UNM - PNM STATEWIDE MATHEMATICS CONTEST. February 2, 2019 Second Round Three Hours

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1 UNM - PNM STTEWIDE MTHEMTICS CONTEST February, 09 Second Round Three Hours. certain bacterium splits into two identical bacteria every second. flask containing one such bacterium gets filled with the bacteria completely in 09 seconds. Now, how long would it take the flask to get filled completely by the bacteria if it initially contained two bacteria? nswer: 08 seconds. Solution: 08 seconds. If we start with one bacterium, after one second we have two bacteria, which is the case we want. It took 09 seconds when starting from one bacterium, which means starting from two bacteria must have taken 08 seconds.. famous YouTuber, MathematicianJanet, is paid two thousand dollars every month as part of her sponsorship money. In addition, she makes nine dollars every time someone watches one of her videos on Youtube. If MathematicianJanet earned 6, 07 dollars in the last three months, how many of her videos were watched during those three months? (Note: YouTuber is a person who posts videos on Youtube.com. Other people can then watch these videos.) nswer: 3. Solution: She earned 6, 07 6, 000 0, 07 from videos watched. Total videos watched were 0, 07/ Iris visited her grandparents during the summer vacation. Last summer, she saw six baskets with same number of eggs in her grandparents house. Her grandma told her that if she took out fifty eggs from each basket, then the total number of the remaining eggs of the six baskets is equal to the number of eggs of the initial two baskets. Her grandma asked Iris if she knew how many eggs each basket has initially. Can you help Iris answer this question? nswer: 75. Solution: Iris grandma took out eggs which is equal to the number of eggs of the initial four baskets. So initially each basket has 75 eggs.. Let {a n } be a sequence of real numbers with the following properties: a. a n+3 a n + 3. a n+ a n +. Calculate a 09. nswer: a Solution: Notice that lso we have Then we have a 09 a 07 + a 05 + a a n a n+ a n +. a 09 a 08 + a 07 + a

2 5. Let x, y, z be positive real numbers. Suppose x + y + z xyz, what is the minimum value of u x yz + y zx + z xy? nswer: u min 3. Solution: Since x, y, z > 0 and x + y + z xyz, so xy + xz + yz. Notice that by the geometric-arithmetic inequality, we have lso x yz + y xz z. Thus x + y xy, x + z xz, y + z yz. u x yz + y xz + z xy x + y + z x + y + ( z + xy + xz + ) yz ( 3 xy + xz + ) yz 3. The above equalities holds if and only if x y z and xy + xz + yz, i.e., x y z 3. Thus u min Suppose p is a prime number and the number of positive factors of p + 7 is no more than 0. What are the possible values of p? nswer: p, 3. Solution: When p, p , the total number of factors of 75 is 3 6. When p 3, p , the total number of factors of 80 is 5 0. When p > 3, p + 7 (p )(p + ) + 7. Since p is an odd number, (p )(p + ) is divisible by. Since p is not divisible by 3, (p )(p + ) must be divisible by 3. Thus p + 7 is divisible by. Let p + 7 m, where m. If m is divisible by a prime number greater than 3, then the total number of factors of p + 7 is at least 6. If m is divisible by 3, then the total number of factors of p + 7 is at least 3. If m k, k, then the total number of factors of p + 7 is ( + k) 6. We thus conclude that p, 3 are the only possible values. 7. Helen wants to approximate the perimeter of a circle but has forgotten the formula. Her friend Stephen suggests the following approximation to the perimeter of the circle: let the circle be inscribed in a n-sided regular polygon (recall: an inscribed circle is the largest possible circle that can be drawn on the inside of the polygon.) Then, approximate the perimeter of the circle by the perimeter of this n-sided regular polygon. What is the approximation to the perimeter of a circle with a diameter of units if Helen uses Stephen s method with n 6?

