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Difference between revisions of "2019 AIME I Problems/Problem 6"
(→‎Solution 6 (5-second PoP))
 
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−
The 2019 Aime I takes place on March 13, 2019.
+
==Problem==
 +
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)==
 +
Let <math>\angle MKN=\alpha</math> and <math>\angle LNK=\beta</math>. Let <math>P</math> be the project of <math>L</math> onto line <math>NK</math>. Note <math>\angle KLP=\beta</math>.
 +
 
 +
Then, <math>KP=28\sin\beta=8\cos\alpha</math>.
 +
Furthermore, <math>KN=\frac{65}{\sin\alpha}=\frac{28}{\sin\beta} \Rightarrow 65\sin\beta=28\sin\alpha</math>.
 +
 
 +
Dividing the equations gives
 +
<cmath>\frac{65}{28}=\frac{28\sin\alpha}{8\cos\alpha}=\frac{7}{2}\tan\alpha\Rightarrow \tan\alpha=\frac{65}{98}</cmath>
 +
 
 +
Thus, <math>MK=\frac{MN}{\tan\alpha}=98</math>, so <math>MO=MK-KO=\boxed{090}</math>.
 +
 
 +
==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 <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
 +
 
 +
==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 <math>P</math> be the intersection of <math>LO</math> and <math>KN</math> as shown above. Note that <math>m\angle KPL = 90^{\circ}</math> as given in the problem. Since <math>\angle KPL \cong \angle KLN</math> and <math>\angle PKL \cong \angle LKN</math>, <math>\triangle PKL \sim \triangle LKN</math> by AA similarity. Similarly, <math>\triangle KMN \sim \triangle KPO</math>. Using these similarities we see that
 +
<cmath>\frac{KP}{KL} = \frac{KL}{KN}</cmath>
 +
<cmath>KP = \frac{KL^2}{KN} = \frac{28^2}{KN} = \frac{784}{KN}</cmath>
 +
and
 +
<cmath>\frac{KP}{KO} = \frac{KM}{KN}</cmath>
 +
<cmath>KP = \frac{KO \cdot KM}{KN} = \frac{8\cdot KM}{KN}</cmath>
 +
Combining the two equations, we get
 +
<cmath>\frac{8\cdot KM}{KN} = \frac{784}{KN}</cmath>
 +
<cmath>8 \cdot KM = 28^2</cmath>
 +
<cmath>KM = 98</cmath>
 +
Since <math>KM = KO + MO</math>, we get <math>MO = 98 -8 = \boxed{090}</math>.
 +
 
 +
Solution by vedadehhc
 +
 
 +
==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>).
 +
 
 +
As <math>\triangle KOL \sim \triangle KHP</math> (as <math>LO \parallel PH</math>, using the fact that <math>H</math> is the orthocenter), we may let <math>OH = 8k</math> and <math>LP = 28k</math>.
 +
 
 +
Then using similarity with triangles <math>\triangle KLH</math> and <math>\triangle KMP</math> we have
 +
 
 +
<cmath>\frac{28}{8+8k} = \frac{8+8k+HM}{28+28k}</cmath>
 +
 
 +
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 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
 +
<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 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 <math>KLMN</math> is inscribed in the circle with diameter <math>\overline{KN}</math> and <math>XOMN</math> is inscribed in the circle with diameter <math>\overline{ON}</math>. Furthermore, <math>(XLN)</math> is tangent to <math>\overline{KL}</math>. Then, <cmath>KO\cdot KM=KX\cdot KN=KL^2\implies KM=\frac{28^2}{8}=98,</cmath>and <math>MO=KM-KO=\boxed{090}</math>.
 +
 
 +
(Solution by TheUltimate123)
 +
 
 +
If you're wondering why <math>KX \cdot KN=KL^2,</math> it's because PoP on <math>(XLN)</math> or by <math>KX \cdot KN=KX \cdot (KX+XN)=KX^2+KX \cdot XN=KX^2+LX^2=KL^2</math> (last part by geometric mean theorem / similarity).
 +
 
 +
Note: the "semicircle" circumscribing points XOMN is not a semicircle. That is just there to tell you that X, O, M, N are indeed concyclic, so ignore the subtlety of the diagram that makes O seems slightly off the marks than it should be.
 +
 
 +
==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 <math>L</math> to <math>NK</math> point <math>P</math>. Let <math>PO=x</math>. Now, we have that <math>KP=\sqrt{64-x^2}</math> by the Pythagorean Theorem. Once again by Pythagorean, <math>LO=\sqrt{720+x^2}-x</math>. Using Power of a Point, we have
 +
 
