Difference between revisions of "2019 AIME I Problems/Problem 6"

Revision as of 13:01, 15 March 2019

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

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

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

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

Thus, $MK=\frac{MN}{\tan\alpha}=98$ , so $MO=MK-KO=\boxed{90}$ .

Solution (Similar triangles)

(writing this, don't edit) $[asy] size(250); real h = sqrt(98^2+65^2); real l = sqrt(h^2-28^2); pair K = (0,0); pair N = (h, 0); pair M = ((98^2)/h, (98*65)/h); pair L = ((28^2)/h, (28*l)/h); pair P = ((28^2)/h, 0); pair O = ((28^2)/h, (8*65)/h); draw(K--L--N); draw(K--M--N--cycle); draw(L--M); label("K", K, SW); label("L", L, NW); label("M", M, NE); label("N", N, SE); draw(L--P); label("P", P, S); dot(O); label("O", shift((1,1))*O, NNE); label("28", scale(1/2)*L, W); label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE); [/asy]$

First, let $P$ be the intersection of $LO$ and $KN$ . 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, $\triangle KMN \sim \triangle KPO$ .

Solution 2 (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)

Video Solution

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

@@ Line 2: / Line 2: @@
 In convex quadrilateral <math>KLMN</math> side <math>\overline{MN}</math> is perpendicular to diagonal <math>\overline{KM}</math>, side <math>\overline{KL}</math> is perpendicular to diagonal <math>\overline{LN}</math>, <math>MN = 65</math>, and <math>KL = 28</math>. The line through <math>L</math> perpendicular to side <math>\overline{KN}</math> intersects diagonal <math>\overline{KM}</math> at <math>O</math> with <math>KO = 8</math>. Find <math>MO</math>.
+==Solution 1==
+Let <math>\angle MKN=\alpha</math> and <math>\angle LNK=\beta</math>. Note <math>\angle KLP=\beta</math>.
+Then, <math>KP=28\sin\beta=8\cos\alpha</math>.
+Furthermore, <math>KN=65/\sin\alpha=28/\sin\beta \Rightarrow 65\sin\beta=28\sin\alpha</math>.
+Dividing the equations gives
+<cmath>\frac{7}{2}\tan\alpha=\frac{65}{28}\Rightarrow \tan\alpha=\frac{65}{98}</cmath>
+Thus, <math>MK=\frac{MN}{\tan\alpha}=98</math>, so <math>MO=MK-KO=\boxed{90}</math>.
 ==Solution (Similar triangles)==
 (writing this, don't edit)

2019 AIME I (Problems • Answer Key • Resources)
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