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Thread: A classic geometry problem

  1. #1
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    A classic geometry problem

    Challenge Question:

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


    Moderator edit: This is now an approved challenge question.
    Last edited by mr fantastic; Apr 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 $\displaystyle ABC$ is an isosceles triangle with base $\displaystyle BC$ , the included angle of $\displaystyle A $ is $\displaystyle 20$ degree . A point $\displaystyle D$is marked off on side $\displaystyle AC $ such that $\displaystyle AD = BC $ . What is the angle of $\displaystyle ABD $ ?
    Dear simplependulum,

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

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

    By the law of sines,

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

    $\displaystyle BC=2aSin10$

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

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

    By the law of sines,

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

    $\displaystyle BC=2aSin10$

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

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

    By the law of sines,

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

    $\displaystyle BC=2aSin10$

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

    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

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

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

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

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

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

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

    $\displaystyle \tan(x) = \tan(10) $

    $\displaystyle x = 10 $

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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

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

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

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

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

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

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

    $\displaystyle \tan(x) = \tan(10) $

    $\displaystyle x = 10 $


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

    Tonio
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  6. #6
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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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    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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    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 $\displaystyle DE \parallel BC $ such that $\displaystyle B,E $ are on the same side of $\displaystyle AC$ and $\displaystyle DE = AB $ .
    Think of why $\displaystyle 20^o $ is used but not the others .
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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 $\displaystyle \triangle ABC$ build a regular $\displaystyle 18$-gon such that $\displaystyle A$ is the center of it
    (which is possible due to the fact that $\displaystyle AB=AC$, $\displaystyle \angle BAC = 20^{\circ}$).
    Now, we are not assuming $\displaystyle AD=BC$. We'll come back to it later.
    Continue AB until it gets to another vertex, call it $\displaystyle J$, of the regular polygon (I'll leave it to you to verify the fact that it is indeed getting there), and draw a line $\displaystyle BS$ as shown in the attached figure. Next, draw $\displaystyle JT$ as shown, and call $\displaystyle M$ the intersection of $\displaystyle JT$ with $\displaystyle BS$. 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 $\displaystyle \triangle JMS$ it follows that $\displaystyle [JM]=2[JS]$, and in triangle $\displaystyle \triangle JMB$ we have $\displaystyle [JM]=2[AD]$. Therefore $\displaystyle [JS]=[AD]$, but $\displaystyle [JS]=[BC]$, so $\displaystyle [AD]=[BC]$. Next notice that $\displaystyle \angle JBS=10^\circ$ (again, by considering the circumscribing circle).
    We have thus proven that the conditions $\displaystyle \angle JBS=10^\circ$ and $\displaystyle [AD]=[BC]$ 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 $\displaystyle \triangle ABC$ build a regular $\displaystyle 18$-gon such that $\displaystyle A$ is the center of it
    (which is possible due to the fact that $\displaystyle AB=AC$, $\displaystyle \angle BAC = 20^{\circ}$).
    Now, we are not assuming $\displaystyle AD=BC$. We'll come back to it later.
    Continue AB until it gets to another vertex, call it $\displaystyle J$, of the regular polygon (I'll leave it to you to verify the fact that it is indeed getting there), and draw a line $\displaystyle BS$ as shown in the attached figure. Next, draw $\displaystyle JT$ as shown, and call $\displaystyle M$ the intersection of $\displaystyle JT$ with $\displaystyle BS$. 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 $\displaystyle \triangle JMS$ it follows that $\displaystyle [JM]=2[JS]$, and in triangle $\displaystyle \triangle JMB$ we have $\displaystyle [JM]=2[AD]$. Therefore $\displaystyle [JS]=[AD]$, but $\displaystyle [JS]=[BC]$, so $\displaystyle [AD]=[BC]$. Next notice that $\displaystyle \angle JBS=10^\circ$ (again, by considering the circumscribing circle).
    We have thus proven that the conditions $\displaystyle \angle JBS=10^\circ$ and $\displaystyle [AD]=[BC]$ 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 : $\displaystyle 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 $\displaystyle E $ be a point outside the triangle such that $\displaystyle \Delta ADE $ is an equilateral triangle so $\displaystyle \angle BAE = 20^o + 60^o = 80^o $ . It is easy to find that $\displaystyle \Delta ABE $ is congruent to $\displaystyle \Delta ABC $ , $\displaystyle AB = BE$ . Note that $\displaystyle D $ is inside the $\displaystyle \Delta ABE $ . Now consider $\displaystyle \Delta ABD $ and $\displaystyle \Delta EBD $ , we have $\displaystyle AD = DE $ , $\displaystyle AB = BE $ and $\displaystyle \angle BAC = \angle BED = 20^o $ . Therefore , $\displaystyle \Delta ABD $ is congruent to $\displaystyle \Delta ABE $ . We can see that $\displaystyle AD $ bisects $\displaystyle \angle ABE $ so $\displaystyle \angle ABD = \frac{20^o}{2} = 10^o $



    Spoiler:

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

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