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

(Solution (Similar triangles))
Line 5: Line 5:
 
(writing this, don't edit)
 
(writing this, don't edit)
 
<asy>
 
<asy>
size(100);
+
size(250);
draw((0,0)--(1,0));
+
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);
 
</asy>
 
</asy>
  

Revision as of 08:53, 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); [/asy]

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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