**stan@hare.DEC (Stanley Rabinowitz)** (08/23/84)

It was recently asked to find the vertex angle of an isosceles triangle that has its orthocenter on its incircle. The cosine of the vertex angle is 1/9. Proof: Call the triangle ABC with vertex angle A and AB=AC=a, BC=b. Let M be the midpoint of BC, so that AM is an altitude. Let the measure of angle CAM be t (pronounced "theta"). Then angle BAM is also t and angle ABM is 90-t. Let P be the center of the inscribed circle. This circle meets BC at M and AC at Q. Let BD be the altitude from B (with D on AC). Let AM meet the incircle at point H. Since we are given that the orthocenter (intersection of the altitudes) lies on the incircle, we must have that BD passes through H. Note further that CQ=CM=b/2, and so AQ=a-b/2. Now the significant point is that right triangles ADH and AMC are similar. Let AH=x and PH=PQ=PM=r, the radius of the inscribed circle. Since AD=a cos 2t, from the similar triangles we have that AD/AH=AM/AC or a cos 2t x+2r (1) -------- = ---- . x a Also, since AM is a secant to the circle, and AQ is a tangent, we have (AH)(AM)=(AQ)^2 or 2 (2) x(x+2r) = (a-b/2) . But from triangle ACM we see that sin t = b/2a, so we can rewrite (2) as 2 (3) x(x+2r) = (a - a sin t) . Substituting the value of x+2r from (1) into (3) gives 2 2 2 (4) a cos 2t = a (1 - sin t) or 2 2 (5) 1 - 2 sin t = (1 - sin t) . Solving this for sin t yields sin t = 2/3 . We are asked to find cos 2t, which we can get from equation (4). We find that 2 cos 2t = (1-2/3) = 1/9. Stanley Rabinowitz UUCP: ...{decvax,ucbvax,allegra}!decwrl!dec-rhea!dec-hare!stan ARPA: stan%hare.DEC@DECWRL.ARPA ENET: {hare,turtle,algol,kobal,golly}::stan USPS: 6 Country Club Lane, Merrimack, NH 03054 (603) 424-2616