2021 AMC 10A Spring 考试题目

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1.

What is the value of (222)(323)+(424)?(2^2-2)-(3^2-3)+(4^2-4)?

11

22

55

88

1212

Answer: D
Solution:

(222)(323)+(424)=26+12=8. \begin{align*} (2^2 - 2) - &(3^2 - 3) + (4^2 - 4) \\ &= 2 - 6 + 12 \\ &= 8. \end{align*}

Thus, D is the correct answer.

2.

Portia's high school has 33 times as many students as Lara's high school. The two high schools have a total of 26002600 students. How many students does Portia's high school have?

600600

650650

19501950

20002000

20502050

Answer: C
Solution:

Let xx be the number of students in Lara's high school. Then Portia's high school has 3x3x students.

Therefore, 3x+x=2600x=650. \begin{align*} 3x + x &= 2600 \\ x &= 650. \end{align*} Then 3x=1950.3x = 1950.

Thus, C is the correct answer.

3.

The sum of two natural numbers is 17,402.17,402. One of the two numbers is divisible by 10.10. If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?

10,27210,272

11,70011,700

13,36213,362

14,23814,238

15,42615,426

Answer: D
Video solution:
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Written solution:

Let xx and yy be the two numbers. WLOG, let xx be divisible by 10.10. Then the units digit of xx is 0.0.

If we erase the units digit, then we are essentially dividing xx by 10.10. The problem statement also gives us that x10=y.\dfrac{x}{10} = y.

Therefore, x10+x=17,40211x10=17,402x=15,820. \begin{align*} \dfrac{x}{10} + x &= 17,402 \\ \dfrac{11x}{10} &= 17,402 \\ x &= 15,820. \end{align*} Then xx10=14,238. x - \dfrac{x}{10} = 14,238.

Thus, D is the correct answer.

4.

A cart rolls down a hill, travelling 55 inches the first second and accelerating so that during each successive 11-second time interval, it travels 77 inches more than during the previous 11-second interval. The cart takes 3030 seconds to reach the bottom of the hill. How far, in inches, does it travel?

215215

360360

29922992

31953195

32423242

Answer: D
Video solution:
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Written solution:

The distance travelled every second forms an arithmetic sequence: 5,5+7,5+27, 5, 5 + 7, 5 + 2 \cdot 7, \ldots

The sum of an arithmetic sequence is \text{# of terms} \cdot \dfrac{\text{first + last}}{2}. We know the number of terms is 3030 and the first term is 5.5. The last term is 5+297=208.5 + 29 \cdot 7 = 208.

Plugging these values into the expression yields 305+2082=15213=3195. 30 \cdot \dfrac{5 + 208}{2} = 15 \cdot 213 = 3195.

Thus, D is the correct answer.

5.

The quiz scores of a class with k>12k > 12 students have a mean of 8.8. The mean of a collection of 1212 of these quiz scores is 14.14. What is the mean of the remaining quiz scores in terms of k?k?

148k12\dfrac{14-8}{k-12}

8k168k12\dfrac{8k-168}{k-12}

14128k\dfrac{14}{12} - \dfrac{8}{k}

14(k12)k2\dfrac{14(k-12)}{k^2}

14(k12)8k\dfrac{14(k-12)}{8k}

Answer: B
Video solution:
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Written solution:

The sum of the scores of everyone in the class is 8k.8k. The sum of the scores in the collection of 1212 is 1214=168.12 \cdot 14 = 168.

This means that the sum of the scores of everyone not in the collection is 8k168.8k - 168. There are also k12k - 12 people not in the collection. Therefore, the average is 8k168k12. \dfrac{8k - 168}{k - 12}.

Thus, B is the correct answer.

6.

Chantal and Jean start hiking from a trailhead toward a fire tower. Jean is wearing a heavy backpack and walks slower. Chantal starts walking at 44 miles per hour. Halfway to the tower, the trail becomes really steep, and Chantal slows down to 22 miles per hour. After reaching the tower, she immediately turns around and descends the steep part of the trail at 33 miles per hour. She meets Jean at the halfway point. What was Jean's average speed, in miles per hour, until they meet?

1213\dfrac{12}{13}

11

1312\dfrac{13}{12}

2413\dfrac{24}{13}

22

Answer: A
Video solution:
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Written solution:

Let 2d2d be the distance from the fire tower, where d>0.d > 0.

