Monday, January 22, 2007

2007:. 41

Since there are so many truly able mathematicians among my readers, I've decided to give them something to keep busy with! This set will start at the hardest and work backwards.

Difficulty: hard.
Standard instructions for this series: No calculator allowed. Express answers in reduced form. Rationalize denominators. Radicals must be reduced. All numbers are base ten unless otherwise specified. Do not approximate radicals or π. Leave such answers as 1025π or √39, for example. Source: UVM Math Contest
Answer is here.

Sunday, January 21, 2007

2007. 38

Daily, they'll get harder.  How hard? Try this one:

Difficulty: hard. Especially when you aren't using a calculator.
However, this does mean that the problem is solvable that way and stuff will cancel when you realize the method to use. Of course, when you don't see the method, it's frustrating as hell.

Standard instructions for this series: No calculator allowed. Express answers in reduced form. Rationalize denominators. Radicals must be reduced. All numbers are base ten unless otherwise specified. Do not approximate radicals or π. Leave such answers as 1025π or √39, for example. Source: UVM Math Contest

Answer is here.

Saturday, January 20, 2007

2007: 20, 21, and 22.

Problem 20

Difficulty: Easy, but fun. A new way to look at averages and the effect of an addition on the average, depending on n.

Problem 21

Difficulty: Linear programming.Any bets this will somehow work out to an integer?

Problem 22


Difficulty: Easy. Once you've got the details right.

Standard instructions for this series: No calculator allowed. Express answers in reduced form. Rationalize denominators. Radicals must be reduced. All numbers are base ten unless otherwise specified. Do not approximate radicals or π. Leave such answers as 1025π or √39, for example. Source: UVM Math Contest

answer 20, answer 21 and answer 21.

Friday, January 19, 2007

2007: 19 and 23

Problem 19


Difficulty: easy. Notice we didn't specify which color.

Problem 23

Difficulty: Very clever geometry question.

Standard instructions for this series: No calculator allowed. Express answers in reduced form. Rationalize denominators. Radicals must be reduced. All numbers are base ten unless otherwise specified. Do not approximate radicals or π. Leave such answers as 1025π or √39, for example. Source: UVM Math Contest

answer 19 and answer 23.

Thursday, January 18, 2007

2007: 18 and 24

Problem 2*3²

Difficulty: easy-ish. A little trigonometric - algebraic manipulation and "Surprise!"

Problem 2³*3

Difficulty: Easy. At least once in every test, in some form or another, the creators work the year into a problem.  This test was for 2007.

Standard instructions for this series: No calculator allowed. Express answers in reduced form. Rationalize denominators. Radicals must be reduced. All numbers are base ten unless otherwise specified. Do not approximate radicals or π. Leave such answers as 1025π or √39, for example. Source: UVM Math Contest

answer 18 and answer 24.

Wednesday, January 17, 2007

2007: 17 and 25.

Day Seventeen


Difficulty: easy-schmeasy, if you subtract. More complicated if you try to add.

Problem 25 = 5².
(I wonder how often you can switch the digits and still write an equivalent expression.)



Difficulty: Medium. More damned logarithms, but there's a neat little prime factorization going on, too.

Standard instructions for this series: No calculator allowed. Express answers in reduced form. Rationalize denominators. Radicals must be reduced. All numbers are base ten unless otherwise specified. Do not approximate radicals or π. Leave such answers as 1025π or √39, for example. Source: UVM Math Contest

answer 17 and answer 25.

Tuesday, January 16, 2007

2007: 16 and 26

Sweet Sixteen

Difficulty: easy-ish. Another SAT-type reasoning question. Fun to watch the students work out the ramifications.

Problem 26

Difficulty: Medium. Is it wrong of me to get really annoyed at this question?  It's not terribly difficult, but it is so ... artificial.

