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

(Solution (Similar triangles))
(Solution (Similar triangles))
Line 17: Line 17:
 
draw(K--M--N--cycle);
 
draw(K--M--N--cycle);
 
draw(L--M);
 
draw(L--M);
label("K", K, W);
+
label("K", K, SW);
 
label("L", L, NW);
 
label("L", L, NW);
 
label("M", M, NE);
 
label("M", M, NE);
label("N", N, E);
+
label("N", N, SE);
 
draw(L--P);
 
draw(L--P);
 
label("P", P, S);
 
label("P", P, S);
Line 29: Line 29:
 
</asy>
 
</asy>
  
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.
+
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>.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 10:54, 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, $\triangle KMN \sim \triangle KPO$.

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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