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Two Triangles Divided (Posted on 2007-12-16) Difficulty: 3 of 5
There are two triangles ABC, and A'B'C'. The bases are AB and A'B'.

Triangle ABC has an altitude drawn from C to AB, meeting the base at point P. This altitude divides the triangle into two unequal right triangles.

Triangle A'B'C' also has a point, P' on its base, with a line segment connecting it to vertex C', but chosen so that angle A'P'C' is 60°, with the resulting triangle A'P'C' though, not being equilateral.

All eight line segments are of integer length, and each triangle has a perimeter less than 50. The bases, AB and A'B', are the longest sides in each of the two respective original triangles, and they differ by 1 unit in length.

  • What are the dimensions of the triangles ABC and A'B'C'?
  • What are the lengths of CP and C'P'?

See The Solution Submitted by Charlie    
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Solution Answer | Comment 2 of 4 |

AB = 21
AC = 10 (or 17)
BC = 17 (or 10)
CP = 8

A'B' = 22
A'C' = 13
B'C' = 13
C'P' = 8


  Posted by DJ on 2007-12-17 17:30:26
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