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 by vedadehhc

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)

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

(Diagram by vedadehhc)

Call the base of the altitude from $L$ to $NK$ point $P$ . Let $PO=x$ . Now, we have that $KO=\sqrt{64-x^2}$ 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)

Video Solution

Video Solution: https://www.youtube.com/watch?v=0AXF-5SsLc8

2019 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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