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Math Help - Another geometry puzzle

  1. #1
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    Another geometry puzzle

    Point P is completely inside equilateral triangle ABC.
    AP = 3, BP = 4, CP = 5.
    Without calculating the area and the side length of the triangle,
    find size of angle APB.
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  2. #2
    Junior Member RaisinBread's Avatar
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    Before I start posting my incredibly long and unpractical (probably far from optimal) solution, is the answer 5pi over 6?
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  3. #3
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    Quote Originally Posted by RaisinBread View Post
    Before I start posting my incredibly long and unpractical (probably far from optimal) solution, is the answer 5pi over 6?
    No; answer is an integer; hint: over 100 degrees.
    If you're involved in something "long and complicated", then STOP
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  4. #4
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    It is an old problem , my idea is to reflect  P across the three sides  BC , CA , AB , let the reflection be  A' , B' , C' respectively .

    I find that  \Delta AB'C' ~,~ \Delta BC'A' ~,~ \Delta CA'B' are  30^o - 120^o - 30^o triangles . Then  \Delta A'B'C' has the sides ratio  3:4:5 so it is right-angled .

     \angle APB = \angle AC'B = 30^o + 90^o + 30^o = 150^o which it is also the answer given by RaisinBread .

    The second solution is more elegant , but not by me .
    The idea is to rotate  \Delta APC by  60^o , by sending  C to  B ,  P to  P' . Then ,  \Delta AP'P is equilateral and  PP' = 3 so  \angle BPP' = 90^o . Therefore ,  \angle APB = 60^o + 90^o  = 150^o
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  5. #5
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    Yes.
    QUOTE=simplependulum;633536]I  \angle APB = \angle AC'B = 30^o + 90^o + 30^o = 150^o which it is also the answer given by RaisinBread .
    [/QUOTE]
    Sorry RaisinBread; didn't realise you were giving me radians.
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  6. #6
    Junior Member RaisinBread's Avatar
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    It's all good, I'm happy to have found the answer! I see the solution is really simple, but believe it or not, I solved the whole thing algebraically with functions and intersections.

    In short, I did this:

    1. I represented the triangle in the x-y plane, such that one side is resting on the x axis, and that the summit opposed to that side is on the y axis. I can thus make any equilateral triangle with the following equations:
    y={\sqrt{3}} x+a, y=-{\sqrt{3}} x+a and y=0, where a is any real number.
    Another geometry puzzle-1.jpg

    2. Let the left summit be A, the top be B and the right be C. I centered 3 circles of respective radii 3, 4 and 5 on these summits.
    Another geometry puzzle-2.jpg
    I know I have to find the point a (upon which all my equations depend) at which the three circles intersect at a single point, that point being P.

    3. I knew that three non concentric circles could only intersect at the intersection of their radical axis, so i isolated the radical axis of the circles with center A and C, using trigonometry, which gave me the x coordinate of where the circles would intersect.

    I then proceeded to solve quadratic equations to find at which y did the circles around A and B intersected, and then the same thing with the circles around B and C. I got two solutions each time, but I could see that in both case, the lowest intersection would be the one where the three circles intersected.
    Another geometry puzzle-3.jpg
    (These are the x and y coordinates of the point P, provided the two y are equal)

    I then found the value of a by conditioning that the two y coordinates be equal, and isolated a=\frac {\ {\sqrt{36{\sqrt{3}}+75}}}{2}

    4. Putting these results in the graph I had a confirmation that I was on the right track; Another geometry puzzle-4.jpg. Now all I needed to do was find the equations of the line from A to P, and the line from P to B, which was easy to do now since I had the value of a. I then changed these slopes into polar coordinates to find what angle they make with the y axis, subtracted the angle values and eureka, got 5pi over 6 radians!
    Another geometry puzzle-victory.jpg

    Very long and impractical solution, but I can't say I didn't have fun doing it!
    Last edited by RaisinBread; March 26th 2011 at 06:15 AM.
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  7. #7
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    Nice...you made me tired just reading it!!
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  8. #8
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    That's amazing that you solved for an exact value using functions and intersections...you make my head feel dizzy...

    Well done, I went the philosophical route simplependulum took, knowing that it is an equilateral triangle.
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