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        riddles >> medium >> Area of Triangle
        
(Message started by: ThudanBlunder on Jan 20^th, 2009, 4:25pm)

Title: Area of Triangle
Post by ThudanBlunder on Jan 20^th, 2009, 4:25pm

Find the area of the green equilateral triangle below.

Title: Re: Area of Triangle
Post by balakrishnan on Jan 21^st, 2009, 12:40am

[hide]45+169/4*http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3~ 118.179[/hide]
Each side of the triangle is of length [hide]http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(169+60http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3)[/hide]

Title: Re: Area of Triangle
Post by JiNbOtAk on Jan 21^st, 2009, 9:52pm

How did you get that bala ? My attempt at using the sine and cosine rule resulted in messy equations, which somehow I don't think is the correct approach.

Title: Re: Area of Triangle
Post by towr on Jan 22^nd, 2009, 8:48am

I think you can solve it by tiling the plane; you get equilateral triangles with sides 5, 12 and 13, and a right-angled triangle with sides 5,12,13.

[hide]1/2 of each equilateral, plus 3/2 of the right-angled one; and since the two smaller equilaterals add to the larger, you get 13²*(sqrt(3)/4) + 3/2*30, which simplifies to what balakrishnan gave.[/hide]

Title: Re: Area of Triangle
Post by balakrishnan on Jan 23^rd, 2009, 10:09am

Towr,
It would be nice if you can draw a figure of your deduction.
My method: It is very easy to deduce that : if a,b,c are the distances of a point P(interior to the triangle) from the vertices of an equilateral triangle, then the side s of the triangle is:
s=[a²+b²+c²+http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(6[a²b²+b²c²+c² a² ]-3[a⁴+b⁴+c⁴] )]/2
which simplifies to c²+abhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3 when c²=a²+b²

Title: Re: Area of Triangle
Post by ThudanBlunder on Jan 23^rd, 2009, 10:47am

Also Equation 16 on this (http://mathworld.wolfram.com/EquilateralTriangle.html) page.

Title: Re: Area of Triangle
Post by towr on Jan 23^rd, 2009, 11:19am

on 01/23/09 at 10:09:08, balakrishnan wrote:

It would be nice if you can draw a figure of your deduction.

Unfortunately I'm on a different computer for the weekend. But what I did was first take six of the triangles with the 5 lines coming together at a point. Now the distance between the endpoints of the length 5 line piece in adjacent triangles must also be 5.
And then you do similarly for 12 and 13. So you get 3 hexagons and put those together.

I'll see if I can put the figure up sunday, unless someone else wants to do it first.

Title: Re: Area of Triangle
Post by Barukh on Jan 24^th, 2009, 12:46am

Let me try, towr - if I got you right. ;)

Absolutely brilliant solution! :D

Title: Re: Area of Triangle
Post by towr on Jan 24^th, 2009, 2:44am

Yup, that's what I got. Except my figure didn't look half as nice and clean :P

Title: Re: Area of Triangle
Post by Grimbal on Jan 24^th, 2009, 9:18am

I would trust towr thought of the following.

Title: Re: Area of Triangle
Post by Grimbal on Jan 24^th, 2009, 9:21am

Hey, I was sure I checked for answers before replying.

Title: Re: Area of Triangle
Post by ecoist on Jan 24^th, 2009, 10:10am

I wholeheartedly agree that towr's solution is brilliant! However, I interpreted his solution based on his use of the word "tiling". Consider a tiling of the plane by equilateral triangles. Those triangles pointing straight up are copies of the green triangle. Those pointing staight down are copies of the green triangle rotated clockwise 60 degrees. Then the resulting segments of lengths 5, 12, and 13, from all triangles form the sides of a plane tiling by hexagons (each hexagon with parallel opposite sides of lengths 5, 12, and 13). Hence the area of this hexagon, which is easy to compute, is twice the area of the green triangle![img][/img]

Title: Re: Area of Triangle
Post by ecoist on Jan 24^th, 2009, 6:53pm

Another way to look at this problem. The side length s of the green triangle equals the minimum value of the sum of the distances from a point to each of the vertices of the triangle with side lengths 5, 12, and 13. Is this the idea behind balakrishnan's solution?

