Question

Let ABC be an acute angled triangle, AD be the bisector of angleBAC with D on BC and BE be the altitude from B on AC. If angleCED > π/n where n is a natural number. Find the least value of n​

Answer 1

Answer:

Since the angle at B has to be acute, we have the constraint that |AE|.|EC|<|EB|2

Proof: Consider if the angle at B is 90° - by similar triangles, |AE||EB|=|EB||EC|⟹|AE|.|EC|=|EB|2

If B is outside of the circle with AC as its diameter, then the altitude BE intersects the circle at B’ and |AE|.|EC|=|EB′|2<|EB|2 , and the angle at B is acute. If B is inside the circle, then we can extend EB to intersect the circle at B’, and |AE|.|EC|=|EB′|2>|EB|2 , and the angle at B is obtuse.

If we fix BE, then we can notice that for a given A, the larger EC is, the smaller the angle DEC will be. So the smallest value of the angle DEC occurs when |EC|=|EB|2|AE| (that is, when the angle at B is 90°). Also, from the angle bisector theorem, |BD||DC|=|AB||AC| .

I got somewhat stuck here. By fixing BE and moving A and C with a right-angle ∠ABC , you can see that the angle ∠BAC varies between 0 and π2 , and by playing around with this, you can see that the angle ∠ADE can be adjusted to be larger, smaller, or the same as the half-angle ∠DAC , and further that ∠CED=∠CAD+∠ADE . We can see that the angle ∠CED is always larger than ∠DAC . If we move A close to E , the angle ∠CED is larger than, but gets closer to, π4 . As A gets further away from E, and C gets closer, the angle ∠CED gets larger, and approaches π2 . However, I have not figured out any way to show that this is monotonic in |AE| (in fact, I’m not sure it is). I believe that the angle CED is bounded below by π4 , but I have not been able to prove it