3 nswer: / 3. Solution: The regular hexagon can be divided into congruent right-triangles. One such triangle C is indicated in the figure below. The perimeter of the hexagon is x where x is the length of the line segment. We are given that length of line segment C is. lso note that triangle CD is an equilateral triangle with length of each side equal to twice of length of line segment, that is equal to x. Thus, length of line segment C is x. y the Pythagorean theorem on triangle C, x + (x) Solving this gives x / 3. The perimeter of the hexagon is then / 3. D C C x x 8. Four points are placed randomly on a circle. What is the probability that the quadrilateral formed by these points contains the center of the circle? nswer: /. Solution: Let us label the points as,,c and D. Define the clockwise semi-circle (CSC) of a point as the part of the circle starting from the point and ending at its antipode while traversing the circle in the clockwise direction. Then the probability of the quadrilateral not containing the center is Probability of, C, D being contained in the clockwise semi-circle of + Probability of, C, D being contained in the clockwise semi-circle of + Probability of,, D being contained in the clockwise semi-circle of C + Probability of,, C being contained in the clockwise semi-circle of D Each of these probabilities is the same. Let us compute one of them, say, Probability of, C, D being contained in the clockwise semi-circle of (Prob. of in CSC of ) (Prob. of C in CSC of ) (Prob. of D in CSC of ) (/) (/) (/) /8 Therefore, the probability of the of the quadrilateral not containing the center is (/8) /. nd hence the probability of the quadrilateral containing the center is / /. 9. This question deals with equally-spaced points. Such points arise in many mathematical, engineering and computer science applications. We define the n + equally-spaced points x i ih where h /n. 3

4 (a) Prove that where n! n (n ) (n ). (b) Establish the bound that for any x [0, ] if 0 j n then (j + )!(n j)! n! x x i hn+ n!, i0 where n i0 z i z z z n. You may use the fact that for x j x x j+ we have x x j x x j+ h () Solution: (a) First note that this equality is obviously true for j 0. For 0 < j n, we will show that, The left hand side simplifies to, (j + )!(n j)! n! (j + )!(n j)! n! (j + )!(n j)! (n)(n ) (n j + )(n j)! (j + )(j)(j ) ()() (n)(n ) (n j + ) ( ) ( ) ( ) j + j n n n j + Since 0 < j n, we have 0 < (j+c) (n +c) for ( j) c. For such a c we have (j+c)/(n +c) and hence, ( ) ( ) ( ) (j + )!(n j)! j + j ()() () n! n n n j + lternate solution to part (a) Part (a) can also be shown using induction. gain note that this equality is obviously true for j 0. For rest we use induction. For n we have!!. Now let this be true for n, that is, (k + )!(n k)! n! for 0 k n. Consider j n. Let k j and hence 0 k n Then, (j + )!(n + j)! (j + )(k + )!(n + k )! (j + )(k + )!(n k)! (j + )n! (n + )n! (n + )! (b) Since x [0, ] we have x j x x j+ for some integer 0 j n. Using () we have j j+ x x i x x i x x i i0 i0 ij ij+ x x i h j x x i i0 ij+ x x i From x j x x j+, we have x x i { (x j+ x i ), i < j (x i x j ), i > j + ()

5 This implies, i0 x x i h j (x j+ x i ) i0 ij+ (x i x j ) h j hj h n (j+)+ (j i + ) i0 ij+ (i j) where we used the fact that x j+ x i (j + )h ih (j + )h and x i x j (i j)h. Simplifying further, x x i hn+ (j + )!(n j)! i0 Finally, use the result from part(a) to finish the proof. 0. Here we have two circles C, C with centers at O, O. C, C intersect at two points,. P, E are two points in C ; Q, F are two points in C. The line EF is a tangent line of C and C. The line P Q is parallel to the line EF. The line P E intersects the line QF at the point R. Show that P R QR. P O Q O E F R Proof: Let us make a parallel line of EF passing through the point. Suppose this line intersects the circle O at G, the circle O at H, the line P E at X, the line F Q at Y, the line EO at M and the line F O at M. Let Z be the intersection point of the line R and P Q, N be the intersection point of EO and P Q and N be the intersection point of F O and P Q. Let EO intersects the circle O at L and F O intersects the circle O at L. 5

6 L L P N Z N O Q O M G X M E F Y H R Let the radius of C be r and the radius of C be r. Let the length of the line segment EM F M be a and the length of the line segment EN F N be b. Notice that the line EF is tangent to the circle C, hence L E is perpendicular to the line EF. y our assumptions, GH is parallel to EF, thus the angle GM E is a right angle. lso the triangle GEL is a triangle inscribed in a circle with one of the sides being the diameter. Thus the angle EGL is a right triangle with the hypothenuse being the diameter. Since the triangle GEM is similar to the triangle GEL, we have GE EM EL ar. Similarly since the triangle EP N is similar to the triangle EP L, we have Notice that EP EN EL br. EX EP a b, we have EX a b br. Since the triangle XP is similar to the triangle EXG, we have Similarly we have Thus So we obtain X P EX a EG b. Y a Q b. X P Y Q. X Y P Q. Notice that the line GH is parallel to the line P Q, thus we have X P Z R RZ Y ZQ. 6

7 That implies Together, we have X P Z Y ZQ. P P Z Q ZQ. That shows P Z QZ which implies P R QR. 7

SOLUTIONS FOR IMO 2005 PROBLEMS

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