A classic geometry problem

**simplependulum** · April 3rd 2010, 06:43 AM

Challenge Question:

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

Moderator edit: This is now an approved challenge question.

**Sudharaka** · April 3rd 2010, 07:08 AM

Originally Posted by simplependulum

Triangle $ABC$ is an isosceles triangle with base $BC$ , the included angle of $A$ is $20$ degree . A point $D$ is 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???

**tonio** · April 3rd 2010, 01:19 PM

Originally Posted by Sudharaka

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

**simplependulum** · April 4th 2010, 06:20 AM

Originally Posted by Sudharaka

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:

**tonio** · April 4th 2010, 07:31 AM

Originally Posted by simplependulum

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:

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

Tonio

**Wilmer** · April 4th 2010, 08:50 AM

Originally Posted by tonio

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

It's a CHALLENGE question, Tonio. See edit on initial post.

**tonio** · April 4th 2010, 12:28 PM

Originally Posted by Wilmer

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

**Wilmer** · April 4th 2010, 05:56 PM

Originally Posted by tonio

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 ?

**simplependulum** · April 4th 2010, 10:29 PM

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:

**simplependulum** · April 4th 2010, 10:32 PM

Originally Posted by Wilmer

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 .

**Unbeatable0** · April 5th 2010, 02:42 AM

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.

**simplependulum** · April 5th 2010, 07:23 PM

Originally Posted by Unbeatable0

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: