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

(removed solution 1 - unclear explanation, and very similar to solution 3 (now solution 2))
(Video Solution 2)
 
(9 intermediate revisions by 7 users not shown)
Line 1: Line 1:
==Problem 6==
+
==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>.
 
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>.
  
Line 102: Line 102:
  
 
(Solution by TheUltimate123)
 
(Solution by TheUltimate123)
 +
 +
If you're wondering why <math>KX \cdot KN=KL^2,</math> it's because <math>KX \cdot KN=KX \cdot (KX+XN)=KX^2+KX \cdot XN=KX^2+LX^2=KL^2</math> (last part by similarity).
  
 
==Solution 6 (Alternative PoP)==
 
==Solution 6 (Alternative PoP)==
Line 132: Line 134:
 
(Diagram by vedadehhc)
 
(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>KO=\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  
+
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>(KO)(OM)=(LO)(OQ)</cmath> (<math>Q</math> is the intersection of <math>OL</math> with the circle <math>\neq L</math>)
Line 143: Line 145:
  
 
(Solution by RootThreeOverTwo)
 
(Solution by RootThreeOverTwo)
 +
==Solution 7 (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>.
  
 
==Video Solution==
 
==Video Solution==
 
Video Solution:
 
Video Solution:
 
https://www.youtube.com/watch?v=0AXF-5SsLc8
 
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==
 
==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 18:08, 14 March 2021

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 $\angle LNK=\beta$. 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 (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 3 (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 4 (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 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)

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

Solution 6 (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)

Solution 7 (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, $OM = 98-8 = \boxed{90}$.

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 (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS