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

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Because <math>\angle KLN = \angle KMN = 90^{\circ}</math>, <math>KLMN</math> is a cyclic quadrilateral. Hence, by Power of Point, <cmath>KO\cdot KM = KL^2 \implies KM=\dfrac{28^2}{8}=98 \implies MO=98-8=\boxed{090},</cmath> as desired.
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Because <math>\angle KLN = \angle KMN = 90^{\circ}</math>, <math>KLMN</math> is a cyclic quadrilateral. Hence, by Power of Point, <cmath>KO\cdot KM = KL^2 \implies KM=\dfrac{28^2}{8}=98 \implies MO=98-8=\boxed{090}</cmath> as desired.
  
 
~Mathkiddie
 
~Mathkiddie

Latest revision as of 19:11, 19 January 2024

Problem

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 (Trig)

Let $\angle MKN=\alpha$ and $\angle LNK=\beta$. Let $P$ be the project of $L$ onto line $NK$. 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 2 (Cyclic Quads, PoP)

[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]

Because $\angle KLN = \angle KMN = 90^{\circ}$, $KLMN$ is a cyclic quadrilateral. Hence, by Power of Point, \[KO\cdot KM = KL^2 \implies KM=\dfrac{28^2}{8}=98 \implies MO=98-8=\boxed{090}\] as desired.

~Mathkiddie

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)

If you're wondering why $KX \cdot KN=KL^2,$ it's because PoP on $(XLN)$ or by $KX \cdot KN=KX \cdot (KX+XN)=KX^2+KX \cdot XN=KX^2+LX^2=KL^2$ (last part by similarity).

Solution 7 (Alternative PoP)

[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]

(Diagram by vedadehhc)

Call the base of the altitude from $L$ to $NK$ point $P$. Let $PO=x$. Now, we have that $KP=\sqrt{64-x^2}$ by the Pythagorean Theorem. Once again by Pythagorean, $LO=\sqrt{720+x^2}-x$. Using Power of a Point, we have

\[(KO)(OM)=(LO)(OQ)\] ($Q$ is the intersection of $OL$ with the circle $\neq L$)

\[8(MO)=(\sqrt{720+x^2}-x)(\sqrt{720+x^2}+x)\]

\[8(MO)=720\]

\[MO=\boxed{090}\].

(Solution by RootThreeOverTwo) \[\]

Remark: Length of OQ

Since $P$ is on the circle’s diameter, $QP = LP = \sqrt{720+x^2}$. So, $OQ = PQ + PO = x + \sqrt{720+x^2}+x)$. ~diyarv

Solution8 (just one pair of 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] Note that since $\angle KLN = \angle KMN$, quadrilateral $KLMN$ is cyclic. Therefore, we have \[\angle LMK = \angle LNK = 90^{\circ} - \angle LKN = \angle KLP,\]so $\triangle KLO \sim \triangle KML$, giving \[\frac{KM}{28} = \frac{28}{8} \implies KM = 98.\] Therefore, $OM = 98-8 = \boxed{90}$.

Solution 9 (Pythagoras Bash)

[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]

By Pythagorean Theorem, $KM^2+65^2 = KN^2 = 28^2 + LN^2$. Thus, $LN^2 = KM^2 + 65^2 - 28^2$.

By Pythagorean Theorem, $KP^2 + LP^2 = 28^2$, and $PN^2 + LP^2 = LN^2$.

\[PN^2 = (KN - KP)^2 = (\sqrt{KM^2 + 65^2} - KP)^2\]

It follows that \[(\sqrt{KM^2 + 65^2} - KP)^2 + LP^2 = KM^2 + 65^2 - 28^2\]

\[KM^2 + 65^2 - 2\sqrt{KM^2 + 65^2}(KP) + KP^2 + LP^2 = KM^2 + 65^2 - 28^2\]

Since $KP^2 + LP^2 = 28^2$, \[KM^2 + 65^2 - 2\sqrt{KM^2 + 65^2}(KP) + 28^2 = KM^2 + 65^2 - 28^2\]

\[-2\sqrt{KM^2 + 65^2}(KP) = -2 \times 28^2\]

\[KP = \frac{28^2}{\sqrt{KM^2 + 65^2}}\]

$\angle OKP = \angle NKM$ (it's the same angle) and $\angle OPK = \angle KMN = 90^{\circ}$. Thus, $\triangle KOP \sim \triangle KNM$.

Thus,

\[\frac{KO}{KN} = \frac{KP}{KM}\]

\[\frac{8}{\sqrt{KM^2 + 65^2}} = \frac{\frac{28^2}{\sqrt{KM^2 + 65^2}}}{KM}\]

Multiplying both sides by $\sqrt{KM^2 + 65^2}$:

\[8 = \frac{28^2}{KM}\]

\[KM = 98\]

Therefore, $OM = 98-8 = \boxed{90}$

~ Solution by adam_zheng

Video Solution

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

Video Solution 2

https://youtu.be/I-8xZGhoDUY

~Shreyas S

Video Solution 3

https://www.youtube.com/watch?v=pP3cih_8bg4

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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