In an isoceles triangle, if the altitudes intersect on the inscribed circle, then the cosine of the vertical angle A is?
maybe it's just me, but i don't know what "vertical" angle of a triangle means. i'd look it up if i wasn't so lazy
EDIT: ok, looked it up. it seems vertical angle has to be in reference to a given side, so i still don't know. are we talking the angle vertical to the side that is (perhaps) of a distinct length? that would make sense, maybe
Here's a sketch. I think you just have to prove that O, I and A are aligned... looks like it's always the case.
O represents the orthocenter (that is to say the intersection of the altitudes), center of the circumscribed circle c and I the center of the inscribed circle d.
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The inscribed circle was drawn randomly, as well as the position of O on it. I then took a random point on circle c, named A.
Then, I drew the tangents from A to the inscribed circle.
One of this tangents intersects the circumscribed circle in B.
Then I drew the circle e of center A and of radius AB, since ABC is an isosceles triangle.
C must be on this circle e and also on the circumscribed circle c. AC must also be tangent to the inscribed circle d. And C is on these three elements iff A, O and I are aligned.
This doesn't give the solution, but maybe it'll help... ?
Ok, so geometry isn't my strong suit, but let me submit this anyway. maybe if someone could remind me how does the altitude of an isosceles triangle relate to the other sides. also, i have yet to use the fact that the triangle is inscribed in a circle.
but i found that $\displaystyle \cos A = 2 - \frac 12 \bigg( \frac ab \bigg)^2$ and $\displaystyle \cos A = \frac {2d^2}{b^2} - 1$
here d is the height, b is the side opposite to angle B (one of the other angles)