ARML 1999 Tryout Solutions

Solution 1:  Let's take a quick look at the graph of the numerator, .  It is a parabola that crosses the x-axis at -1 and -4, so for positive x it looks roughly like this:

But the thing we want to minimize is this divided by x.  How do we picture that?  For a point (x, f(x)) on the graph, the ratio is just the slope of the line from the origin to the point.  For example, for the point shown, is given by the slope of the dashed line.

Now we see what to do.  The minimum slope will be given by the line that is just tangent to the parabola.  So let's find the tangent point.
Well, if a line L is tangent to a parabola P whose leading term is  , then    units over from the tangent point, the line will be    units below the parabola.
In this case, and considering the origin, which is 4 units below the parabola,  .  So  , i.e. the origin is 2 units over from the tangent point.  So the tangent point is at (2, 18), and the minimum slope is 9.

Solution 2:  Expanding the fraction, we get .  The minimum value of    is given immediately by the arithmetic-geometric mean inequality, applied to the means of and x:

				with equality iff
				"	x=2

So the minimum value of is 9.  (And this minimum is attained at , as we can check.)

The length of the altitude to side S is given by the twice the area divided by the length of S.  We're given the lengths, and we need the area, so it's time to use Heron's formula:

We'll get the shortest altitude by dividing 2A by the longest side, 40:  
12 is a nice simple answer!  Maybe there's a nice picture to go along with it?
Yes, the 12 and 13 make a 5-12-13 triangle, and on the other side we get a 12-35-37 triangle:

If the four primes are all odd, then their sum will be even while their product is odd.  But an even number cannot divide an odd number, so this case cannot be the case.  Well, the only prime that isn't odd is 2, so 2 must be one of the four primes, and in fact it must be the least one since there is no smaller prime. (1 is not a prime, by definition.)

We sketch a picture to get an idea what the problem is talking about:

We see that our picture is not to scale, for example BD is supposed to be longer than AB, but we won't worry about that, since this looks like a problem for Stewart's Theorem anyway:



We want to find AC, which is 2x, or .

Solution 1:  We can make a table where, given the first two rolls, we note how many possibilities for the third roll will result in a sequence satisfying the given condition:

		First Roll
		1	2	3	4	5	6
Second       Roll	1	0	0	0	0	0	0
	2	4	0	0	0	0	0
	3	3	3	0	0	0	0
	4	2	2	2	0	0	0
	5	1	1	1	1	0	0
	6	0	0	0	0	0	0

We see there are a total of 20 cases where each throw is greater than the previous, out of a total of possibilities, so the probability is .

Solution 2:  Out of the 216 equally likely outcomes, there are    ways for the three rolls to be three different, increasing numbers.  So the probability is .

Drawing a diagram, we see that the trapezoid is symmetric and must be an isosceles trapezoid.

If  h  is the distance between and , then thinking of the areas in terms of a base times this height (over two), we see that CD:AB =  4:5.
So let's say the radius of the circle is 5x.  Then AB is 10x, and CD is 8x.

Drawing a vertical line through the middle,we see that  h is 3x.
So the total area of the trapezoid must be times 3x, or .  And since we know the total area is 270, x must be .
We are trying to find BC, so let's drop an altitude from C to make an x--3x--x triangle.
So BC is 10.

First let's draw a diagram to see what the problem is talking about:

Thinking about the relationship of the area of to , we imagine D and E as being able to slide back and forth along and .  The area of is proportional to the product of AD and AE.  (In fact, it is given by the formula .)  So, to have half the area of , we want (AD)(AE) to be half of  (AB)(AC).
We are given that AD is of AB, so we want AE to be of AC to make the product be .
So AE is of AC, and AE:EC is 4:1.

