## Tuesday, January 1, 2002

Problem 01: 4/5
Problem 02: 8 pounds
3/4 of the gas weighs 18 pounds. Gas = 24 pounds. Thus tank is 8
Problem 03: 30
$\frac{1}{(\frac{1}{5}-\frac{1}{6})} = \frac{1}{\frac{1}{30}} = 30$
Problem 04: 99%
Problem 05: 20°
BAD = 20° arcAC = 140° arcACE = 180°,
thus arcCE=40° and CAE = 20°
BAC = 60° thus DAE = 60° - 20° - 20° = 20°
Problem 06: 30°
Problem 07: -2+4i
Problem 08: 2008
$f(2002)=f(3*\dfrac{2002}{3})=\dfrac{1}{1+\dfrac{2002}{3}}=\dfrac{3}{2005}$
Problem 09: 36
4 years * 3 days * 3 months
Problem 10: 25%
$\dfrac{1}{\dfrac{4}{3}a}+\dfrac{1}{\dfrac{4}{3}b}=\dfrac{3}{4}(\dfrac{1}{a}+\dfrac{1}{b})$
Problem 11: -2 or 17
Cancel x+2, then multiply out both sides and simplify. Remember the x+2 when done.
Problem 12: $\frac{11}{21}$
The number of 3-element subsets with 3 evens: C(4,3) = 4
The number of 3-element subsets with 2 odd, 1 evens: C(5,2)*4 = 40
Problem 13: -1
The terms pattern out as 1, 2, -1, 2
Problem 14: $\frac{25}{6}$
(10-x)(0.4)+x(1.0)=0.65(10)
Problem 15: $\sqrt{6}$
Square the second expression.
Problem 16: $\frac{2}{3}$
Half-angle formula for sin divided by the half-angle formula for cos.
Problem 17: $\sqrt{3}$ or $\frac{1}{9}$
OMG, this was a pain. I'll post if requested.
Problem 18: 44
Problem 19: 12
Problem 20: -1
Problem 21: 26
Problem 22: $\frac{9}{2}$; 4.5
Problem 23: 11
Problem 24: 100°; length = $\dfrac{10\pi}{9}$
Problem 25: 9
Problem 26: 34
Problem 27: $6\sqrt{3}$
Problem 28: $\frac{29}{2}$; 14.5
Problem 29: $\frac{4}{9}$
Problem 30: $(\frac{3}{2}, \frac{5}{2})$
Problem 31: 264
Problem 32: 6π
Problem 33: d = 9/5 miles
Problem 34: n = 1, 2
Problem 35: 105 ways
Problem 36: 1001/334
Problem 37: 7/2
Pretty hefty polynomial expansion. I'll post it if requested.
Update: I'll post the "trick" (which I got from the test creators; my method was hairy):
Make the substitution $x=3^t-9$ and $y=9^t-3$
Then the student should notice that the equation becomes $x^3 + y^3 = (x+y)^3$

This is only possible if the middle terms of the expansion cancel each other.
So, $3x^2y-3xy^2 = 0$, and factoring gives us $3xy(x+y)=0$ and three corresponding results.
$x = 3^t-9 = 0$ which means that t = 2
$y = 9^t-3 = 0$ which means that t = 0.5
and $9^t+3^t-12 = 0$ and t = 1
Final answer is the sum of the three values is $\dfrac{7}{2}$
Problem 38: 22/3
Law of Cosines, solve for cos A: $cosA = \frac{11}{21}$, then plug back in for side AD.
Problem 39: -2
f(2002) = f(1001) + 1 = f(500) = f(250) + 1 = f(125) + 2 = f(62) + 1
= f(31) + 2 = f(15) + 1 = f(7) = f(3) - 1 = f(1) - 2 = -2
Problem 40: 4/3
FH is the altitude.
By similar triangles, AH/FH = 4/1 and HE/FH = AE/1.
Add the equations: AH/FH + HE/FH = 4 + AE; AE/FH = 4 +AE
Swap: FH = AE /(4 + AE)
1/6 = 1/2*AE*(FH) = 1/2 AE*[AE/(4+AE)]
Algebraic manipulation and x = AE: 3x² - x - 4 = (3x - 4)(x + 1)
AE is either 4/3 or -1.
Problem 41: 74
3 squares = 50. Original triangle + other easy triangle = 12.
Angle α in the upper left so the obtuse angle is 180°-α. Sin(α) = 4/5 = sin(180°-α), making the area of the upper obtuse triangle A = ½*3*5*4/5 = 6. Similar logic gives us the area of the lower triangle A = ½*4*5*3/5 = 6.
Total 62 + 6 + 6 = 74.

1. Interesting. I used the sine rule for the second part of problem 38, thinking that it would at least avoid me having to solve a quadratic, but of course triangle ABD is isosceles, so everything cancels very nicely in the second application of the cosine rule. A nice question.

2. In problem 40, where, exactly is H?

3. FH is the altitude of AFE, FH is perpendicular to AE.