Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2015 UNCO Math Contest II Answer Key"

(Created page with "1) <math>17</math> 2) <math>3\pi</math> 3) <math>P(x) = 5x^2−3x+17</math> 4) <math>28</math> 5) <math>8</math> 6) <math>1197</math> 7) (a) One possible example i...")
 
 
Line 5: Line 5:
  
  
−
3) <math>P(x) = 5x^2−3x+17</math>
+
3) <math>P(x)=5x^2-3x+17</math>
  
  
Line 17: Line 17:
  
  
−
7) (a) One possible example is a square box with large top and bottom and small height, with the top and bottom colored gold and the four sides blue. If the top and bottom are squares of side <math>s</math>, then the total area colored gold is <math>2 × s^2</math> . If the height of the box is <math>h</math>, then the total area colored blue is <math>4 × h × s</math>. It is easy to select s and h to make <math>4 × h × s</math> equal to half of <math>2 × s^2</math>. Set <math>4 × h × s = s^2</math> or <math>4 × h = s</math>. Choose, say, <math>h = 1</math> and <math>s = 4</math>.
+
7) (a) One possible example is a square box with large top and bottom and small height, with the top and bottom colored gold and the four sides blue. If the top and bottom are squares of side <math>s</math>, then the total area colored gold is <math>2 \times s^2</math> . If the height of the box is <math>h</math>, then the total area colored blue is <math>4\times h\times s</math>. It is easy to select s and h to make <math>4\times h\times s</math> equal to half of <math>2\times s^2</math>. Set <math>4\times h\times s=s^2</math> or <math>4\times h=s</math>. Choose, say, <math>h=1</math> and <math>s=4</math>.
 +
 
 
(b) Triangulate each face by connecting the point of tangency on the face to each vertex of the
 
(b) Triangulate each face by connecting the point of tangency on the face to each vertex of the
 
face. Look at a gold face and one of the triangles into which it has been cut. The gold triangle
 
face. Look at a gold face and one of the triangles into which it has been cut. The gold triangle
Line 27: Line 28:
  
  
−
8) <math>\frac{157}{(18 × 17 × 16} = \frac{157}{4896}</math>
+
8) <math>\frac{157}{(18 \times 17 \times 16)} = \frac{157}{4896}</math>
  
  
Line 34: Line 35:
  
 
10) (a) <math>14</math> (b) <math>132</math>
 
10) (a) <math>14</math> (b) <math>132</math>
−
(c) <math>\frac{1}{n + 1}\choose{2n}{n}=\choose{2n}{n}-\choose{2n}{n-1}=\choose{2n}{n}-\choose{2n}{n+1}=\frac{1}{2n+1}\choose{2n+1}{n}=\frac{(2n)!}{(n+1)!n!}</math>
+
 
 +
(c) <math>\frac{1}{n + 1} \binom{2n}{n}</math>
 +
<math>= \binom{2n}{n}- \binom{2n}{n-1}= \binom{2n}{n}- \binom{2n}{n+1}=\frac{1}{2n+1} \binom{2n+1}{n}=\frac{(2n)!}{(n+1)!n!}</math>
  
 
BONUS  <math>462</math>
 
BONUS  <math>462</math>

Latest revision as of 00:19, 6 November 2015

1) $17$


2) $3\pi$


3) $P(x)=5x^2-3x+17$


4) $28$


5) $8$


6) $1197$


7) (a) One possible example is a square box with large top and bottom and small height, with the top and bottom colored gold and the four sides blue. If the top and bottom are squares of side $s$, then the total area colored gold is $2 \times s^2$ . If the height of the box is $h$, then the total area colored blue is $4\times h\times s$. It is easy to select s and h to make $4\times h\times s$ equal to half of $2\times s^2$. Set $4\times h\times s=s^2$ or $4\times h=s$. Choose, say, $h=1$ and $s=4$.

(b) Triangulate each face by connecting the point of tangency on the face to each vertex of the face. Look at a gold face and one of the triangles into which it has been cut. The gold triangle meets a blue triangle along the edge where the two faces meet. Consider the two triangles and the radii that go from the tangent points on the two faces (=the apexes of the triangles) to the center of the sphere. By symmetry, the two triangles are congruent. This means that for each triangle colored gold, there is a congruent triangle colored blue. Therefore, the total area colored blue is at least as large as the total area colored gold.


8) $\frac{157}{(18 \times 17 \times 16)} = \frac{157}{4896}$


9) (a) $264$ (b) $5700$


10) (a) $14$ (b) $132$

(c) $\frac{1}{n + 1} \binom{2n}{n}$ $= \binom{2n}{n}- \binom{2n}{n-1}= \binom{2n}{n}- \binom{2n}{n+1}=\frac{1}{2n+1} \binom{2n+1}{n}=\frac{(2n)!}{(n+1)!n!}$

BONUS $462$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_UNCO_Math_Contest_II_Answer_Key&oldid=72778"