superprismatic Posted August 27, 2009 Report Share Posted August 27, 2009 Consider the triangle with vertices at (0,0), (a,0), and (b,c) with 0 < b < a. Consider a point (x,y) inside this triangle. Shoot perpendiculars from (x,y) to each side of the triangle. This divides the triangle into 3 quadrilaterals. Find (x,y) such that these quadrilaterals all have the same area. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 28, 2009 Report Share Posted August 28, 2009 I think I got it! x=a/2 y=c/3 Because each quadrilateral has two right angles (created by the perpendiculars) you can cut them in half into 6 equal right triangles. Meaning thier areas all equal a*c/12. I used that to solve for y and then substituted the answer for y and solved for x. Let me know if I am wrong in any way. If I am not mistaken (x,y) would also create equal aread triangles if lines where drawn to the corners of the original triangle and it is the balance point for the triangle. a*c/12= a*y/4 c/12=y/4 c/3=y a*c/12=x*y/2 a*c/12=x*c/6 a/12=x/6 a/2=x Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 28, 2009 Report Share Posted August 28, 2009 (edited) I think I got it! x=a/2 y=c/3 Because each quadrilateral has two right angles (created by the perpendiculars) you can cut them in half into 6 equal right triangles. Meaning thier areas all equal a*c/12. I used that to solve for y and then substituted the answer for y and solved for x. Let me know if I am wrong in any way. If I am not mistaken (x,y) would also create equal aread triangles if lines where drawn to the corners of the original triangle and it is the balance point for the triangle. a*c/12= a*y/4 c/12=y/4 c/3=y a*c/12=x*y/2 a*c/12=x*c/6 a/12=x/6 a/2=x I may be wrong but my geometry is very rusty... Assuming that the quadrilaterals can be cut in equal halves by line joing the circumcenter (where perpendiculars meet) to the side indirectly assumes that the circumcenter and the incenter (where angle bisectors meet) are the same points. This would be a very specific case: an equilateral triangle only! Which is why you got x = a/2 and y = c/3. Now it remains to be seen whether the 3 quadrilaterals can have equal areas if and only if the triangle is an equilateral triangle! Edited August 28, 2009 by DeeGee Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 28, 2009 Report Share Posted August 28, 2009 For starters, Redsnuff’s answer does not work (in my view,) since I am not convinced that the triangles he constructs have to be equal in area. As a simple counter-example to his solution, take a = 6, b = 4, c = 3. The point (3,1) does not work. Secondly, I think that for any triangle for which all angles are acute, and many others, there must be a solution to Superprismatic’s problem. Take a random point inside the triangle. One of the quadrilaterals will be smallest. Move the point away from its vertex until it has an equal area to the largest of the other quadrilaterals. There will now still be a smallest quadrilateral. Repeat the procedure ad infinitum. A point will be found to equilibrize the situation. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 28, 2009 Report Share Posted August 28, 2009 I can't find a way too solve this without it getting rather messy. I'm sure there is a simple way to solve it. Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted August 28, 2009 Author Report Share Posted August 28, 2009 (edited) Consider the triangle with vertices at (0,0), (a,0), and (b,c) with 0 < b < a. Consider a point (x,y) inside this triangle. Shoot perpendiculars from (x,y) to each side of the triangle. This divides the triangle into 3 quadrilaterals. Find (x,y) such that these quadrilaterals all have the same area. Sorry, I misstated the problem. In order to guarantee being able to shoot perpendiculars, the triangle needs to be acute. So, the problem should read: Consider an acute triangle with vertices at (0,0), (a,0), and (b,c) with 0 < b < a. Consider a point (x,y) inside this triangle. Shoot perpendiculars from (x,y) to each side of the triangle. This divides the triangle into 3 quadrilaterals. Find (x,y) such that these quadrilaterals all have the same area. Edited August 28, 2009 by superprismatic Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted August 30, 2009 Report Share Posted August 30, 2009 (edited) Take the sides in order given. Three distances from (x, y) to the sides are d1, d2, d3. Distance from point (x0, y0) to line containing points (x1, y1) and (x2, y2) is This gives d1, d2 and d3. Three points of intersection are p1, p2, p3. Point of intersection (x0, y0) of perpendicular from point (x, y) to line containing points (x1, y1) and (x2, y2) is x0 = [(y-y1) + mx1 +x/m]/[m+1/m] y0 = y1 + m(x0-x) where m = (y2-y1)/(x2-x1) Apply from (x, y) to the three sides of the triangle to get p1, p2 and p3. Lines from (x, y) to the vertices