Hardest math question I have ever seen O.O

Test Preparation SAT Preparation
1

<p>It's from October 12th, 2005 PSAT.</p>

<p><a href="http://oi54.tinypic.com/f54kyq.jpg%5B/url%5D"&gt;http://oi54.tinypic.com/f54kyq.jpg&lt;/a&gt;&lt;/p>

<p>The figure above shows a circle with radius r and centre P and an arc of length 6. The two radii shown are extended 3 units outside the circle. There is an arc of lenght x, which is part of a larger circle (not shown) also centred at P. What must x equal?
A) 9
B) 12
C) r+3
D) pi(r+3)/3
E)(6r+18)/r</p>

<p>The answer is E.</p>

<p>Why?</p>

2

<p>Ok here we go:
*If you can’t solve this math problem with mathematical method, you should set up your own values for both x and r.
*And that’s it.</p>

3

<p>I actually got it right, because I knew other answers were wrong.</p>

<p>But I don’t <em>know</em> why E is the correct answer and that’s killing me…especially a day before SAT and especially when I want 800 in math.</p>

4

<p>Oh…nevermind…I got it, my bad.</p>

<p>If anyone’s interested…2PIr/a=6 (where a is 360/angle)
a=2PIr/6=PIr/3
x=2pi(r+3)/a=6(r+3)/r=E(the answer)</p>

5

<p>Let the angle be theta
then the arc length: L = theta * radius</p>

<p>It is obvious that we can have two equations:

  1. 6 = theta * r
  2. x = theta * (r+3)</p>

<p>Eq 1) gives you: theta = 6/r and put theta in Eq 2) yields:
x= (r+3) * 6/r = (6r + 18)/r</p>

6

<p>Easy Way: Using ratios</p>

<p>when radius is r, arc length is 6</p>

<p>when radius is r+3, arc length is x</p>

<p>r:6=(r+3):x</p>

<p>6(r+3) = xr</p>

<p>x = (6(r+3))/r = (6r+18)/r</p>

<p>Bam</p>

7

<p>^
cool</p>

<p>10char</p>