Plane Geometry Question: Distinction

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Discussion Overview

The discussion revolves around a geometric problem involving an equilateral triangle PQR and a point A on side QR, where the relationship between the lengths PA and QA is to be proven. The focus is on applying geometric principles, particularly the cosine rule and Pythagorean theorem, to establish the equation PA^2 = 7QA^2.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant requests assistance in proving that PA^2 = 7QA^2 given RA = 2QA.
  • Another participant questions the original poster's approach by asking how they have attempted to prove the statement.
  • A hint is provided to apply the cosine rule on triangle PQA to find the relationship between PA and QA.
  • A participant suggests using an arbitrary equilateral triangle and demonstrating the relationship arithmetically, arguing that all triangles PQA are congruent.
  • One participant expresses frustration over perceived sarcasm and clarifies their understanding of the cosine rule, stating they have attempted to solve the problem but did not arrive at a relevant answer.
  • Another participant proposes an alternative method using the midpoint of RQ and the Pythagorean theorem to relate the segments to QA.
  • A participant presents a calculation using the cosine rule, arriving at the conclusion PA^2 = 7QA^2, and suggests that the discussion about sarcasm should end.

Areas of Agreement / Disagreement

Participants express differing views on the approach to solving the problem, with some advocating for the cosine rule and others suggesting alternative methods. There is no consensus on the best method, and the discussion remains unresolved regarding the most effective approach to the proof.

Contextual Notes

Some participants indicate they have made attempts to solve the problem but have not reached a satisfactory conclusion. There are references to the need for effort in problem-solving before receiving direct assistance.

aek
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PQR is an equilateral triangle, and A is a point on QR such that RA = 2 QA. Prove that PA^2 = 7QA^2

If someone can help, i'd really appreciate it. Thanks
 
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aek said:
PQR is an equilateral triangle, and A is a point on QR such that RA = 2 QA. Prove that PA^2 = 7QA^2

If someone can help, i'd really appreciate it. Thanks
How have you attempted to prove this statement ?
 
aek said:
PQR is an equilateral triangle, and A is a point on QR such that RA = 2 QA. Prove that PA^2 = 7QA^2

If someone can help, i'd really appreciate it. Thanks

HINT : Apply cosine rule on triangle PQA.
 
Thanks. . .

Why are you answering my question with a question? Just abit ironic that's all. Thanks for the hint hyper, it was helpful.
 
aek said:
Why are you answering my question with a question? Just abit ironic that's all. Thanks for the hint hyper, it was helpful.

I can't even tell if that's sarcasm. In any case, it was I that gave the hint, not hyper.

The policy of this forum is not to do homework or just solve the problem for the poster unless it is evident some effort or at least some serious thinking has been put into the solution by the poster. So I can understand why hyper posted what he did.

Have you solved the problem yet ? If you're still finding difficulty, apply the Cosine Rule as I suggested and see what you get. If stuck, post here with working, then you'll get help.
 
You can choose any arbitrary equilateral triangle PQR and let a point A on QR such that RA = 2QA. Show arithmatically (with numbers) that PA^2 = 7QA^2. Then proceed to argue that all triangles PQA are congruent, and therefore any proportions wrt its lengths (PA^2 = 7QA^2) must remain constant.
 
yeah, it was sarcasm

yep i got the point,
but i too know that you have to use the cosine rule and I've tried and couldn't find a relevant answer. If you want proof that i tried, i could scan my work but you being so trustful that wouldn't be necassary :P
 
Here is an alternative but longer way to do it that avoids trig. Let M be the midpoint of RQ. Since PQR is equilateral, you know (it's a theorem) that PM must be perpendicular to RQ. Then you can use Pythagoras's theorem a couple of times to find how the different segments relate to QA.
 
aek said:
yep i got the point,
but i too know that you have to use the cosine rule and I've tried and couldn't find a relevant answer. If you want proof that i tried, i could scan my work but you being so trustful that wouldn't be necassary :P

[tex]PQ = QR = 3QA[/tex]

[tex]PA^2 = PQ^2 + QA^2 - 2(PQ)(QA)\cos 60^o[/tex]

[tex]PA^2 = 9QA^2 + QA^2 - 2(3QA)(QA)(\frac{1}{2})[/tex]

[tex]PA^2 = 7QA^2[/tex]

Perhaps now the sarcasm can end.
 
  • #10
perhaps it may. well thanks for the help, even though it took some time but truly, i really do appreciate it.
 

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