bonanova Posted January 1, 2014 Report Share Posted January 1, 2014 (edited) Place an arbitrary point P inside an isosceles equilateral triangle. From P, draw lines to the vertices and perpendiculars to the sides. Six smaller triangles now touch at P. Prove that the sums of areas of the two alternating sets of three triangles are equal. Edited January 1, 2014 by bonanova Correction Quote Link to comment Share on other sites More sharing options...

0 k-man Posted June 23, 2014 Report Share Posted June 23, 2014 In the picture below, the black lines are drawn as per the OP. The red lines are drawn through the point P parallel to the sides of the triangle. The red lines are dividing the triangle into 3 equilateral triangles (B,D,F) and 3 parallelograms (A,C,E). The black lines divide those triangles and parallelograms in half proving that the sum of blue shaded areas equals the sum of the white areas (every blue area marked with a letter has an equal white area marked with the same letter). Now let's look at triangles formed by black lines. Starting from left corner going clockwise (A+B) + (C+D) + (E+F) = (B+C) + (D+E) + (F+A). 1 Quote Link to comment Share on other sites More sharing options...

0 plasmid Posted January 1, 2014 Report Share Posted January 1, 2014 Do you mean equilateral instead of isosceles?Consider an isosceles triangle where the two equal sides are much longer than the other one, and move the point P very close to a vertex that doesn't connect the two sides of equal length.Four of the six sub-triangles will be vanishingly small, and for the two non-tiny subtriangles orange > blue. Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted January 1, 2014 Author Report Share Posted January 1, 2014 Do you mean equilateral instead of isosceles?Consider an isosceles triangle where the two equal sides are much longer than the other one, and move the point P very close to a vertex that doesn't connect the two sides of equal length.isosceles.jpgFour of the six sub-triangles will be vanishingly small, and for the two non-tiny subtriangles orange > blue. Quite right. Equilateral. I changed the OP. Thanks. Quote Link to comment Share on other sites More sharing options...

0 plasmid Posted June 22, 2014 Report Share Posted June 22, 2014 First consider the triangle where P lies exactly in the center. Label the four subdivisions of the triangle A-F. They are obviously all the same size and the sum of the areas (A+C+E) equals the sum (B+D+F). Now consider what happens if you move point P up or down the line down the center of the triangle – the line between A and F and between C and D. There will still be symmetry about that line, so the areas of A and F are equal, the areas of B and E are equal, and the areas of C and D are equal if point P is anywhere along that line, so (A+C+E) still equals (B+D+F). By symmetry, if you were to start with P in the center and were to move it along any of the other two lines going through the middle (the line between B and C and between F and E, or the line between A and B and between D and E) then the sum of areas remains equal. So placing P at any point on the dashed lines in the figure will lead to subdivisions with area (A+C+E) = (B+D+F). Also notice that any point in the triangle can be reached by picking a point #1 on one of the dotted lines, a point #2 on another dotted line, and moving partway from point #1 to point #2. Next we'll consider what happens if you were to place P somewhere between two points on different dotted lines. I'll refer to the subdivisions of the triangle as A1, B1, C1, etc. when P is at point #1 and A2, B2, etc. when P is at point #2. And I'll refer to the base of a triangle (the side that's in common with the large equilateral triangle) as bA1, bB1, … bA2, bB2, etc. and the height of a triangle (the line perpendicular to the base and going toward P) as hA1, hC1, hE1, hA2, hC2, hE2. Notice that hA1 would be the same as hB1, hC1=hD1, hA2=hB2, etc. When P is at point #1 the area of A1 is bA1*hA1 / 2, and when P is at point #2 the area of A2 is bA2*hA2 / 2. If P has moved some fraction f of the way from point #1 to point #2, then the area of Af would be ((1-f)bA1 + fbA2) * ((1-f)hA1 + fhA2) / 2 = (1-f)^2 (bA1*hA1) / 2 + (f-f^2) (bA1*hA2) / 2 + (f-f^2) (bA2*hA1) / 2 + f^2 (bA2*hA2) / 2 Notice that that has terms bA1*hA1 and bA2*hA2 which are interpretable and we know based on the equality when P is at point #1 or #2 that those terms by themselves will lead to equal sums of alternating triangles, but it also has terms bA1*hA2 and bA2*hA1 which will require some sorting out. For the moment, I'll define the term deltaA as (bA1*hA2)+(bA2*hA1) so the equation for the area becomes (1-f)^2 (bA1*hA1) / 2 + f^2 (bA2*hA2) / 2 + (f-f^2) deltaA / 2 In order to generate a useful equation with those mixed terms, you can start off by writing an equation saying that the sum of areas (A1+C1+E1) equals (B1+D1+F1). bA1*hA1 + bC1*hC1 + bE1*hE1 = bB1*hA1 + bD1*hC1 + bF1*hE1 Combine terms that are multiplied by a common height (bA1-bB1)/hA1 + (bC1-bD1)/hC1 + (bE1-bF1)/hE1 = 0 We know that this is also true for the triangle with P at point #2, so we can write (bA1-bB1)/hA1 + (bC1-bD1)/hC1 + (bE1-bF1)/hE1 = 0 (bA2-bB2)/hA2 + (bC2-bD2)/hC2 + (bE2-bF2)/hE2 = 0 If we add those two left-hand sides together, we get ( hA2[bA1-bB1] + hA1[bA2-bB2] ) / hA1hA2 + .... = 0 (bA1*hA2 - bB1*hA2 + bA2*hA1 - bB2*hA1) / hA1hA2 + .... = 0 (deltaA - deltaB) / hA1hA2 + …. = 0 Since the right hand side of that equation is zero, the delta terms cancel out and the sum of areas remains equal as P moves from point #1 to point #2. Quote Link to comment Share on other sites More sharing options...

