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March 16th, 2015, 03:27 AM  #1 
Member Joined: Aug 2013 Posts: 40 Thanks: 3  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. 
March 16th, 2015, 03:47 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587  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 $180^\circ$ ?

March 16th, 2015, 03:50 AM  #3 
Member Joined: Aug 2013 Posts: 40 Thanks: 3 
Sorry, the angles are: theta, pi/2 +theta and pi/2  2theta. Sorry about the mistake. Eva 
March 16th, 2015, 06:21 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,919 Thanks: 2202 
I get approximately 0.41064866 : 0.70532433 : 1 or 0.45141623 : 0.72570811 : 1. Where did you get this problem from? Last edited by skipjack; March 16th, 2015 at 10:40 AM. 
March 16th, 2015, 06:46 AM  #5 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  It seems like there are three possible orders: $\theta<\pi/22\theta<\pi/2+\theta$, $\theta=\pi/22\theta<\pi/2+\theta$, and $\pi/22\theta<\theta<\pi/2+\theta.$ The middle case gives $\theta=\pi/6$. The first case has $(\pi/22\theta)\theta=(\pi/2+\theta)(\pi/22\theta)$ 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.

March 16th, 2015, 09:38 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,919 Thanks: 2202 
Aren't you putting the angles in arithmetic progression rather than the sides?

March 16th, 2015, 08:59 PM  #7 
Member Joined: Aug 2013 Posts: 40 Thanks: 3 
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. 
March 17th, 2015, 01:33 AM  #8 
Global Moderator Joined: Dec 2006 Posts: 20,919 Thanks: 2202 
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.

March 17th, 2015, 02:24 AM  #9 
Member Joined: Aug 2013 Posts: 40 Thanks: 3 
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 
March 17th, 2015, 07:32 AM  #10 
Global Moderator Joined: Dec 2006 Posts: 20,919 Thanks: 2202 
In ascending order of length, the sides are ad, 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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