Program Output for April 2016 Puzzler:

Set of all triangles whose sides are integer values and
(arbitrarily) less than 11 and whose areas match another triangle.
Sorted by area
Triangle(side lengths = 2, 2, 3;	 perimeter=7, area=1.984313)
Triangle(side lengths = 1, 4, 4;	 perimeter=9, area=1.984313)
Triangle(side lengths = 2, 6, 7;	 perimeter=15, area=5.562149)
Triangle(side lengths = 3, 4, 4;	 perimeter=11, area=5.562149)
Triangle(side lengths = 2, 7, 7;	 perimeter=16, area=6.928203)
Triangle(side lengths = 4, 4, 4;	 perimeter=12, area=6.928203)
Triangle(side lengths = 2, 8, 8;	 perimeter=18, area=7.937254)
Triangle(side lengths = 4, 4, 6;	 perimeter=14, area=7.937254)
Triangle(side lengths = 2, 9, 10;	 perimeter=21, area=8.181534)
Triangle(side lengths = 4, 5, 8;	 perimeter=17, area=8.181534)
Triangle(side lengths = 2, 9, 9;	 perimeter=20, area=8.944272)
Triangle(side lengths = 3, 6, 7;	 perimeter=16, area=8.944272)
Triangle(side lengths = 3, 8, 10;	 perimeter=21, area=9.921567)
Triangle(side lengths = 4, 5, 6;	 perimeter=15, area=9.921567)
Triangle(side lengths = 5, 5, 6;	 perimeter=16, area=12.000000)
Triangle(side lengths = 5, 5, 8;	 perimeter=18, area=12.000000)
Triangle(side lengths = 4, 7, 9;	 perimeter=20, area=13.416408)
Triangle(side lengths = 4, 7, 7;	 perimeter=18, area=13.416408)

 Since Matthew knows that the perimeter = 11 and that Kristen
 is unable to solve it initially, the solution must be in the set of
 triangles (above) that have an area equivalent to another triangle.
 The only integral triangle that has an area in common with another
 and a perimeter of 11 is:

 Triangle(side lengths = 3, 4, 4;	 perimeter=11, area=5.562149)

 Hence, this is the solution.  Since Kristen knows the area,
 she initially knows the solution must be either:

 Triangle(side lengths = 3, 4, 4;	 perimeter=11, area=5.562149) or
 Triangle(side lengths = 2, 6, 7;	 perimeter=15, area=5.562149)

 Once Matthew declares he knows the answer after learning that
 Kristen also couldn't initially determine it, Kristen can now deduce
 the solution because, of her 2 choices with the given area, perimeters
 of 11 and 15 in the above set, there is a single triangle with a
 perimeter of 11 but 2 with a 15 perimeter. She realizes that if the
 perimeter was 15  Matthew couldn't declare he knew the answer.  Hence
 now she is also sure of the solutions.