# 2019 AIME I Problems/Problem 6

## Problem 6

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 $MO$.

## Solution 1 (Trig)

Let $\angle MKN=\alpha$ and $\angle LNK=\beta$. Note $\angle KLP=\beta$.

Then, $KP=28\sin\beta=8\cos\alpha$. Furthermore, $KN=\frac{65}{\sin\alpha}=\frac{28}{\sin\beta} \Rightarrow 65\sin\beta=28\sin\alpha$.

Dividing the equations gives $$\frac{65}{28}=\frac{28\sin\alpha}{8\cos\alpha}=\frac{7}{2}\tan\alpha\Rightarrow \tan\alpha=\frac{65}{98}$$

Thus, $MK=\frac{MN}{\tan\alpha}=98$, so $MO=MK-KO=\boxed{090}$.

## Solution 2 (Similar triangles)

$[asy] size(250); real h = sqrt(98^2+65^2); real l = sqrt(h^2-28^2); pair K = (0,0); pair N = (h, 0); pair M = ((98^2)/h, (98*65)/h); pair L = ((28^2)/h, (28*l)/h); pair P = ((28^2)/h, 0); pair O = ((28^2)/h, (8*65)/h); draw(K--L--N); draw(K--M--N--cycle); draw(L--M); label("K", K, SW); label("L", L, NW); label("M", M, NE); label("N", N, SE); draw(L--P); label("P", P, S); dot(O); label("O", shift((1,1))*O, NNE); label("28", scale(1/2)*L, W); label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE); [/asy]$

First, let $P$ be the intersection of $LO$ and $KN$ as shown above. Note that $m\angle KPL = 90^{\circ}$ as given in the problem. Since $\angle KPL \cong \angle KLN$ and $\angle PKL \cong \angle LKN$, $\triangle PKL \sim \triangle LKN$ by AA similarity. Similarly, $\triangle KMN \sim \triangle KPO$. Using these similarities we see that $$\frac{KP}{KL} = \frac{KL}{KN}$$ $$KP = \frac{KL^2}{KN} = \frac{28^2}{KN} = \frac{784}{KN}$$ and $$\frac{KP}{KO} = \frac{KM}{KN}$$ $$KP = \frac{KO \cdot KM}{KN} = \frac{8\cdot KM}{KN}$$ Combining the two equations, we get $$\frac{8\cdot KM}{KN} = \frac{784}{KN}$$ $$8 \cdot KM = 28^2$$ $$KM = 98$$ Since $KM = KO + MO$, we get $MO = 98 -8 = \boxed{090}$.

## Solution 3 (Similar triangles, orthocenters)

Extend $KL$ and $NM$ past $L$ and $M$ respectively to meet at $P$. Let $H$ be the intersection of diagonals $KM$ and $LN$ (this is the orthocenter of $\triangle KNP$).

As $\triangle KOL \sim \triangle KHP$ (as $LO \parallel PH$, using the fact that $H$ is the orthocenter), we may let $OH = 8k$ and $LP = 28k$.

Then using similarity with triangles $\triangle KLH$ and $\triangle KMP$ we have

$$\frac{28}{8+8k} = \frac{8+8k+HM}{28+28k}$$

Cross-multiplying and dividing by $4+4k$ gives $2(8+8k+HM) = 28 \cdot 7 = 196$ so $MO = 8k + HM = \frac{196}{2} - 8 = \boxed{090}$. (Solution by scrabbler94)

## Solution 4 (Algebraic Bashing)

First, let $P$ be the intersection of $LO$ and $KN$. We can use the right triangles in the problem to create equations. Let $a=NP, b=PK, c=NO, d=OM, e=OP, f=PC,$ and $g=NC.$ We are trying to find $d.$ We can find $7$ equations. They are $$4225+d^2=c^2,$$ $$4225+d^2+16d+64=a^2+2ab+b^2,$$ $$a^2+e^2=c^2,$$ $$b^2+e^2=64,$$ $$b^2+e^2+2ef+f^2=784,$$ $$a^2+e^2+2ef+f^2=g^2,$$ and $$g^2+784=a^2+2ab+b^2.$$ We can subtract the fifth equation from the sixth equation to get $a^2-b^2=g^2-784.$ We can subtract the fourth equation from the third equation to get $a^2-b^2=c^2-64.$ Combining these equations gives $c^2-64=g^2-784$ so $g^2=c^2+720.$ Substituting this into the seventh equation gives $c^2+1504=a^2+2ab+b^2.$ Substituting this into the second equation gives $4225+d^2+16d+64=c^2+1504$. Subtracting the first equation from this gives $16d+64=1504.$ Solving this equation, we find that $d=\boxed{090}.$ (Solution by DottedCaculator)