We want to find tan C.  Well, .
If we don't remember the tangent-of-a-sum formula, we can easily work it out from sine-of-a-sum and cosine-of-a-sum.  Anyway, trying not to forget the minus sign:

After drawing the figure, we see that in order to find AC, it will be useful to consider DF, so let's let .  Since is the angle bisector, we have .  Also, since is a miniature version of , is similarly the angle bisector of , and is also .  Multiplying out the denominators of these equations, we get , and .  Solving these yields , so , or 29.4.

What a weird equation!  Let's multiply it out and see if we can regroup the terms to match some identities.

(sin 3+sin )(sin 3-sin )=sin(3-)

(2sin 2 cos )(2sin cos 2)=sin 2

sin 2=0                  or      4cos sin cos 2=1

sin 4=1

So the possible values for are:

Let's see, they make an angle of almost 60°, and they'll do that next at around five minutes to one.  But what will be the time exactly?  Let's just find the next time the hands will overlap (around 1:05), and then exactly ten minutes before that, the hands should make exactly the same angle as they do now at ten minutes past noon.
The hands meet 11 times every 12 hours, so after meeting at noon, they'll next meet hours later.  Subtracting off the ten minuts that have already passed since noon and the ten final minutes before the hands meet again, we see we will have to wait hours, which simplifies to hours, or or 45 minutes.  That seems about right.

This looks like a job for the Law of Cosines.



What?  You mean x has to be zero?  Does that make sense?  Why yes, it does, it means the triangle has to be equilateral, which certainly satisfies the stated problem!  So the area is .

As each student knows, there must have been at least 10 failures.  Could there have been only 10?  No, because even a failing student saw 10 failures other than their own!  So there must have been at least 11 failures.  Could there have been only 11?  Yes, if each failing student saw only the other ten, and each passing student saw only ten of the eleven failures.  So there might have only been 11.

This looks like a job for the Law of Cosines again.  Let x be the cosine of the angle that appears twice.
    These numbers are so big!  Let's take out a factor of 9.



Ok, now we can find BC:

Given values for x and y, clearly at most one value of z can work.
Let's start at the beginning:

So the possible triples are:  (1,1,1), (1,2,2), (2,3,6), (2,4,4), (3,3,3)

Mod 9, a number is the same as the sum of its digits. (This is because powers of 10 look just like powers of 1 in mod 9 land.)  So we need to find all the ways to pick four numbers from that set that add up to a multiple of nine.
To get a sum of 9, we must use three 2s and a 3.  There are 4 four digit solutions of this form.
To get a sum of 18, we can use two 4s and two 5s (this gives 6 solutions), or three 5s and a 3 (giving 4 more solutions).
No other multiple-of-nine sums are possible.
So in total there are 4 + 6 + 4 = 14  such numbers.

Consider what happens if we rotate the whole figure by 180° so that the parallelogram winds up where it started.   will have rotated 180°, so its final position (shown here as a dotted line) must be parallel to its original position.  But wait, the area between and the dotted line is exactly the discrepancy in area between the two parts of the parallelogram, which we are told is zero!  So must in fact be the same place as the dotted line, so CF must be the same as AE, namely 3, so DF must be 5.

Across	Down
1. A perfect cube that is 40 more       than a perfect square	1. Number of real solutions between 0 and 2 pi to the equation 2(sin 100x)(cos 16x) = cos 16x
5. One Fibonacci number followed by another Fibonacci number
6. A Fibonacci number	3. The digits are in strictly increasing order
	4. A perfect square

Well!  This looks like 7 problems in one, so we'd better work on it sensibly if we want to finish in time.  Only 1 Down and 1 Across seem to be "stand-alone" problems that don't depend much on anything else, so let's do them first.
1 Down:  This equation is true if either    or  .
The solutions to are .  There are 32 of them.
The solutions to are .  There are 200 of them.
Is there any overlap between these two lists?  No, because everything in the second list will have a factor of 3 in the denominator even after it is completely reduced, and nothing in the first list will.  So there are a total of 232 solutions to this equation!

