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General Discussion => Other Discussions => Math and Science => Topic started by: Sorunome on November 20, 2013, 03:52:49 pm

Title: Some math problem
Post by: Sorunome on November 20, 2013, 03:52:49 pm

So yeah, i stumbled upon this math problem some time ago, yes, I did solve it eventually, but it was so much fun I thought I'd share it with you guys :)

(http://img.ourl.ca//mathproblem.png)
Note how one of those two angles is outside of the inner triangle.

Title: Re: Some math problem
Post by: ruler501 on November 20, 2013, 04:15:07 pm

Spoiler For my quick attempt:

0+s t_1sin(25)=s-s t_2cos(25)
t_1sin(25)=1-t_2cos(25)
(square so I'd assume that t_1=t_2=t since that really simplifies the work)
\frac{1}{sin(25)+cos(25)}=t
\theta=arctan(\frac{1-\frac{1}{sin(25)+cos(25)}sin(25)}{1-\frac{1}{sin(25)+cos(25)}cos(25)}

EDIT: didn't do angles correctly so have to redo it.

EDIT2:

Spoiler For more complicated:

$0+s t_1sin(25)=s-s t_2cos(20)$

$t_1sin(25)=1-t_2cos(20)$

$1-t_1cos(25)=t_2sin(20)$

$t_2sin(20)+t_1cos(25)=1=t_1sin(25)+t_2cos(20)$

$t_2(sin(20)-cos(20))=t_1(sin(25)-cos(25))$

$t_2=\frac{sin(25)-cos(25)}{sin(20)-cos(20)}t_1$

$t_1sin(25)=1-\frac{sin(25)-cos(25)}{sin(20)-cos(20)}t_1cos(20)$

$t_1=\frac{1}{sin(25)+cos(20)\frac{sin(25)-cos(25)}{sin(20)-cos(20)}}$

$x=\frac{sin(25)}{sin(25)+cos(20)\frac{sin(25)-cos(25)}{sin(20)-cos(20)}}$

$y=\frac{sin(20)(sin(25)-cos(25))}{(sin(20)-cos(20))(sin(25)+cos(20)\frac{sin(25)-cos(25)}{sin(20)-cos(20)})}$

$\theta=arctan(\frac{1-x}{1-y})$

Title: Re: Some math problem
Post by: Adriweb on November 20, 2013, 04:27:30 pm

About 43°

(http://i.imgur.com/EKVFNbr.png)

But anyway... closest "good" value seems to be 45°.

And anyway² : original drawing doesn't look so "good" geometry-wise, so.... :P

Title: Re: Some math problem
Post by: Keoni29 on November 20, 2013, 05:29:01 pm

I did not calculate or make notes but I found the path to the solution. It is just a matter of calculating angle after angle after angle.

Title: Re: Some math problem
Post by: Sorunome on November 20, 2013, 05:45:07 pm

Quote from: Keoni29 on November 20, 2013, 05:29:01 pm

I did not calculate or make notes but I found the path to the solution. It is just a matter of calculating angle after angle after angle.

Nope, it's not, keep in mind that those 25° angles aren't both of inside of that triangle.

If you found an angle after angle solution I'd be happy to hear because I failed to find one after tons of trying out ;)

Title: Re: Some math problem
Post by: Keoni29 on November 20, 2013, 05:58:05 pm

But you can... it's an awful lot of angles before you get to it though :)

Title: Re: Some math problem
Post by: Sorunome on November 20, 2013, 06:16:09 pm

Quote from: Keoni29 on November 20, 2013, 05:58:05 pm

But you can... it's an awful lot of angles before you get to it though :)

Would you mind showing me how, please? Because I did indeed try out first of all only angle completion, because it looks like such a task but it didn't work out at all for me, that way.

Title: Re: Some math problem
Post by: Runer112 on November 20, 2013, 06:31:34 pm

I am strongly of the opinion that this cannot be solved simply by filling in angle values using geometric rules. Although the answer does come out to an integer, adjusting one of the starting angles slightly to a different integer results in the answer coming out as irrational, which would thus not be obtainable by application of simple geometric rules. So I believe trigonometry is required. Which I applied to get my solution, which I will withhold for now.