Standard instructions for this series: No calculator allowed. Express answers in reduced form. Rationalize denominators. Radicals must be reduced. All numbers are base ten unless otherwise specified. Do not approximate radicals or π. Leave such answers as 1025π or √39, for example. Source: UVM Math Contest

answer 16 and answer 26.

Monday, January 15, 2007

2007: 15 and 27

Day Fifteen ... We're almost there.


Difficulty: pretty easy. Kids will get wrapped up in the P(A) and if-then, but it's an example of a problem that is much easier if you imagine the grid of all possibilities. Definitely a candidate for the SAT.

Problem 27

Difficulty: Medium. It's funny how easily the kids get thrown by an infinite series.

Standard instructions for this series: No calculator allowed. Express answers in reduced form. Rationalize denominators. Radicals must be reduced. All numbers are base ten unless otherwise specified. Do not approximate radicals or π. Leave such answers as 1025π or √39, for example. Source: UVM Math Contest

answer 15 and answer 27.

Sunday, January 14, 2007

2007: 14 and 28

Day Fourteen
Difficulty: easy-ish. It's similar to a lot of SAT questions in that you don't actually solve for a, b or c but use them all in concert to find some other value.

Problem 28

Difficulty: Medium. Advanced Logarithmic manipulation.

Standard instructions for this series: No calculator allowed. Express answers in reduced form. Rationalize denominators. Radicals must be reduced. All numbers are base ten unless otherwise specified. Do not approximate radicals or π. Leave such answers as 1025π or √39, for example. Source: UVM Math Contest

answer 14 and answer 28.

Saturday, January 13, 2007

2007: 13 and 29

Day Thirteen
Difficulty: easy-ish. Mostly, it's an interpretation question.

I love this next problem. So very cool.
Problem 29
i.e., find the shaded area.
Difficulty: Medium.

Standard instructions for this series: No calculator allowed. Express answers in reduced form. Rationalize denominators. Radicals must be reduced. All numbers are base ten unless otherwise specified. Do not approximate radicals or π. Leave such answers as 1025π or √39, for example. Source: UVM Math Contest

answer 13 and answer 29.

Friday, January 12, 2007

2007, 12 and 30

Day Twelve

Difficulty: easy-ish. Does your student remember his trig identities?

Problem 30

Difficulty: Medium.

Standard instructions for this series: No calculator allowed. Express answers in reduced form. Rationalize denominators. Radicals must be reduced. All numbers are base ten unless otherwise specified. Do not approximate radicals or π. Leave such answers as 1025π or √39, for example. Source: UVM Math Contest

answer 12 and answer 30.

Thursday, January 11, 2007

2007: 11 and 31

First, an easy one.
And then something more complicated, though personally, I'm not a fan of probability questions.


Standard instructions for this series: No calculator allowed. Express answers in reduced form. Rationalize denominators. Radicals must be reduced. All numbers are base ten unless otherwise specified. Do not approximate radicals or π. Leave such answers as 1025π or √39, for example. Source: UVM Math Contest

answer 11 and answer 31.

Wednesday, January 10, 2007

2007. 10 and 32

Day Ten
Difficulty: easy-ish.

Problem 32:
Difficulty: Tricky. It seems easy, but then tries to kneecap you.

Standard instructions for this series: No calculator allowed. Express answers in reduced form. Rationalize denominators. Radicals must be reduced. All numbers are base ten unless otherwise specified. Do not approximate radicals or π. Leave such answers as 1025π or √39, for example. Source: UVM Math Contest

answer 10 and answer 32.

Tuesday, January 9, 2007

2007: 9 and 33

Day Nine.
Difficulty: easy-ish.

Number 33: 

Difficulty: Hard.
(or as the kids said, OMFG)
It's actually more complicated than hard.

Standard instructions for this series: No calculator allowed. Express answers in reduced form. Rationalize denominators. Radicals must be reduced. All numbers are base ten unless otherwise specified. Do not approximate radicals or π. Leave such answers as 1025π or √39, for example. Source: UVM Math Contest

answer 9 and answer 33.

