Problem
Consider an obtuse angled triangle ABC in which the difference between the largest and the smallest angle is π/2 and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on a circle of radius 1. Let a be the area of the triangle ABC. Then find the value of (64a)².
Solution
Since we need to find the area of the triangle, we’ll start with using one of the standard formulas.
Here’s one: Δ = 2R²sinAsinBsinC.
We used this one since the value of R is given. Plus, some relation between the angles is known, and the one given between the sides can be converted to angles as well.
So, the idea is the use these relations to find the angles and then the area. Or, we could also end up find the values of their trigonometric ratios, which can be substituted in the above formula. Let’s see how it goes.
Assuming A < B < C, we can express the relation between the angles as C = A + π/2.
Then, using the angle sum property, we’ll get B = π – (A + C) = π/2 – 2A
We’ve found all angles in terms of A. But that’s not enough, since the area will be an expression involving A.
That is, Δ = 2sinAsin(π/2 – 2A)sin(π/2 + A) = 2sinAcos2AcosA = sin2Acos2A. A nice looking expression, but we still need the value of A.
The only hope we have is the given relation between the sides.
Since A < B < C, we have a < b < c. So, the relation between the sides can be expressed as 2b = a + c.
Let’s use the sine rule to convert this to a relation between angles. We’ll get 2sinB = sinA + sinC.
Now, using the angle relations that we got above, this relation changes to 2sin(π/2 – 2A) = sinA + sin(π/2 + A). On simplifying, we’ll get 2cos2A = sinA + cosA.
Using cos2A = cos²A – sin²A or (cosA + sinA)(cosA – sinA), the above relation changes to 2(cosA + sinA)(cosA – sinA) = sinA + cosA.
Now, we can cancel out the non-zero term sinA + cosA, since A is acute (as C is obtuse). So, we get the simplified relation as cosA – sinA = 1/2.
At this point, you might be tempted solve this equation and find the value of A. We could do that, but it might not be necessary.
If we square both sides, we’ll get sin2A = 3/4 (skipped a couple of steps here). Using this, cos2A will come out to be √7/4. And, we’re done!
The required area a will be sin2Acos2A or 3√7/16. This gives the value of (64a)² as (64 x 3√7/16)².
This simplifies to (4 x 3√7)² or 16 x 9 x 7 or 1008.
Comments
A fairly simple problem. If you have the formulas handy, then there won’t be much trouble solving this one. Remember to use the sine rule to interconvert relations between sides and angles, whenever necessary. Second, simplify the expressions / equations as much as possible. Finally, see if the given relations can be converted to the required ones, using trigonometric identities – we saved a fair bit of effort by not solving that equation.