Anyone up for two hard math problems?

Test Preparation SAT Subject Test Preparation
1

<li><p>Points A and B are on the parabola y = 4x2 + 7x - 1 and the origin is the midpoint of AB. What is the length of AB?
A) 2/5 B) 5 + 2/ /2 C) 5 + /2 D) 13 E) 114</p></li>
<li><p>A truncated cone has horizontal bases with radii 18 and 2. A sphere is tangent to the top, bottom and lateral surface of the truncated cone. What is the radius of the sphere?
A) 6 B) 4/5 C) 9 D) 10 E) 6/3</p></li>
</ol>

<p>I’m not sure if there is a forum on this site that accommodates questions like these or not, so I just stuck it here. These are two sample questions of the AMC 12. So no, not homework, and no, not pertinent to the SAT Subject Math. </p>

<p>According to the answer key, the answer to number 7 is E and the answer to number 8 is A.</p>

<p>Looking at the graph, I don’t see how 114 is even possible as an answer to number 7. Doing it analytically, I got 5root2 (7.07).</p>

<p>As for number 8, I came up with a system of equations, but with 5 equations. Though, theoretically, it should work, I’m too lazy to solve a 5 variable system. I’m sure there is an easier way to do the problem; I doubt they would make you use 5 variables to solve a problem…</p>

2

<p>Okay, I solved #7 and I got 5 * sqrt(2) as well. Here's my process...someone please find a flaw: </p>

<p>Let M be the midpoint defined as (0,0)</p>

<p>Let A = (x, 4x^2 + 7x - 1 )
Let B = (z, 4z^2 + 7z - 1) </p>

<p>(x+z)/2 = 0 </p>

<p>(4x^2 + 4z^2 + 7x + 7z - 2)/2 = 0 </p>

<p>Since the first equation is x + z = 0
x = -z, which we can use to perform a substitution: </p>

<p>(4x^2 + 4x^2 + 7x - 7x -2) = 0
8x^2 - 2 = 0
4x^2 - 1 = 0
x = +- 1/2 </p>

<p>So now the points we have for AB are:
(0.5, 3.5), (-0.5, -3.5) </p>

<p>Distance of AB = sqrt(50) = 5 * sqrt(2).</p>

<p>Btw, I love these sorts of problems, even if I can't solve them. They use Math Level 2 and beyond concepts with reasoning skills greater than those needed for SAT I. They're great! :P</p>

3

<p>Yeah, that is exactly what I did to solve #7. The answer 114 is actually absolutely absurd because the graph is a parabola opening up a little bit below and a little bit to the left of the origin. To have AB be a length of 114 would mean that one of the y values would have to be A LOT lower than the minimum of the graph.</p>

<p>So yeah, pretty sure that question is just messed up and 5root2 is the answer.</p>

<p>Still pretty stumped for number 8..</p>

4

<p>What's a truncated cone?</p>

5

<p>A cone that has part of its top lopped off.</p>

6

<p>Basically imagine a round 3D trapezoid with a sphere touching the top, the sides, and the bottom. Or just imagine a 2D trapezoid with a circle touching the top, the bottom, and the sides. I've figured out the answer through my calculator using nefarious (guess and check) activities but I can't figure out a clean-cut solution for the problem and its killing me :P</p>

7

<p>I think this is what #8 looks like</p>

<p><a href="http://i206.photobucket.com/albums/bb26/Iam3elliott/untitled.jpg%5B/url%5D"&gt;http://i206.photobucket.com/albums/bb26/Iam3elliott/untitled.jpg&lt;/a&gt;&lt;/p>

8

<p>Wow, finally figured it out and its really simple. Tangents are equal when they are from the same point and connected to a point on a circle. Thus, one of the legs of the trapezoid (as seen on ThisCouldBeHeaven's pic) has a length of 18 + 2 = 20. Then you use the pythagorean theorem to find the height of the trapezoid. 20^2 - 16^2 = h^2. The height of the trapezoid is 12 and it follows that the radius of the circle is 6. And here I was thinking up of all sorts of crazy methods and going no where 0.o That would be an awesome SAT question.</p>

9

<p><em>smacks head</em></p>

<p>Thanks for that!</p>

10

<p>This would be a level 5 at the very least. More like a level 7 or 8 SAT math question.</p>

11

<p>
[QUOTE]
Wow, finally figured it out and its really simple.

[/QUOTE]

It's only simple once you figure it out.</p>

12

<p>I didn't know the point to tangent rule... interesting...</p>

<p>It's faster if you see 20 and 16 and assume 3, 4, 5, especially if you don't have a calculator.</p>

13

<p>
[quote]
Tangents are equal when they are from the same point and connected to a point on a circle. Thus, one of the legs of the trapezoid (as seen on ThisCouldBeHeaven's pic) has a length of 18 + 2 = 20

[/quote]
</p>

<p>Wait, wait. Maybe I'm not understanding what you said, but I don't think I know of this relation. Can you elaborate?</p>

14

<p>I think he means this:</p>

<p><a href="http://i206.photobucket.com/albums/bb26/Iam3elliott/untitled2.jpg%5B/url%5D"&gt;http://i206.photobucket.com/albums/bb26/Iam3elliott/untitled2.jpg&lt;/a&gt;&lt;/p>

<p>I wasn't familiar with it either, but it makes sense.</p>

15

<p>Ah yes, I remember that theorem from Geometry now. Although, is there any proof/theorem that proves that the circle is tangent to both bases at the exact mid point of the bases? While it seems intuitive due to the symmetry of the trapezoid, I can't seem to think of a way to prove that true and it's necessary information to use the tangent theorem.</p>