Title: Re: Area of Triangle
Post by balakrishnan on Jan 25^th, 2009, 4:28am

No, My solution just takes the cosine of the angles between each of the segments of lenghts 5,12 and 13 and uses the fact that
coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup2.gif(A)+coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup2.gif(B)+coshttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup2.gif(C)-2cos(A)cos(B)cos(C)=1 when A+B+C=2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif
This is very easy to simplify in case of ahttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup2.gif+bhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup2.gif=chttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup2.gif

Title: Re: Area of Triangle
Post by Grimbal on Jan 26^th, 2009, 7:56am

on 01/24/09 at 18:53:29, ecoist wrote:

This is an interesting result. But I fail to see 1. how you derive this and 2. how it can be used to actually compute s.

Title: Re: Area of Triangle
Post by ecoist on Jan 26^th, 2009, 9:43am

I saw this at cut-the-knot, but couldn't find the exact reference. Erect equilateral triangles on each side of an arbitrary triangle ABC. Label the third vertex of the equilateral triangle erected on BC by A'. Similarly label vertices B' and C'. Then the segments AA', BB', and CC' are all of equal length, and are concurrent at a point p. That common length is also the sum |pA|+|pB|+|pC|. I have no idea how to compute s from this, but balakrishnan seems to know how to do it.

To give you an idea of what is going on here, choose a point p inside ABC and construct the segments pA, pB, and pC. Now rotate the triangle BpC about B until BC coincides with BA'. Then, the segments Ap, pp', and p'A', where p' is the image of p in the rotation, form a polygonal line from A to A' of the same lengths as those of pA, pB, and pC. Since the shortest path from A to A' is a straight line, it follows that the sum of the distances from p to the vertices of ABC is minimized when p is chosen as the point of concurrency of AA', BB', and CC'.

Title: Re: Area of Triangle
Post by balakrishnan on Jan 26^th, 2009, 9:51am

If ABC is the triangle(here right-angled) with sides a,b,c and P is the point inside, then:
Position of P such that it minimizes the sum of the distances is a point that subtends an angle of 120 degrees at all the 3 sides. This simplifies to
p^2+q^2+pq=a^2
p^2+r^2+pr=b^2
r^2+q^2+rq=c^2
where p,q and r are the distances of P from A,B,C respectively.

The above solves to
(p+q+r) = sqrt((a^2 + b^2 + c^2 + sqrt(3*(a+b+c)(a+b-c)(a-b+c)(-a+b+c) ))/2)

substituting a=12,b=5 and c=13 gives the required answer

Title: Re: Area of Triangle
Post by balakrishnan on Jan 26^th, 2009, 10:15am

You might get more insights upon solving this (http://projecteuler.net/index.php?section=problems&id=143) problem : :P

Title: Re: Area of Triangle
Post by ecoist on Jan 26^th, 2009, 8:47pm

Sorry, guys, I goofed. As balakrishnan's link points out, the minimum sum of distances from p to the vertices of a triangle ABC only works for triangles, all of whose angles are less or equal 120 degrees. However, the concurrency of the segments pA, pB, and pC still holds for arbitrary triangles, including degenerate ones, and their lengths are still all the same. A problem on this site shows this (clever geometry or mundane complex number calculation does the job), but I haven't yet located the link.

Title: Re: Area of Triangle
Post by JohanC on Jan 27^th, 2009, 8:49am

on 01/26/09 at 09:43:26, ecoist wrote:

I saw this at cut-the-knot, but couldn't find the exact reference. ...

Weird, Cut-the-knot (http://www.cut-the-knot.org/Generalization/fermat_point.shtml) itself now points in its 7th solution about the Fermat point to Grimbal at the wu::forum. Alexander Bogomolny only points to riddles::medium, so we would need Towr's help to find out which thread is referred to. Or maybe we are talking about a cyclic reference?

Title: Re: Area of Triangle
Post by ThudanBlunder on Jan 28^th, 2009, 12:53am

on 01/27/09 at 08:49:49, JohanC wrote:

It is here (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1203290327;).