Then Chantal hiked for d4+d2+d3=13d12 \dfrac{d}{4} + \dfrac{d}{2} + \dfrac{d}{3} = \dfrac{13d}{12} hours.

If Jean travelled dd miles in 13d12\dfrac{13d}{12} hours, then his speed was d÷13d12=1213 d \div \dfrac{13d}{12} = \dfrac{12}{13} miles per hour.

Thus, A is the correct answer.

7.

Tom has a collection of 1313 snakes, 44 of which are purple and 55 of which are happy. He observes that

• all of his happy snakes can add,

• none of his purple snakes can subtract, and

• all of his snakes that can't subtract also can't add.

Which of these conclusions can be drawn about Tom's snakes?

Purple snakes can add.

Purple snakes are happy.

Snakes that can add are purple.

Happy snakes are not purple.

Happy snakes can't subtract.

Answer: D
Video solution:
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Written solution:

Note that the third condition ensures that purple snakes can't add.

We also know that all happy snakes can add, which means that happy snakes can't be purple as well.

Thus, D is the correct answer.

8.

When a student multiplied the number 6666 by the repeating decimal, 1.a b a b=1.a b,\underline{1}.\underline{a} \ \underline{b} \ \underline{a} \ \underline{b}\ldots=\underline{1}.\overline{\underline{a} \ \underline{b}}, where aa and bb are digits, he did not notice the notation and just multiplied 6666 times 1.a b.\underline{1}.\underline{a} \ \underline{b}. Later he found that his answer is 0.50.5 less than the correct answer. What is the 22-digit number a b?\underline{a} \ \underline{b}?

1515

3030

4545

6060

7575

Answer: E
Video solution:
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Written solution:

We can express 1.a b\underline{1}.\overline{\underline{a} \ \underline{b}} as an infinite geometric sum: 1.a b=1+.a b+.00 a b+ \underline{1}.\overline{\underline{a} \ \underline{b}} = 1 + .\underline{a} \ \underline{b} + .00 \ \underline{a} \ \underline{b} + \cdots We can therefore use the formula for the sum of a geometric sum: S=first term1ratio=.a b11100 S = \dfrac{\text{first term}}{1 - \text{ratio}} = \dfrac{.\underline{a} \ \underline{b}}{1 - \dfrac{1}{100}} =10099(.a b)=a b99. = \dfrac{100}{99} \left(. \underline{a}\ \underline{b}\right) = \dfrac{\underline{a}\ \underline{b}}{99}. We also know that 1.a b=1+a b100 1. \underline{a}\ \underline{b} = 1 + \dfrac{\underline{a}\ \underline{b}}{100}

Then 66(1+a b100)+.5=66(1+a b99)66100a b+.5=6699a b1150a b=.5a b=75. \begin{align*} 66\left(1 + \dfrac{\underline{a}\ \underline{b}}{100}\right) + .5 &= 66\left(1 + \dfrac{\underline{a}\ \underline{b}}{99}\right) \\ \dfrac{66}{100}\underline{a}\ \underline{b} + .5 &= \dfrac{66}{99}\underline{a}\ \underline{b} \\ \dfrac{1}{150}\underline{a}\ \underline{b} &= .5 \\ \underline{a}\ \underline{b} &= 75. \end{align*}

Thus, E is the correct answer.

9.

What is the least possible value of (xy1)2+(x+y)2(xy-1)^2+(x+y)^2 for real numbers xx and y?y?

00

14\dfrac{1}{4}

12\dfrac{1}{2}

11

22

Answer: D
Video solution:
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Written solution:

Expanding, we get x2y22xy+1+x2+2xy+y2=x2y2+x2+y2+1. \begin{gather*} x^2y^2 - 2xy + 1 + x^2 + 2xy + y^2 \\ = x^2y^2 + x^2 + y^2 + 1. \end{gather*} Note that every square must be non-negative. Therefore, the minimum value is when all the terms except 11 are 0,0, making the sum 1.1.

This is attainable when x=y=0.x = y = 0.

Thus, D is the correct answer.

10.

Which of the following is equivalent to (2+3)(22+32)(24+34)(2+3)(2^2+3^2)(2^4+3^4)(28+38)(216+316)(2^8+3^8)(2^{16}+3^{16})(232+332)(264+364)?(2^{32}+3^{32})(2^{64}+3^{64})?

