Find the position vector of ##Q##

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The discussion focuses on finding the position vector of point Q in a geometry problem, specifically an O level question worth two marks. The original poster used similarity to approach the problem but seeks a clearer or simpler method, suspecting that reflection might be involved. They mention that triangles OPC and QPB are similar with a ratio of 3, and that triangle QOA is also similar to these triangles. The poster is looking for insights or alternative methods to solve the problem effectively. The conversation highlights the challenges of understanding the marking scheme and finding a straightforward solution.
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Homework Statement
see attached (highlighted in Red)
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O level question; i used similarity would appreciate an easier approach for 2 marks.

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The ms solution (approach) is not clear to me. Here it is;



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My approach; using similarity

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Any insight welcome its a 2 mark question- cannot seem to find easier way though i suspect reflection.
 
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anuttarasammyak said:
...still not getting it. You mind giving some reason or thought. Thks.
 
Triangle OPC and QPB are similar with ratio 3.
 
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anuttarasammyak said:
Triangle OPC and QPB are similar with ratio 3.
And ΔQOA is similar to each of those two.
 
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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