Expected Value Question : 2024 Mathcounts State Sprint #24 level 1.5

Problem. Dennis rolls three fair six-sided dice, obtaining a, b, c ∈ {1,…,6}. Find \[ \mathbb{E}\!\bigl[\,|a-b|+|b-c|+|c-a|\,\bigr]. \]

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Try the question first before scrolling down to read the solution. 

Solution.

Step 1 — Linearity of expectation.

\[ \mathbb{E}\!\bigl[\,|a-b|+|b-c|+|c-a|\,\bigr] \;=\; \mathbb{E}[|a-b|]+\mathbb{E}[|b-c|]+\mathbb{E}[|c-a|] \;=\; 3\,\mathbb{E}[|a-b|]. \]

Step 2 — Expected absolute difference of two dice.

Let \(X = |a-b|\). Its distribution is

\[ \begin{array}{c|cccccc} d & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \Pr(X=d) & \tfrac{6}{36} & \tfrac{10}{36} & \tfrac{8}{36} & \tfrac{6}{36} & \tfrac{4}{36} & \tfrac{2}{36} \end{array} \] \[ \mathbb{E}[|a-b|] =\frac{1}{36}\bigl(0\cdot6 + 1\cdot10 + 2\cdot8 + 3\cdot6 + 4\cdot4 + 5\cdot2\bigr) =\frac{70}{36} =\frac{35}{18}. \]

Step 3 — Final answer.

\[ \mathbb{E}\!\bigl[\,|a-b|+|b-c|+|c-a|\,\bigr] = 3 \times \frac{35}{18} = \boxed{\tfrac{35}{6}}. \]

Dimensional Change questions I:

Questions written by Willie, a volunteer.  Answer key and detailed solutions below.

1a. There is a regular cylinder, which has a height equal to its radius. If the radius and height are both increased by 50%, by what % does the total volume of the cylinder increase?

1b. If the radius and height are both decreased by 10%, by what % does the total volume of the cylinder decrease?

1c. If the radius is increased by 20% and the height is decreased by 40%, what % of the volume of the original cylinder does the volume of the new cylinder represent?

1d. If the radius is increased by 40% and the height is decreased by 20%, what % of the volume of the original cylinder does the volume of the new cylinder represent?

1e. If the height is increased by 125%, what % does the radius need to be decreased by for the volume to remain the same?

2. If the side of a cube is increased by 50%, by what % does the total surface area of the cube increase?

3a. If the volume of a cube increases by 72.8%, by what % does the total surface area of the cube increase?

3b. By what % did the side length of the cube increase?

4. You have a collection of cylinders, all having a radius of 5. The first cylinder has a height of 2, the second has a height of 4, the third a height of 6, etc. The last cylinder has a height of 50. What is the sum of the volumes of all the cylinders (express your answer in terms of pi)?

Answer key: (Each question should not take you more than 30 seconds to solve if you really understand the concepts involved.)

 

 

1a.  The volume of a cylinder is πr²x h (height). The radius itself will be squared and the height stays at constant ratio. The volume will increased thus (1.5)³ - 1³ -- the original 100% of the volume = 2.375

=237.5%

1b.  Like the previous question: 1³ - 0.9³ [when it's discount/percentage decrease, you use the 100% or 1 - the discount/decrease percentage] = 0.271 =  27.1% decrease

1c.  1.2² [100% + 20% increase = 1.2] x 0.6 [100% -40% = 0.6] = 0.864  or  

86.4% of the original volume

1d.  1.4² [100% + 40% increase = 1.4] x 0.8 [100% -20% = 0.8] = 1.568 = 156.8% of the original volume

1e.  When the height of a cylinder is increased 125%, the total volume is is 225% of the original cylinder, or 9/4.

Since the radius is used two times (or squared), it has to decrease 4/9^1/2 = 2/3 for the new cylinder to have the same volume as the old one. [9/4 times 4/9 = 1 or the original volume.]

1 - (2/3) = 1/3 = 0.3 = 33.3%

2. Surface area is 2-D so 1.5² - 1 = 1.25 = 125% increase

3a. If a volume of a cube is increased by 72.8 percent, it's 172.8% or 1.728 of the original volume. Now you are going from 3-D (volume) to 2-D (surface area). 1.728^2/3 = 1.44 or 44% increase. [Don't forget to minus 1 (the original volume) since it is asking you the percentage increase.]

3b. From surface area, you can get the side increase by using 1.44^1/2 = 1.2, so 20% increase.

Or you can also use 1.728^1/3 = 1.2;  1.2 - 1 = 20%

4. The volume of a cylinder is πr²x h . (2 + 4 + 6 + ...50) x 5²π = (25 x 26) x 25π =16250π

Dimentional Change Questions III: Similar Shapes

There are numerous similar triangle questions on Mathcounts.

Here are the basics:

If two triangles are similar, their corresponding angles are congruent and their corresponding sides will have the same ratio or proportion.

Δ ABC and ΔDEF are similar. \(\frac{AB}{DE}\) = \(\frac{AC}{DF}\) = \(\frac{BC}{EF}\)= their height ratio = their perimeter ratio.







Once you know the linear ratio, you can just square the linear ratio to get the area ratio and cube the linear ratio to get the volume ratio. 

Practice Similarity of Triangles here.  Read the notes as well as work on the practice problems.  There is instant feedback online. 

Other practice sheets on Similar Triangles                                                        

Many students have trouble solving this problem when the two similar triangles are superimposed. 

Just make sure you are comparing smaller triangular base with larger triangular base and smaller triangular side with corresponding larger triangular side, etc... In this case:

\(\frac{BC}{DE}\)= \(\frac{AB}{AD}\) = \(\frac{AC}{AE}\)




Questions to ponder (Solutions below)

#1: Find the area ratio of Δ ABC to trapezoid BCDE to DEGF to FGIH. You can easily get those ratios using similar triangle properties. All the points are equally spaced and line \(\overline{BC}\)// \(\overline{DE}\) // \(\overline{FG}\) // \(\overline{HI}\). 

#2: Find the volume of the cone ABC to Frustum BCDE to DEGF to FGIH. Again, you can use the similar cone, dimensional change property to easily get those ratios.Same conditions as the previous question.

 

Answer key: 

#1:

 #2:

3 - D Geometry and harder AMCs, AIME links

For Mathcounts advanced group -- harder  Mathcounts State/National problem on 3-D geometry.                      

From Mathcounts Mini :

Using Similarity to Solve Geometry Problems

Area and Volume 

Relationship Between A Box's Dimensions, Volume and Surface Area

AMC, AIME videos from AoPS

Harder/Hardest AMC-10, 12 and AIME problems explained by Mrs. Rusczyk