3127+21273^{127} + 2^{127}

3127+2127+23633^{127} + 2^{127} + 2 \cdot 3^{63}+3263 + 3 \cdot 2^{63}

312821283^{128} - 2^{128}

3128+21283^{128} + 2^{128}

51275^{127}

Answer: C
Video solution:
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Written solution:

Notice that if we multiply by 32=1,3 - 2 = 1, we end up with a bunch of difference of squares.

(32)(2+3)=3222(3222)(22+32)=3424 \begin{gather*} (3 - 2)(2 + 3) = 3^2 - 2^2 \\ (3^2 - 2^2)(2^2 + 3^2) = 3^4 - 2^4\\ \vdots \end{gather*} This ends up giving us a final value of 31282128.3^{128} - 2^{128}.

Thus, C is the correct answer.

11.

For which of the following integers bb is the base-bb number 2021b221b2021_b - 221_b not divisible by 3?3?

33

44

66

77

88

Answer: E
Video solution:
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Written solution:

We can express this expression in base 1010 using the definition of bases: 2b3+2b+12b22b1 2b^3 + 2b + 1 - 2b^2 - 2b - 1 =2b32b2=2b2(b1).= 2b^3 - 2b^2 = 2b^2(b - 1).

For this to be divisible by 3,3, either bb or b1b - 1 must be divisible by 3.3.

The only answer choice that satisfies neither of these conditions is 8.8.

Thus, E is the correct answer.

12.

Two right circular cones with vertices facing down as shown in the figure below contains the same amount of liquid. The radii of the tops of the liquid surfaces are 33 cm and 66 cm. Into each cone is dropped a spherical marble of radius 11 cm, which sinks to the bottom and is completely submerged without spilling any liquid. What is the ratio of the rise of the liquid level in the narrow cone to the rise of the liquid level in the wide cone?

1:11:1

47:4347:43

2:12:1

40:1340:13

4:14:1

Answer: E
Video solution:
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Written solution:

Let the heights of the narrow wide cones be h1h_1 and h2h_2 respectively.

Then 1332πh1=1362πh2, \dfrac{1}{3} \cdot 3^2 \pi h_1 = \dfrac{1}{3} \cdot 6^2 \pi h_2, which gives us h1h2=4. \dfrac{h_1}{h_2} = 4. Also note that 3h1\dfrac{3}{h_1} and 6h2\dfrac{6}{h_2} must remain constant due to similar triangles.

After the marble is added to each cone, let 3x3x be the radius of the narrow cone and 6y6y be that of the wide cone.

Then the height of the narrow cone is h1xh_1x and that of the wide cone is h2yh_2y due to similar triangles.

Equating the final volumes, we get 13(3x)2π(h1x)=13(6y)2π(h2y), \dfrac{1}{3} (3x)^2 \pi (h_1x) = \dfrac{1}{3} (6y)^2 \pi (h_2y), which gives us h1x3=4h2y3. h_1x^3 = 4h_2y^3. We know that h1h2=4,\dfrac{h_1}{h_2} = 4, so this equation gives us that x=y.x = y.

The desired ratio is h1xh1h2yh2=h1(x1)h2(y1)=4. \dfrac{h_1x - h_1}{h_2y - h_2} = \dfrac{h_1(x - 1)}{h_2(y - 1)} = 4.

Thus, E is the correct answer.

13.

What is the volume of tetrahedron ABCDABCD with edge lengths AB=2,AB = 2, AC=3,AC = 3, AD=4,AD = 4, BC=13,BC = \sqrt{13}, BD=25,BD = 2\sqrt{5}, and CD=5?CD = 5?

33

232\sqrt{3}

44

333\sqrt{3}

66

Answer: C
Video solution:
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Written solution:

Analyzing the side lengths, we can realize that ACD and ABC \triangle ACD \text{ and } \triangle ABC are both right triangles.

Therefore, we can treat ACD\triangle ACD as the base of the tetrahedron and AB\overline{AB} as the altitude.

With this setup, we can use the formula for the volume of a pyramid, yielding 133422=4. \dfrac{1}{3} \cdot \dfrac{3 \cdot 4}{2} \cdot 2 = 4.

