Problem:
What is the number of ordered triples (a,b,c) of positive integers, with aβ€bβ€cβ€9, such that there exists a (non-degenerate) triangle β³ABC with an integer inradius for which a,b, and c are the lengths of the altitudes from A to BC,B to AC, and C to AB, respectively? (Recall that the inradius of a triangle is the radius of the largest possible circle that can be inscribed in the triangle.)
Answer Choices:
A. 2
B. 3
C. 4
D. 5
E. 6
Solution:
Let x,y, and z be the lengths of BC,AC, and AB, respectively. Let r be the inradius of β³ABC. Then
Dividing by Area(β³ABC) gives (a1β+b1β+c1β)r=1, so
a1β+b1β+c1β=r1β
Because a,b,cβ€9, it follows that
r1β=a1β+b1β+c1ββ₯91β+91β+91β=31β
implying rβ€3. It is then possible to find the solutions (a,b,c) by examining cases based on the value of r.
item If r=1, then, because aβ€bβ€c, either a=2 or a=3. If a=2, then b1β+c1β=21β. Then, because bβ€c, either b=3 or b=4. If b=3, then c=6; and if b=4, then c=4. If a=3, then it must be that b=c=3. So the solutions in this case are (2,3,6),(2,4,4), and (3,3,3).
If r=2, then, because aβ€bβ€c, it follows that a=3,a=4,a=5, or a=6. If a=3, then b1β+c1β=61β, which has no solutions because 91β+91β>61β. If a=4, then b1β+c1β=41β. In this case, b=c=8 is the only solution. If a=5, then b1β+c1β=103β, which gives no solutions. If a=6, then it follows that b=c=6. So the solutions in this case are (4,8,8) and (6,6,6).
If r=3, then (a,b,c)=(9,9,9) is the only solution because if a<9, then c>9.
Finally, the altitude lengths must be checked to ensure that these lengths give dimensions for a valid triangle. In the (2,3,6) case, the side lengths of the triangles become (3t,2t,t) for some t, which does not form a triangle. In the (2,4,4) and (4,8,8) cases, the side lengths of the triangles become (2t,t,t) for some t, which again does not form a triangle. The rest of the cases, namely (3,3,3), (6,6,6), and (9,9,9), do produce valid triangles because if all of the altitudes have length h, then it is possible to form an equilateral triangle with side length 323ββh. Thus there are (B)3β ordered triples satisfying the given conditions.