1992 Sydney Grammar School Paper: Triangle geometry problem (1 Viewer)

asianese · Jul 6, 2012

8c) Last question
(c) The lengths of the sides of a triangle forman arithmetic progression and the largest angle of the triangle exceeds the smallest by 90◦. Find the ratio of the lengths of the sides.

What I've got so far:

a, b, c are in arithmetric series, so

$c-b=b-a \\a+c=2b$

And if the largest angle exceeds the smallest by 90, then

$\gamma = 90+\alpha\\ $In a triangle$ \;\; \alpha+\beta+\gamma=180 \\ \alpha+\beta+90+\alpha=180 \\ \beta=90-2\alpha$

Using sine rule

$\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{c}{\sin\gamma}=\frac{c}{\cos\alpha}\\ $Taking$ \;\; \frac{a}{\sin\alpha}=\frac{c}{\cos\alpha},\\ \frac{a}{c}=\tan\alpha$

$$Taking$ \;\; \frac{a}{\sin\alpha}=\frac{b}{\sin\beta}\\ \frac{2b-c}{\sin\alpha}=\frac{b}{\sin\beta}\\ \frac{2b}{\sin\alpha}-\frac{b}{\sin\beta}=\frac{c}{\sin\alpha}\\ $Using$ \;\; \beta=90-2\alpha , \\ b(\frac{2}{\sin\alpha}-\frac{1}{\sin{(90-2\alpha)}}=\frac{c}{\sin\alpha}\\ b(\frac{2}{\sin\alpha}-\frac{1}{\cos{2\alpha}})=\frac{c}{\sin\alpha}\\ \frac{b}{c}=\frac{\cos{2\alpha}}{2\cos{2\alpha}-\sin\alpha}$

$$Taking$ \;\; \frac{a}{\sin\alpha}=\frac{b}{\sin\beta}\\ \frac{a}{b}=\frac{\sin \alpha}{\sin \beta}=\frac{\sin \alpha}{\cos 2\alpha}$
Any help is appreciated. I don't know where I'm going with this lol. I generally hate geometry but this seemed like a fun problem.

Fus Ro Dah · Jul 6, 2012

Assuming your working is correct, a/b = tan alpha and a/c = tan alpha, so a/b = a/c, so b = c.

If b = c, then b/c = 1.

Sub b/c = 1 into your ratio for b/c then solve for the unique solution of alpha, then sub back into the expressions for a/b and a/c, then get your ratios.

asianese · Jul 6, 2012

bleakarcher · Jul 6, 2012

a/b=sin(alpha)/cos(2*alpha) *

bleakarcher · Jul 6, 2012

check your final line, beta=pi/2-2*alpha not pi/2-alpha

bleakarcher · Jul 6, 2012

I just did part of the problem and got that the ratio of the largest side to the smallest is [4+sqrt(7)]/3.

asianese · Jul 6, 2012

Oh shoot..Will redo later

seanieg89 · Jul 6, 2012

bleakarcher said:
I just did part of the problem and got that the ratio of the largest side to the smallest is [4+sqrt(7)]/3.

This is correct. The question is just using the sine rule and then solving a quadratic.

asianese · Jul 6, 2012

I can't seem to get the quadratic..Is it a quadratic in one of the sides?

1992 Sydney Grammar School Paper: Triangle geometry problem (1 Viewer)

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