+501 Two rays with common endpoint O form a 30-degree angle. Point A lies on one ray, point B on the other ray, and AB=1. What is the maximum possible length of OB?
What I did so far, is I let AO=a and BO=b, and I used the Pythagorean theorem on AOD and AOB. I got b = sqrt(1-a^2/4)+a*sqrt(3)/2, but I'm not sure how to proceed from there.
+33659 As follows:
Let side OA (opposite angle B) be b, and side OB (opposite angle A) be a.
From the cosine rule we have: \(1^2=a^2+b^2-2ab\frac{\sqrt3}{2}\) remembering that \(cos(30\deg)=\frac{\sqrt3}{2}\).
Rewrite as a quadratic in a: \(a^2-\sqrt3ba+b^2-1=0\)
Solve for a in terms of b: \(a=\frac{\sqrt3}{2}b+\frac{\sqrt{(2+b)(2-b)}}{2}\) (This solution will give the largest value of a)
Differentiate a wrt b to get: \(\frac{da}{db}=\frac{\sqrt3}{2}-\frac{b}{2\sqrt{(2+b)(2-b)}}\)
Set this to zero and solve for b to get: \(b=\sqrt3\)
Substitute this back into the expression for a above to get: \(a=2\)
(Corrected for earlier brain fade!!!).
+501 How did you get that equation that would give you the largest value of a?
Also, why can't OB = 2, for example?
Consider a circle, radius 1 , with its centre on the line OB .
Call the centre b' and start with b' at O.
Move b' away from O.
The maximum distance Ob' will happen when the line OA is tangential to the circle.
When that happens the line from OA to OB will be at right angles to OA.
Ob' will then be the hypotenuse of a 30 deg right-angled triangle.
In that triangle, 1/Ob' = sin(30 deg) =1/2.
So 0b' = 2.