EDIT: Here's a modified picture with some suggestions which hint to how I solved the problem with trigonometry and algebra.

Spoiler For Approach hint:

(http://img.ourl.ca/hint-1.png)

EDIT 2: And here's the worked out solution. Each step is nested in a spoiler, so only peek as much as you need to!

Spoiler For Step 1:

(http://img.ourl.ca/step1.png)

Spoiler For Step 2:

(http://img.ourl.ca/step2.png)

Spoiler For Step 3 (final):

(http://img.ourl.ca/step3-3.png)

Title: Re: Some math problem
Post by: Sorunome on November 21, 2013, 09:31:01 am

Yup, i did what runer did :)

Title: Re: Some math problem
Post by: Chockosta on November 21, 2013, 04:26:42 pm

Hey, I was passing by and I just saw this topic.
Runner's solution works, but using analysis for such a problem is sad.

I found this nice solution in 5 minutes :

- Clearly PCA=PAD=20°
- Let O be the circle that passes through P, C, A
- Since C and A are on O and PCA=PAD, DA is tangent to O in A (corollary of inscribed angles theorem)
- DA and AB are perpendicular so AB passes through the center of O
- Since C and A are on O and PCD=PAC, DC is tangent to O in C
- DC and CB are perpendicular so CB passes through the center of O
- Hence the center of O is the intersection of AB and CB, B
- Since B is the center of O and P, A belong to O, PBA=2*PCA=40°

I hope you can understand despite my poor English

Title: Re: Some math problem
Post by: Legimet on November 21, 2013, 04:38:10 pm

I think Chockosta's solution is the best, since it doesn't use trig.

Title: Re: Some math problem
Post by: SpiroH on November 22, 2013, 09:00:52 am

Nice one Chockosta! Here a little drawing showing your solution:
(http://imageshack.us/a/img5/2952/5qs4.png)    (http://imageshack.us/a/img818/1925/t9fa.png)
1. Assume B is the center of the drawn circle (ABCD is a square!).
2. BP and BA are both radii of the circle (have equal lengths) => ABP triangle is isosceles!
3. Meaning, angles BPA=PAB=70º => ABP = 180 - 140 = 40º!

Title: Re: Some math problem
Post by: Sorunome on November 22, 2013, 09:04:28 am

OH, i get it now, that is a very elegant solution, great job! :)

Title: Re: Some math problem
Post by: Chockosta on November 22, 2013, 02:00:59 pm

Thanks for those drawings, but they only show the easy part of the proof.

The interesting part is that B is the center of the drawn circle, which is true if and only if PAC=PCD.
Here are some drawings :

Spoiler For Spoiler:

(http://www.pixenli.com/images/1385/1385146290011778700.png)

The inscribed angle theorem says that the blue angle remains the same when you move B.
(This is a well-known and useful result)
The corollary of this theorem says that it is the same as the angle with the tangent in C.
(To understand this, keep in mind that the blue angle doesn't change when B is getting close to C)

Spoiler For Spoiler:

(http://www.pixenli.com/images/1385/1385146384087833100.png)

In this drawing you can notice that the purple angles are acting the same way as mentionned before.
Therefore, you can say that AD is tangent to the circle in A.
Moreover, you can do the same with the blue angles. So CD is tangent to the circle in C.

Spoiler For Spoiler:

(http://www.pixenli.com/images/1385/1385146589089257400.png)

Since ABCD is a square, CB is perpendicular to CD in C. But CD is tangent to the circle in C.
Therefore CB is a radius of the circle.
You can do the same and find that AB is also a radius.
The two radii meet in B, so B is the center of the circle.

The conclusion is now quite obvious, see SpiroH's drawings
I personnally did it with the inscribed angle theorem :

Spoiler For Spoiler:

(http://upload.wikimedia.org/wikipedia/commons/thumb/8/87/Inscribed_angle_theorem.svg/250px-Inscribed_angle_theorem.svg.png)

You can apply this to PAC and PBC and find out that PBC=2*PAC, so PBC=40°

Title: Re: Some math problem
Post by: Sorunome on November 22, 2013, 04:39:45 pm

Great job Chockosta, would have never come up with that solution :)