Difference between revisions of "2019 AIME I Problems/Problem 6"
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− | ==Problem | + | ==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>. | ||
− | ==Solution (Similar triangles)== | + | ==Solution 1 (Trig)== |
− | ( | + | Let <math>\angle MKN=\alpha</math> and <math>\angle LNK=\beta</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 (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 3 (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 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,</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 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 <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 <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)== | ||
+ | |||
+ | <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) | ||
+ | ==Solution 7 (just one pair of similar triangles)== | ||
<asy> | <asy> | ||
− | size( | + | 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> | </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
Contents
- 1 Problem
- 2 Solution 1 (Trig)
- 3 Solution 2 (Similar triangles)
- 4 Solution 3 (Similar triangles, orthocenters)
- 5 Solution 4 (Algebraic Bashing)
- 6 Solution 5 (5-second PoP)
- 7 Solution 6 (Alternative PoP)
- 8 Solution 7 (just one pair of similar triangles)
- 9 Video Solution
- 10 Video Solution 2
- 11 Video Solution 3
- 12 See Also
Problem
In convex quadrilateral side is perpendicular to diagonal , side is perpendicular to diagonal , , and . The line through perpendicular to side intersects diagonal at with . Find .
Solution 1 (Trig)
Let and . Note .
Then, . Furthermore, .
Dividing the equations gives
Thus, , so .
Solution 2 (Similar triangles)
First, let be the intersection of and as shown above. Note that as given in the problem. Since and , by AA similarity. Similarly, . Using these similarities we see that and Combining the two equations, we get Since , we get .
Solution by vedadehhc
Solution 3 (Similar triangles, orthocenters)
Extend and past and respectively to meet at . Let be the intersection of diagonals and (this is the orthocenter of ).
As (as , using the fact that is the orthocenter), we may let and .
Then using similarity with triangles and we have
Cross-multiplying and dividing by gives so . (Solution by scrabbler94)
Solution 4 (Algebraic Bashing)
First, let be the intersection of and . We can use the right triangles in the problem to create equations. Let and We are trying to find We can find equations. They are and We can subtract the fifth equation from the sixth equation to get We can subtract the fourth equation from the third equation to get Combining these equations gives so Substituting this into the seventh equation gives Substituting this into the second equation gives . Subtracting the first equation from this gives Solving this equation, we find that (Solution by DottedCaculator)
Solution 5 (5-second PoP)
Notice that is inscribed in the circle with diameter and is inscribed in the circle with diameter . Furthermore, is tangent to . Then, and .
(Solution by TheUltimate123)
If you're wondering why it's because (last part by similarity).
Solution 6 (Alternative PoP)
(Diagram by vedadehhc)
Call the base of the altitude from to point . Let . Now, we have that by the Pythagorean Theorem. Once again by Pythagorean, . Using Power of a Point, we have
( is the intersection of with the circle )
.
(Solution by RootThreeOverTwo)
Solution 7 (just one pair of similar triangles)
Note that since , quadrilateral is cyclic. Therefore, we have so , giving Therefore, .
Video Solution
Video Solution: https://www.youtube.com/watch?v=0AXF-5SsLc8
Video Solution 2
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.