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Here's the question from the 2014 CEMC Fermat contest, Q23

I added some annotations to the diagram

I labelled the two points T and S and $\overline{TS}$ = $x$

Since $$\triangle PQR \sim \triangle PTS \text{ and } QR = PR = 125$$ $$\therefore TS = PS = x$$ I then used Heron's formula to find the area of $\triangle PQR$, which was 7500

$3Area_{QTSR} = Area_{PTS}$, so $Area_{QTSR} = 1875$ and $Area_{PTS} = 5625$

$$Area_{PTS} = \sqrt{s(s-a)(s-b)(s-c)}$$ $$5625 = \sqrt{(75+x)(75)(75)(x-75)}$$ $$x = \sqrt{11250}$$

$Area_{QTSR} = \frac{(125+x)}{2}h = 1875$

After substituting for x and solving for h I got an answer around 16.229. However, the answer was around 16.07

This was their solution

I'm not sure where my solution is wrong and I haven't been able to figure out why.

John Qu
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1 Answers1

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In your expression for the area of PTS, if I'm understanding right you've assumed that the length PT is 150 units, but it isn't; that's PQ and PT is shorter. So your value of $x$ is coming out slightly too small, which gives you a value of $h$ that's slightly too big.

A useful general technique for this sort of situation where you have one answer you know is right and one you know is wrong: go through the wrong one step by step using the information you have from the right one, and see where the first wrong statement is.

In this case, the other solution gives the same area as you for PTS, so everything's OK up to there; your next line is just Heron's formula, which is fine; what about the next line? Well, the other solution tells you that $x=\frac{\sqrt3}2\cdot125$, and you can plug that into the formula and see that it doesn't give 5625. So something's wrong there that wasn't wrong before, so somehow it must not be true that $s,s-a,s-b,s-c=75+x,75,75,x-75$. And contemplating that for a minute should lead to enlightenment.

Gareth McCaughan
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