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 19:11, 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.