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 1 (Trig)== | ==Solution 1 (Trig)== | ||
− | Let <math>\angle MKN=\alpha</math> and <math>\angle LNK=\beta</math>. Note <math>\angle KLP=\beta</math>. | + | 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>. | Then, <math>KP=28\sin\beta=8\cos\alpha</math>. | ||
Line 14: | Line 13: | ||
Thus, <math>MK=\frac{MN}{\tan\alpha}=98</math>, so <math>MO=MK-KO=\boxed{090}</math>. | Thus, <math>MK=\frac{MN}{\tan\alpha}=98</math>, so <math>MO=MK-KO=\boxed{090}</math>. | ||
− | ==Solution 2 (Similar triangles)== | + | ==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> | <asy> | ||
size(250); | size(250); | ||
Line 54: | Line 83: | ||
Solution by vedadehhc | Solution by vedadehhc | ||
− | ==Solution | + | ==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>). | 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>). | ||
Line 65: | Line 94: | ||
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) | 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 | + | ==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, and g=NC.</math> We are trying to find <math>d.</math> We can find <math>7</math> equations. They are | + | 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=c^2,</cmath> | ||
<cmath>4225+d^2+16d+64=a^2+2ab+b^2,</cmath> | <cmath>4225+d^2+16d+64=a^2+2ab+b^2,</cmath> | ||
Line 77: | Line 106: | ||
(Solution by DottedCaculator) | (Solution by DottedCaculator) | ||
− | ==Solution | + | ==Solution 6 (5-second PoP)== |
<asy> | <asy> | ||
Line 104: | Line 133: | ||
(Solution by TheUltimate123) | (Solution by TheUltimate123) | ||
− | ==Solution | + | 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 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 <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> | + | <cmath>KM = 98</cmath> |
− | < | + | Therefore, <math>OM = 98-8 = \boxed{90}</math> |
− | + | ~ Solution by adam_zheng | |
− | |||
==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 20:11, 19 January 2024
Contents
- 1 Problem
- 2 Solution 1 (Trig)
- 3 Solution 2 (Cyclic Quads, PoP)
- 4 Solution 3 (Similar triangles)
- 5 Solution 4 (Similar triangles, orthocenters)
- 6 Solution 5 (Algebraic Bashing)
- 7 Solution 6 (5-second PoP)
- 8 Solution 7 (Alternative PoP)
- 9 Solution8 (just one pair of similar triangles)
- 10 Solution 9 (Pythagoras Bash)
- 11 Video Solution
- 12 Video Solution 2
- 13 Video Solution 3
- 14 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
. Let
be the project of
onto line
. Note
.
Then, .
Furthermore,
.
Dividing the equations gives
Thus, , so
.
Solution 2 (Cyclic Quads, PoP)
Because ,
is a cyclic quadrilateral. Hence, by Power of Point,
as desired.
~Mathkiddie
Solution 3 (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 4 (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 5 (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 6 (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 PoP on
or by
(last part by similarity).
Solution 7 (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)
Remark: Length of OQ
Since is on the circle’s diameter,
. So,
.
~diyarv
Solution8 (just one pair of similar triangles)
Note that since
, quadrilateral
is cyclic. Therefore, we have
so
, giving
Therefore,
.
Solution 9 (Pythagoras Bash)
By Pythagorean Theorem, . Thus,
.
By Pythagorean Theorem, , and
.
It follows that
Since ,
(it's the same angle) and
. Thus,
.
Thus,
Multiplying both sides by :
Therefore,
~ Solution by adam_zheng
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.