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Math Help - A classic geometry problem

  1. #1
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    A classic geometry problem

    Challenge Question:

    Triangle ABC is an isosceles triangle with base BC , the included angle of  A is 20 degree . A point Dis marked off on side  AC such that  AD = BC . What is the angle of  ABD ?


    Moderator edit: This is now an approved challenge question.
    Last edited by mr fantastic; April 4th 2010 at 03:37 AM. Reason: Re-opened thread now that rules for posting here have been complied with.
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    Quote Originally Posted by simplependulum View Post
    Triangle ABC is an isosceles triangle with base BC , the included angle of  A is 20 degree . A point Dis marked off on side  AC such that  AD = BC . What is the angle of  ABD ?
    Dear simplependulum,

    If you draw the traingle as indicated you will see that,

    Take, A\hat{B}D=x\Rightarrow{D\hat{B}C}=80-x

    By the law of sines,

    \frac{BC}{Sinx}=\frac{a}{Sin(160-x)}

    BC=2aSin10

    \frac{2aSin10}{Sinx}=\frac{a}{Sin(160-x)}

    Can you do it from here???
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    Quote Originally Posted by Sudharaka View Post
    Dear simplependulum,

    If you draw the traingle as indicated you will see that,

    Take, A\hat{B}D=x\Rightarrow{D\hat{B}C}=80-x

    By the law of sines,

    \frac{BC}{Sinx}=\frac{a}{Sin(160-x)}

    BC=2aSin10

    \frac{2aSin10}{Sinx}=\frac{a}{Sin(160-x)}

    Can you do it from here???


    Hmmm... I can't see clearly what the continuation is without getting into heavy stuff (probably power series or something: I'm not sure), but it perhaps doesn't matter: the problem's title says geometry and, at least over here, this means the problem must be solved using exclusively tools from geometry, without any trigonometry, calculus, etc.
    I wonder what the OP originally meant.

    Tonio
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    Quote Originally Posted by Sudharaka View Post
    Dear simplependulum,

    If you draw the traingle as indicated you will see that,

    Take, A\hat{B}D=x\Rightarrow{D\hat{B}C}=80-x

    By the law of sines,

    \frac{BC}{Sinx}=\frac{a}{Sin(160-x)}

    BC=2aSin10

    \frac{2aSin10}{Sinx}=\frac{a}{Sin(160-x)}

    Can you do it from here???

    I first saw this problem in a paper of IMO preliminary selection contest but the suggested solution uses trigonometry . Of course , as i said it classic , there is actually a solution ( collected from a book )which only uses the properties of isosceles , equilateral triangle , similar triangles and parallel lines .

    Below is the trigonometric method which is similar to the contest 's .

    Spoiler:
    we have

     \frac{ \sin(x+20) }{\sin(x) } = \frac{1}{2\sin(10)}

     \cos(20) + \cot(x) \sin(20) = \frac{1}{2\sin(10)}

     \cot(x) \sin(20) = \frac{1- 2\sin(10)\cos(20)}{2\sin(10)}

     \cot(x) \sin(20) = \frac{1- (\sin(30) - \sin(10))}{2\sin(10)}

     \cot(x) \sin(20) = \frac{\sin(30) + \sin(10)}{2\sin(10)}

     \cot(x) \sin(20) = \frac{2 \sin(20)\cos(10)}{2\sin(10)}

     \tan(x) = \tan(10)

     x = 10

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  5. #5
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    Quote Originally Posted by simplependulum View Post
    I first saw this problem in a paper of IMO preliminary selection contest but the suggested solution uses trigonometry . Of course , as i said it classic , there is actually a solution ( collected from a book )which only uses the properties of isosceles , equilateral triangle , similar triangles and parallel lines .

    Below is the trigonometric method which is similar to the contest 's .

    Spoiler:
    we have

     \frac{ \sin(x+20) }{\sin(x) } = \frac{1}{2\sin(10)}

     \cos(20) + \cot(x) \sin(20) = \frac{1}{2\sin(10)}

     \cot(x) \sin(20) = \frac{1- 2\sin(10)\cos(20)}{2\sin(10)}

     \cot(x) \sin(20) = \frac{1- (\sin(30) - \sin(10))}{2\sin(10)}

     \cot(x) \sin(20) = \frac{\sin(30) + \sin(10)}{2\sin(10)}

     \cot(x) \sin(20) = \frac{2 \sin(20)\cos(10)}{2\sin(10)}

     \tan(x) = \tan(10)

     x = 10


    Ok, and what's the pure-geometric solution, please?

    Tonio
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    Quote Originally Posted by tonio View Post
    Ok, and what's the pure-geometric solution, please?
    Tonio
    It's a CHALLENGE question, Tonio. See edit on initial post.
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    Quote Originally Posted by Wilmer View Post
    It's a CHALLENGE question, Tonio. See edit on initial post.

