+1911 Triangle $MNO$ is an isosceles triangle with $MN = NO = 25\;\text{cm}$. A line segment, drawn from the midpoint of $\overline{MO}$ perpendicular to $\overline{MN}$, intersects $\overline{MN}$ at point $P$ with $NP:PM = 2:1$. What is the length of the altitude drawn from point $N$ to $\overline{MO}$? Express your answer in simplest radical form.
+129881 Call the midpoint of MO = Q
Call the altitude A
NP = (2/3)(25) = 50/3
PM = 25/3
QN^2 = [ A^2 - (50/3)^2 ]
QN^2 = [ MQ^2 - (25/3)^2 ]
A^2 - (50/3)^2 = MQ^2 - (25/3)^2
A^2 - 2500/9 = MQ^2 - 625/9
MQ^2 = A^2 - 1875/9
MN ^2 = MQ^2 + A^2
25^2 = A^2 - 1875/9 + A^2
625 = 2A^2 - 1875/9
5625 = 18A^2 - 1875
18A^2 = 5625 + 1875
18A^2 = 7500
A^2 = (7500 / 18) = ( 1250/3)
A = sqrt (1250/3) = sqrt ( 625 * 2 / 3) = 25 sqrt (2/3) = 25sqrt (6) / 3