Monday, January 8, 2007

2007. 8 and 34

Day Eight will make you think a second ... or ten.
Difficulty: easy.

Difficulty: Hard-ish
(if they don't know the 3D Pythagorean theorem and have to work it out.)

Standard instructions for this series: No calculator allowed. Express answers in reduced form. Rationalize denominators. Radicals must be reduced. All numbers are base ten unless otherwise specified. Do not approximate radicals or π. Leave such answers as 1025π or √39, for example. Source: UVM Math Contest

answer 8 and answer 34.

Sunday, January 7, 2007

2007. 7 and 35

Problem 7
Difficulty: so easy the kids kept second-guessing themselves. ;-)

Problem 35
Difficulty: Hard (last page)

Standard instructions for this series: No calculator allowed. Express answers in reduced form. Rationalize denominators. Radicals must be reduced. All numbers are base ten unless otherwise specified. Do not approximate radicals or π. Leave such answers as 1025π or √39, for example. Source: UVM Math Contest

answer 7 and answer 35.

Saturday, January 6, 2007

2007: 6 and 36

Day Six will make you think a bit.

Difficulty: easy - ish.

Difficulty: Hard (last page)

Standard instructions for this series: No calculator allowed. Express answers in reduced form. Rationalize denominators. Radicals must be reduced. All numbers are base ten unless otherwise specified. Do not approximate radicals or π. Leave such answers as 1025π or √39, for example. Source: UVM Math Contest

answer 6 and answer 36.

Friday, January 5, 2007

2007. 5 and 37

Day Five, Number 5
Difficulty: easy.
and Problem 37 because 38 was already posted and I didn't want to lose any of the comments.

Difficulty: Hard. (Last page)

Standard instructions for this series: No calculator allowed. Express answers in reduced form. Rationalize denominators. Radicals must be reduced. All numbers are base ten unless otherwise specified. Do not approximate radicals or π. Leave such answers as 1025π or √39, for example. Source: UVM Math Contest

answer 5 and answer 37.

Thursday, January 4, 2007

2007. 4 and 39

Problem (4)
Difficulty: first page.

Problem 39
Note: Remainder.
Difficulty: Hard. (last page)

Standard instructions for this series: No calculator allowed. Express answers in reduced form. Rationalize denominators. Radicals must be reduced. All numbers are base ten unless otherwise specified. Do not approximate radicals or π. Leave such answers as 1025π or √39, for example. Source: UVM Math Contest
Yeah, I know, it should read "larger" and "smaller" but I didn't have the energy to fix it.


answer 4 and answer 39.

Wednesday, January 3, 2007

2007. 3 and 40

Day Three
Difficulty: easy.

I'm going to save my self some time and just include the difficult one with the easy one every day.
 Please enjoy this mathematical tidbit.

Difficulty: Hard.
Standard instructions for this series: No calculator allowed. Express answers in reduced form. Rationalize denominators. Radicals must be reduced. All numbers are base ten unless otherwise specified. Do not approximate radicals or π. Leave such answers as 1025π or √39, for example. Source: UVM Math Contest

answer 3 and answer 40.

Tuesday, January 2, 2007

2007. 2

Day two.
Difficulty: easy.
Standard instructions for this series: No calculator allowed. Express answers in reduced form. Rationalize denominators. Radicals must be reduced. All numbers are base ten unless otherwise specified. Do not approximate radicals or π. Leave such answers as 1025π or √39, for example. Source: UVM Math Contest
Answer is here.

Monday, January 1, 2007

2007: 1

Day one. Question 1
Difficulty: easy.
Standard instructions for this series: No calculator allowed. Express answers in reduced form. Rationalize denominators. Radicals must be reduced. All numbers are base ten unless otherwise specified. Do not approximate radicals or π. Leave such answers as 1025π or √39, for example. Source: UVM Math Contest
Answer is here.

