# A classic geometry problem

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

Originally Posted by simplependulum
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???
• 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, $\displaystyle A\hat{B}D=x\Rightarrow{D\hat{B}C}=80-x$

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

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

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

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

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

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

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

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

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

$\displaystyle x = 10$

• 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

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

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

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

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

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

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

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

$\displaystyle x = 10$

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

Tonio
• 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 $\displaystyle DE \parallel BC$ 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 .
• 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 $\displaystyle \triangle ABC$ 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.
• 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 $\displaystyle \triangle ABC$ 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:

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