Thus, C is the correct answer.

14.

All the roots of the polynomial z610z5+Az4+Bz3z^6-10z^5+Az^4+Bz^3+Cz2+Dz+16+Cz^2+Dz+16 are positive integers, possibly repeated. What is the value of B?B?

88-88

80-80

64-64

41-41

40-40

Answer: A
Video solution:
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Written solution:

By Vieta's formulas, we get that the product of the roots is 1616 and that their sum is 10.10.

Given that all the roots are positive integers, we can see that the roots are 1,1,2,2,2,2. 1, 1, 2, 2, 2, 2.

The function is therefore just (z1)2(z2)4=(z22z+1) (z - 1)^2(z - 2)^4 = (z^2 - 2z + 1) (z48z3+24z232z+16). (z^4 - 8z^3 + 24z^2 - 32z + 16).

Calculating just the z3z^3 term, we get 32z348z38z3=88z3. -32z^3 - 48z^3 - 8z^3 = -88z^3.

Thus, A is the correct answer.

15.

Values for A,B,C,A,B,C, and DD are to be selected from {1,2,3,4,5,6}\{1, 2, 3, 4, 5, 6\} without replacement (i.e. no two letters have the same value). How many ways are there to make such choices so that the two curves y=Ax2+By=Ax^2+B and y=Cx2+Dy=Cx^2+D intersect?

(The order in which the curves are listed does not matter; for example, the choices A=3,A=3,B=2,B=2,C=4,C=4, D=1D=1 is considered the same as the choices A=4,A=4,B=1, B=1, C=3,C=3, D=2.D=2.)

3030

6060

9090

180180

360360

Answer: C
Video solution:
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Written solution:

Setting the equations equal to each other, we get Ax2+B=Cx2+D Ax^2 + B = Cx^2 + D x2(AC)=DB x^2(A - C) = D - B x2=DBAC0 x^2 = \dfrac{D - B}{A - C} \geq 0 since squares are non-negative.

This means DBD - B and ACA - C must both have the same sign.

If we choose two distinct values for (A,C)(A, C) and (B,D),(B, D), there are 22 ways to arrange them such that the numerator and denominator both have the same sign.

We have to divide by 2,2, however, since the two curves are not considered distinct.

Therefore, the total number of tuples is 12(62)(42)2=90. \dfrac{1}{2} \binom{6}{2} \binom{4}{2} \cdot 2 = 90.

Thus, C is the correct answer.

16.

In the following list of numbers, the integer nn appears nn times in the list for 1n200.1 \leq n \leq 200.1,2,2,3,3,3,4,4,4,4,1, 2, 2, 3, 3, 3, 4, 4, 4, 4,,200,200,,200 \ldots, 200, 200, \ldots , 200What is the median of the numbers in this list?

100.5100.5

134134

142142

150.5150.5

167167

Answer: C
Solution:

The total number of numbers in the list is 1+2++200 1 + 2 + \cdots + 200=2002012=20100. = \dfrac{200 \cdot 201}{2} = 20100.

We want to find the median kk such that k(k+1)2\dfrac{k(k + 1)}{2} is near 201002.\dfrac{20100}{2}.

Multiplying by 2,2, we want k(k+1)k(k + 1) near 20100.20100. Note that 20100142.\sqrt{20100} \approx 142.

Plugging in k=142k = 142 yields 12142143=10153. \dfrac{1}{2} \cdot 142 \cdot 143 = 10153.

10153142<10050,10153 - 142 < 10050, which shows that 142142 is our desired median (142142 is the 1004910049th and 1005010050th number).

Thus, C is the correct answer.

17.

Trapezoid ABCDABCD has ABCD,BC=CD=43,\overline{AB}\parallel\overline{CD},BC=CD=43, and ADBD.\overline{AD}\perp\overline{BD}. Let OO be the intersection of the diagonals AC\overline{AC} and BD,\overline{BD}, and let PP be the midpoint of BD.\overline{BD}.

Given that OP=11,OP=11, the length of ADAD can be written in the form mn,m\sqrt{n}, where mm and nn are positive integers and nn is not divisible by the square of any prime. What is m+n?m+n?

6565

132132

157157

194194

215215

Answer: D
Video solution:
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Written solution:

First, we can show that BPCBDA.\triangle BPC \sim \triangle BDA.

