Question

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.

Answer 1

[asy]
pair M,N,O;
M = (0,0);
N=equator(M,25);
O = (50,0);
draw(M--N--O--cycle);
label("$M$",M,SW);
label("$N$",N,W);
label("$O$",O,SE);
pair P = 2*N/3+M/3;
draw(P--foot(P,M,O));
label("$P$",P,S);
label("$50$",M--O);
label("$25$",(M--N),W);
[/asy]

Let the altitude from $N$ to $\overline{MO}$ intersect $\overline{MO}$ at point $Q$. Since $\triangle MNO$ is an isosceles triangle with $MN=25$, we have $MQ=25/2$. Therefore, we see that $NMOQ$ is a right trapezoid. Let the foot of the perpendicular from $P$ to $\overline{MQ}$ be $R$. Since $\overline{MN}\parallel\overline{OQ}$, we have $\triangle NPQ\sim\triangle MRO$. Then \[\frac{NP}{PM} = \frac{NO}{OM}\quad\Rightarrow \quad\frac{2}{1} = \frac{25}{OM}\quad \Rightarrow\quad OM =
\frac{25}{2}.\]
[asy]
pair M,N,O,R;
M = (0,0);
N=equator(M,25);
O = (25/2,0); // (50+0)/2 = 50/2
R = foot((2*N+M)/3,M,O);
draw((2*N+M)/3--R);
draw(M--N--O--cycle);
label("$M$",M,SW);
label("$N$",N,W);
label("$O$",O,SE);
label("$Q$",foot(N,M,O),N);
label("$P$", (2*N+M)/3,S);
label("$R$",R,E);
label("$50$",(M--N),W);
label("$25/2$",(M--O),S);
[/asy]

We know that $MN = 25$, and we want to find $AQ$. Using Pythagorean Theorem, \[OQ = \sqrt{MN^2+OM^2} = \sqrt{25^2+\left(\frac{25}{2}\right)^2} = \frac{25}{2}\sqrt{5}.\] Now, observe that since $\triangle MNQ\sim\triangle OQM$ (by AA), we have \[\frac{MN}{NQ} = \frac{OQ}{MQ}\quad\Rightarrow \quad\frac{25}{NQ} = \frac{OQ}{25/2}\quad\Rightarrow \quad NQ = \frac{625}{2\cdot 25\sqrt{5}} = \frac{25\sqrt5}{2}.\] Since $\triangle MNQ$ is a $45-45-90$ triangle, we now know that $QO = NQ = \frac{25\sqrt5}{2}$. Therefore, \[AQ = OQ+OQ = \frac{25}{2}\sqrt5 + \frac{25}{2}\sqrt5 = \boxed{25\sqrt5}.\]