# Difference between revisions of "2014 AIME II Problems/Problem 14"

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==SOLUTION== | ==SOLUTION== | ||

As we can see, | As we can see, | ||

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<math>M</math> is the midpoint of <math>BC</math> and <math>N</math> is the midpoint of <math>HM</math> | <math>M</math> is the midpoint of <math>BC</math> and <math>N</math> is the midpoint of <math>HM</math> | ||

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<math>AHC</math> is a <math>45-45-90</math> triangle, so <math>\angle{HAB}=15^\circ</math>. | <math>AHC</math> is a <math>45-45-90</math> triangle, so <math>\angle{HAB}=15^\circ</math>. | ||

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<math>AHD</math> is <math>30-60-90</math> triangle. | <math>AHD</math> is <math>30-60-90</math> triangle. | ||

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<math>AH</math> and <math>PN</math> are parallel lines so <math>PND</math> is <math>30-60-90</math> triangle also. | <math>AH</math> and <math>PN</math> are parallel lines so <math>PND</math> is <math>30-60-90</math> triangle also. | ||

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Then if we use those informations we get <math>AD=2HD</math> and | Then if we use those informations we get <math>AD=2HD</math> and | ||

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<math>PD=2ND</math> and <math>AP=AD-PD=2HD-2ND=2HN</math> or <math>AP=2HN=HM</math> | <math>PD=2ND</math> and <math>AP=AD-PD=2HD-2ND=2HN</math> or <math>AP=2HN=HM</math> | ||

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Now we know that <math>HM=AP</math>, we can find for <math>HM</math> which is simpler to find. | Now we know that <math>HM=AP</math>, we can find for <math>HM</math> which is simpler to find. | ||

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We can use point <math>B</math> to split it up as <math>HM=HB+BM</math>, | We can use point <math>B</math> to split it up as <math>HM=HB+BM</math>, | ||

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We can chase those lengths and we would get | We can chase those lengths and we would get | ||

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<math>AB=10</math>, so <math>OB=5</math>, so <math>BC=5\sqrt{2}</math>, so <math>BM=\dfrac{1}{2} \cdot BC=\dfrac{5\sqrt{2}}{2}</math> | <math>AB=10</math>, so <math>OB=5</math>, so <math>BC=5\sqrt{2}</math>, so <math>BM=\dfrac{1}{2} \cdot BC=\dfrac{5\sqrt{2}}{2}</math> | ||

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Then using right triangle <math>AHB</math>, we have HB=10 sin (15∘) | Then using right triangle <math>AHB</math>, we have HB=10 sin (15∘) | ||

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So HB=10 sin (15∘)=<math>\dfrac{5(\sqrt{6}-\sqrt{2})}{2}</math>. | So HB=10 sin (15∘)=<math>\dfrac{5(\sqrt{6}-\sqrt{2})}{2}</math>. | ||

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And we know that <math>AP = HM = HB + BM = \frac{5(\sqrt6-\sqrt2)}{2} + \frac{5\sqrt2}{2} = \frac{5\sqrt6}{2}</math>. | And we know that <math>AP = HM = HB + BM = \frac{5(\sqrt6-\sqrt2)}{2} + \frac{5\sqrt2}{2} = \frac{5\sqrt6}{2}</math>. | ||

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Finally if we calculate <math>(AP)^2</math>. | Finally if we calculate <math>(AP)^2</math>. | ||

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<math>(AP)^2=\dfrac{150}{4}=\dfrac{75}{2}</math>. So our final answer is <math>75+2=77</math>. | <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{77}</math> | <math>m+n=\boxed{77}</math> | ||

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Thank you. | Thank you. | ||

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-Gamjawon | -Gamjawon |

## Revision as of 06:48, 31 March 2014

## 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