Let DBC=α.\angle DBC = \alpha. Then DCB=1802α\angle DCB = 180 - 2\alpha since DCB\triangle DCB is isosceles.

Since ABCD,\overline{AB} \parallel \overline{CD}, we get that ABD=α.\angle ABD = \alpha. Then since BPC\triangle BPC and BDA\triangle BDA are right triangles, they are similar.

Using this fact, we get 2=BDBP=ABAC=AB43. 2 = \dfrac{BD}{BP} = \dfrac{AB}{AC} = \dfrac{AB}{43}. From this, we get that AB=86.AB = 86.

We also get that ABOCDO\triangle ABO \sim \triangle CDO from parallel sides and vertical angles.

Therefore 2=ABCD=BOOD=BP+11BP11. 2 = \dfrac{AB}{CD} = \dfrac{BO}{OD} = \dfrac{BP + 11}{BP - 11}. Solving, we get BP=33BP = 33 and BD=66.BD = 66.

Using the Pythagorean Theorem on ADB,\triangle ADB, we get that AD=862662=4190. AD = \sqrt{86^2 - 66^2} = 4\sqrt{190}.

Thus, D is the correct answer.

18.

Let ff be a function defined on the set of positive rational numbers with the property that f(ab)=f(a)+f(b)f(a\cdot b)=f(a)+f(b) for all positive rational numbers aa and b.b. Suppose that ff also has the property that f(p)=pf(p)=p for every prime number p.p. For which of the following numbers xx is f(x)<0?f(x) < 0?

1732\dfrac{17}{32}

1116\dfrac{11}{16}

79\dfrac{7}{9}

76\dfrac{7}{6}

2511\dfrac{25}{11}

Answer: E
Video solution:
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Written solution:

Note that for any number of the form pep^e where pp is prime, f(pe)=ef(p)=ep. f(p^e) = ef(p) = ep. This can be seen by applying the function property multiple times.

Also note that f(a)=f(abb)=f(ab)+f(b), \begin{align*}f(a) &= f\left(\dfrac{a}{b} \cdot b\right) \\&= f\left(\dfrac{a}{b}\right) + f(b),\end{align*} which gives us f(ab)=f(a)f(b). f\left(\dfrac{a}{b}\right) = f(a) - f(b).

We can know calculate each answer choice one-by-one.

f(1732)=f(17)f(32) f\left(\dfrac{17}{32}\right) = f(17) - f(32) =1752=7. = 17 - 5 \cdot 2 = 7.

f(1116)=f(11)f(16)=1142=3 \begin{gather*} f\left(\dfrac{11}{16}\right) = f(11) - f(16) \\ = 11 - 4 \cdot 2 = 3 \end{gather*}

f(79)=f(7)f(9)=723=1 \begin{gather*} f\left(\dfrac{7}{9}\right) = f(7) - f(9) \\ = 7 - 2 \cdot 3 = 1 \end{gather*}

f(76)=f(7)f(6)=7f(2)f(3)=2 \begin{gather*} f\left(\dfrac{7}{6}\right) = f(7) - f(6) \\ = 7 - f(2) - f(3) = 2 \end{gather*}

f(2511)=f(25)f(11)=2511=1 \begin{gather*} f\left(\dfrac{25}{11}\right) = f(25) - f(11) \\ = 2 \cdot 5 - 11 = -1 \end{gather*}

Thus, E is the correct answer.

19.

The area of the region bounded by the graph of x2+y2=3xy+3x+yx^2+y^2 = 3|x-y| + 3|x+y| is m+nπ,m+n\pi, where mm and nn are integers. What is m+n?m + n?

1818

2727

3636

4545

5454

Answer: E
Solution:

We can case on the signs of xyx - y and x+y.x + y.

Case 1:1:xy=xy, |x - y| = x - y,x+y=x+y|x + y| = x + y

Substituting and simplifying, we get x26x+y2=0(x3)2+y2=32. \begin{gather*} x^2 - 6x + y^2 = 0 \\ (x - 3)^2 + y^2 = 3^2. \end{gather*} This is a circle with radius 33 centered at (3,0).(3, 0).

