2019 AIME I Problems/Problem 6: Difference between revisions

Revision as of 19:12, 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 (Simplicity)

Note that $KLMN$ is cyclic with diameter $KN$ since $\angle KLN = \angle KMN = \frac{\pi}{2}$ . Also, note that we have $\triangle KML \sim \triangle KLO$ by SS similarity.

We see this by $\angle LKM = \angle OKL$ and $\angle KLO = \angle KML$ . The latter equality can be seen if we extend $LP$ to point $L'$ on $(KLMN)$ . We know $LK = KL'$ from which it follows $\angle KLO = \angle KML$ .

Let $MO = x$ . By $\triangle KML \sim \triangle KLO$ we have

$\frac{KL}{KO} = \frac{KM}{KL} \Rightarrow \frac{28}{8} = \frac{x+8}{28}.$

$98 = x + 8 \Rightarrow x = \boxed{090}.$

- gregwwl

Solution 2 (Trig)

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

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

Dividing the equations gives $\frac{65}{28}=\frac{28\sin\alpha}{8\cos\alpha}=\frac{7}{2}\tan\alpha\Rightarrow \tan\alpha=\frac{65}{98}$

Thus, $MK=\frac{MN}{\tan\alpha}=98$ , so $MO=MK-KO=\boxed{090}$ .

Solution 3 (Similar triangles)

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

First, let $P$ be the intersection of $LO$ and $KN$ as shown above. 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, $\triangle KMN \sim \triangle KPO$ . Using these similarities we see that $\frac{KP}{KL} = \frac{KL}{KN}$ $KP = \frac{KL^2}{KN} = \frac{28^2}{KN} = \frac{784}{KN}$ and $\frac{KP}{KO} = \frac{KM}{KN}$ $KP = \frac{KO \cdot KM}{KN} = \frac{8\cdot KM}{KN}$ Combining the two equations, we get $\frac{8\cdot KM}{KN} = \frac{784}{KN}$ $8 \cdot KM = 28^2$ $KM = 98$ Since $KM = KO + MO$ , we get $MO = 98 -8 = \boxed{090}$ .

Solution by vedadehhc

Solution 4 (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)

Solution 5 (5-second PoP)

$[asy] size(8cm); pair K, L, M, NN, X, O; K=(-sqrt(98^2+65^2)/2, 0); NN=(sqrt(98^2+65^2)/2, 0); L=sqrt(98^2+65^2)/2*dir(180-2*aSin(28/sqrt(98^2+65^2))); M=sqrt(98^2+65^2)/2*dir(2*aSin(65/sqrt(98^2+65^2))); X=foot(L, K, NN); O=extension(L, X, K, M); draw(K -- L -- M -- NN -- K -- M); draw(L -- NN); draw(arc((K+NN)/2, NN, K)); draw(L -- X, dashed); draw(arc((O+NN)/2, NN, X), dashed); draw(rightanglemark(K, L, NN, 100)); draw(rightanglemark(K, M, NN, 100)); draw(rightanglemark(L, X, NN, 100)); dot("$K$", K, SW); dot("$L$", L, unit(L)); dot("$M$", M, unit(M)); dot("$N$", NN, SE); dot("$X$", X, S); [/asy]$ Notice that $KLMN$ is inscribed in the circle with diameter $\overline{KN}$ and $XOMN$ is inscribed in the circle with diameter $\overline{ON}$ . Furthermore, $(XLN)$ is tangent to $\overline{KL}$ . Then, $KO\cdot KM=KX\cdot KN=KL^2\implies KM=\frac{28^2}{8}=98,$ and $MO=KM-KO=\boxed{090}$ .

(Solution by TheUltimate123)

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 (Trig)==
+==Solution 1 (Simplicity) ==
+Note that <math>KLMN</math> is cyclic with diameter <math>KN</math> since <math>\angle KLN = \angle KMN = \frac{\pi}{2}</math>. Also, note that we have <math>\triangle KML \sim \triangle KLO</math> by SS similarity.
+We see this by <math>\angle LKM = \angle OKL</math> and <math>\angle KLO = \angle KML</math>.
+The latter equality can be seen if we extend <math>LP</math> to point <math>L'</math> on <math>(KLMN)</math>. We know <math>LK = KL'</math> from which it follows <math>\angle KLO = \angle KML</math>.
+Let <math>MO = x</math>. By <math>\triangle KML \sim \triangle KLO</math> we have
+<cmath>\frac{KL}{KO} = \frac{KM}{KL} \Rightarrow \frac{28}{8} = \frac{x+8}{28}.</cmath>
+<cmath>98 = x + 8 \Rightarrow x = \boxed{090}.</cmath>
+- gregwwl
+==Solution 2 (Trig)==
 Let <math>\angle MKN=\alpha</math> and <math>\angle LNK=\beta</math>. Note <math>\angle KLP=\beta</math>.
@@ Line 13: / Line 28: @@
 Thus, <math>MK=\frac{MN}{\tan\alpha}=98</math>, so <math>MO=MK-KO=\boxed{090}</math>.
-==Solution 2 (Similar triangles)==
+==Solution 3 (Similar triangles)==
 <asy>
 size(250);
@@ Line 53: / Line 68: @@
 Solution by vedadehhc
-==Solution 3 (Similar triangles, orthocenters)==
+==Solution 4 (Similar triangles, orthocenters)==
 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>).
@@ Line 64: / Line 79: @@
 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)
-==Solution 4 (Algebraic Bashing)==
-First, let <math>P</math> be the intersection of <math>LO</math> and <math>KN</math>. We can use the right triangles in the problem to create equations. Let <math>a=NP, b=PK, c=NO, d=OM, e=OP, f=PC, and g=NC.</math> We are trying to find <math>d.</math> We can find <math>7</math> equations. They are
-<cmath>4225+d^2=c^2,</cmath>
-<cmath>4225+d^2+16d+64=a^2+2ab+b^2,</cmath>
-<cmath>a^2+e^2=c^2,</cmath>
-<cmath>b^2+e^2=64,</cmath>
-<cmath>b^2+e^2+2ef+f^2=784,</cmath>
-<cmath>a^2+e^2+2ef+f^2=g^2,</cmath>
-and <cmath>g^2+784=a^2+2ab+b^2.</cmath>
-We can subtract the fifth equation from the sixth equation to get <math>a^2-b^2=g^2-784.</math> We can subtract the fourth equation from the third equation to get <math>a^2-b^2=c^2-64.</math> Combining these equations gives <math>c^2-64=g^2-784</math> so <math>g^2=c^2+720.</math> Substituting this into the seventh equation gives <math>c^2+1504=a^2+2ab+b^2.</math> Substituting this into the second equation gives <math>4225+d^2+16d+64=c^2+1504</math>. Subtracting the first equation from this gives <math>16d+64=1504.</math> Solving this equation, we find that <math>d=\boxed{090}.</math>
-(Solution by DottedCaculator)
 ==Solution 5 (5-second PoP)==

2019 AIME I (Problems • Answer Key • Resources)
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