Question

In convex quadrilateral $KLMN,$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM},$ side $\overline{KL}$ is perpendicular to diagonal $\overline{LN},$ $MN = 65,$ and $KL = 28.$ The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8.$ Find $MO.$

Answer

Answer: 90

Concepts:similarity right triangle altitude

Difficulty rating: 2600

Solution:

Let $F$ be the foot of the perpendicular from $L$ to $\overline{KN},$ so $O$ lies on segment $LF.$ In right triangle $KLN$ (right angle at $L$ ), the altitude $LF$ to the hypotenuse gives the geometric mean relation $KF \cdot KN = KL^2 = 28^2 = 784.$

Triangles $KFO$ and $KMN$ share angle $K,$ and $\angle KFO = 90^\circ = \angle KMN,$ so they are similar. Hence $\frac{KF}{KM} = \frac{KO}{KN},$ that is, $KO \cdot KM = KF \cdot KN = 784.$ With $KO = 8$ this gives $KM = 98,$ so $\begin{aligned} MO &= KM - KO \\ &= 98 - 8 = 90. \end{aligned}$

2019 AIME I Problem 6

Answer: 90

Difficulty rating: 2600

Solution:

Problem 6 in Other Years