MathLabs

Problem 5

Let ABC be a triangle and X an interior point of ABC. Show that at least one of the angles XAB, XBC, XCA is less than or equal to 30 degrees.
Step 4 of 4: The three-angle bound contradicts the assumption
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x=A−30∘, y=B−30∘, z=C−30∘,x+y+z=90∘,sin⁡xsin⁡ysin⁡z≤18x=A-30^\circ,\ y=B-30^\circ,\ z=C-30^\circ,\quad x+y+z=90^\circ,\quad \sin x\sin y\sin z\le\frac18
Detailed analysis

Set x=A−30 degrees, y=B−30 degrees, z=C−30 degrees; their sum is 90 degrees. Since sin x sin y is at most (1−cos(x+y))/2=(1−sin z)/2, the product is at most one half of (1−sin z)sin z, which is at most 1/8. This contradicts the strict inequality above, so at least one of alpha,beta,gamma is at most 30 degrees.