A circle is inscribed in triangle ABC (AC=AB). The point of heights’ intersection (orthocenter) (Fig. 1) belongs to this circle. Find the angles of triangle ABC (Fig. 2).
or, perhaps, vice-versa
... in an isosceles triangle, AC=BC, the quotient of AC:AB is 3:4, thus, show that the orthocenter lies on the incircle ...
Probably, and presumably, in the very first post, the very first sentence should read like:
"A circle is inscribed in triangle ABC (AC=AB)" ... rather than AC=AC
Liudmyla Hetmanenko, any clarification please.
The circumcircle of ABC, and heights from C and A can be drawn for a "geometric" solution
plus
https://www.google.com/search?q=r%2FR+%2B+1+%3D+cos+A+%2B+cos+B+%2B+cos+C
Using Steftcho's notation we have:
angle BCH=A/2 (both equal to 90-B). In triangle CPH we have tan(BCH)=HP/CP=(2r)/(a/2) and it results
tan(A/2)=(4r)/a=(4r)/(2RsinA)=(4r)/(4Rsin(A/2)cos(A/2)) =>
(sinA/2)/(cosA/2)=r/(RsinA/2 cosA/2) =>(sin A/2)^2=r/R
But from sin A/2=sqrt((p-b)(p-c)/bc), and similarly, we obtain
r=4Rsin A/2 sin B/2 sin C/2
Now, B being equal to C, it results sin A/2 =4(sin B/2)^2 = 2(1-cos B) and with
B=(180-A)/2=90-A/2
=> sinA/2=2(1-sin A/2) => sin A/2 =2/3
but all this proof is an almost same approach of what Peter and Steftcho proven !
Again a very nice problem.
The answer is alpha=arccos(2/3). Can be simple calculated with r=A/s with A the area of the triangle and s is half the circumference of the triangle.
The problem becomes more interesting when angle ACB is not equal to angle ABC.
Enno Diekema
Your question intrigued me, and I'd like to share what I found. There is another property of the semiperimeter of a triangle: if the intersection point of the altitudes lies on the inscribed circle.
Let the ABC triangle’s heights intersect the inscribed circle at the points of X, Y, and T (Fig. 1). The ABC triangle’s orthocenter is on the inscribed circle then and only then, if
△XYT~△ABC.
And therefore,
S_(△XYT)=r^2/R^2 S_(△ABC).
Liudmyla Hetmanenko Very nice the extension! But the proof of statement(i.e. the extension) "The ABC triangle’s orthocenter is on the inscribed circle if and only if △XYT~△ABC" is quite simple, only an elementary exercise in using inscriptible quadrilaterals. Clearly, solving the initial problem is more difficult!