    Yes, I know but the OP said there exists a pure-geometric solution though his uses trigonometry. I'm asking for the pure-geometric one.

    Tonio
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  8. #8
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    Quote Originally Posted by tonio View Post
    Yes, I know but the OP said there exists a pure-geometric solution though his uses trigonometry. I'm asking for the pure-geometric one.
    Tonio
    Don't think so: he gave the trig solution because it's evident;
    BUT the challenge is to find the geometric solution; right, simplep ?
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  9. #9
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    I shouldn't give the solution now but if you really want the pure-geometric solution , i think it is okay to give the hints :


    The solution has three additional lines but i am going to give you only one of the lines because this is exactly the key to this problem while the others are not very important .

    Hints:

    Spoiler:
    Draw  DE \parallel BC such that  B,E are on the same side of  AC and  DE = AB  .
    Think of why  20^o is used but not the others .
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  10. #10
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    Quote Originally Posted by Wilmer View Post
    Don't think so: he gave the trig solution because it's evident;
    BUT the challenge is to find the geometric solution; right, simplep ?

    Any method is welcome . The geometric solution just brings us more happiness .

    BTW , my initial solution is the use of complex number . haha .
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    A figure is added in the attachment to make the proof easier to read.

    On the given triangle \triangle ABC build a regular 18-gon such that A is the center of it
    (which is possible due to the fact that AB=AC, \angle BAC = 20^{\circ}).
    Now, we are not assuming AD=BC. We'll come back to it later.
    Continue AB until it gets to another vertex, call it J, of the regular polygon (I'll leave it to you to verify the fact that it is indeed getting there), and draw a line BS as shown in the attached figure. Next, draw JT as shown, and call M the intersection of JT with BS. All the angles shown in the figure can easily be calculated by considering equal arcs on the circumscribing circle of the regular polygon. Thus, from triangle \triangle JMS it follows that [JM]=2[JS], and in triangle \triangle JMB we have [JM]=2[AD]. Therefore [JS]=[AD], but [JS]=[BC], so [AD]=[BC]. Next notice that \angle JBS=10^\circ (again, by considering the circumscribing circle).
    We have thus proven that the conditions \angle JBS=10^\circ and [AD]=[BC] are obtained simultaneously, from which the result follows.
    Attached Thumbnails Attached Thumbnails A classic geometry problem-mhfgeometry.png  
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  12. #12
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    Quote Originally Posted by Unbeatable0 View Post
    A figure is added in the attachment to make the proof easier to read.

    On the given triangle \triangle ABC build a regular 18-gon such that A is the center of it
    (which is possible due to the fact that AB=AC, \angle BAC = 20^{\circ}).
    Now, we are not assuming AD=BC. We'll come back to it later.
    Continue AB until it gets to another vertex, call it J, of the regular polygon (I'll leave it to you to verify the fact that it is indeed getting there), and draw a line BS as shown in the attached figure. Next, draw JT as shown, and call M the intersection of JT with BS. All the angles shown in the figure can easily be calculated by considering equal arcs on the circumscribing circle of the regular polygon. Thus, from triangle \triangle JMS it follows that [JM]=2[JS], and in triangle \triangle JMB we have [JM]=2[AD]. Therefore [JS]=[AD], but [JS]=[BC], so [AD]=[BC]. Next notice that \angle JBS=10^\circ (again, by considering the circumscribing circle).
    We have thus proven that the conditions \angle JBS=10^\circ and [AD]=[BC] are obtained simultaneously, from which the result follows.

    Wow , again ! A unbeatable method !

    The key to this problem is : this isosceles triangle has the property that the difference of its base angle and the other angle is equal to 60 :  80^o - 20^o = 60^o

    I would like to share my three solutions :

    Spoiler:

    Note that

    Draw such that are on the same side of and .

    Since , , ,

    is congruent to and

    .


    Consider and , therefore , is an equilateral triangle .

    Then , is an isosceles triangle with base angle , so .



    Spoiler:


    Let  E be a point outside the triangle such that  \Delta ADE is an equilateral triangle so  \angle BAE = 20^o + 60^o = 80^o . It is easy to find that  \Delta ABE is congruent to  \Delta ABC ,  AB = BE . Note that  D is inside the  \Delta ABE . Now consider  \Delta ABD and  \Delta EBD , we have  AD = DE ,  AB = BE and  \angle BAC = \angle BED = 20^o . Therefore , \Delta ABD is congruent to  \Delta ABE . We can see that  AD bisects  \angle ABE so  \angle ABD = \frac{20^o}{2} = 10^o



    Spoiler:

    Let  E be a point inside the  \Delta ABC such that \Delta BCE is an equilateral triangle ,  \angle ABE = 80^o - 60^o = 20^o . It is easy to prove that  ABED is an isosceles trapezium .  \angle ABD = \angle ABE = \frac{20^o}{2} = 10^o .

    Last edited by simplependulum; July 18th 2010 at 09:06 PM.
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