Two vertices of an isosceles triangle are (1,2) and (4,6). The inradius of the triangle is $\frac{3}{2}$. Find the maximum possible area for the triangle.
My work, for the two possible structures of this problem, has been the elaboration of setting unknown x, and solve equations(sometimes 4th degree polynomials, and you have to guess/test its roots without a calculator), but i never got a neat answer.
The solution goes:
there are two possible structures. In the first, the side of length 5 is the base, and then the median from the opposite vertex is 3 times the in radius in length, so the area is $3*(\frac{3}{2}*\frac{5}{2})= \frac{45}{4}$. In the second, the side of length 5 is one of two equal sides, and then the base must be $ 2*(\sqrt{(25-9(3/2)^2)})= \sqrt{19} < 5$, so the maximum is $\frac{45}{4}$.
I don't think the solution is ever logically right. Any ideas?
The solution is wrong. It has confused the incentre with the centroid.
The incentre, or centre of the incircle, is where the angle bisectors meet.
The centroid, which is the intersection of the medians, is one-third of the way along the medians. The medians connect each vertex with the midpoint of the opposite side.
A median is only the same as an angle bisector when the two adjacent sides are equal. The three medians only intersect at the incentre when all three sides are equal.