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

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(Solution (Similar triangles): add alternate solution)
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First, let <math>P</math> be the intersection of <math>LO</math> and <math>KN</math>. Note that <math>m\angle KPL = 90^{\circ}</math> as given in the problem. Since <math>\angle KPL \cong \angle KLN</math> and <math>\angle PKL \cong \angle LKN</math>, <math>\triangle PKL \sim \triangle LKN</math> by AA similarity. Similarly, <math>\triangle KMN \sim \triangle KPO</math>.
 
First, let <math>P</math> be the intersection of <math>LO</math> and <math>KN</math>. Note that <math>m\angle KPL = 90^{\circ}</math> as given in the problem. Since <math>\angle KPL \cong \angle KLN</math> and <math>\angle PKL \cong \angle LKN</math>, <math>\triangle PKL \sim \triangle LKN</math> by AA similarity. Similarly, <math>\triangle KMN \sim \triangle KPO</math>.
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==Solution 2 (Similar triangles, orthocenters)==
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Extend <math>KL</math> and <math>NM</math> past <math>L</math> and <math>M</math> respectively to meet at <math>P</math>. Let <math>H</math> be the intersection of diagonals <math>KM</math> and <math>LN</math> (this is the orthocenter of <math>\triangle KNP</math>).
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As <math>\triangle KOL \sim \triangle KHP</math> (as <math>LO \parallel PH</math>, using the fact that <math>H</math> is the orthocenter), we may let <math>OH = 8k</math> and <math>LP = 28k</math>.
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Then using similarity with triangles <math>\triangle KLH</math> and <math>\triangle KMP</math> we have
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<cmath>\frac{28}{8+8k} = \frac{8+8k+HM}{28+28k}</cmath>
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Cross-multiplying and dividing by <math>4+4k</math> gives <math>2(8+8k+HM) = 28 \cdot 7 = 196</math> so <math>MO = 8k + HM = \frac{196}{2} - 8 = \boxed{090}</math>. (Solution by scrabbler94)
  
 
==Video Solution==
 
==Video Solution==

Revision as of 13:57, 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 (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

See Also

2019 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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