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Topic: Area of Triangle (Read 1173 times) |
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ThudnBlunder
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Find the area of the green equilateral triangle below.
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balakrishnan
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Re: Area of Triangle
« Reply #1 on: Jan 21st, 2009, 12:40am » |
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45+169/4* 3~ 118.179
Each side of the triangle is of length (169+603)
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| « Last Edit: Jan 21st, 2009, 12:46am by balakrishnan » |
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JiNbOtAk
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Re: Area of Triangle
« Reply #2 on: Jan 21st, 2009, 9:52pm » |
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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.
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towr
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Re: Area of Triangle
« Reply #3 on: Jan 22nd, 2009, 8:48am » |
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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.
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.
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| « Last Edit: Jan 22nd, 2009, 9:04am by towr » |
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balakrishnan
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Re: Area of Triangle
« Reply #4 on: Jan 23rd, 2009, 10:09am » |
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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+(6[a2b2+b2c2+c2 a2 ]-3[a4+b4+c4] )]/2
which simplifies to c2+ab3 when c2=a2+b2
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ThudnBlunder
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Re: Area of Triangle
« Reply #5 on: Jan 23rd, 2009, 10:47am » |
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Also Equation 16 on this page.
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towr
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Re: Area of Triangle
« Reply #6 on: Jan 23rd, 2009, 11:19am » |
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on Jan 23rd, 2009, 10:09am, balakrishnan wrote:| It would be nice if you can draw a figure of your deduction. |
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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.
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Barukh
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Let me try, towr - if I got you right. 
Absolutely brilliant solution! 
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towr
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Re: Area of Triangle
« Reply #8 on: Jan 24th, 2009, 2:44am » |
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Yup, that's what I got. Except my figure didn't look half as nice and clean 
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Grimbal
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I would trust towr thought of the following.
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Grimbal
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Re: Area of Triangle
« Reply #10 on: Jan 24th, 2009, 9:21am » |
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Hey, I was sure I checked for answers before replying.
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ecoist
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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]
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| « Last Edit: Jan 27th, 2009, 2:43pm by ecoist » |
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ecoist
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Re: Area of Triangle
« Reply #12 on: Jan 24th, 2009, 6:53pm » |
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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?
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| « Last Edit: Jan 24th, 2009, 8:19pm by ecoist » |
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balakrishnan
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Re: Area of Triangle
« Reply #13 on: Jan 25th, 2009, 4:28am » |
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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
cos (A)+cos(B)+cos(C)-2cos(A)cos(B)cos(C)=1 when A+B+C=2
This is very easy to simplify in case of a+b=c
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| « Last Edit: Jan 25th, 2009, 4:29am by balakrishnan » |
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Grimbal
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Re: Area of Triangle
« Reply #14 on: Jan 26th, 2009, 7:56am » |
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on Jan 24th, 2009, 6:53pm, ecoist wrote:| 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. |
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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.
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ecoist
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Re: Area of Triangle
« Reply #15 on: Jan 26th, 2009, 9:43am » |
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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'.
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balakrishnan
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Re: Area of Triangle
« Reply #16 on: Jan 26th, 2009, 9:51am » |
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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
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balakrishnan
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Re: Area of Triangle
« Reply #17 on: Jan 26th, 2009, 10:15am » |
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You might get more insights upon solving this problem :
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ecoist
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Re: Area of Triangle
« Reply #18 on: Jan 26th, 2009, 8:47pm » |
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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.
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JohanC
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Re: Area of Triangle
« Reply #19 on: Jan 27th, 2009, 8:49am » |
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on Jan 26th, 2009, 9:43am, ecoist wrote:| I saw this at cut-the-knot, but couldn't find the exact reference. ... |
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Weird, Cut-the-knot 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?
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ThudnBlunder
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Re: Area of Triangle
« Reply #20 on: Jan 28th, 2009, 12:53am » |
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on Jan 27th, 2009, 8:49am, JohanC wrote:
Weird, Cut-the-knot 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? |
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