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March 16th, 2015, 03:27 AM   #1
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Sides of a triangle in arithmetic sequence

Hi all,

if anyone has a few moments to help me, I would be grateful.

If an obtuse triangle has three sides in arithmetic sequence, with angles, Theta, Pi + theta and Pi - 2theta, what are the ratio of the sides?

Many thanks,
Eva

Last edited by skipjack; March 17th, 2015 at 07:38 AM.
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March 16th, 2015, 03:47 AM   #2
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Originally Posted by evaeva View Post
Hi all,

if anyone has a few moments to I help i would be grateful.

If an obtuse triangle has three sides in arithmetic sequence, with angles, Theta, Pi +theta and Pi - 2theta, what are the ratio of the sides?

many thanks
Eva
If $\theta > 0$ is one angle of the triangle, how can the same triangle have another angle equal to $\pi+\theta$ ? Wouldn't that angle be greater than $180^\circ$ ?
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March 16th, 2015, 03:50 AM   #3
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Sorry,

the angles are: theta, pi/2 +theta and pi/2 - 2theta.

Sorry about the mistake.

Eva
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March 16th, 2015, 06:21 AM   #4
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I get approximately 0.41064866 : 0.70532433 : 1 or 0.45141623 : 0.72570811 : 1.

Where did you get this problem from?
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Last edited by skipjack; March 16th, 2015 at 10:40 AM.
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March 16th, 2015, 06:46 AM   #5
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Originally Posted by evaeva View Post
the angles are: theta, pi/2 +theta and pi/2 - 2theta.
It seems like there are three possible orders: $\theta<\pi/2-2\theta<\pi/2+\theta$, $\theta=\pi/2-2\theta<\pi/2+\theta$, and $\pi/2-2\theta<\theta<\pi/2+\theta.$ The middle case gives $\theta=\pi/6$. The first case has $(\pi/2-2\theta)-\theta=(\pi/2+\theta)-(\pi/2-2\theta)$ or $\theta=\pi/12$. The last case is similar to the first. Find the last case and check if (1) the angles add up to $\pi$ and (2) if the triangle is obtuse in each of the three cases.
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March 16th, 2015, 09:38 AM   #6
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Aren't you putting the angles in arithmetic progression rather than the sides?
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March 16th, 2015, 08:59 PM   #7
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Hi skipjack,

it comes from an Australian yr 12 Trial HSC exam.

The answer is sqrt 7 - 1: sqrt 7 : sqrt 7 + 1.

It uses the sine and cosine rules. I was wondering whether there's a simpler way to deduce this. The algebra gets pretty long.

Regards

Last edited by skipjack; March 17th, 2015 at 01:38 AM.
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March 17th, 2015, 01:33 AM   #8
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That's equivalent to my second answer above, so they missed my first answer. I used the sine rule. As you haven't posted the "pretty long" method, I can't tell whether I can find a simpler trigonometric method. It's unlikely that there's a simpler method that doesn't use trigonometry.
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March 17th, 2015, 02:24 AM   #9
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Thanks skip,

I have attached the solution I have, agree the sine rule is the best, but was hoping for something less cumbersome.

can you tell me where I have missed the second solution?

kind regards
Eva
Attached Images
File Type: jpg 1.1.jpg (17.9 KB, 6 views)
File Type: jpg 1.2.jpg (18.3 KB, 4 views)
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March 17th, 2015, 07:32 AM   #10
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In ascending order of length, the sides are a-d, a and a+d, but there are two ways to put the corresponding opposite angles in ascending order, depending on the value of θ.

If the angles, in ascending order, are θ, $\pi$/2 - 2θ, and $\pi$/2 + θ, the sine rule and the fact that the lengths of the sides form an arithmetic progression lead easily to the equation
cos(θ) + sin(θ) = 2cos(2θ), squaring which gives 1 + sin(2θ) = 4cos²(2θ) = 4(1 - sin²(2θ)).

Hence 4sin²(2θ) + sin(2θ) - 3 = 0, i.e. (sin(2θ) + 1)(4sin(2θ) - 3) = 0.

As sin(2θ) = -1 is an extraneous solution, sin(2θ) = 3/4.

Hence (after some algebra) cos(2θ) = √7/4, cos(θ) = (√7 + 1)/4 and sin(θ) = (√7 - 1)/4.
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