2008 OHMIO Individual Questions and Solutions

1.When sending text messages on my cell phone, the letter/number groupings are 2:abc, 3:def, 4:ghi, 5:jkl, 6:mno, 7:pqrs, 8:tuv, 9:wxyz. To type/text the letter r requires pressing the 7-key three times (once for p, twice for q, thrice for r, and four times for s), etc. In total, how many times must those number keys be pressed in sending the entire alphabet as a text message?

Answer: 56

Solution: There are many solutions using simple arithmetic. One approach: keys 2,3,4,5,6,8 get pressed 1+2+3 = 6 times each, and keys 7,9 get pressed 1+2+3+4 = 10 times each, for a total of (6)(6)+(2)(10)=56.

2. Joe drove 495 miles from A to B, with an average speed of 55 mph, then drove 495 miles back from B to A. If his average speed was 60 mph for the entire trip, find his average speed for the trip from B to A.

Answer: 66 mph

Solution: His time for the entire trip was t = = = 16.5 hrs, and his time from A to B was t = = = 9 hrs, so from A to B took 7.5 hrs. So, his average speed from B to A was r = = = 66 mph. log 2 x 3 3. Find the only real solution to  2   lne ex .

Answer: e2, or x = e2.

1 l o g 2 x Solution: We can simplify both sides to get 2 2  3 e x . This simplifies further to 1  2 3 . Raising both sides to the sixth power, we get x  e x 3 2 2 2 x = e x , from which x = e , since x ≠ 0.

4. When I was half as old as I will be twenty years from now, I was three years older than my sister. Twenty years ago I was twice my sister's age. How old am I now?

Answer: 26 (years old)

Solution: Let x be "my" present age. By the last part of the first sentence, the sister's age is x–3. By the second sentence, x–20 = 2(x–23), so x = 26. (The first part of the first sentence doesn't say much, other than < x, based on the past tense of the word "was"!) 5. Suppose that f(x) is a polynomial function which is one-to-one, that f -1(-1) = 0 and f -1(1) = -1. Which of the following is a consequence? (Indicate by letter.) A: The maximum output of f(x) among inputs in [-1,0] must be -1. B: The minimum output of f(x) among inputs in [-1,0] must be 0. C: The maximum output of f(x) among inputs in [0,1] must be -1. D: The minimum output of f(x) among inputs in [0,1] must be 0.

Answer: C

Solution: Since f(-1) = 1, option A is untrue. Since f(0) = -1, options B and D are untrue. By the process of elimination, the answer is C. To see that C is in fact true, however, read on. Since f is one-to-one, it must be always increasing or always decreasing as traced from left to right. Since f(-1) = 1 and f(0) = -1, we see that f decreases from output 1 at x = -1 to output -1 to its right at x = 0. Thus f is decreasing. Therefore the maximum output of f(x) over interval [0,1] occurs at its left endpoint x = 0, at which the output is f(0) = -1.

6. A right triangle has hypotenuse of length and area 123. Find the sum of the lengths of its two legs.

Answer: 50 Solution: Letting a,b be its leg lengths, a 2+b2 = 2008 and ab = 123, so 2ab = 492. Thus (a+b)2 = a2+2ab+b2 = 2008+492 = 2500, so a+b = 50.

7. In the sequence 75, 27, 275 there are seven digits (including repeat occurences). How many digits are there in the sequence 1, 4, 9, 16, 25, ..., 10000 ?

Answer: 358

Solution: There are three 1-digit numbers, nine with 2 or fewer digits (since 9 2 is the largest perfect square smaller than 100), thirty-one with 3 or fewer digits (since 31 2 is the largest perfect square smaller than 1000, what with 32 2 being 1024), ninety-nine with four or fewer digits (since 99 2 is the largest perfect square smaller than 10000), and one with 5 digits or more, namely 10000 itself. So, the number of digits is (3)(1)+(9–3)(2)+(31–9)(3)+(99–31)(4)+(1)(5) = 358.

8. A solid cylinder has a circular base with radius 4 feet and has height 6π feet. A string is to wind around the cylinder once, starting at a point P on its base and ending at the point on the cylinder's top directly above P. How many feet long is the shortest possible such string?