1 Across:  We now know this starts with 2, so let's list all the cubes from 2000 to 2999:

12×12×12 = 144×12 = 1440 + 288 = 1728	Ooops, too low!
13×13×13 = 169×13 = 1690 + 507 = 2197	Subtracting 40, no perfect square can end in 7
14×14×14 = 196×14 = 1960 + 784 = 2744	Is 2704 a perfect square?  Perhaps ?  Yes.

We see that the cubes are increasing so rapidly that the next will be over 3000.
We have probably spent several minutes working on this so far, so it's a good thing we've already solved half the puzzle!

1           2	2           7	3           4	4           4
5     3
6     2

Hmm, what should we do next?  Probably we should make a list of Fibonacci numbers to help with the remaining Across problems!
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, ...

Great, there's only one four-digit one that starts with 2, so 6 Across must be 2584.
So now 4 Down is a perfect square over 400 ending in 4, so it must be .

1           2	2           7	3           4	4           4
5     3			8
6     2	5	8	4

Now, 5 Across ends with a Fibonacci number that ends in 8, which we see from our list must be 8 itself, so the first part must be a 3-digit Fibonacci number starting with 3, which we see must be 377.

1           2	2           7	3           4	4           4
5     3	7	7	8
6     2	5	8	4

Now we can immediately double-check 3 Down, and we see that 2 Down is 25×31, so that checks out too, so we probably got it all right!

For starters, we should solve the equation for either n or x, in terms of the other.

Solution 1:  Solving for n, we find , so the largest x that leads to an integer n is x = 13.

Solution 2:  Solving for x, we find , so the largest value for x, given that is an integer, is 13, attained when n is also 13.

Let's look at a vertical plane that passes through the center of the upper basketball and the centers of two diagonally opposite lower basketballs, so their points of tangency will also be in the plane.

Looking at the triangle formed by the three centers, we know that the base is (by considering a diagonal from the square of four basketballs on the floor).  We also know that each of the upper sides is 25, so this is an isosceles right triangle of height .  So the height above the floor of the center of the upper basketball is

Let .  Look at :  it is .  Well, that's almost the same as , but actually it's 4 less.  So , and .

Are both solutions possible?  Well, there are two possible values for a: one just under 5, and the reciprocal of that.  These will lead to different values of , one a little bit more under 5, and the other being almost -5.  This matches what we found, so both and are indeed possible values.

Imagine the parallel being able to move back and forth, from the largest side of the triangle to the opposite point and back.  The trapezoid's perimeter will change linearly with the height of the parallel, from 30 when the parallel is at the bottom to 42 when the parallel is at the top.  The given perimeter of 39 is of the way from 30 to 42, so the parallel must be of the way from the bottom to the top, so the area of the little triangle must be of the area of the 13-14-15 triangle, which is 84.  So its area must be .

How on earth are we supposed to factor that?!?  Well, wait a minute, 27 is , so we can write it as .  Aha!  This is a sum of two cubes, which we know how to factor:

Well, 103 is prime, so that must be the answer.  Can we easily factor 9709?  It's not divisible by 3 or 5, but it is divisible by 7:  Since 1001=7·11·13, we see that .  If 1387 has a factor over 100, then it will have a corresponding factor under 14, which we can quickly check that it doesn't.  So we have confirmed that 103 is the only 3-digit prime factor of 1000027.

Drawing a picture, we see there are two possibilities for the third circle: a big "outside" one and a small "inside" one.  Our pictures aren't accurate enough to tell whether the resulting radii are between 1 and 9, so let's find the possible radii.  

Looking at the right triangles whose hypotenuses connect the circles' centers, we should be able to wring some equations out of the picture.

The sum of the two lower lengths (or their difference, if the lower triangle hangs out to the left as in the case for the bigger "outside" possibility) must be 12, so we have:

So in fact the "outside" possibility gives a radius of exactly 9, and therefore must be eliminated as not being between 1 and 9, leaving the unique solution .