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        riddles >> medium >> Area of Quadrilateral?
        
(Message started by: THUDandBLUNDER on Jul 25^th, 2005, 4:47am)

Title: Area of Quadrilateral?
Post by THUDandBLUNDER on Jul 25^th, 2005, 4:47am

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?

Title: Re: Area of Quadrilateral?
Post by Presley on Jul 25^th, 2005, 8:30am

Since there are no numbers, I'm going to take a wild stab in the dark and say:

[hide]The area of the quadrilateral ADXE is the area of the triangle ABC minus a, b, and c.[/hide]

That just seems too obvious, so I'm not really sure.

Title: Re: Area of Quadrilateral?
Post by THUDandBLUNDER on Jul 25^th, 2005, 9:28am

on 07/25/05 at 08:30:18, Presley wrote:

Since there are no numbers, I'm going to take a wild stab in the dark and say:

[hide]The area of the quadrilateral ADXE is the area of the triangle ABC minus a, b, and c.[/hide]

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.

Title: Re: Area of Quadrilateral?
Post by Eigenray on Jul 25^th, 2005, 12:56pm

[hide]ac(a+2b+c)/(b²-ac)[/hide]?

Title: Re: Area of Quadrilateral?
Post by THUDandBLUNDER on Jul 25^th, 2005, 2:05pm

on 07/25/05 at 12:56:25, Eigenray wrote:

[hide]ac(a+2b+c)/(b²-ac)[/hide]?

Yes.
That a highly-educated guess, was it? :)

Title: Re: Area of Quadrilateral?
Post by Eigenray on Jul 25^th, 2005, 2:58pm

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).

Title: Re: Area of Quadrilateral?
Post by Barukh on Jul 27^th, 2005, 10:40am

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.