Problem 1. At 8 AM Black Widow and Hawkeye began to move towards each other from two cities. They were planning to meet at the midpoint between two cities, but because Black Widow was driving 100 mi/h faster than Hawkeye, they met at the point that is located 120 miles from the midpoint. When they met Black Widow said ”If I knew that you drive so slow I would have started one hour later, and then we would have met exactly at the midpoint”. Find the distance between cities. Solution.
Figure 1
Let A and B be cities, C be the point where Black Widow and Hawkeye met, and M be the midpoint of AB. Denote the length of AM by x mi and and the speed of Hawkeye by v mi/h. Thus, the speed of Black Widow is v + 100 mi/h. Black Widow covered the distance of x + 120 mi and Hawkeye the distance of x − 120 mi. Setting their travel times equal we get the following equation:
x + 120 x − 120 = . v + 100 v 1 2 If Black Widow started one hour later they both would cover x mi. Therefore, x x 1 + = . v + 100 v We get the following system of equations: x+120 x−120 v+100 = v x x . 1 + v+100 = v Then, v(x + 120) = (v + 100)(x − 120) . v(v + 100) + xv = x(v + 100) Simplifying these equations, we get the following system: 100x = 240v + 12000 . v2 + 100v = 100x After substitution, we get v2 − 140v − 12000 = 0. Thus, √ √ v = 70 ± 4900 + 12000 = 70 ± 16900 = 70 ± 130. Therefore v = 200, x = 600. Answer: the distance between cities is 1200 mi. 3 x−1 x−2 Problem 2. Solve the inequality: x−2 ≤ x−1. Solution. x − 1 x − 2 − ≤ 0 x − 2 x − 1 (x − 1)2 − (x − 2)2 ≤ 0 (x − 1)(x − 2) 2x − 3 ≤ 0 (x − 1)(x − 2) x − 3/2 ≤ 0 (x − 1)(x − 2)
Figure 2
−∞ < x < 1 or 1.5 ≤ x < 2
Answer: (−∞, 1) ∪ [1.5, 2). 4 Problem 3. Solve the equation: (x − y − z)2 + (2x − 3y + 2z + 4)2 + (x + y + z − 8)2 = 0.
Solution. x − y − z = 0 2x − 3y + 2z + 4 = 0 x + y + z − 8 = 0 After adding the first and the third equations we get x = 4. Thus,
y + z = 4 3y − 2z = 12 Multiplying the first equation by 2 and adding to the rre- sult the second equation we get
5y = 20. Thus,
y = 4, z = 0.
Answer: x=4, y=4, z=0. 5 Problem 4. Three camps are located in the vertices of an equilateral triangle. The roads connecting camps are along the sides of the triangle. Captain America is inside the triangle and he needs to know the distances between camps. Being able to see the roads he has found that the sum of the shortest distances from his location to the roads is 50 miles. Can you help Captain America to evaluate the distances between the camps. Solution. Denote the locations of cities by A, B, and C, and the location of Captain America by P . Let h1, h2, and h3 be heights of the triangles AP B, BPC, and AP C.
Figure 3
h1 + h2 + h3 = 50
Let |AB| = x, then 6
xh xh xh Area(AP B) = 1 , Area(BPC) = 2 , Area(AP C) = 3 . 2 2 2 The height of h the triangle ABC equals √ x 3 h = px2 − (x/2)2 = 2 . √ xh x2 3 Therefore, Area(ABC) = 2 = 4 . Since Area(AP B)+Area(BPC)+Area(AP C) = Area(ABC) we get √ xh xh xh x2 3 1 + 2 + 3 = . 2 2 2 4 Therefore, √ x(h + h + h ) x2 3 1 2 3 = . 2 4 Thus, √ 50x x2 3 = 2 4 and √ 100 100 3 x = √ = . 3 3
Answer:√ the distances between the camps equal 100 3 3 miles. 7 Problem 5. N regions are located in the plane, every pair of them have a non-empty overlap. Each region is a connected set, that means every two points inside the region can be connected by a curve all points of which belong to the region. Iron Man has one charge remaining to make a laser shot. Is it possible for him to make the shot that goes through all N regions? Solution. Choose an arbitrary straight line l and project all regions on that line. The projections are intervals and each pair of intervals has a non-empty intersection. Denote the left endpoins of intervals by a1, a2, . . . , aN , and the right endpoints by b1, b2, . . . , bN . Then for every pair of endpoints ai, bi, aj, bj,
ai ≤ bj, aj ≤ bi.
Indeed, since intervals [ai, bi] and [aj, bj] have non-empty intersection, there exists a point ci,j such that
ai ≤ ci,j ≤ bi, aj ≤ ci,j ≤ bj.
Figure 4
Therefore, there exists a point c that belongs to all inter- vals. Thus the straight line passing through the point c perpendicular to lthe line l intersects all regions. 8 Problem 6. Numbers 1, 2,..., 100 are randomly di- vided in two groups 50 numbers in each. In the first group the numbers are written in increasing order and denoted a1, a2, . . . , a50. In the second group the numbers are writ- ten in decreasing order and denoted b1, b2, . . . , b50. Thus, a1 < a2 < ··· < a50 and b1 > b2 > ··· > b50. Evaluate
|a1 − b1| + |a2 − b2| + ··· + |a50 − b50|.
Solution. Divide the numbers a1, a2, . . . , a50 in two groups: less than or equal to 50 and greater than 50. Let the first group contains n numbers. Thus,
a1 < a2 < ··· < an ≤ 50 < an+1 < an+2 < ··· < a50.
Now, let us divide the numbers b1, b2, . . . , b50 in two groups: less than or equal to 50 and greater than 50. Then the first group contains 50 − n numbers, and the second group n numbers. Thus,
b1 > b2 > ··· > bn > 50 ≥ bn+1 > bn+2 > ··· > b50. Therefore,
|a1 − b1| + |a2 − b2| + ··· + |a50 − b50|
= (b1 −a1)+... (bn −an)+(an+1 −bn+1)+···+(a50 −b50). In the last sum all terms with plus are greater than 50 and all terms with minus are less than or equal to 50 (absolute values). Thus the last sum equals (51 + 52 + ··· + 100) − (1 + 2 + ··· + 50) 50(51 + 100) 50(1 + 50) = − = 2500. 2 2 Answer: 2500.