Difference between revisions of "2019 AIME I Problems/Problem 6"
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− | The | + | ==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>. 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> | ||
+ | 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: | ||
+ | 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}} |
Revision as of 18:08, 14 March 2021
Contents
[hide]- 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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.