0 plasmid Posted June 22, 2014 Report Share Posted June 22, 2014 Drat. I just looked over that again and saw that I divided instead of multiplying when manipulating that equation after introducing the deltaA term, so it won't work. 1 Quote Link to comment Share on other sites More sharing options...

0 plasmid Posted June 22, 2014 Report Share Posted June 22, 2014 (edited) First consider the triangle where P lies exactly in the center. Label the four subdivisions of the triangle A-F. They are obviously all the same size and the sum of the areas (A+C+E) equals the sum (B+D+F).Now consider what happens if you move point P up or down the line down the center of the triangle the line between A and F and between C and D. There will still be symmetry about that line, so the areas of A and F are equal, the areas of B and E are equal, and the areas of C and D are equal if point P is anywhere along that line, so (A+C+E) still equals (B+D+F). By symmetry, if you were to start with P in the center and were to move it along any of the other two lines going through the middle (the line between B and C and between F and E, or the line between A and B and between D and E) then the sum of areas remains equal.So placing P at any point on the dashed lines in the figure will lead to subdivisions with area (A+C+E) = (B+D+F). Also notice that any point in the triangle can be reached by picking a point #1 on one of the dotted lines, a point #2 on another dotted line, and moving partway from point #1 to point #2. Next we'll consider what happens if you were to place P somewhere between two points on different dotted lines. To make things easier later on, define points #1 and #2 by drawing a line through the desired ultimate location of point P and parallel to the nearest edge of the equilateral triangle. For example, if you wanted to place P somewhere within the area of B in the figure above, you could place point #1 on the line between F and A and point #2 on the line between B and C, so #1 P and #2 form a line parallel to the side of the large equilateral triangle touching A and B.I'll refer to the subdivisions of the triangle as A1, B1, C1, etc. when P is at point #1 and A2, B2, etc. when P is at point #2. And I'll refer to the base of a triangle (the side that's in common with the large equilateral triangle) as bA1, bB1, bA2, bB2, etc. and the height of a triangle (the line perpendicular to the base and going toward P) as hA1, hC1, hE1, hA2, hC2, hE2. Notice that hA1 would be the same as hB1, hC1=hD1, hA2=hB2, etc.When P is at point #1 the area of A1 is bA1*hA1 / 2, and when P is at point #2 the area of A2 is bA2*hA2 / 2. If P has moved some fraction f of the way from point #1 to point #2, then the area of Af would be((1-f)bA1 + fbA2) * ((1-f)hA1 + fhA2) / 2=(1-f)^2 (bA1*hA1) / 2+ (f-f^2) (bA1*hA2) / 2+ (f-f^2) (bA2*hA1) / 2+ f^2 (bA2*hA2) / 2Notice that that has terms bA1*hA1 and bA2*hA2 which are interpretable and we know based on the equality when P is at point #1 or #2 that those terms by themselves will lead to equal sums of alternating triangles, but it also has terms bA1*hA2 and bA2*hA1 which will require some sorting out. I'll define the term deltaA as (bA1*hA2)+(bA2*hA1) so the equation for the area becomes(1-f)^2 (bA1*hA1) / 2+ f^2 (bA2*hA2) / 2+ (f-f^2) deltaA / 2In order to generate a useful equation with those mixed terms, you can start off by writing an equation saying that the sum of areas (A1+C1+E1) equals (B1+D1+F1) and (A2+C2+E2) equals (B2+D2+F2).bA1*hA1 + bC1*hC1 + bE1*hE1 = bB1*hA1 + bD1*hC1 + bF1*hE1bA2*hA2 + bC2*hC2 + bE2*hE2 = bB2*hA2 + bD2*hC2 + bF2*hE2Combine terms that are multiplied by a common height(bA1-bB1)hA1 + (bC1-bD1)hC1 + (bE1-bF1)hE1 = 0(bA2-bB2)hA2 + (bC2-bD2)hC2 + (bE2-bF2)hE2 = 0For simplicity in the upcoming section, I'll rename those terms in the equationsGH + IJ + KL = 0MN + OP + QR = 0Note that(G+M) (H+N) + (I+O) (J+P) + (K+Q) (L+R)= (GH + GN + MH + MN) + (IJ + IP + OJ + OP) + (KL + KR + QL + QR)= (GH + IJ + KL) + (MN + OP + QR) + (GN + MM + IP + OJ + KR + QL)= 0 + 0 + (GN + MM + IP + OJ + KR + QL)= (bA1-bB1)hA2 + (bA2-bB2)hA1 + (bC1-bD1)hC2 + (bC2-bD2)hC1 + (bE1-bF1)hE2 + (bE2-bF2)hE1= bA1*hA2 + bA2*hA1 - bB1*hA2 - bB2*hA1 + ....= deltaA - deltaB + deltaC - deltaD + deltaE - deltaFThe term at the beginning of that last block (G+M) (H+N) + (I+O) (J+P) + (K+Q) (L+R) is precisely the lengths you would get if you were to consider an equilateral triangle that's twice as large as the original triangle and with the point P moved halfway from #1 to #2. Remember that points #1 and #2 are positioned such that they form a line parallel to one edge of the large equilateral triangle and are on the two dotted lines that lie farthest apart. That means that the triangle described by (G+M) (H+N) + (I+O) (J+P) + (K+Q) (L+R) has its point P lying halfway between #1 and #2. That means it lies on the dotted line between those two. That means that, since it lies on a dotted line, the sum of its alternating sub-triangles must be equal. That means that(G+M) (H+N) + (I+O) (J+P) + (K+Q) (L+R) = 0And that means thatdeltaA - deltaB + deltaC - deltaD + deltaE - deltaF = 0And going back to the equation of the area of a subtriangle after moving point P some fraction of the way between points #1 and #2, that means that the sum of areas for all intermediate points are equal. QED. Edit: added the figure back Edited June 23, 2014 by plasmid 1 Quote Link to comment Share on other sites More sharing options...

