Source: NCTM Mathematics Teacher, September 2006

**Solution**

Points , and are the vertices of the right triangle . The length of hypotenuse equals

The slope of equals

Hypotenuse intersects the -axis at point . Using and we express the slope of as

Thus, intersects the -axis at . Observe that the four vertical lines , and intersect the -axis at equal intervals. By the Proportionality theorem, if parallel lines intersect two transversals, then they divide the transversals proportionally. Given that the four vertical lines are parallel and divide the transversal -axis into thirds, they also divide the transversal into thirds.

**Answer**: units

**Alternative solution**

Using points and we express the slope of as

The triangle with vertices at and is a right triangle.

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Source: NCTM Mathematics Teacher, September 2006

**Solution
**It helps to build a deck of cards made out of cutout papers and simulate a few draws in order to understand the mechanics of the game. For example, we discover that having the card numbered in third position does not guarantee a win, unless the card numbered is not in first position.

An indirect approach to solve the problem is to figure out how many ways can we lose the game? If the card numbered is in positions , or from the top, we will definitely lose because there is no way for it to bubble up to the top in draws. The probability for each of these seven events equals . We will also lose if the card numbered is in third position and the card numbered is in first position. The probability for this event equals .

The probability for losing equals

Hence, the probability for winning equals

**Answer**:

**Alternative solution**

A direct approach to solve the problem is to figure out how many ways can we win the game? If the card numbered is in first or second positions, we will win for sure. The probability for each of these two events equals . If the card numbered is in third position and the card numbered is not in first position, we will also win. The probability for this event equals .

The probability of winning equals

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Source: NCTN Mathematics Teacher, September 2006

**Solution**

The above figure shows the ground distance traveled by the rolling wheel of radius feet in revolutions.

feet

The diameter of the wheel equals feet.

**Answer**: feet

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(a)

(b)

(c)

(d)

(c)

Source:NCTM Mathematics Teacher, September 2006

**Solution**

Suppose and

(a)

(b)

(c)

(d)

(e)

(a),(d),(e) have the smallest value

Suppose and

(a)

(b)

(c)

(d)

(e)

(b) and (d) have the smallest value

In all cases (d) has the smallest value.

**Answer**: (d)

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Source: NCTM Mathematics Teacher, September 2006

**Solution**

Let represent the volume of the ball and represent the volume of the cube

Let represent the radius of the ball and represent the length of the side of the cube

and

Multiply both sides by

Divide both sides by

Let represent the surface area of the ball and represent the surface area of the cube

and

Thus, is to as is to or is to .

**Answer**:

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(a)

(b)

(c)

(d)

(e)

Source: NCTM Mathematics Teacher, September 2006

**Solution
**Heron’s Area Formula: area = where and are the lengths of the sides.

(a)

area =

(b)

area =

(c)

area =

(d)

area =

(e)

area =

Triangle (b) has the largest area equal to square units.

**Answer**: (b)

**Alternate solution
**

Since , triangle is a right triangle. All other triangles are either acute or obtuse. Area of triangle is the largest area.

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Source: NCTM Mathematics Teacher, September 2006

**Solution
**

Area of disc =

Area of square =

**Answer**: square centimeters

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Source: NCTM Mathematics Teacher, September 2006

**Solution**

We construct products of positive consecutive integers and see which ones if any can be expressed as a power of .

is the only such product and there is no other product because as the integers grow bigger and bigger, more and more prime numbers are introduced into the products making impossible.

yields the first solution and yields the second solution .

**Answer**:

**Alternative solution 1
**We construct successive positive powers of and see which ones if any can be factored into a pair of factors that are different by .

is the only such power.

**Alternative solution 2
**

Solving the quadratic equation by completing the square for

and

**Alternative solution 3**

The figure shows the graphs of and of . They intersect at two points one of which has integer coordinates leading to .

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Source: NCTM Mathematics Teacher, September 2006

**Solution**

The graph of is a circle centered at the origin and radius = . The graph of is a hyperbola with vertices at and . The two graphs intersect at four points. There are four real solutions that satisfy the two equations.

**Answer**:

**Alternate solution**

Add the two equations

————————

Substitute the value of into the first equation

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Source: NCTM Mathematics Teacher, September 2006

**Solution**

There are two types of liars, the Mon-Tue-Wed liar and the Thu-Fri-Sat liar. We need to determine who is who? Clearly, Kerry is lying when he said that “I lie on Sundays” since both of them tell the truth on Sundays. Therefore, his statement “I lie on Saturdays” is false; he tells the truth on Saturdays and so Kerry is the Mon-Tue-Wed liar and Kelly is the Thu-Fri-Sat liar.

Since we know Kerry is lying, the conversation can only take place on Monday, Tuesday, or Wednesday.

Suppose it was Monday. Since Kelly is telling the truth on Mondays, the statement “I will lie tomorrow” (Tuesday) is true, which characterizes her as a Tue. liar. But that is not possible, because she is a Thu-Fri-Sat liar.

Suppose it was Tuesday. Since Kelly is telling the truth on Tuesdays, the statement “I will lie tomorrow” (Wednesday) is true, which characterizes her as a Wed. liar. But that is not possible, because she is a Thu-Fri-Sat liar.

Suppose it was Wednesday. Since Kelly is telling the truth on Wednesdays, the statement “I will lie tomorrow” (Thursday) is true, which characterizes her as a Thu. liar. There is no contradiction here, she is indeed a Thu-Fri-Sat liar.

**Answer**: Wednesday

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