SOLUTION: H is the orthocenter of acute triangle ABC and the extensions of AH, BH, and CH intersect the circumcircle of traingle ABC at A prime, B prime and C prime. We know angle AHB : angl

Question 1210573: H is the orthocenter of acute triangle ABC and the extensions of AH, BH, and CH intersect the circumcircle of traingle ABC at A prime, B prime and C prime. We know angle AHB : angle BHC : angle CHA = 2 : 5 : 8. Find angle AprimeBprimeCprime in degrees.
Answer by ikleyn(53617) (Show Source): You can put this solution on YOUR website!
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H is the orthocenter of acute triangle ABC and the extensions of AH, BH, and CH intersect the circumcircle
of triangle ABC at A', B' and C'. We know angle AHB : angle BHC : angle CHA = 2 : 5 : 8. Find angle A'B'C' in degrees.
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This "problem" is SELF-CONTRADICTORY and describes a situation which NEVER may happen in Euclidian geometry.

Indeed, in an acute triangle, the orthocenter (the intersection point of three altitudes) is always inside
the triangle. So, the angles of visibility to any side of an acute triangle from the orthocenter
must be less than 180 degrees each.

Let's look at the given angles. We have 2 + 5 + 8 = 15 equal parts, so each part is 360/15 = 24 degrees.

Then angle AHB = 2*24 = 48 degrees, angle BHC = 5*24 = 120 degrees and angle CHA = 8*24 = 192 degrees.

We see that angle CHA of 192 degrees is greater than 180 degrees, which, as we discussed above, may not happen.

So, the posed problem is a FAKE.

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