When Does This Point Not Exist?

First draw a circle with centre O, Now since ABC is acute angled, therefore distance between BC & A is greater than R, so draw a mildly visible diameter, then draw BC in one of the semicircles, parallel to this diameter, till here no generality is lost... Now pick any point as A on the other semicircle & draw a circle C2 with A & O as diameter it will be intersected by AB at P & by AC at Q..., since Angle AOX = 90, X shall lie on this circle..., now drop a perpendicular from O to BC, let i intercept BC at M, now M will be the midpoint join A to M..., now if angle CAM > 40, X will lie inside triangle AMC and if angle CAM <40, then X will lie inside triangle AMB but arc PQ is available in both AMC & AMB, therefore X was possible!! Then the only case left, is to make CAM = 40, now our X must lie on the diameter AO of C2, which is not allowed as the angle wont be 90, but as CAM = 40, it is now the angle bisector aswell as the median, so our triangle ABC is isoceles..., BAC = 80, ABC = 50, N/2 = 25...,

If the answer is $50^{\circ}$ , then how does one explain this?

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Clearly the $X$ shown in the diagram fits all conditions.

I edited the problem to state "X has to be in the triangle." @Sreejato Bhattacharya is that okay?

P.S the statement $X\ne A$ is not needed. Also, the fact that the triangle is acute is also not needed.

We would like to construct X, ignoring the first condition, where it cannot equal A or O. Draw circle with diameter AO, $\omega$ . If $\angle AXO=90$ , then X must lie on $\omega$ . Furthermore, if $\angle BAX = \angle CAM$ then X must lie on the symmedian of A. Denote X can equal O when O lies on the symmedian of A. This restricts X to a single point. If X is to not exist, then X must coincide with O. Since O is the intersection of the perpendicular bisectors, O lies on the perpendicular bisector of BC. Also, O lies on the symmedian of A in this case. I shall now prove that M' must be M in this situation. Denote $\angle MAM'$ as $\gamma$ . From basic angle chasing, we find that $\angle AMB = \angle AM'B = 90 - \frac{\gamma} {2}$ , meaning that M is M'. Thus the midpoint of BC lies on the angle bisector of A, meaning ABC is isoceles with $AB= AC$ . Thus $\angle ABC = 50$ and $\boxed{ \frac{N} {2} =25}$ .

Suppose that $X$ exists, we will follow through these steps.

1.) From $\angle BAX = \angle CAM$ and $\overline{AM}$ is a median line, therefore $\overline{AX}$ is a symmedian line.

2.) Since $M$ is a midpoint of $\overline{BC}$ , therefore $\overline{OM} \perp \overline{BC}$ , and from $\angle AXO = 90^{\circ}$ , therefore, $\overline{AX}$ is a portion of an altitude.

3.) I claim that the isogonal conjugate of orthocenter (intersected by $\overline{AX}$ ) is the circumcircle (intersected by $\overline{AO}$ ), so $\angle BAX = \angle CAO$ .

4.) Finally, $\angle CAM = \angle CAO$ , which means $A,O,M$ are collinear.

The only possible case is $\triangle ABC$ is an isosceles where $\angle ABC = \angle ACB$ (and yes $X$ violates $X \neq O$ even though everything is correct), giving $\boxed{\angle ABC = 50^{\circ}}$ .