'07 District 1

[nsguy1350]

I got a 304, missed 4, skipped 4. Broke 300! Now I need to keep this up and start breaking 359 :p
Today was slightly better than yesterday, considering I broke all of my goals today, compared to none yesterday (Yesterday, 1 question off in Calc, 1 in Math, then 1 and 2 in NS. Today, broke by 2 in NS, again 2 in NS, then 3 in Calc, and 1 in Math).

Questions:
16. I got B. 11.3" instead of A. Where did I go wrong?
37. How am I supposed to know this?
22. I # at optimization... didn't get this one or the Calc one. Help please?
29. Should I use Brahmagupta's formula, the formula I never wanted to learn? Seems sort of tedious, but I don't see any other option. Unless I'm being dumb with the geometry again.
48. I had ideas, but I failed on this probability.
51. I used a couple methods and neither worked. Linear regression does not work. What should I do?

The other 2 mistakes were me being dumb again... one, I forgot an absolute value (#4), and another, I didn't see the less than or equal, instead doing less than (#50.)

[redhawk2015]

16. Imagine drawing [unparseable or potentially dangerous latex formula] where [unparseable or potentially dangerous latex formula] intersects [unparseable or potentially dangerous latex formula]. Then, [unparseable or potentially dangerous latex formula] is a 30-60-90 right triangle. [unparseable or potentially dangerous latex formula] is opposite the 30, so [unparseable or potentially dangerous latex formula], or half of 13.

22. Looking at where the fences must be placed, [unparseable or potentially dangerous latex formula], with [unparseable or potentially dangerous latex formula] and [unparseable or potentially dangerous latex formula]. So, [unparseable or potentially dangerous latex formula]. The area is maximized when [unparseable or potentially dangerous latex formula]. So, [unparseable or potentially dangerous latex formula]

29. Don't bother with area inside the quadrilateral - draw a 6 by 8 box around it that touches the top left and bottom right points and subtract the areas not within the shape. Doing this, I get [unparseable or potentially dangerous latex formula]

[nsguy1350]

Thanks. On #16, I used the Law of Cosines to get XY = 6.505... . Why is that wrong?
And on 22, where did you get the 3h and 3l? There are only two vertical fences, and one horizontal, I thought.

[redhawk2015]

The inner horizontal fence is another [unparseable or potentially dangerous latex formula], and the broken-up vertical fences add up to create another [unparseable or potentially dangerous latex formula].

[nsguy1350]

Oh, I see.
I understood the question as saying the borders were not part of the fencing for some reason. That would've made a tough problem! Thanks.

[redhawk2015]

#37 and #41 I have no idea...#51 I would use linear regression, but if it also didn't work, I'm not sure what else to do.

[nsguy1350]

Oh, I figured out 37.
You have 12 vertices. Imagine a prism, two flat faces, and then sides that connect each face.
12 vertices for 2 sides means 6 vertices per side, for a hexagon.
Hexagon means 6 lateral sides!

[redhawk2015]

I noticed that too...I was thinking about that earlier and realized how ridiculously easily the problem actually is...

[Salvadoramerican]

[nsguy1350]

I think he read my question incorrectly, I did not ask for 41. I did graph the problem.