Case 2:2:xy=yx,|x - y| = y - x,x+y=x+y|x + y| = x + y

As above, we get x2+y26y=0x2+(y3)2=32. \begin{gather*} x^2 + y^2 - 6y = 0 \\ x^2 + (y - 3)^2 = 3^2. \end{gather*} This is a circle with radius 33 centered at (0,3).(0, 3).

Case 3:3: xy=xy,|x - y| = x - y,x+y=xy|x + y| = -x - y

Again, we get x2+y2+6y=0x2+(y+3)2=32. \begin{gather*} x^2 + y^2 + 6y = 0 \\ x^2 + (y + 3)^2 = 3^2. \end{gather*} This is a circle with radius 33 centered at (0,3).(0, -3).

Case 4:4: yx=xy,|y - x| = x - y,x+y=xy|x + y| = -x - y

Finally, we get x2+6x+y2=0(x+3)2+y2=32. \begin{gather*} x^2 + 6x + y^2 = 0 \\ (x + 3)^2 + y^2 = 3^2. \end{gather*} This is a circle with radius 33 centered at (3,0).(-3, 0).

These circles form the following graph:

The 44 semicircles form 22 full two circle with radius 33 for a total area of 18π.18\pi.

The middle square also contributes 62π=36π6^2 \pi = 36 \pi to the total area. This means that the area of the region is 54π.54 \pi.

Thus, E is the correct answer.

20.

In how many ways can the sequence 1,2,3,4,51,2,3,4,5 be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?

1010

1818

2424

3232

4444

Answer: D
Video solution:
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Written solution:

Note that all the sequences are symmetric about 3.3. x1,x2,x3,x4,x5 x_1, x_2, x_3, x_4, x_5 is a valid sequence if and only if 6x1,6x2,6x3, 6 - x_1, 6 - x_2, 6 - x_3,6x4,6x5 6 - x_4, 6 - x_5 is a valid sequence.

Therefore, we can just count all the sequences that begin with 1,1, 2,2, 31,31, and 32.32.

1,3,2,5,41, 3, 2, 5, 4

1,4,2,5,31, 4, 2, 5, 3

1,4,3,5,21, 4, 3, 5, 2

1,5,2,4,31, 5, 2, 4, 3

1,5,3,4,21, 5, 3, 4, 2

2,1,4,3,52, 1, 4, 3, 5

2,1,5,3,42, 1, 5, 3, 4

2,3,1,5,42, 3, 1, 5, 4

2,4,1,5,32, 4, 1, 5, 3

2,4,3,5,12, 4, 3, 5, 1

2,5,1,4,32, 5, 1, 4, 3

2,5,3,4,12, 5, 3, 4, 1

3,1,4,2,53, 1, 4, 2, 5

3,1,5,2,43, 1, 5, 2, 4

3,2,4,1,53, 2, 4, 1, 5

3,2,5,1,43, 2, 5, 1, 4

This shows that there are 1616 valid sequences starting with the above numbers. Doubling this yields the total number of sequences, 32.32.

Thus, D is the correct answer.

21.

Let ABCDEFABCDEF be an equiangular hexagon. The lines AB,CD,AB, CD, and EFEF determine a triangle with area 1923,192\sqrt{3}, and the lines BC,DE,BC, DE, and FAFA determine a triangle with area 3243.324\sqrt{3}. The perimeter of hexagon ABCDEFABCDEF can be expressed as m+np,m +n\sqrt{p}, where m,n,m, n, and pp are positive integers and pp is not divisible by the square of any prime. What is m+n+p?m + n + p?

4747

5252

5555

5858

6363

Answer: C
Solution:

Let P,P, Q,Q, R,R, X,X, Y,Y, and ZZ be the points at ABundefinedCDundefined,\overleftrightarrow{AB}\cap\overleftrightarrow{CD},CDundefinedEFundefined,\overleftrightarrow{CD}\cap\overleftrightarrow{EF},EFundefinedABundefined,\overleftrightarrow{EF}\cap\overleftrightarrow{AB},BCundefinedDEundefined,\overleftrightarrow{BC}\cap\overleftrightarrow{DE},DEundefinedFAundefined,\overleftrightarrow{DE}\cap\overleftrightarrow{FA}, and FAundefinedBCundefined,\overleftrightarrow{FA}\cap\overleftrightarrow{BC}, respectively.

Since ABCDEFABCDEF is equiangular, all of its interior angles are 720÷6=120.720^{\circ} \div 6 = 120^{\circ}.

