Minnesotata State University, Mankato

The 46th Annual High School Mathematics Contest Problems and

Solutions

Contents

1. Answer key 1
2. Problems and Solutions 1
1. Answer key
Problem 1 2 3 4 5 6 7 8 9 10
number
Answer b e e d c
a c d d c
Problem 11 1 13 14 115 17 18 1 20
number 2 6 9
Answer c a b b d d a a c b
Problem 21 2 23 24 25 2 27 28 2 30
number 2 6 9
Answer d c a c b e e b c b
2. Problems and Solutions

1. If x + y = 7 and xy = 6 then is

.
Answer-b) Observe that

2. If then is

c) 2017 d) 2020 e) 4037.

Answer-e) Observe that

which implies

3. A drawer contains 10 socks and 5 of them are red. When three socks are drawn at random,
what is the probability that all three of them are red?

.
Answer-e) The probability is
2

.
4. If x and y are positive numbers that satisfy

( 2 − y2 = 24 x
x2 + y2 = 26
then x + y is
a) 3 b) 4 c) 5 d) 6 e) none of these.
Answer-d) Observe that

.
5. The larger circle of radius 9 centered at C is tangent to the smaller circle of radius 4 centered
at D. If the circles are tangent to the line l at A and B, then what is the area of the triangle
∆ABC?

C
•
•D

• • l
A B

a) 36 b) 47 c) 54 d) 67 e) none of these.
Answer-c) Appying the Pythagorean theorem to the triangle ∆CD0D

C
•
0• •D
D

• • l
A B
we find that

p
AB = DD0 = (9 + 4)2 − (9 − 4)2 = 12.

The area of the triangle is

6. If θ is a solution of the equation

cos(50◦) + cos(70◦) = cosθ◦
then cos(9θ◦) is

a) 0 e) 1.
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Answer-a) Using the addition formulas

(
cos70 = cos(60 + 10) = cos60cos10 − sin60sin10 cos50 =
cos(60 − 10) = cos60cos10 + sin60sin10 we find that

cos(50) + cos(70) = 2cos60cos10 = cos10 =⇒ θ = ±10 mod 360.

The answer is cos(9θ) = cos(±90) = 0.

7. There are four boxes full of $ 10, $ 20 $ 50, $ 100 bills

$ 100 bills $ 50 bills $ 20 bills $ 10 bills

You can choose your boxes and take 2,3,4, and 5 bills from the chosen boxes
respectively. What is the largest amount of money you can get? (For example, if you
take 2 bills from the $100-box, 3 bills from the $ 50-box, 4 bills from the $ 20- box
and 5 bills from §10-box, you would get $480.)
a) $480 b) $560 c) $780 d) $850 e) $920.
Answer-c) By using the greedy algorithm, we find the largest sum is
5 · 100 + 4 · 50 + 3 · 20 + 2 · 10 = 780.
8. If x and y are positive numbers satisfying
log3(x2y2) = 2log3 y + 6
then x is
a) 1 b) 3 c) 9 d) 27 e) cannot be determined
Answer-d) [ANT:E: MMC 2017-15] Observe that log 3(x2y2) = 2log2 y +
2 =⇒ 2log3 x = 6 =⇒ x = 27.

9. Suppose that 4ABC is a triangle, that M is the midpoint of AC, and the

segments AM, MC, MB and AB all have length 1. Find the area of 4ABC.

e) none of these.
Answer-d) There are a variety of ways to solve this, but they all revolve around the
fact that the condition AM = MC = BM forces the triangle to have a right angle at
∠ABC. This is a theorem that is sometimes seen in high school geometry courses. It
can also be derived using inequality arguments. Here it is also possible to notice
that 4AMB is equilateral and 4BMC is isosceles, and then calculate the angle values
based on the angle sum for a triangle. Once you have established that

√4ABC is right, the Pythagorean theorem can√ be used show that BC has length
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3, and so the area of the triangle must be 3 · 1 · (1/2).

10. If sinx = 2cosx, then what is (sinx)(cosx) ?

Answer-c) Observe that

.
with same parity on signs. sin .

11. If the sum of two of the roots of x3 − ax2 + bx − c = 0 is zero, then ab − c is

a) −2 b) −1 c) 0 d) 1 e) 2.
Answer-c) Let α,β,γ be the roots of x3 −ax2 +bx+c = 0 and assume α+β = 0. Then
α+β+γ=a γ=a
 
αβ + βγ + γα = b =⇒ αβ = b =⇒ ab = c.
 
αβγ = c αβγ = c
The answer is ( c ).

12. Suppose that the 25 squares below are filled with the integers 1 through 25 such a way that
the sums of integers in each row and column are the same. What is the common sum?

a) 65 b) 72 c) 83 d) 96 e) none of these.
Answer-a) Let x1,x2,..,x25 be integers in the magic square from left to right and from top to
bottom. Let s be the common sum. Then
5s = (x1+x2+x3+x4+x5)+···+(x21+x22+x23+x24+x25) = 1+2+···+25 = 25·13 and so the sum is s =
5 · 13 = 65.

