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Trapezing The Parabola (Posted on 2007-12-01) Difficulty: 3 of 5
The two diagonals PR and QS of a trapezoid PQRS intersect at the point (1, 0) and PS is parallel to QR. All the four vertices of the trapezoid lie on the parabola y2 = 4x and PR = QS = 100/9.

Determine the area of the trapezoid PQRS.

See The Solution Submitted by K Sengupta    
Rating: 3.0000 (1 votes)

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Solution numerical solution via computer | Comment 1 of 3

To appease my conventionality I turned the parabola upright to y=x^2/4, rather than x=y^2/4.

The diagonals must have some absolute value of slope, say a. One has positive and the other negative, slope. Let's take the one with positive slope, y = ax+1.

Let p be the x value (remembering it's an upright parabola, not sideways as in the statement) of the right end of the lower (shorter) of the parallel sides of the trapezoid, and q be the x value of the right end of the higher (longer) parallel side.

Then q^2/4 = aq+1 and p^2/4 = 1-ap. The lower y value is then 1-ap and the higher is 1+aq.

The following program uses the quadratic formula to solve for p and q, and then y1 and y2 for a given value of a (the slope of the diagonal), and then iteratively solves so that the diagonal is 100/9:

DEFDBL A-Z
stp = 1
bgn = 1: fin = 5
DO
  FOR a = bgn TO fin STEP stp
    p = (-4 * a + SQR(16 * a * a + 16)) / 2
    q = (4 * a + SQR(16 * a * a + 16)) / 2
    PRINT USING "####.#######"; a; p; q;
    y1 = 1 - a * p
    y2 = 1 + a * q
    d = SQR((y1 - y2) ^ 2 + (p + q) ^ 2)
    PRINT USING "####.#######"; y1; y2; d
    IF d > 100 / 9 THEN EXIT FOR
    prevA = a
    rsp$ = INKEY$
    IF rsp$ = CHR$(27) THEN EXIT FOR
  NEXT
  stp = stp / 4
  bgn = prevA
  fin = a
LOOP UNTIL rsp$ = CHR$(27)
area = (y2 - y1) * (p + q)
PRINT area

The solution found is:

  slope of                                                     len of
    diag         p = x1      q = x2       y1           y2       diag    
   1.3333333   0.6666667   6.0000000   0.1111111   9.0000000  11.1111111
  

  area ~= 59.25925925925926
 
 as one base is twice p, the other base is twice q and the height is y2-y1.
 
The area looks like 59 + 259/999 = 59 + 7/27 = 1600/27.

This is a correction from my first post, as I had not solved the quadratic correctly.

Edited on December 2, 2007, 2:03 pm
  Posted by Charlie on 2007-12-02 13:51:13

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