define six right triangles whose areas are t1, t2, t3, t4, t5 and t6. t1 = .5 x d3 x dist [p3, (0, 0)] - where dist is distance between the two points. t2 = .5 x d1 x dist [p1, (0, 0)] t3 = .5 x d1 x dist [p1, (a, 0)] t4 = .5 x d2 x dist [p2, (a, 0)] t5 = .5 x d2 x dist [p2, (b, c)] t6 = .5 x d3 x dist [p3, (b, c)] The areas of the three quadrilaterals are A1 = t1+t2 A2 = t3+t4 A3 = t5+t6. Algebraic: Plug all the formulas together from the first spoiler, and set A1 = A2 = ac/6 [= 1/3 of the triangle area.] Solve for x and y. I'll leave that for someone else: Tag team, Bushindo? Numeric: For an initial x and y Calculate A1 and A2. Iterate. I did the numeric approach for a specific case of a=12, b=9, c=8. Triangle area is .5 x 12 x 8 = 48 Solution: x = 6.4815, y = 2.4815 gives A1 = A2 = 16. Edit: As an aside, this disproves redsnuff's His solution would be x=6 [slightly too small] and y=2.6667 [slightly large]. Edited August 31, 2009 by bonanova Add reference in 2nd spoiler to redsnuff's post Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 30, 2009 Report Share Posted August 30, 2009 I am working a different approach to that of Bonanova, one which does not require iteration. I hope that we will find eventual agreement, though. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted August 30, 2009 Report Share Posted August 30, 2009 I gave the equations to calculate of A1(x,y) and A2(x,y). Equating them to A/3 lets you solve for x,y. Algebraically, you get a closed form expression. Numerically, you get specific values. In this problem, algebraic is a bit clumsy; but the equations are there. Are there other approaches? Perhaps the equal area condition is stationary point that presents a simpler equation to solve. Don't know, but I'm interested to see what you come up with. I've missed the Aha! solution often enough. Quote Link to comment Share on other sites More sharing options...
0 bushindo Posted August 31, 2009 Report Share Posted August 31, 2009 (edited) I gave the equations to calculate of A1(x,y) and A2(x,y). Equating them to A/3 lets you solve for x,y. Algebraically, you get a closed form expression. Numerically, you get specific values. In this problem, algebraic is a bit clumsy; but the equations are there. Are there other approaches? Perhaps the equal area condition is stationary point that presents a simpler equation to solve. Don't know, but I'm interested to see what you come up with. I've missed the Aha! solution often enough. Alright, tag team. Here's an idea that may reduce the computational load. Here's an image of a triangle. Let the triangle on the right be the original triangle, and let the blue point be (x,y) such that the 3 quadrilaterals created are equal. Let a be the angle in degrees that the triangle makes at the origin. Note that a is not the length of the side, non-bold a is the length of a side (bad choice of notation, I know, but I didn't think of that when I drew the graphs). Note that solving for the area of the two lower pink triangles on the right hand side is easy. The main idea here is to rotate the entire triangle so that another side is flat along the x axis. Then we can solve the another set of pink triangles easily. Recall that the rotational matrix that rotates vectors a degrees counterclockwise is cos( [b]a[/b] ) -sin( [b]a[/b] ) sin( [b]a[/b] ) cos( [b]a[/b] ) We rotate the entire triangle by (180 - a) degrees counterclockwise so that another side is now flat along the x axis. The new coordinate for the blue center point is (x*cos( 180 - a ) - y sin( 180 - a ), x*sin( 180 - a ) + y*cos( 180 - a ) ) From here we can compute two new sub-triangles easily (denoted by pink triangles on left hand side). To do the last rotation, we'll need to do a translation first, and then a rotation. But the idea remains the same. Of course, once we have expressions for the areas of the 6 sub-triangles, we can solve for x and y. So, this seems to be less cumbersome by a smidge. Tag team, someone else? Edited August 31, 2009 by bushindo Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 31, 2009 Report Share Posted August 31, 2009 (edited) I attach my approach, which very probably is equivalent to that of Bonanova's and Bushindo's (indubitably correct.) I may have committed some trivial inaccuracies, but the method speaks for itself. Would Superprismatic please come out of hiding and spread his pearls of wisdom on the problem. I certainly cannot find an "aha" approach, as BN calls it. P.S. I have just found one error, though it does not nullify the approach. Brainden Question.doc Edited August 31, 2009 by jerbil Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 31, 2009 Report Share Posted August 31, 2009 I have just amended the silly errors occurring in my last post. Brainden Question.doc Quote Link to comment Share on other sites More sharing options...
