Difference between revisions of "2014 AIME II Problems/Problem 14"
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==Solution 3== | ==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}.</cmath> | 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}.</cmath> | ||
− | Since <math>\triangle{HCA}</math> is a 45-45-90 | + | Since <math>\triangle{HCA}</math> is a 45-45-90<math> |
− | <cmath>HC=\frac{5\sqrt2+5\sqrt6}{2}</cmath> <cmath>HM=\frac{5\sqrt6}{2}</cmath> <cmath>HN=\frac{5\sqrt6}{4}</cmath> We know that <math>\triangle{AHD}\simeq \triangle{PND}< | + | <cmath>HC=\frac{5\sqrt2+5\sqrt6}{2}</cmath> </math>MC=\frac{BM}{2},<math> <cmath>HM=\frac{5\sqrt6}{2}</cmath> <cmath>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}< | + | Thus, </math>(AP)^2=\dfrac{150}{4}=\dfrac{75}{2}<math>. So our final answer is </math>75+2=77<math>. |
− | <math>m+n=\boxed{077} | + | </math>m+n=\boxed{077}$ |
== See also == | == See also == |
Revision as of 02:47, 4 March 2018
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
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 Since is a 45-45-90MC=\frac{BM}{2},\triangle{AHD}\simeq \triangle{PND}$and are 30-60-90. Thus, <cmath>AP=2 \cdot HN=\frac{5\sqrt6}{4}.</cmath>
Thus,$ (Error compiling LaTeX. Unknown error_msg)(AP)^2=\dfrac{150}{4}=\dfrac{75}{2}75+2=77m+n=\boxed{077}$
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 |
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