Answer: 10π , or 10 π ft

Solution: The circumference is 2π r = 8 π ft. A piece of paper 6 π ft high and 8 π ft wide could barely wrap around the cylinder, with a seam running from P straight up to the cylinder's top. The string's path "traces out" a curve along the paper from one corner of the paper to the opposite corner, so the shortest possible string length is the diagonal length of the paper, which is = 10π ft. 9 . Despite laws concerning exponents and logs, the functions f(x) = 2 ln(x) and g(x) = ln(x2) are not quite the same. For the function h(x) = f(|g(x)|), list all the real numbers x not in the domain of h.

Answer: -1, 1, and 0. Or, x = -1, 1, and 0.

Solution: The domain of f is the interval 0 < x < ∞ . Thus, so long as g(x) is defined and not zero, |g(x)| will be in the domain of f, making f(|g(x)|) defined. So, for x to avoid being in the domain of h, it must be that either g(x) = 0 or g(x) is undefined. Now, g(x) is undefined only when x = 0. And g(x) = 0 only when x 2 = 1, which happens when x is -1 or 1. 10. Find the number c for which the graph of y = has the line y = c as a horizontal asymptote.

Answer: -251, or y = -251

Solution: For "rational functions" such as this one, to find the limit of y as x approaches either ∞ or - ∞ , it is correct to ignore all terms other than the "dominant" terms on top and bottom, since the resulting simpler function of y will have the same "end behavior" as the original. In this case, that simpler function is y = = -251. Or, one can more carefully evaluate 2 0 0 8 x 3  1 4 9 2 x 2 l i m 3 (and similarly the limit as x approaches -∞ ) by dividing on top and x  1 8 1 2  8 x 1 4 9 2 2 0 0 8  2 0 0 8  0 3 l i m x   2 5 1 bottom by x to get x  1 8 1 2 . Or, one can use L'Hospital's rule  8 0  8 x 3 2 0 0 8 x 3  1 4 9 2 x 2 6 0 2 4 x 2  2 9 8 4 x 2 9 8 4 repeatedly to get l i m 3  l i m 2  l i m  2 5 1   2 5 1 . x  1 8 1 2  8 x x   2 4 x x   2 4 x A b o r d e r i s s h a d e d a s s h o w n a r o u n d a 3 , 4 , 5 r i g h t t r i a n g l e , c o n s i s t i n g o f p o i n t s o u t s i d e t h e t r i a n g l e a t d i s t a n c e 1 o r l e s s 11. f r o m a t l e a s t o n e p o i n t o f t h e t r i a n g l e . H o w m a n y s q u a r e u n i t s o f a r e a a r e s h a d e d ?

Answer: 12+π , or 12+ π sq. units

Solution: The region decomposes into three rectangular pieces of combined area 5+4+3 - 12 and three sectors which together form a circle of radius 1, of area π r2 = π .

12. An object's position s(t) along a coordinate line after t seconds of travel is given by s(t) 4 3 2 = at +bt +ct +dt+e, measured in feet. Given that a≠ 0 and that this object's average velocity over the interval 0 ≤ t ≤ 2 equals its instantaneous velocity at time t = 1, find Answer: -4 Solution: The instantaneous velocity function is v(t)= 4at 3+3bt2+2ct+d, so the instantaneous velocity at t = 1 is 4a+3b+2c+d. The average velocity is = = 8a+4b+2c+d. So, 8a+4b+2c+d = 4a+3b+2c+d, or 4a = -b, so = -4. 13. A hyperbola centered at (0,0) has y = ± x as its asymptotes. If its foci are at (0,±3

Answer: (0,±) Solution: The hyperbola has equation of the form – = 1, where a,b > 0. For such a hyperbola, the distance to the foci is , the vertices are at (0,±a), and the asymptotes have equations y = ± x. From the asymptote information, = , so a = b. From the focal point information, a2+b2 = 32 = 9. Substituting b for a, we have (b)2+b2 = 9, or 3b2 = 9, so b = and a = b = . So, the vertices are (0,±).

