Difference between revisions of "2014 AIME II Problems/Problem 14"
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Here's a solution that doesn't need sin 15. | Here's a solution that doesn't need sin 15. | ||
− | As above, get to AP=HM. 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>AB</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>AB</math>. Therefore <math>MH:AO=MC:CO</math>, or <math>MH=\dfrac{5\sqrt3}{\sqrt2}</math>, and we're done. |
== See also == | == See also == |
Revision as of 08:59, 27 September 2014
Contents
[hide]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
http://www.artofproblemsolving.com/Wiki/images/5/59/AOPS_wiki.PNG ( This is the diagram.)
Solution
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
Then using right triangle , we have HB=10 sin (15∘)
So HB=10 sin (15∘)=.
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 sin 15.
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.
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.