• Aug 2nd 2008, 08:42 AM
Matt Westwood
Take a rectangle ABCD.

Draw the line AC dividing the rectangle diagonally in two.

In triangle ABC, inscribe a circle. Let its centre be E.

Draw perpendiculars from E to AB and BC, meeting CD at F and DA at G

What is the area of ABCD divided by the area of EFDG?

How neat is that?
• Aug 10th 2008, 04:36 AM
Coffee Cat
I think it makes a pretty fine problem
http://farm4.static.flickr.com/3167/...722c9d.jpg?v=0

Let's say the length of rectangle ABCD is L, and the width is W. If r is the radius of circle E, then the length of rectangle EFDG is (L-r) and the width is (W-r).

We can split triangle ABC into three smaller triangles: AEC, AEB and BEC. Each has an altitude of r.

http://farm4.static.flickr.com/3192/...70c17f.jpg?v=0

area(ABC) = area(AEC) + area(AEB) + area(BEC)

$
\frac {LW} {2} = \frac {r \sqrt {L^2 + W^2} }{2} + \frac {Lr} {2} + \frac {Wr} {2}
$

$
r = \frac {LW} {L + W + \sqrt {L^2 + W^2} }
$

So the area of rectangle EFDG is

$
A_{EFDG} = (L - \frac {LW} {L + W + \sqrt {L^2 + W^2} } ) (W - \frac {LW} {L + W + \sqrt {L^2 + W^2} } )
$

Whoa. So I suppose unless we are provided with the L/W ratio, we will not have a pretty answer for this problem.
• Aug 10th 2008, 05:27 AM
Matt Westwood
There's a neater way of doing it. Hint: use Pythagoras.
• Aug 10th 2008, 06:00 AM
Quick
Quote:

Originally Posted by Matt Westwood
Take a rectangle ABCD.

Draw the line AC dividing the rectangle diagonally in two.

In triangle ABC, inscribe a circle. Let its centre be E.

Draw perpendiculars from E to AB and BC, meeting CD at F and DA at G

What is the area of ABCD divided by the area of EFDG?

How neat is that?

I cheated and didn't do the algebra. Although making the incenter took a good deal of time (Worried) so I wouldn't recommend using the java [draw] [/draw] tags.

Anyway, you can drag points "A" and "C" around and you'll find that the ratio is always 2 :)
• Aug 10th 2008, 06:06 AM
Coffee Cat
To Quick: Amazing! How did you make this fine applet?

To Matt Westwood: Still thinking about it ^^

Edit: Ah, I'm retiring for the night. I'm so curious about the answer. ^^
• Aug 10th 2008, 06:12 AM
Quick
Quote:

Originally Posted by Coffee Cat
To Quick: Amazing! How did you make this fine applet?

To Matt Westwood: Still thinking about it ^^

I did this:

[draw]Point(20,60) [label ('A')]; UnitPoint(1,230) [hidden]; Line(1,2) [hidden]; Point(230,240) [label ('C')]; Perpendicular (3, 4) [hidden]; Intersect (3, 5) [label('B')]; Perpendicular (3, 1) [hidden]; Perpendicular (5, 4) [hidden]; Intersect (7, 8) [label('D')]; {10} Segment (1, 6); Segment (6, 4); Segment (4, 9); Segment (9, 1); Segment (1,4); {15} Angle(4, 1, 6, 10, 30, 'hi') [hidden]; Angle(1, 6, 4, 0, 0, 'hi') [hidden]; Angle(6, 4, 1, 0, 0, 'hi') [hidden]; Calculate(0, 10, '18', 'A360/ 3.14159 *') (15) [hidden]; Calculate(0, 0, '19', 'A360/ 3.14159 *') (16) [hidden]; {20} Rotation/MeasuredAngle (4, 1, 18)[hidden]; Rotation/MeasuredAngle (1, 6, 19) [hidden];{22} Line (1, 20) [hidden]; Line (6, 21) [hidden]; {24} Intersect (22,23) [label('E')]; Perpendicular (10,24) [color(128, 0, 128)]; Intersect (25,10) [hidden]; Circle (24,26); {28} Perpendicular (11,24) [color(128, 0, 128)]; Intersect (28,13) [label('G')]; Intersect (25,12) [label('F')];{31} Polygon (1, 6, 4, 9) [color(250,250,0), layer(1)]; Polygon (9, 29, 24, 30) [color(0,250,0), layer(2)]; {33} Area (31, 10, 20, 'ABCD = '); Area (32, 10, 40, 'EFDG = '); Calculate(10, 270, 'ABCD/EFGD = ', 'AB /') (33, 34);[/draw]

(Surprised) That's a lot for something simple. Most of the time the coding is a lot easier... Anyway, here is a tutorial.