## Wednesday, December 17, 2014

### 2012 Mathcounts School Round Harder Questions

Check out Mathcounts here, the best competition math program for middle school students.

2012 Mathcounts Sprint: #29: The difference of the squares of two distinct positive numbers is equal to twice the square of their difference. What is the ratio of the smaller number to the larger? Express your answer as a common fraction?
Solution:
x2 - y2 = 2(x -y)2 $$\rightarrow$$ (x - y)(x + y) = 2 (x -y)(x - y)
Divide both sides by (x - y) and you have x + y = 2 x - 2y
Move x to the right and y to the left and change their sign. x = 3y
$$\dfrac {y} {x}=\dfrac {1} {3}$$

2012 Mathcounts Sprint #30:The area of a particular regular hexagon is $$x^{3}$$ square units, where x is the measure of the distance from the center of the hexagon to the midpoint of a side. What is the side length of the hexagon?
Solution I:
#30:  Use 30-60-90 degree angle ratio and you can get the side length of the regular hexagon being$$\dfrac {2x} {\sqrt {3}}$$.
The area of the hexagon is $$\dfrac {2x} {\sqrt {3}}\times x$$ (the height) $$\times \dfrac {1} {2}\times 6$$ = x3
so $$x =\dfrac {6} {\sqrt {3}}$$or $$2\sqrt {3}$$
Side length is $$\dfrac {2x} {\sqrt {3}}$$
Plug in and you get x = 4

Solution II:
Let the side length be s and use the formula for finding the area of an equilateral triangle and you have:
$$\dfrac {s*x } {2}\times 6=3x*s=x^{3}$$ [given]$$\rightarrow$$ s = $$\dfrac {x^{2}} {3}$$
Use the formula for area of an equilateral triangle times 6 or $$\rightarrow$$$$\dfrac {3\sqrt {3}} {2}s^{2}$$
$$\left( \dfrac {x ^{2}} {3}\right) ^{2}\times \dfrac{3\sqrt{3}}{2}=x^{3}$$ $$\rightarrow$$$$x =\dfrac {6} {\sqrt {3}}$$
Plug into the s = $$\dfrac {x^{2}} {3}$$and you get s = 4

2012 Mathcounts Target: #8: A 4 inch × 4 inch square is surrounded by a border consisting of all points in the plane of the square that are within 1.5 inches of the square and not in the square. In square inches, what is the area of the border? Express your answer as a decimal to the nearest tenth.
Solution:

There would be four rectangular strips with area that is 1.5 times 4 x 4 = 24 inch2 .
The corner combined will be a whole circle with a radius of 1.5 inch, so its area is (1.5)2 times π.

You can practice more problems here

2012 School team round harder problems
If you are aiming higher, you should use school team round as target round questions for more practices.

Hints or detailed solutions :
#1: dimensional change question -- As long as two polygons are similar, all the 1-D similar length ratio will stay constant so
the bases, the heights, the diagonals, the perimeter will still be 5 : 4 ratio.
You should be able to answer this in seconds if you truly understand the concept.

#4: another dimensional change question : 1 - 1. 5 x $$\left( 0.8\right) ^{2}$$ = 0.04 = 4%

#8 : triangular number , but don't overlook the part that 3 lines, three intersections so it's 1 + 2 = 3, etc... so
6 lines, it's 1 + 2 + 3 + 4 + 5 = 15 intersections.

#9 : You only need to concern about line MQ, the other information is irrelevant.

More practices on radicals form Regents prep

## Monday, December 15, 2014

### 2013 AMC 10 A and B : Solutions to the Harder Problems

From AoPS videos :

2013 AMC-10 B #23  It's a good idea to be able to get all those split-side lengths, height to the hypotenuse fast and right. Those are basic skills.

Also, you should be able to get the area, in-radius and others fast as well of 13-14-15 and 10-17-21
triangles.

2013 AMC 10 A Math Jam from AoPS

2013 AMC 10 B Math Jam from AoPS