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Topic: Area of Quadrilateral? (Read 2058 times) |
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ThudnBlunder
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In the triangle below, if EXB has area a, CXB has area b, and DXC has area c, what is the area of the quadrilateral ADXE?
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Presley
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Re: Area of Quadrilateral?
« Reply #1 on: Jul 25th, 2005, 8:30am » |
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Since there are no numbers, I'm going to take a wild stab in the dark and say:
The area of the quadrilateral ADXE is the area of the triangle ABC minus a, b, and c.
That just seems too obvious, so I'm not really sure.
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ThudnBlunder
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Re: Area of Quadrilateral?
« Reply #2 on: Jul 25th, 2005, 9:28am » |
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on Jul 25th, 2005, 8:30am, Presley wrote:Since there are no numbers, I'm going to take a wild stab in the dark and say:
The area of the quadrilateral ADXE is the area of the triangle ABC minus a, b, and c.
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Oh, you stuck your neck out with that answer! And modestly hid it, too! 
But I was hoping for a formula in terms of a, b, and c.
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| « Last Edit: Jul 25th, 2005, 9:30am by ThudnBlunder » |
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Eigenray
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Re: Area of Quadrilateral?
« Reply #3 on: Jul 25th, 2005, 12:56pm » |
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ac(a+2b+c)/(b2-ac)?
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| « Last Edit: Jul 25th, 2005, 12:58pm by Eigenray » |
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ThudnBlunder
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Re: Area of Quadrilateral?
« Reply #4 on: Jul 25th, 2005, 2:05pm » |
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on Jul 25th, 2005, 12:56pm, Eigenray wrote:
Yes.
That a highly-educated guess, was it? 
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Eigenray
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Re: Area of Quadrilateral?
« Reply #5 on: Jul 25th, 2005, 2:58pm » |
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I suck at geometry, so I coordinatized. WLOG (via a linear transformation, which multiplies all areas by a constant), assume that B is a right angle, and AB=BC=1. If x is angle XBC, and y is angle XCB, one may solve for the coordinates of E, X, and D, and compute the areas
2(b+c)=1/(1+cot x),
2(a+b)=tan y,
2b=1/(cot x + cot y).
Solving the first two for cot x, cot y, plugging into the third, and rearranging one finds
2b(a+b)(b+c)/(b2-ac)=1,
and since this has dimensions of area, it must in general be equal to 2K, where K is the area of the whole triangle. Thus
K-(a+b+c)=ac(a+2b+c)/(b2-ac).
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Barukh
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Re: Area of Quadrilateral?
« Reply #6 on: Jul 27th, 2005, 10:40am » |
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Here’s another approach.
Introduce additional notations: p = AD/AC, q = AE/AB, e = area of triangle ABX, d = area of triangle AXC, K = area of ABC (following Eigenray’s notation).
The first two relations between the quantities are almost obvious: (why?)
(a+b)/K = 1-q, (c+b)/K = 1-p. (1)
Two more relations are less obvious, but use the same idea: (why?)
d/b = q/(1-q), e/b = p/(1-p). (2)
And, of course,
b + d + e = K (3)
Adding relations (2) and using (3), we get: K/b = p/(1-p) + q/(1-q) + 1, which after re-arrangement gives:
b/K = (1-p)(1-q)/(1-pq). (4)
Plugging (4) into (1), we obtain expressions for a/K and c/K and note that ac/b2 = pq. Therefore, from (4), the final result is:
b/K = (a+b)(b+c)(b2–ac)/b2,
and Eigenray’s formula for K follows. I’m sure this second part of the derivation can be further simplified, but don’t see how.
T&B, thanks for the nice problem! By the way, I doubt that it belongs to Easy section.
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