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        riddles >> medium >> Hardy Perennial
        
(Message started by: Barukh on Nov 21st, 2003, 12:17am)
Title: Hardy Perennial
Post by Barukh on Nov 21st, 2003, 12:17am
Triangle ABC is isosceles. All the angles' measures are in degrees. Find the angle marked '?'
Title: Re: Hardy Perennial
Post by aero_guy on Nov 22nd, 2003, 12:52pm
OK, I solved it, though I had to bust out the law of sines.  Trickier than it initially looks.  I notice now that there are similar triangles involved.  The question is how would I have known that initially without going through the mess?
Title: Re: Hardy Perennial
Post by Dudidu on Nov 23rd, 2003, 10:03am
Barukh hi,
Could it be [hide]30[smiley=suptheta.gif][/hide].
If it does then you can read the following explanation:
[hide]You can look at the attached file to see a drawing of how I got it[/hide] ;D
Title: Re: Hardy Perennial
Post by aero_guy on Nov 23rd, 2003, 2:39pm
Dudidu, that is exactly what I got and exactly the way I got it.  What I noticed is that the middle triangle is similar (geometrically speaking) to the one that contains it and the one to its lower left.  I was wondering if there was a I way I could have found this without using the law of sines (as we both did) to find the angle.
Title: Re: Hardy Perennial
Post by THUDandBLUNDER on Nov 23rd, 2003, 3:43pm

Quote:
I was wondering if there was a I way I could have found this without using the law of sines (as we both did) to find the angle.

Hint:[hide]Draw a line BD such that D lies on AC and angle ABD = 20 degrees.[/hide]
Title: Re: Hardy Perennial
Post by Barukh on Nov 24th, 2003, 5:55am

on 11/23/03 at 14:39:20, aero_guy wrote:
Dudidu, that is exactly what I got and exactly the way I got it.

The following question is for both you and Dudidu: how did you solve the sines equation?


Quote:
I was wondering if there was a I way I could have found this without using the law of sines (as we both did) to find the angle.

I haven't found the way to follow THUDandBLUNDER's hint yet  >:(, but here's another way: [hide] Consider the regular 18-gon[/hide].
Title: Re: Hardy Perennial
Post by Lightboxes on Nov 24th, 2003, 7:47am

Quote:
You can look at the attached file to see a drawing of how I got it
\
I'm confused...
::[hide]
If the 70 degrees (both of them are) are created with two lines that intersect, then the remainding angles of the intersecting lines are as follows (I think):
360 - (70+70) = 220.
Then 220 / 2 - 110.
That means that the triangle to the left has thre angles now.  80+30+110??

But also (110-x) + 50 + ? = 180.
That leads to ? = 50.
Then the very top small triangle is 20+50+? = 180.  Then ? =110.  Then 110 + x + 80 = 180?
Unless these are not truely lines and bend as they intersect?  I think this is the case.
[/hide]::
Title: Re: Hardy Perennial
Post by aero_guy on Nov 24th, 2003, 8:10am
He messed up, it should be 40 not 80.  You find the 40 specifically because you know the 110 and the 30.

Barukh, doesn't Dudido's picture show you?

T and B, there you go, now that is as nice and simple as it should be.  A little bit of mirrored symmetry and you have it.  Still doesn't use similar triangles, but I guess that was a red herring.
Title: Re: Hardy Perennial
Post by Barukh on Nov 24th, 2003, 8:47am

on 11/24/03 at 08:10:28, aero_guy wrote:
Barukh, doesn't Dudido's picture show you?

Well, what I've got from Dudidu's picture - after simplification - is: [hide]sin(20+x) / sin(x) = 2sin(50)[/hide]. Is there a way to solve it without guessing?
Title: Re: Hardy Perennial
Post by aero_guy on Nov 24th, 2003, 2:31pm
Using the equations he has at the side gives [hide]sin(x)/sin(160-x)=sin(40)/sin(80)[/hide].

Now I bust out my old analytic geometry book (I very often use the tables at the back but almost never look at the rest) we see that sin(A+B)=sin(A)*cos(B)+cos(A)*sin(B).

Combine the two and you should be able to solve explicitly for x.  This is of course the ineligant way of doing it.
Title: Re: Hardy Perennial
Post by Dudidu on Nov 27th, 2003, 1:00am

Quote:
I'm confused...
Lightboxes you should be, I made a writing mistake. As aero_guy indicated it should have been 40 and not 80 (you can also see that in the equations I suggested, I used 40 and not 80).

Quote:
Using the equations he has at the side gives...
This is exectly what I got and as aero_guy indicated (again) it can be solved (inteligently) by known trigonometry equalities or (simply) by a calculator ::).
Title: Re: Hardy Perennial
Post by THUDandBLUNDER on Jan 4th, 2004, 6:42am
http://mathcircle.berkeley.edu/BMC4/Handouts/geoprob.pdf
Title: Re: Hardy Perennial
Post by ThudanBlunder on Dec 8th, 2007, 10:17am
Here is a nice animation (http://agutie.homestead.com/files/LangleyProblem.html).


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