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

Revision as of 08:51, 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) $[asy] size(250); real h = sqrt(98^2+65^2); real l = sqrt(h^2-28^2); pair K = (0,0); pair N = (h, 0); pair M = ((98^2)/h, (98*65)/h); pair L = ((28^2)/h, (28*l)/h); pair P = ((28^2)/h, 0); pair O = ((28^2)/h, (8*65)/h); draw(K--L--N); draw(K--M--N--cycle); draw(L--M); label("K", K, W); label("L", L, NW); label("M", M, NE); label("N", N, E); draw(L--P); label("P", P, S); dot(O); label("O", shift((1,1))*O, NNE); label("28", scale(1/2)*L, W); label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE); [/asy]$

First, let $P$ be the intersection of $LO$ and $KN$ . Note that $m\angle KPL = 90^{\circ}$ as given in the problem.

Video Solution

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

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

Revision as of 08:51, 15 March 2019

Contents

Problem 6

Solution (Similar triangles)

Video Solution

See Also

@@ Line 26: / Line 26: @@
 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 <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.
 ==Video Solution==