Difference between revisions of "2014 AIME II Problems/Problem 14"
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</asy> | </asy> | ||
− | ==Solution== | + | ==Solution 1== |
As we can see, | As we can see, | ||
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-Gamjawon | -Gamjawon | ||
− | Solution 2. | + | ==Solution 2==. |
Here's a solution that doesn't need <math>\sin 15^\circ</math>. | Here's a solution that doesn't need <math>\sin 15^\circ</math>. | ||
As above, get to <math>AP=HM</math>. As in the figure, let <math>O</math> be the foot of the perpendicular from <math>B</math> to <math>AC</math>. Then <math>BCO</math> is a 45-45-90 triangle, and <math>ABO</math> is a 30-60-90 triangle. So <math>BO=5</math> and <math>AO=5\sqrt{3}</math>; also, <math>CO=5</math>, <math>BC=5\sqrt2</math>, and <math>MC=\dfrac{BC}{2}=5\dfrac{\sqrt2}{2}</math>. But <math>MO</math> and <math>AH</math> are parallel, both being orthogonal to <math>BC</math>. Therefore <math>MH:AO=MC:CO</math>, or <math>MH=\dfrac{5\sqrt3}{\sqrt2}</math>, and we're done. | As above, get to <math>AP=HM</math>. As in the figure, let <math>O</math> be the foot of the perpendicular from <math>B</math> to <math>AC</math>. Then <math>BCO</math> is a 45-45-90 triangle, and <math>ABO</math> is a 30-60-90 triangle. So <math>BO=5</math> and <math>AO=5\sqrt{3}</math>; also, <math>CO=5</math>, <math>BC=5\sqrt2</math>, and <math>MC=\dfrac{BC}{2}=5\dfrac{\sqrt2}{2}</math>. But <math>MO</math> and <math>AH</math> are parallel, both being orthogonal to <math>BC</math>. Therefore <math>MH:AO=MC:CO</math>, or <math>MH=\dfrac{5\sqrt3}{\sqrt2}</math>, and we're done. | ||
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+ | ==Solution 3== | ||
+ | Break our diagram into 2 special right triangle by dropping an altitude from <math>B</math> to <math>AC</math> we then get that <cmath>AC=5+5\sqrt{3}, BC=5\sqrt{2}.HC=\frac{5\sqrt2+5\sqrt6}{2}, HM=\frac{5\sqrt6}{2}, HN=\frac{5\sqrt6}{4}.</cmath> We know that <math>\triangle{AHD}\simeq \triangle{PND}</math> and are 30-60-90. Thus, <cmath>AP=2 \cdot HN=\frac{5\sqrt6}{4}.</cmath> | ||
+ | |||
+ | Thus, <math>(AP)^2=\dfrac{150}{4}=\dfrac{75}{2}</math>. So our final answer is <math>75+2=77</math>. | ||
+ | |||
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+ | <math>m+n=\boxed{077}</math> | ||
== See also == | == See also == |
Revision as of 03:42, 4 March 2018
Contents
Problem
In , and . Let and be points on the line such that , , and . Point is the midpoint of the segment , and point is on ray such that . Then , where and are relatively prime positive integers. Find .
Diagram
Solution 1
As we can see,
is the midpoint of and is the midpoint of
is a triangle, so .
is triangle.
and are parallel lines so is triangle also.
Then if we use those informations we get and
and or
Now we know that , we can find for which is simpler to find.
We can use point to split it up as ,
We can chase those lengths and we would get
, so , so , so
We can also use Law of Sines:
Then using right triangle , we have
So .
And we know that .
Finally if we calculate .
. So our final answer is .
Thank you.
-Gamjawon
==Solution 2==. Here's a solution that doesn't need .
As above, get to . As in the figure, let be the foot of the perpendicular from to . Then is a 45-45-90 triangle, and is a 30-60-90 triangle. So and ; also, , , and . But and are parallel, both being orthogonal to . Therefore , or , and we're done.
Solution 3
Break our diagram into 2 special right triangle by dropping an altitude from to we then get that We know that and are 30-60-90. Thus,
Thus, . So our final answer is .
See also
2014 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.