This means that the exterior angles are all 60,60^{\circ}, making all the small triangles equilateral triangles. This also shows that PQR\triangle PQR and XYZ\triangle XYZ are also equilateral.

We know from the problem statement that [PQR]=34PQ2=1923, [PQR] = \dfrac{3}{4} \cdot PQ^2 = 192\sqrt{3}, [XYZ]=34YZ2=3243. [XYZ] = \dfrac{3}{4} \cdot YZ^2 = 324\sqrt{3}. Solving gives us that PQ=163PQ = 16\sqrt{3} and YZ=36.YZ = 36.

Finally, the perimeter of ABCDEFABCDEF is AB+BC+CD+DE+EF AB + BC + CD + DE + EF +FA=AZ+PC+CD+DQ + FA = AZ + PC + CD + DQ +YF+FA=(YF+FA+AZ) + YF + FA = (YF + FA + AZ) +(PC+CD+DQ)=YZ + (PC + CD + DQ) = YZ +PQ=36+163. + PQ = 36 + 16\sqrt{3}.

With this, we have (36+16+3=55).(36 + 16 + 3 = 55).

Thus, C is the correct answer.

22.

Hiram's algebra notes are 5050 pages long and are printed on 2525 sheets of paper; the first sheet contains pages 11 and 2,2, the second sheet contains pages 33 and 4,4, and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the notes. When Hiram comes back, he discovers that his roommate has taken a consecutive set of sheets from the notes and that the average (mean) of the page numbers on all remaining sheets is exactly 19.19. How many sheets were borrowed?

1010

1313

1515

1717

2020

Answer: B
Video solution:
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Written solution:

Let the sheets the roommate took be aa through b.b. This is the same as taking pages 2a12a - 1 through 2b.2b.

The sum of the pages taken is (2a1+2b)(2b2a+2)2. \dfrac{(2a - 1 + 2b)(2b - 2a + 2)}{2}. The sum of the pages remaining is 19(50(2b2a+2)). 19(50 - (2b - 2a + 2)). The sum of these equals the sum of all the pages, 50512.\dfrac{50 \cdot 51}{2}.

Therefore (2a1+2b)(2b2a+2)2+ \dfrac{(2a - 1 + 2b)(2b - 2a + 2)}{2} +19(50(2b2a+2))=50512.19(50 - (2b - 2a + 2)) = \dfrac{50 \cdot 51}{2}.

Then (2a+2b39)(ba+1) (2a + 2b - 39)(b - a + 1) =50132=2513. = \dfrac{50 \cdot 13}{2} = 25 \cdot 13.

We can set 2a+2b39=25 2a + 2b - 39 = 25 and ba+1=13 b - a + 1 = 13 to get a+b=32 a + b = 32 and ba=12. b - a = 12. This yields b=22 and a=12. b = 22 \text{ and } a = 12.

The desired answer is 2210+1=13.22 - 10 + 1 = 13.

Thus, B is the correct answer.

23.

Frieda the frog begins a sequence of hops on a 3×33 \times 3 grid of squares, moving one square on each hop and choosing at random the direction of each hop-up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she "wraps around" and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops "up", the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square.

Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?

916\dfrac{9}{16}

58\dfrac{5}{8}

34\dfrac{3}{4}

2532\dfrac{25}{32}

1316\dfrac{13}{16}

Answer: D
Solution:

Let MM denote the center, EE an edge, and CC a corner.

The only ways Frieda can reach a corner in 44 or less moves are EC,EEC,EEEC, and EMEC. EC, EEC, EEEC, \text{ and } EMEC.

We have to find the probability of each of these cases happening.

Case 1:EC1 : EC

On the first hop, Frieda necessarily moves to an E.E. From there, Frieda has a 12\dfrac{1}{2} chance of going to a C.C. This gives us a probability of 112=12 1 \cdot \dfrac{1}{2} = \dfrac{1}{2} for this case.

Case 2:EEC2 : EEC

Again, Frieda moves to an EE on her first move. On the second move, however, she has a 14\dfrac{1}{4} chance of wrapping around to another E.E. Then there is a 12\dfrac{1}{2} chance of going to a CC from the E.E. This is a probability of 11412=18. 1 \cdot \dfrac{1}{4} \cdot \dfrac{1}{2} = \dfrac{1}{8}.

