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Circle in hypotenuse (Posted on 2003-11-14) Difficulty: 3 of 5
Prove that if you draw a right triangle and then draw a circle with its center on the hypotenuse's midpoint such that it intersects at least 1 vertex, it will in fact intersect all three.

See The Solution Submitted by Antonio    
Rating: 2.5000 (4 votes)

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re: solution | Comment 6 of 13 |
(In reply to solution by SilverKnight)

Now that we have established that the midpoint is at (X/2, Y/2) we can then break it down

Pythagorean theorem states that X² + Y² = H²

Therefor the radius of the circle is (√(X² + Y²))/2

We also know that the secondary hypoteneuse from midpoint to vertex of right angle will have a distance of √((X/2)² + (Y/2)²)

Through equating these two above equations results in a true statement, we know that a cicle will pass through all vertexes.
  Posted by Roger on 2003-11-16 08:12:52

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