Answers to 2007

Problem 01: -14/29 just plug, crank, and pull.
Problem 02: 5/3
Conjugates are the red herring here.
(4√2 + √2)÷(4√2 - √2) = 5√2 ÷ 3√2 =5/3
Problem 03: 2
Draw a vertical line down the middle and the students will quickly see that the two halves are 2x2 and the red portion is 1+1 = 2
Problem 04: 112
Choose three consecutive integers: x-1, x, x+1
(x+1)³ - x³ = 666 + x³ - (x-1)³
x³ + 3x² + 3x +1 - x³ = 666 + x³ - x³ + 3x² - 3x + 1
Stuff cancels all over the place: 6x = 666; x = 111; Answer is 112
Problem 05: 1
Problem 06: 17
(x+2)² = x² + 8²
x² + 4x + 4 = 64 + x²
4x = 60; x=15; hypotenuse = 17. Even if you missed the Pythagorean triple.
Problem 07: 15
Problem 08: 36
Problem 09: 117
Problem 10: 59
Problem 11: 2
9x + 2(3x+2) = 243
32x + 18(3x) = 243
complete the square for U = 3x
u² + 18u + 81 = 243 + 81 = 324 = 18²
u + 9 = +- 18
3x = 9, -27 .... 2 is only real answer.
Problem 12: 13/14
Problem 13: 930
Problem 14: 140√2
ab = 28; ac = 20; bc=70
a²b²c² = 28*20*70
abc = √(4*7*4*5*7*2*5) = 140√2
Problem 15: 6/31
In these simple dice rolling questions, I like to imagine the full sample space of 36 possibilities. We are told that 6 is not an option and that occurs 5 times, leaving 31 total.
11 12 13 14 15 16
21 22 23 24 25 26
31 32 33 34 35 36
41 42 43 44 45 46
51 52 53 54 55 56
61 62 63 64 65 66
Thus, P(7) = 6/31
Problem 16: 48
Problem 17: 62
Build a rectangle around the figure and subtract the four triangles.
10*10 - ½(4*7 + 3*7 + 3*3 + 6*3) = 62
Problem 18: 1/5
Problem 19: 6/55
Problem 20: 16
Problem 21: 10
Problem 22: 1/13
Problem 23: 8
Problem 24: 4013
Problem 25: 6
log144(3a * 2b) = c
144c = 3a * 2b
24c * 32c = 2b * 3a
b=4c and a=2c; answer 6
Problem 26: a=3, b=2/13
Problem 27: 1/6
Problem 28: 49
Problem 29: 5π/8
Problem 30: 300
Problem 31: 64/175
Problem 32: 18√3
ABC is a 30-60-90, with B=90o. AC is 12, split 3,3, and 6. If the altitude is BN, then AN=3 and BN=3√3, making the area = 18√3
Problem 33: 3
Problem 34: 339
4a+4b+4c=124, so a+b+c=31. Square it
a²+ab+ac+ba+b²+bc+ca+cb+c²=961
2ab + 2ac + 2bc = 622
a² + b² + c² = 339, which is also d²
Problem 35: 250
Assume first that p is odd. The list of integers is symmetric:
x-2, x-1, x, x+1, x+2; 5x = 57
In general, then px = 57, making p a power of 5.
p=5, x=56. 5 terms
p=25, x=55. 25 terms
p=125, x=54 = 625 terms.
p=625, x=53. Impossible because x was the "middle" number and we can't have 312 on either side.

