12 Favorite Problems

Carol Love

EMAT 6600

Link here for GSP file of these problems.

1. Spot’s doghouse has a regular hexagonal base that measures on yard on each side. Spot is tethered to a vertex with a two-yard rope. What is the area, in square yards, of the region outside the doghouse that Spot can reach? (100 Favorite Calendar Problems #22)

The area of the reachable area is two-thirds of the two-yard circle plus two one-sixth sectors of the one-yard circles.

_{} square yards

2. Carl’s Cone: Make a pattern for a sawed off cone that is 12 inches tall, has a base diameter of 24 inches, and a top diameter of 8 inches.

First, *x* is the missing additional height of the original triangle. To
find *x*, use a proportion with similar
triangles.

_{}

Now, *b* is the inner radius of the pattern. To find *b*, use the Pythagorean Theorem.

_{}

Next, *l* = *a* + *b* is the outer radius of the patter. To
find *l*, use a proportion with similar
triangles.

_{}

Now, construct the concentric
circles with radii *b* and *l* (_{} and _{}).

The length of the arc along the
larger circle must be equal to the circumference of the bottom of the cone, or
24π. The circumference of the larger circle is _{}. The central angle, *θ*,
intersecting the arc must be in the same proportion to 2π as the length of
the arc is to the circumference of the larger circle.

_{}

Therefore, the pattern is as
follows, with an inner radius of _{}, an outer radius of _{}, and a intercepted arc of 199.7º.

3. Oil Tank Problem: A cylindrical oil tank with a radius of 18” and a depth of 48” is filled such that there are 10” of oil in the tank. What volume of oil is in the tank?

The volume of the entire tank is _{} in^{3}. To
find the volume of the remaining oil, construct a cross-section of the tank.

To find *θ*, the central angle intersecting the arc where the oil
contacts the tank, we need the cosine ratio.

_{}

The oil fills the area of the
sector that is 2*θ* less the area
of the isosceles triangle. The area of the 2*θ*
sector is

_{}

The area of the isosceles triangle
depends on *x*.

_{}

So, the area of the isosceles triangle is

_{}

Next, the area of the oil is _{} square inches.
Multiplying by the depth, the volume of oil remaining is _{} cubic inches. Finally,
since there are 0.004329004 cubic inches in a gallon, there are _{}gallons of oil remaining.

4. Each
of the small circles in the figure has radius 1. The innermost circle is
tangent to the six circles that surround it, and each of those circles is
tangent to the large circle and to its small-circle neighbors. Find the area of
the shaded region, *yellow* below. (100
Favorite Calendar Problems #23)

The area of each small circle is _{}.

Because the diameters of three
small circles form the diameter of the large circle, the diameter of the large
circle is 6, making the radius of the large circle 3. The area of the large
circle is, therefore, _{}. The yellow area is the area of the large circle less the
combined area of the seven smaller circles, or _{}.

5. Pick
a point *P* on the inside of a unit
square. Connect *P* with the vertices
of the base of the square. What is the probability that the triangle so
completed will be an acute triangle?

The triangle is a right triangle
when *P* is on the 3 non-base sides of
the square and when *P* is on the
circle. Only angle *P* can be obtuse,
and it is such when *P* is within the
semicircular region. The probability of the triangle being acute can be found
using an area model and finding _{}. The area outside the semicircle is found by subtracting the
area of the semicircular region from the area of the square. The area of the
semicircular region I is _{}, so the area outside the semicircle is _{}. Therefore the probability of an acute triangle is _{}.

6. A circle’s diameter is divided in the ratio of 3 to 4, and semicircles are drawn on each segment, as shown. What is the ratio of the upper area to the lower area? (100 Favorite Calendar Problems #32)

The upper area is one-half the
large circle (diameter 7*x*) plus
one-half of a circle of diameter 4*x*
less one-half of a circle with diameter 3*x*.

_{}

The lower area is one-half the
large circle plus one-half of a circle with diameter 3*x* less one-half of a circle of diameter 4*x*.

_{}

Therefore, the ratio of the upper area to the lower area is

_{}

7. Magic Square: Arrange the numbers 1 through 9 in a 3 by 3 array – a Magic Square – such that the sum of any row, column, or the two diagonals is the same.

8 |
1 |
6 |

3 |
5 |
7 |

4 |
9 |
2 |

This result is unique in that the only other solutions involve rotations or reflections of this solution. The five must be used in the greatest number of sums, and must therefore be placed in the center square. The 4-5-6 combination must be used on the diagonal and determines the sum of the rows, columns, and diagonals. The 1 and 9 are the upper and lower limits of the numbers, and must be used in a row or column, and together. The higher number should sum with the 4 and the lower with the 6.

Other 3 by 3 magic squares can be found as linear combinations of this original magic square. For instance, a magic square with 21 at the center is a 4(original square) + 1, and yields

33 |
5 |
25 |

13 |
21 |
29 |

17 |
37 |
9 |

using the numbers 5, 9, 13, 17,
21, 25, 29, 33, 37, which are of the form 4*x*
+ 1, where *x* is the original numbers
1 – 9.

A different magic square with 21
at the center can be formed using *x* +
16, and the numbers 17 – 25, as shown.

24 |
17 |
22 |

19 |
21 |
23 |

20 |
25 |
18 |

Here is a 4 by 4 magic square using 1 – 16.

1 |
2 |
15 |
16 |

13 |
14 |
3 |
4 |

12 |
7 |
10 |
5 |

8 |
11 |
6 |
9 |

8. The figure is composed of twenty-five unit squares. What is the area of the unshaded region?

The unshaded region is composed
of 4 triangles. The area of the top triangle is _{}, the area of the right-side triangle is_{}, the area of the bottom triangle is _{}, and the area of the left-side triangle is _{}. The area of the unshaded region is the sum of these four
areas, or 4 + 4 + 6 + 6 = 20 square units.

9. Let
*C* be a cube where the length in
inches of its long diagonal is the same as its volume in cubic inches. What is
the length in inches or each side? (100 Favorite Calendar Problems #16)

The length of the longest diagonal
is _{}. The volume of the cube is *s*^{3}. So,

_{}

10. A strip of uniform width is cut from three sides of a square region, leaving a rectangular region with are that is one-half that of the square region. Determine the quotient of the square length divided by the strip’s width. (100 Favorite Calendar Problems #20)

The area of the square is s^{2},
so the area of the rectangular region is _{}. The dimensions of the rectangular region are *s* – *w*
and *s* – 2*w*. Therefore the area of the rectangular region is _{}, and

_{}

So, either _{}, which leads to

_{}

or _{}, which leads to

_{}

Since the strip is cut from the
square, the width of the strip must be shorter than the side of the square,
leading to the choice of _{}.

11. If the perimeter of an isosceles triangle is 36 cm, and the altitude to its base is 12 cm, what is the triangle’s area? (100 Favorite Calendar Problems #25)

Checking to ensure the perimeter is correct,

2*x* + (18 – *x*) + (18 – *x*) = 2*x* + 36 – 2*x* = 36.

Now, *x* can be found using the Pythagorean Theorem.

_{}

So, the area of the triangle is _{} cm^{2}.

12. A square is
inscribed in a semicircle as shown. Find the ratio of *s* to *a*. (100 Favorite
Calendar Problems #90)

The radius, *r*, of the semicircle is found using the Pythagorean Theorem.

_{}

And *a* = *r* – *s*, so

_{}

Finally, _{}