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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.

shreyas1  Oct 3, 2018
 #1
avatar+94087 
+2

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

Melody  Oct 4, 2018

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