Let P and P’ denote points inside rectangles ABCD and A’B'C’D', respectively. If PA = a+b, PB=a+c, PC = c+d, PD =b+d, P’A=ab, P’B'=ac, P’C'=cd, prove that P’D'=bd.
(Pi Mu Epsilon Journal – 4(Spring 1967)258, proposed by Stanley Rabinowitz)
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*Let sides AB and CD are the horizontal sides, and sides BC and AD are the vertical sides.
*Draw 2 perpendicular lines (vertical and horizontal) inside rectangle ABCD such that their intersection is point P. It follows that the the vertical line segment is perpendicular to sides AB and CD, and the horizontal line segment is perpendicular to sides BC and AD.
*Let:
point a1 – intersection of the v.l.segment and side AB
point a2 - intersection of the h.l.segment and side BC
point a3 – intersection of the v.l.segment and side CD
point a4 – intersection of the h.l.segment and side AD
*Let: u=distance bet. A and a1= distance bet. D and a3
w=distance bet. B and a1= distance bet. C and a3
x=distance bet. A and a4= distance bet. B and a2
y=distance bet. D and a4= distance bet. C and a2
*Then, by Pythagorean theorem
(PA)2 = x2 + u2, (PB)2 = w2 + x2 , (PC)2 = w2 + y2, (PD)2 = y2 + u2
Since (y2 + u2) + (w2 + x2) = (x2 + u2) + (w2 +y2).
Therefore, (PD)2 + (PB)2 = (PA)2 + (PC)2,
(b+d)2 + (a+c)2 = (a+b)2 + (c+d)2,
(ac)2 + (bd)2 = (ab)2 + (cd)2 – - – (1)
*Same procedure to A’B'C’D',
(P’D')2 = (P’A')2 + (P’C')2 – (P’B')2,
(P’D')2 = (ab)2 + (cd)2 – (ac)2
*Substitute (1):
(P’D')2 = [(ac)2 + (bd)2] – (ac)2
Therefore, P’D’ = bd
Quote: “As long as algebra and geometry have been separated, their progress have been slow and their uses limited; but when these two sciences have been united, they have lent each mutual forces, and have marched together towards perfection” - Joseph-Louis Lagrange
Nice thinking! Have to remember this page!
Thanks!