Assume p is even.
x and x+1 are the middle two terms.
x-3, x-2, x-1, x, x+1, X+2, x+3, x+4 : 8 terms => 8x + 4
In general, p terms = px + .5p = p(x+1/2).
Since x is integer, we need a 2: q(2x+1) = 57.
5(2x+1)=57: x=56, means ten terms.
25(2x+1)= 57: x=1562, 50 terms
125(2x+1)=57: x=310, 250 terms
625(2x+1)=57: x=62, impossible for x to be the middle.
Problem 36: 46
logab + 10logba = 7
logab + 10/logab = 7
U + 10/U = 7
U² + 10 = 7U
U² - 7U + 10 = 0 ... (U-5)(U-2) = 0 .... logab = 2, 5
a5 = b has three candidates in range: 2,32; 3,243; 4;1024
since 45² = 2025, a² = b has 43 candidates in range. 2² = 4 through 44² = 1836
46 candidates.
Problem 37: 19
Problem 38: a=9 and b=11
Two equations:
386a = 272b
146a=102b
Starting with the second: 1a² + 4a + 6 = 1b² + 0b + 2
a² + 4a + 4 = b²
(a+2)² = b² gives us a+2 = b
Substitute into equation 1
3a² + 8a + 6 = 2b² + 7b + 2
3a² + 8a + 6 = 2(a+2)² + 7(a+2) + 2
expand
3a² + 8a + 6 = 2a² + 8a + 8 + 7a + 14 + 2
3a² + 8a + 6 = 2a² + 15a + 24
a2 - 7a - 18 = 0
(a+2)(a-9) gives us a = -2, 9
a = -2 is impossible so we're left with a=9 and thus b=11
back to number 38.
Problem 39: 224x³
Simplify by canceling an x³ and then by substituting u = x³.  Now for some long division. Yay!

      u666 +2u663 +3u660 + u657 + ...
-----------------------
u3-1) u669 + u666 + u663 + u660 + ...
      u669 - u666
      -----------
            2u666 + u663
            2u666 -2u663
            ------------
                  3u663 + u660etc.

So each column adds one and projecting down the line gets you to a remainder of 224x³
Hang on. I messed that up. Must revisit.

      u668 + u667 + u666 + u657 + ...
.   -----------------------
u-1) u669 +  0    + 0 +   u666 + 0 + 0 + u663 + u660 + ...
      u669 - u668
      -----------
            u668 + 0
            u668 - u667
            ------------
                  u667 + u666
                  u667 - u666
                 ------------
                       2u666 + 0
 
When you are finished, you will have 224/u-1 but we can't leave it there. We removed a x³ from numerator and denominator so the last bit is really 224x³ / (x6-x3) so the remainder is really 224x³.

Good thing I wasn't taking the test for real.
Problem 40: 3/2
d² = x² +(2x)² +(5-x)² + (2x)² + (5-2x)² + (5-x)² + (5-2x)² + x²
Expand all that and cancel/clean up
sum of d² = 20x² - 30x + 50
min when d(sum)=0
40x - 60 = 0
x = 3/2
Kids get in trouble here because they want to differentiate the square root or get off the idea that the sum of the distances must be at a minimum.
Problem 41: 150
Cleaning up Jonathan's comment a bit:
The semiperimeter, S = (5+6+7)/2 = 9.
Heron's:
A = √(9*4*3*2) = √216 = 6√6.
To get Altitude (height):
Area = ½*5*h = 6√6 so h = 12/5√6
Use CA = 6 and h to find that CPh = 6/5.
Using h and Pythagoras:
q1 = (1/5)² + h²
q2 = (4/5)² + h²
q3 = (9/5)² + h²
q4 = (13/5)² + h²
Sum:
4(12/5*√6)² + (1+16+81+196)/25.
(4*144*6 + 1+16+81+196)/25 = 3750/25
= 150

Eric P said:
First find cos(C) by law of cosines
Cos(C)=[6²-7²-5²]/(-2*7*5)=19/35
q1² = 49 +1 -2(7)(1)(19/35)
q2² = 49 +4 -2(7)(2)(19/35)
q3² = 49 +9 -2(7)(3)(19/35)
q4² = 49 +16-2(7)(4)(19/35)
Lining it up this way and using some quick mental math tricks makes the addition simpler: SUM=4(49)+30-2(7)(10)(19/35)=150

I think I have to side with Eric on this one.