i don't know how to continue ( the answer is 2 : 3)

b. sorry, I'm clueless...

thx
 
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songoku said:

Homework Statement


P is a point on triangle ABC such that area PAB : area PBC : area PCA = 2 : 3 : 5
You mean in triangle ABC don't you?

a. Find AQ : QC
Q? Where did Q come from? Do you mean P?

b. Show that BP = 0.5 BQ

Homework Equations





The Attempt at a Solution


a. i found that AQ : QC = area QAB : area QBC, but i don't know how to continue ( the answer is 2 : 3)

b. sorry, I'm clueless...

thx
 
oh sorry

yes P is a point inside the triangle and the straight line BP intersects the side AC at point Q.

thx
 
i tried and got stuck at the same problem...

anyone can help?
 
Hmm, I think the trick here is to keep using that fact that gave you AQ : QC = area QAB : area QBC, that is, for triangles with the same altitude but different bases, the ratio of their areas is just the ratio of their bases. The corresponding statement for same bases but different altitudes is also useful. I thought I had a solution but I made some pretty strong assumptions and have forgotten all about high school geometry, sorry I can't help further.
 
snipez90 said:
Hmm, I think the trick here is to keep using that fact that gave you AQ : QC = area QAB : area QBC, that is, for triangles with the same altitude but different bases, the ratio of their areas is just the ratio of their bases. The corresponding statement for same bases but different altitudes is also useful. I thought I had a solution but I made some pretty strong assumptions and have forgotten all about high school geometry, sorry I can't help further.
Generally snipez90 is correct, but I can't explain it well without a diagram, I‘ll try to make a picture later.
Hint:Try constructing similar triangles
 
Here it is.
 

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i think the term 'altitude' is the line that perpendicular to the opposite side.

Is BQ perpendicular to AC ?

thx
 
songoku said:
i think the term 'altitude' is the line that perpendicular to the opposite side.

Is BQ perpendicular to AC ?

thx

No, BQ is not necessarily perpendicular to AC
 
  • #10
nice work

thx a lot all
 
  • #11
Coordinate geometry
 

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  • #12
Words have spreaded
 

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  • #13
wow, you use geometry coordinate again to solve triangle problem.

awesome !

thx a lot
 

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