14. Two balloons are being inflated, each at cubic units per minute, each keeping a spherical shape. One stays centered at (0,0,0), the other at (4,4,2). Right now each of them has radius 1. In how many minutes will they first touch each other? [Hint: a 3 sphere of radius r has volume π r .] Answer: 52, or 52 minutes Solution: The distance between their centers is = 6, so they first touch when each has radius 3. At that point, each has volume 36π . Currently, each has volume π . So, letting t be the number of minutes of inflation remaining, π + t = 36 π . Multiplying by 3/ π to simplify, we have 4+2t = 108, so t = 52.

15. Among the integers from 1,000 to 9,999 inclusive, what percentage of them have no two digits in a row being the same? For example, 8083 qualifies, but 3770 does not. Round your answer to the nearest percent.

Answer: 73, or 73%

Solution: The 1st digit can be any of 9 choices (not 0). For each choice, the 2nd digit can be any of 9 choices (not the same as the previous digit), and so on. So, there are 9 4 such numbers out of the 9,000 numbers from 1,000 to 9,999. That's a fraction = = 72.9%, rounding to 73%. 16. Find the slope of the tangent line to f(x) = at x= . Hint: simplify before differentiating. Answer: –28

Solution: By sin2A = 2sinAcosA, we simplify and cancel to get f(x) = = =

= 8 sin7x. So, f´(x) = 56cos7x, so m = 56cos(7π /6) = 56(-/2) = -28. A T r i a n g l e  A B C h a s A B = A C , w i t h B C = 7 . A l i n e 17. t h r o u g h A a n d a p o i n t D o n t h e o p p o s i t e s i d e   D s p l i t s t h e t r i a n g l e a s s h o w n . F i n d c o s  . C B 3 4 Answer: 2/3 Solution: Let x = AB = AC. Applying the Law of Sines to the two smaller triangles, using  as the measure of CDA and π – for the measure of BDA, we have = = = = , so = , and cos  = 2/3 (since sin  ≠ 0). [Comment: it isn't obvious that a triangle fitting the description in the problem exists in the first place! But it does... draw a base of length 7 from B to C, and select A along the perpendicular bisector of the base, just far enough away from the base so that CAD has measure 3cos-1(2/3), or roughly 2.5 radians.]

18. Two particles start at the same time at (1,0) and travel counterclockwise around x 2+y2 = 1, one traveling twice as fast as the other. What is the x-coordinate of the slower particle at the first instant when its x-coordinate is 50% larger than that of the other particle? Answer: , or +

Solution: When the first particle has traveled t units of distance or central angle, the slow point is at (cos t, sin t) and the fast one is at (cos 2t, sin 2t). We solve cos t = cos 2t. By trig identity, this is 2 cos t = 3(2cos 2t–1), or 6cos2t – 2 cos t – 3 = 0. Solving this as a quadratic in terms of cos t, we have cos t = = . This event happens first before either particle leaves the first quadrant, before the faster particle's x-coordinate decreases to 0, so the answer is .

19. The four corners of a square are P(1,1), Q(1,-1), R(-1,-1), S(-1,1). A point T is selected at random in the square (in the usual "random by area" sense). What is the probability that Q,R,T form the vertices of an obtuse triangle?

Answer: π /8

Solution: Consider the semicircle inside the square, with Q,R the ends of its diameter. If T is on the circumference, the QTR will be 90°. If T is outside the semicircle, the angle at T is acute. If T is inside, the angle is obtuse. The angles at Q,R cannot be obtuse. So, the triangle is obtuse when T is inside the semicircle of radius 1. Thus the probability is = . 20. 2008 can be expressed in the form a b(cd+ef), where a,b,c,d,e,f > 1 are primes. Find the product abcdef for any such expression.

Answer: 540

Solution: Note that 2008 has prime factorization 2 3(251). Thus a must be a divisor of 2008, so must be 2 or 251. Since b > 1, we have a = 2. Since b is prime and at most 3, we have b = 2 or 3. So, c d+ef = 251 or 502. The only possible values for c d, ef between 4 and 502 are 22=4, 23=8, 25=32, 27=128, 32=9, 33=27, 35=243, 52=25, 53=125, 72=49, 73=343, 112=121, 132=169, 172=289, 192=361. The only pair of these with sum 251 or 502 is 243 and 8, so cd and ef are cd and ef in either order. Thus abcdef = (2)(3)(3)(5) (2)(3) = 540.