 +
<cmath>(KO)(OM)=(LO)(OQ)</cmath> (<math>Q</math> is the intersection of <math>OL</math> with the circle <math>\neq L</math>)
 +
 
 +
<cmath>8(MO)=(\sqrt{720+x^2}-x)(\sqrt{720+x^2}+x)</cmath>
 +
 
 +
<cmath>8(MO)=720</cmath>
 +
 
 +
<cmath>MO=\boxed{090}</cmath>.
 +
 
 +
(Solution by RootThreeOverTwo)
 +
<cmath> </cmath>
 +
===Remark: Length of OQ===
 +
Since <math>P</math> is on the circle’s diameter, <math>QP = LP = \sqrt{720+x^2}</math>. So, <math>OQ = PQ + PO = x + \sqrt{720+x^2}+x)</math>.
 +
~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 <math>\angle KLN = \angle KMN</math>, quadrilateral <math>KLMN</math> is cyclic. Therefore, we have <cmath>\angle LMK = \angle LNK = 90^{\circ} - \angle LKN = \angle KLP,</cmath>so <math>\triangle KLO \sim \triangle KML</math>, giving <cmath>\frac{KM}{28} = \frac{28}{8} \implies KM = 98.</cmath> Therefore, <math>OM = 98-8 = \boxed{90}</math>.
 +
 
 +
==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, <math>KM^2+65^2 = KN^2 = 28^2 + LN^2</math>. Thus, <math>LN^2 = KM^2 + 65^2 - 28^2</math>.
 +
 
 +
By Pythagorean Theorem, <math>KP^2 + LP^2 = 28^2</math>, and <math>PN^2 + LP^2 = LN^2</math>.
 +
 
 +
<cmath>PN^2 = (KN - KP)^2 = (\sqrt{KM^2 + 65^2} - KP)^2</cmath>
 +
 
 +
It follows that
 +
<cmath>(\sqrt{KM^2 + 65^2} - KP)^2 + LP^2 = KM^2 + 65^2 - 28^2</cmath>
 +
 
 +
<cmath>KM^2 + 65^2 - 2\sqrt{KM^2 + 65^2}(KP) + KP^2 + LP^2 = KM^2 + 65^2 - 28^2</cmath>
 +
 
 +
Since <math>KP^2 + LP^2 = 28^2</math>,
 +
<cmath>KM^2 + 65^2 - 2\sqrt{KM^2 + 65^2}(KP) + 28^2 = KM^2 + 65^2 - 28^2</cmath>
 +
 
 +
<cmath>-2\sqrt{KM^2 + 65^2}(KP) = -2 \times 28^2</cmath>
 +
 
 +
<cmath>KP = \frac{28^2}{\sqrt{KM^2 + 65^2}}</cmath>
 +
 
 +
<math>\angle OKP = \angle NKM</math> (it's the same angle) and <math>\angle OPK = \angle KMN = 90^{\circ}</math>. Thus, <math>\triangle KOP \sim \triangle KNM</math>.
 +
 
 +
Thus,
 +
 
 +
<cmath>\frac{KO}{KN} = \frac{KP}{KM}</cmath>
 +
 
 +
<cmath>\frac{8}{\sqrt{KM^2 + 65^2}} = \frac{\frac{28^2}{\sqrt{KM^2 + 65^2}}}{KM}</cmath>
 +
 
 +
Multiplying both sides by <math>\sqrt{KM^2 + 65^2}</math>:
 +
 
 +
<cmath>8 = \frac{28^2}{KM}</cmath>
 +
 
 +
<cmath>KM = 98</cmath>
 +
 
 +
Therefore, <math>OM = 98-8 = \boxed{90}</math>
 +
 
 +
~ 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
  
−
==Problem 6==
 
−
==Solution==
 
 
==See Also==
 
==See Also==
 
{{AIME box|year=2019|n=I|num-b=5|num-a=7}}
 
{{AIME box|year=2019|n=I|num-b=5|num-a=7}}
 +
 +
[[Category:Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 12:23, 25 November 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 geometric mean theorem / similarity).

Note: the "semicircle" circumscribing points XOMN is not a semicircle. That is just there to tell you that X, O, M, N are indeed concyclic, so ignore the subtlety of the diagram that makes O seems slightly off the marks than it should be.

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)
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