Difference between revisions of "2019 AIME I Problems/Problem 6"
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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 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, and 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) | ||
==Video Solution== | ==Video Solution== |
Revision as of 19:53, 15 March 2019
Contents
Problem 6
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 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)
Video Solution
Video Solution: https://www.youtube.com/watch?v=0AXF-5SsLc8
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.