# A classic geometry problem

• Apr 3rd 2010, 06:43 AM
simplependulum
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 $D$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.
• Apr 3rd 2010, 07:08 AM
Sudharaka
Quote:

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???
• Apr 3rd 2010, 01:19 PM
tonio
Quote:

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
• Apr 4th 2010, 06:20 AM
simplependulum
Quote:

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:
we have

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

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

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

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

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

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

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

$x = 10$

• Apr 4th 2010, 07:31 AM
tonio
Quote:

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:
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
• Apr 4th 2010, 08:50 AM
Wilmer
Quote:

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.
• Apr 4th 2010, 12:28 PM
tonio
Quote:

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
• Apr 4th 2010, 05:56 PM
Wilmer
Quote:

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 ?
• Apr 4th 2010, 10:29 PM
simplependulum
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 .
• Apr 4th 2010, 10:32 PM
simplependulum
Quote:

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 . (Happy)

BTW , my initial solution is the use of complex number . haha .
• Apr 5th 2010, 02:42 AM
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.
• Apr 5th 2010, 07:23 PM
simplependulum
Quote:

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:

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 . $\angle ABD = \angle ABE = \frac{20^o}{2} = 10^o$.