+468 In triangle ABC, BC = 8. The length of median AD is 5. Let M be the largest possible value of AB^2 + AC^2, and let m be the smallest possible value. Find M - m.
+118677 Here is the pic
You are being asked for the difference between the maximum and the minimum values of \(b^2+c^2\)
Note: \(0
| \(x^2+5^2=b^2\\ x^2+25=b^2\\ \) \( (8-x)^2+5^2=c^2\\ 64+x^2-16x+25=c^2\\ x^2-16x+89=c^2\\ \) |
Add
\(x^2+25+x^2-16x+89=b^2+c^2\\ 2x^2-16x+114=b^2+c^2\\ b^2+c^2=2x^2-16x+114\\\)
So we need a maximum and a minimum value for the expression on the right.
Consider the parabola
\(y=2x^2-16x+114\)
This is a concave up parabola. \(16^2-4*2*114 <0\) so all values of y are positive.
The axis of symmetry is \(x=\frac{-b}{2a}=\frac{--16}{2*2}=\frac{16}{4}=4\)
When x=4 \(y=2*4^2-16*4+114=32-64+114=82\)
So the smallest value appears to be 82
The largest value would occur as x approaches 8 or 0 (x cannot actually be 8 or 0)
When x=0 or 8 y=114
so
\(82\le a^2+b^2<114\\~\\ 114-82=32\)
So the difference between the largest and smallest values of b^2+c^2 must be less than 32