13. The remainder when 630 + 830 is divisible by 49 is

a) 0 b) 2 c) 11 d) 12 e) 13.
Answer-b) Using the binomial theorem to

and
 5

we find that
630 + 830 = (a multiple of 49) + 2.
The answer is 2.

14. If a polynomial p(x) of degree ≥ 1 satisfies

p(p(x)) = 5p(x3)
then the degree of p(x) must be
a) 2 b) 3 c) 5 d) 7 e) none of these.
Answer-b) Let n be the degree of p(x). Then
n2 = 3n =⇒ n = 3.

15. What is the remainder when 20 + 21 + 22 + ··· + 299 is divided by 9.

a) 0 b) 1 c) 3 d) 6 e) none of these.
Answer-d) Note that 20 + 21 + 22 + ··· + 299 = 2100 − 1 and 26 ≡ 1 mod 9. The sequence
2n − 1 mod 9 is periodic with period 6 and 100 = 6 · 16 + 4 , 2 100 − 1 ≡ 24 − 1 = 6 mod
9. Elementary pattern search would work as well.

16. What is the number of positive integer solutions (x,y) of the equation
x2 + y2 = 3x + 3y + 4 − 2xy?
a) 0 b) 1 c) 2 d) 3 e) none of these.
Answer-a) Observe that x2+y2 = 3x+3y+4−2xy =⇒ (x+y)2−3(x+y)−4 = 0 =⇒ (x+y+1)
(x+y−4) = 0

There are 3 positive integer solutions (x,y) for x + y = 4.

17. Consider the triangles in the following diagram:

•D

•C

A• •B
6

Suppose that 4ABC is isosceles with AB = AC and 4ABD is isosceles with

AD = BD. Suppose that AD = 9 and BC = 4. Then AB is
√ √

a) 6 b) 62 c) 8 d) 82 e) 9.
Answer-a) Using the isosceles triangle theorem, the angles ∠ABC, ∠ACB and
∠DAB all must be congruent. Therefore we have the similarity 4ABC ∼ 4DAB. This
gives us the equation:

Since AB must be positive, this equation has the unique solution that AB = 6.

18. Find the area of the shaded region formed by a large quarter circle of radius 10 and two
smaller semicircles as shown below.

20

a) 100π−200 b) 200π−100 c) 200π+100 d) 150π+100 e) none of these.

Answer-a) Thinking of

20 20

we find that the area is

19. In how many ways can you walk to a stairway with 7 steps if you can take one or
two steps. (For example, you can walk up a stairway with 3 steps in three different
ways: i) three 1 steps ii) 1 step and then 2 steps and iii) 2 steps and then
1 step.)
a) 15 b) 18 c) 21 d) 27 e) 29.
Answer-c) Let fn be the number of ways to walk to a stairway with n steps. Then fn
satisfies
 7

fn = fn−1 + fn−2
that is, fn is a Fibonacci sequence. Since f1 = 1 and f2 = 2, we find that
f3 = 3 =⇒ f4 = 5 =⇒ f5 = 8 =⇒ f6 = 13 =⇒ f7 = 21.
20. Two points A and B are vertically 3 and 5 ft away from the line l, respectively,
and they are horizontally 10 ft apart. When a point P moves along the line l, what is
the smallest value of AP + PB.
10
•B

A•
5
3

• l
P
√ √ √ √ √

a) 97 b) 2 41 c) 3 43 d) 4 45 e) 185.
Answer-b) [G:N: None] Let A0,B0 be the reflections of A,B with respect to l and let P0 be
the point at which A0B meets l. The sum AP +PB minimizes at P = P0.
10
•B

A•

5
3 • l
P0

A 0•

Applying the Pythagorean theorem to the right triangle above, we find that
√ √
AP0 + P0B = A0B = p10 2 + 82 = 164 = 2 41.

21. The sum of the solutions to

.
8

Answer-d) Let . Then

The sum is .

Remark) If α,β are roots of a quadractic equation ax2 + bx + c = 0 then

22. A student walks home from her school at a speed of miles per hour. She was
greeted by her dog that runs 8 miles per hour when she was 2 miles away from
home. The dog goes back home and then comes back to her and it keeps doing it

What is the total distance the dog runs? (We assume that she and her dog keep the
same paces miles per hour and 8 miles per hour, respectively.)

a)10
Answer-c) Solution I) The total distance she travels is 2 miles, and the ratio of the
speed of her to that of the dog is 3 : 16. The total distance of the dog’s travel is
.

Solution II) For each rendevous point Rn for n ≥ 1, let tn be the time it takes to go from
Rn to Rn+1.
dn
Rn R n +1
• • •
dn +1 Home
Then

and 8tn = dn + dn+1

Eliminating tn, we have

The dog runs

.
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23. If f(x) = −x2 + ax + b has the maximum value 2019 at x = 3, then b is

a) 2010 b) 2015 c) 2018 d) 2020 e) 2024
Answer-a) Completing the square

we find that f(x) has the maximum value . The answer is a = 6 and b =
2010.