## Solution 5 (5-second PoP)

$[asy] size(8cm); pair K, L, M, NN, X, O; K=(-sqrt(98^2+65^2)/2, 0); NN=(sqrt(98^2+65^2)/2, 0); L=sqrt(98^2+65^2)/2*dir(180-2*aSin(28/sqrt(98^2+65^2))); M=sqrt(98^2+65^2)/2*dir(2*aSin(65/sqrt(98^2+65^2))); X=foot(L, K, NN); O=extension(L, X, K, M); draw(K -- L -- M -- NN -- K -- M); draw(L -- NN); draw(arc((K+NN)/2, NN, K)); draw(L -- X, dashed); draw(arc((O+NN)/2, NN, X), dashed); draw(rightanglemark(K, L, NN, 100)); draw(rightanglemark(K, M, NN, 100)); draw(rightanglemark(L, X, NN, 100)); dot("K", K, SW); dot("L", L, unit(L)); dot("M", M, unit(M)); dot("N", NN, SE); dot("X", X, S); [/asy]$ Notice that $KLMN$ is inscribed in the circle with diameter $\overline{KN}$ and $XOMN$ is inscribed in the circle with diameter $\overline{ON}$. Furthermore, $(XLN)$ is tangent to $\overline{KL}$. Then, $$KO\cdot KM=KX\cdot KN=KL^2\implies KM=\frac{28^2}{8}=98,$$and $MO=KM-KO=\boxed{090}$.

(Solution by TheUltimate123)

## Solution 6 (Alternative PoP)

$[asy] size(250); real h = sqrt(98^2+65^2); real l = sqrt(h^2-28^2); pair K = (0,0); pair N = (h, 0); pair M = ((98^2)/h, (98*65)/h); pair L = ((28^2)/h, (28*l)/h); pair P = ((28^2)/h, 0); pair O = ((28^2)/h, (8*65)/h); draw(K--L--N); draw(K--M--N--cycle); draw(L--M); label("K", K, SW); label("L", L, NW); label("M", M, NE); label("N", N, SE); draw(L--P); label("P", P, S); dot(O); label("O", shift((1,1))*O, NNE); label("28", scale(1/2)*L, W); label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE); [/asy]$

Call the base of the altitude from $L$ to $NK$ point $P$. Let $PO=x$. Now, we have that $KP=\sqrt{64-x^2}$ by the Pythagorean Theorem. Once again by Pythagorean, $LO=\sqrt{720+x^2}-x$. Using Power of a Point, we have

$$(KO)(OM)=(LO)(OQ)$$ ($Q$ is the intersection of $OL$ with the circle $\neq L$)

$$8(MO)=(\sqrt{720+x^2}+x)(\sqrt{720+x^2}-x)$$

$$8(MO)=720$$

$$MO=\boxed{090}$$.

(Solution by RootThreeOverTwo)

## Solution 7 (just one pair of similar triangles)

$[asy] size(250); real h = sqrt(98^2+65^2); real l = sqrt(h^2-28^2); pair K = (0,0); pair N = (h, 0); pair M = ((98^2)/h, (98*65)/h); pair L = ((28^2)/h, (28*l)/h); pair P = ((28^2)/h, 0); pair O = ((28^2)/h, (8*65)/h); draw(K--L--N); draw(K--M--N--cycle); draw(L--M); label("K", K, SW); label("L", L, NW); label("M", M, NE); label("N", N, SE); draw(L--P); label("P", P, S); dot(O); label("O", shift((1,1))*O, NNE); label("28", scale(1/2)*L, W); label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE); [/asy]$ Note that since $\angle KLN = \angle KMN$, quadrilateral $KLMN$ is cyclic. Therefore, we have $$\angle LMK = \angle LNK = 90^{\circ} - \angle LKN = \angle KLP,$$so $\triangle KLO \sim \triangle KML$, giving $$\frac{KM}{28} = \frac{28}{8} \implies KM = 98.$$ Therefore, $OM = 98-8 = \boxed{90}$.