Case 3:EEEC3 : EEEC

Using the same logic as above, this case yields a probability of 1141412=132. 1 \cdot \dfrac{1}{4} \cdot \dfrac{1}{4} \cdot \dfrac{1}{2} = \dfrac{1}{32}.

Case $ : EMEC

Finally, this case has a probability of 114112=18. 1 \cdot \dfrac{1}{4} \cdot 1 \cdot \dfrac{1}{2} = \dfrac{1}{8.}

Adding the probabilities together yields a total probability of 12+18+132+18=2532. \dfrac{1}{2} + \dfrac{1}{8} + \dfrac{1}{32} + \dfrac{1}{8} = \dfrac{25}{32}.

Thus, D is the correct answer.

24.

The interior of a quadrilateral is bounded by the graphs of (x+ay)2=4a2(x+ay)^2 = 4a^2 and (axy)2=a2,(ax-y)^2 = a^2, where aa is a positive real number. What is the area of this region in terms of a,a, valid for all a>0?a > 0?

8a2(a+1)2\dfrac{8a^2}{(a+1)^2}

4aa+1\dfrac{4a}{a+1}

8aa+1\dfrac{8a}{a+1}

8a2a2+1\dfrac{8a^2}{a^2+1}

8aa2+1\dfrac{8a}{a^2+1}

Answer: D
Solution:

Note that each of the equations yields two parallel lines.

(x+ay)2=4a2 (x + ay)^2 = 4a^2 results in the two lines x+ay2a=0 x + ay - 2a = 0 and x+ay+2a=0. x + ay + 2a = 0. Both of these lines have a slope of 1a.-\dfrac{1}{a}.

Similarly, (axy)2=a2 (ax-y)^2 = a^2 results in the lines axya=0 ax - y - a = 0 and axy+a=0. ax - y + a = 0. These lines have slope a.a.

Note that each pair of lines is perpendicular to the other pair of lines. This shows that the equations form a rectangle.

Recall that the formula for the distance dd between two parallel lines {Ax+By+C1=0Ax+By+C2=0 \begin{cases} Ax+By+C_1=0 \\ Ax+By+C_2=0 \end{cases} is d=C2C1A2+B2. d = \dfrac{\mid C_2 - C_1 \mid}{\sqrt{A^2 + B^2}}.

Using this formula, we get that the distance between the first pair of lines is 4aa2+1. \dfrac{4a}{\sqrt{a^2 + 1}}. Similarly, the distance between the second pair of lines is 2aa2+1. \dfrac{2a}{\sqrt{a^2 + 1}}.

These are the side lengths of the rectangle. Multiplying yields the area 8a2a2+1. \dfrac{8a^2}{a^2 + 1}.

Thus, D is the correct answer.

25.

How many ways are there to place 33 indistinguishable red chips, 33 indistinguishable blue chips, and 33 indistinguishable green chips in the squares of a 3×33 \times 3 grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally?

1212

1818

2424

3030

3636

Answer: E
Solution:

Let the colors be A,B,A, B, and C.C. Note that we can assign the 33 colors to them in 3!=63! = 6 ways, so we have to multiply by 66 at the end.

Let AA be in the center of the grid.

????A???? \begin{array}{ccc} ? & ? & ? \\ ? & A & ? \\ ? & ? & ? \end{array}

The other AAs have to either be along the diagonal or on the same side.

??A?A?A?? \begin{array}{ccc} ? & ? & A \\ ? & A & ? \\ A & ? & ? \end{array}

A?A?A???? \begin{array}{ccc} A & ? & A \\ ? & A & ? \\ ? & ? & ? \end{array}

The first scenario can happen in 22 ways since there are 22 diagonals. The second has 44 ways since there are 44 sides.

Either way, the positions of the BBs and CCs is fixed.

CBABACACB \begin{array}{ccc} C & B & A \\ B & A & C \\ A & C & B \end{array}

ABACACBCB \begin{array}{ccc} A & B & A \\ C & A & C \\ B & C & B \end{array}

This is a total of 4+2=64 + 2 = 6 ways to arrange the A,B,A, B, and CCs. This gives us a total of 66=366 \cdot 6 = 36 configurations.

Thus, E is the correct answer.