24. Let Sn be the sum of the first n terms of the sequence an. If Sn = 2n for all n ≥ 1, then
what is a2019?
a) 210 b) 220 c) 22018 d) 22019 e) none of these.
Answer-c) Using the general formula an = Sn − Sn−1 for n ≥ 2, we find that
a2019 = 22019 − 22018 = 22018.

25. Let p and q are prime numbers such that

p2 − 2q2 = 1
The number of such pairs (p,q) is
a) 0 b) 1 c) 2 d) 11 e) infinitely many.
Answer-b) It is clear that p > q and that p must be odd because p2 = 1 + 2q2. This, in turn,
implies q must be even because
2q2 = p2−1 = (p+1)(p−1) =⇒ 2q2 = 4m =⇒ q2 = 2m for some integer m.
The only solutions is (p,q) = (3,2).

26. For all values of x, the function f(x) satisfies

f(x + 1) + f(x − 1) = f(x).
If f(0) = 3 and f(1) = 5 then f(2018) + f(2019) + f(2020) is
a) −3 b) 0 c) 5 d) 8 e) none of these.
Answer-e) Observe that

which implies
=⇒ f(x + 2) = −f(x − 1) =⇒ f(x + 3) = −f(x) =⇒ f(x + 6) = f(x). Note that f(−1) =
f(0) − f(1) = −2 and so f(2) = −2. f(2018) + f(2019) + f(2020) = f(2) + f(3) + f(4) =
2 + (−3) + (−5) = −6
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27. Consider the following diagram of shapes in a circle:

A

P
O

B
The points A, B and C are on the circle, O is the center of the circle, and the line
←→
CP is tangent to the circle. Furthermore ∠ABC has measure of 80 degrees. What is the
measures of ∠ACP?
a) 60◦ b) 75◦ c) 90◦ d) 100◦ e) None of these.
Solution-e) The measure is 80 degrees, since ∠ABC and ∠ACP are congruent.
Sometimes this result is proven as an extension of the inscribed angle theorem for
circles. (I.e. if we choose some point X between B and C on the circle then ∠ABC and
∠AXC would be congruent. The tangent angle is sort of a limit case of that). If
students are familiar with that result, then the answer is immediate. If they are not
familiar with that result then they can compare angles in the triangles by using the
fact that 4AOC, 4AOB and 4BOC are isosceles and ∠OCP must be right ←→
(since CP is tangent to the circle).

28. A sequence xn is defined by

x1 = 3 and for all n ≥ 1.

Then x15 is

.
Answer-b) Observe that

The alternating sequence method

gives

.
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29. For integers x and y satisfying xy = 3x + 2y + 23, what is the largest possible value of x − y?
a) 23 b) 25 c) 27 d) 29 e) none of these.
Answer-c) Observe that
xy = 3x + 2y + 23 =⇒ (x − 2)(y − 3) = 29
and so

x − 2 = 1,y − 3 = 29 =⇒ x = 3,y = 32

 x − 2 = 29,y − 3 = 1 =⇒ x = 31,y = 4

 
 xx −− 22 == −−291,y,y−−33==−−291 ==⇒⇒ xx == 127,y,y==−2 26

and so the largest x − y is 27.

30. Two distinct points A and B are on a semicircle with diameter MN and

center C. The point P is on CN and ∠CAP = ∠CBP =

15◦.

A
•

15◦ •B

15◦
◦
75
• •
M C P N
If ∠ACM = 75◦ then ∠BPN equals
a) 55◦ b) 60◦ c) 65◦ d) 70◦ e) 75◦.
Answer-b) Consider the figure below
A
•

◦ •B
R 15
R ◦
15
75◦ θ
• •
M C P N
Observe that
∠CPA = 75 − 15 = 60.
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Let R be the radius of the halfcircle and let θ = ∠CPB. Applying the law of sines to
∆ACP and to ∆CPB, we find that

and
which implies sinθ = sin60 =⇒ θ = 180 − 60 = 120 =⇒ ∠BPN = 60.

Tie breakers. Answer the questions.

a) Factor the following to the product of two polynomials on a,b.
a3 + b3 − ab(a + b).
Solution) Using
a3 + b3 = (a + b)(a2 + ab + b2)
we factor
a3 + b3 − ab(a + b) = (a + b)(a2 + ab + b2) − ab(a + b) = (a + b)(a2 + b2).
b) Use a) or other means to prove that, for all nonnegative real numbers a,b,
a3 + b3 − ab(a + b) ≥ 0.
solution) It is clear from the factoring of b) that
a3 + b3 − ab(a + b) = (a + b)(a2 + b2) ≥ 0.
c) Prove that, for all nonnegative real numbers a,b,c and any positive integer n, an(a − b)(a − c)
+ bn(b − c)(b − a) + cn(c − a)(c − b) ≥ 0.

Solution) Since the inequality is symmetric on a,b,c, we may assume that a ≥ b ≥ c.

Observe that
LHS = an(a − b)(a − c) + bn(b − c)(b − a) + cn(c − a)(c − b)
= (a − b)[an(a − c) − bn(b − c)] + cn(c − a)(c − b) ≥ 0
because an(a − c) ≥ bn(a − c) ≥ bn(b − c) =⇒ an(a − c) − bn(b − c) ≥ 0.