0 bushindo Posted August 31, 2009 Report Share Posted August 31, 2009 That is a great approach. Definitely more elegant than my proposed approach. Well done. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted August 31, 2009 Report Share Posted August 31, 2009 (edited) Solve three quadratic equations. Your two equations have the form . a1x2 + b1xy + c1y2 + d1 = 0. a2x2 + b2xy + c2y2 + d2 = 0. . Where ai, bi, ci, di are expressions involving only a, b, c. Recast these in the standard quadratic form for x: .x2 + B1x + C1 = 0.x2 + B2x + C2 = 0.. where Bi = biy/ai and Ci = [ciy2 + di]/ai Solve 3 and 4 to get expressions for x in terms of a, b, c, and y. Equate them to get a quadratic for y in terms of a, b, c only. Plug y back into either expression for x and you have x, y in terms of a, b, c. Probably will give four values of x, y - only one of which will be inside the triangle. Not all that messy: [1] the process is laid out - just solve 3 quadratics [2] the final expressions probably simplify. Agree with Bushindo - nice approach. Now if I can just figure out why the new site permanently and unalterably makes half my posts bold... Rookie? site admin comment: Well spotted, I have removed the default bold font for span in spoiler. Thanks. Edited August 31, 2009 by rookie1ja site admin added comment/answer Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 1, 2009 Report Share Posted September 1, 2009 Thanks BN for your description as to how to complete the problem most easily. I will definitely do this myself when I have the time, whether you beat me to it or not. Looking at Superprismatic's original post, I note the two conditions (a) The point to be found must be internal to the triangle, (b) 0 < b < a. However, the problem clearly has a solution for some values of c when the triangle is right angled, i.e. b = a. By extension, there must be solutions for some values for b when b > a, so that one angle is obtuse. Inevitably I ask myself how obtuse can I get! Seriously though, to answer this I think an explicit formula needs to be found. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted September 1, 2009 Report Share Posted September 1, 2009 (edited) Thanks BN for your description as to how to complete the problem most easily. I will definitely do this myself when I have the time, whether you beat me to it or not. Looking at Superprismatic's original post, I note the two conditions (a) The point to be found must be internal to the triangle, (b) 0 < b < a. However, the problem clearly has a solution for some values of c when the triangle is right angled, i.e. b = a. By extension, there must be solutions for some values for b when b > a, so that one angle is obtuse. Inevitably I ask myself how obtuse can I get! Seriously though, to answer this I think an explicit formula needs to be found. x = [-b +/- sqrt(b2 - 4ac)] / 2a etc... is an explicit formula, no? It's double valued, which gives you two y values and then two x values for each of them. But it seems impossible for more than one of the four combinations to lie within the triangle [for any a, b, c.] p.s. You will beat me to it. p.p.s. to OP: Doesn't 0 < b < a guarantee an acute triangle? p.p.p.s. To site admin. It was the span tag. Thanks for the fix. site admin: You are welcome Edited September 1, 2009 by bonanova site admin added comment Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted September 1, 2009 Author Report Share Posted September 1, 2009 x = [-b +/- sqrt(b2 - 4ac)] / 2a etc... is an explicit formula, no? It's double valued, which gives you 2 y values and then two x values for each of them. But it seems impossible for more than one of the four combinations to lie within the triangle [for any a, b, c.] p.s. You will beat me to it. p.p.s. to OP: Doesn't 0 < b < a guarantee an acute triangle? p.p.p.s. To site admin. It was the span tag. Thanks for the fix. site admin: You are welcome To guarantee an acute triangle, I think that one must have 0 < b < a as well as c > sqrt(b*[a-b]). If (b,c) is too close to the x-axis the triangle will be obtuse. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted September 1, 2009 Report Share Posted September 1, 2009 To guarantee an acute triangle, I think that one must have 0 < b < a as well as c > sqrt(b*[a-b]). If (b,c) is too close to the x-axis the triangle will be obtuse. Yeah ... didn't think of the short case. The perpendiculars fall outside two of the bases. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 5, 2009 Report Share Posted September 5, 2009 Attached are my most recent thoughts on the matter. It includes an analysis of Bonanova's example as a useful adjunct. Brainden Question.doc Quote Link to comment Share on other sites More sharing options...
Question
superprismatic
Consider the triangle with vertices at
(0,0), (a,0), and (b,c) with 0 < b < a.
Consider a point (x,y) inside this
triangle. Shoot perpendiculars from
(x,y) to each side of the triangle.
This divides the triangle into 3
quadrilaterals. Find (x,y) such that
these quadrilaterals all have the same
area.
Link to comment
Share on other sites
18 answers to this question
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.