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

m (→‎Solution 4(Algebraic Bashing))
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==Solution 1 (Trig)==
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==Solution 1 (Simple)  ==
 +
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>
 +
 
 +
Note: This solution does not use the condition <math>MN=65</math>.
 +
 
 +
- gregwwl
 +
 
 +
==Solution 2 (Trig)==
 
Let <math>\angle MKN=\alpha</math> and <math>\angle LNK=\beta</math>. Note <math>\angle KLP=\beta</math>.  
 
Let <math>\angle MKN=\alpha</math> and <math>\angle LNK=\beta</math>. Note <math>\angle KLP=\beta</math>.  
  
Line 14: Line 30:
 
Thus, <math>MK=\frac{MN}{\tan\alpha}=98</math>, so <math>MO=MK-KO=\boxed{090}</math>.
 
Thus, <math>MK=\frac{MN}{\tan\alpha}=98</math>, so <math>MO=MK-KO=\boxed{090}</math>.
  
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==Solution 2 (Similar triangles)==
+
==Solution 3 (Similar triangles)==
 
<asy>
 
<asy>
 
size(250);
 
size(250);
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Solution by vedadehhc
 
Solution by vedadehhc
  
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==Solution 3 (Similar triangles, orthocenters)==
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==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>).
 
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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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)
 
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)==
+
==Solution 5(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,</math> and <math>g=NC.</math> We are trying to find <math>d.</math> We can find <math>7</math> equations. They are
 
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,</math> and <math>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=c^2,</cmath>
Line 77: Line 93:
 
(Solution by DottedCaculator)
 
(Solution by DottedCaculator)
  
−
==Solution 5(5-second PoP)==
+
==Solution 6(5-second PoP)==
  
 
<asy>
 
<asy>
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(Solution by TheUltimate123)
 
(Solution by TheUltimate123)
  
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==Solution 6  ==
 
−
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>
 
  
−
Note: This solution does not use the condition <math>MN=65</math>.
 
−
 
−
- gregwwl
 
 
==Video Solution==
 
==Video Solution==
 
Video Solution:
 
Video Solution:

Revision as of 20:11, 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 (Simple)

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

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

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

Note: This solution does not use the condition $MN=65$.

- 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(Algebraic Bashing)

First, let $P$ be the intersection of $LO$ and $KN$. We can use the right triangles in the problem to create equations. Let $a=NP, b=PK, c=NO, d=OM, e=OP, f=PC,$ and $g=NC.$ We are trying to find $d.$ We can find $7$ equations. They are \[4225+d^2=c^2,\] \[4225+d^2+16d+64=a^2+2ab+b^2,\] \[a^2+e^2=c^2,\] \[b^2+e^2=64,\] \[b^2+e^2+2ef+f^2=784,\] \[a^2+e^2+2ef+f^2=g^2,\] and \[g^2+784=a^2+2ab+b^2.\] We can subtract the fifth equation from the sixth equation to get $a^2-b^2=g^2-784.$ We can subtract the fourth equation from the third equation to get $a^2-b^2=c^2-64.$ Combining these equations gives $c^2-64=g^2-784$ so $g^2=c^2+720.$ Substituting this into the seventh equation gives $c^2+1504=a^2+2ab+b^2.$ Substituting this into the second equation gives $4225+d^2+16d+64=c^2+1504$. Subtracting the first equation from this gives $16d+64=1504.$ Solving this equation, we find that $d=\boxed{090}.$ (Solution by DottedCaculator)

Solution 6(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

See Also

2019 AIME I (Problems • Answer Key • Resources)
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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