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Math Help - Geometry Marathon

  1. #31
    MHF Contributor red_dog's Avatar
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    \widehat{BDC}=3x=\widehat{BCD}.
    Let AB=DC=a, \ BD=BC=b.
    In \triangle ABD: \ \frac{a}{b}=\frac{\sin 3x}{\sin 2x} (1)
    In \triangle BDC: \ \frac{a}{b}=\frac{\sin 6x}{\sin 3x}=2\cos 3x (2)
    (1) & (2) \Rightarrow \sin 3x=2\sin 2x\cos 3x\Leftrightarrow \sin x+\sin 3x-\sin 5x=0\Leftrightarrow
    \Leftrightarrow \sin x(1-2\cos 4x)=0\Rightarrow \cos 4x=\frac{1}{2}\Rightarrow x=15
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  2. #32
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    Quote Originally Posted by red_dog View Post
    \widehat{BDC}=3x=\widehat{BCD}.
    Let AB=DC=a, \ BD=BC=b.
    In \triangle ABD: \ \frac{a}{b}=\frac{\sin 3x}{\sin 2x} (1)
    In \triangle BDC: \ \frac{a}{b}=\frac{\sin 6x}{\sin 3x}=2\cos 3x (2)
    (1) & (2) \Rightarrow \sin 3x=2\sin 2x\cos 3x\Leftrightarrow \sin x+\sin 3x-\sin 5x=0\Leftrightarrow
    \Leftrightarrow \sin x(1-2\cos 4x)=0\Rightarrow \cos 4x=\frac{1}{2}\Rightarrow x=15
    Please, read the first post.
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  3. #33
    MHF Contributor red_dog's Avatar
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    \widehat{BDC}=\widehat{BCD}=3x\Rightarrow\triangle BDC isosceles.
    Let AB=DC=a, \ BE\perp DC, E\in (DC) and F the midpoint of AB.
    Then FE=DE=EC=\frac{a}{2}\Rightarrow \widehat{DFC}=90
    \widehat{FEC}=2x, \ \widehat{EFC}=\widehat{FCE}=x
    We have \widehat{FBD}=\widehat{FCD}=x\Rightarrow FBCD is cyclic \Rightarrow \widehat{DBC}=\widehat{DFC}=90.
    Then 3x=45\Rightarrow x=15

    (What program do you use for drawing figures?)
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  4. #34
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    Quote Originally Posted by red_dog View Post
    (What program do you use for drawing figures?)
    I think it might be called "Asymptote" but I might be wrong.
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  5. #35
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    Quote Originally Posted by red_dog View Post
    \widehat{BDC}=\widehat{BCD}=3x\Rightarrow\triangle BDC isosceles.
    Let AB=DC=a, \ BE\perp DC, E\in (DC) and F the midpoint of AB.
    Then FE=DE=EC=\frac{a}{2}\Rightarrow \widehat{DFC}=90
    \widehat{FEC}=2x, \ \widehat{EFC}=\widehat{FCE}=x
    We have \widehat{FBD}=\widehat{FCD}=x\Rightarrow FBCD is cyclic \Rightarrow \widehat{DBC}=\widehat{DFC}=90.
    Then 3x=45\Rightarrow x=15

    (What program do you use for drawing figures?)
    Okay, that is correct.

    Post your problem now.

    Actually, drawing with Asymptote it's a lot of complex work, plus, CorelDraw's quality is the best.
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  6. #36
    MHF Contributor red_dog's Avatar
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    Here is the problem (it's not mine).
    Let \triangle ABC with \widehat{A}=\widehat{B} and \widehat{C}=20.
    Let M\in (AC), \ N\in (BC) such that \widehat{ABM}=60, \ \widehat{BAN}=50. Find \widehat{BMN}.

    (Krizalid, maybe you can draw the figure. Thank you)
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  7. #37
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    This is Langley's Problem!!

    I know how to make it, but I'll wait if someone can show another solution!!
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  8. #38
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    Famous problem.
    Discussed Here.
    (But I have an embarassing Trigonometry Solution!)

    But there is a link on page #2 that provides another solution.

    Can I post a problem please? Since I answered red_dog's solution.

    See Here
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  9. #39
    Math Engineering Student
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    Of course you can
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  10. #40
    MHF Contributor red_dog's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Famous problem.
    Can I post a problem please? Since I answered red_dog's solution.
    Yes!
    I didn't know that the problem was already posted.
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  11. #41
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    This one is really really hard. Not mine I found it in a book.

    "Let ABCD EF be a convex polygon with AB=BC=CD and DE=ED=FA so that \angle BCD = \angle EFA = \pi/3. Let G \mbox{ and }H be inside of the hexagon so that \angle AGB = \angle DHE = 2\pi /3. Prove that AG+GB+GH+DH+HE \geq CF. "

    Hints (highlight below)



    1) Reflect through BE

    2)Use Ptolemy's Theorem.
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  12. #42
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    Quote Originally Posted by ThePerfectHacker View Post
    This one is really really hard. Not mine I found it in a book.
    Sorry, I'm gonna have to ask if you can post another problem.

    This is because some people wants to take part (and we dunno what their level is).

    There're some hard problems but not so hard like this one.

    Plus, this is starting, and the level of problems must up slowly.

    In fact, I forgot to mention: Do not post Olympiad problems.
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  13. #43
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    Quote Originally Posted by Krizalid View Post
    Sorry, I'm gonna have to ask if you can post another problem.
    In that case you post a problem for me because I am leaving in like 5 minutes and not going to be here for a day. So that will delay this thread unless you want to wait.

    In fact, I forgot to mention: Do not post Olympiad problems.
    This is an IMO problem. You are right, it is not right to post such hard problems.
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  14. #44
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    Okay, no problem

    This problem looks like hard, but it's not when you find a good auxiliary construction.

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  15. #45
    MHF Contributor red_dog's Avatar
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    We construct the bisector CE, \ E\in BD of \widehat{BCD}. Then \triangle BEC is isosceles.
    Applying bisector's theorem we have \frac{DE}{EB}=\frac{DC}{BC}\Rightarrow \frac{DE}{EC}=\frac{DC}{BC}. (1)
    Also we have \widehat{DEC}=\widehat{DCB}=4x (2)
    (1) & (2) \Rightarrow \triangle DEC\sim\triangle DCB\Rightarrow \frac{DC}{BC}=\frac{DC}{BD}\Rightarrow BC=BD.
    So the triangle BDC is isosceles \Rightarrow \widehat{BDC}=4x\Rightarrow 4x+6x=180\Rightarrow x=18
    Attached Thumbnails Attached Thumbnails Geometry Marathon-triangle.jpg  
    Last edited by red_dog; July 14th 2007 at 09:11 AM.
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