0 plainglazed Posted June 23, 2014 Report Share Posted June 23, 2014 For the general case of an equilateral triangle ABC of side length s, the area = sqrt(3)s^{2}/4 and the sum of the areas of triangles ACK, CBK, BAK = s/2(h_{1}+h_{2}+h_{3}). Equating the two yields h_{1}+h_{2}+h_{3}=sqrt(3)s/2; a constant regardless of the location of K. So s=2(h_{1}+h_{2}+h_{3})/sqrt(3) and the perimeter of the triangle p=6(h_{1}+h_{2}+h_{3})/sqrt(3). here Dm=h_{1}/2 and Dn=h_{1}/2+h_{3 }thus BD=2(Dn)/sqrt(3) or (h_{1}+h_{3}+h_{3})/sqrt(3) similarly CE=(h_{3}+h_{2}+h_{2})/sqrt(3) and AF=(h_{2}+h_{1}+h_{1})/sqrt(3) and BD + CE + AF = 3(h_{1}+h_{2}+h_{3})/sqrt(3) or half of the perimeter. Consequently the sum of the areas of alternating triangles within the equilateral triangle are each 1/2 the total area of the equilateral triangle. Quote Link to comment Share on other sites More sharing options...

0 plainglazed Posted June 23, 2014 Report Share Posted June 23, 2014 très élégant k-man! Quote Link to comment Share on other sites More sharing options...

0 plasmid Posted June 24, 2014 Report Share Posted June 24, 2014 Indeed, beautiful solution, k-man!(Especially compared to all that slogging I was doing. ) Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted June 24, 2014 Author Report Share Posted June 24, 2014 Bowing to the wishes of the majority. Nice job all, and k-man gets the nod. Quote Link to comment Share on other sites More sharing options...

## Question

## bonanova

Place an arbitrary point P inside an

~~isosceles~~equilateral triangle.From P, draw lines to the vertices and perpendiculars to the sides.

Six smaller triangles now touch at P.

Prove that the sums of areas of the two alternating sets of three triangles are equal.

Edited by bonanovaCorrection

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