| Transcript
for the Math
Jam "USAMTS Round 3 Math Jam"
on Jan 11. |
| Math Jam hosted by DPatrick
(Dave Patrick ). |
DPatrick (19:30:19)
Hello and welcome to the third 2006-07 USA Math Talent Search Math Jam.
DPatrick (19:30:27)
Before we get started I would like to take a moment to explain our virtual classroom to those who have not previously participated in a Math Jam or one of our online classes.
DPatrick (19:30:34)
The classroom is
moderated, meaning that students can type into the classroom, but only the moderator (that's me) can choose a comment to drop into the classroom. This helps keep the class organized and on track. This also means that only well-written comments will be dropped into the classroom, so please take time
writing responses that are complete, easy to read, and on-topic. Also, only the moderator can enter into private chats with other users, although due to the size and nature of this Math Jam, it is unlikely that this will happen.
DPatrick (19:30:50)
There will be images in this lecture. The images should appear directly in the classroom window, as in the example below:
DPatrick (19:30:56)
DPatrick (19:30:58)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/a7dcd8010feb2cc00e4a33dc6e5ef2cd.png
DPatrick (19:31:09)
Also, if you click on the link, the image will appear in a separate window. You may have to hold down the Ctrl key while you click on an image link, and/or you may have to disable your popup blocker.
DPatrick (19:31:17)
You can view the 3rd round problems as we discuss them by clicking on the following link:
DPatrick (19:31:21)
http://www.usamts.org/Tests/USAMTSProblems_18_3.pdf
DPatrick (19:31:30)
Let's get started!
DPatrick (19:31:36)
Problem 1/3/18
DPatrick (19:31:40)
DPatrick (19:31:45)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-197460425.gif
DPatrick (19:31:56)
(I apologize in advance if you are color-blind or if you're using a black-and-white display; this is going to be a bit hard to follow.) [img id=em-1]
DPatrick (19:32:19)
DPatrick (19:32:21)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/751f4b201f0b3cb9a40c216de72f6607.png
robertnishihara (19:31:57)
Well each vertex has exactly one of each color
PI-Dimension (19:31:58)
there must be 4 of ea color
DPatrick (19:32:39)
Right. We can observe two important facts right away.
DPatrick (19:32:46)
Every vertex must have exactly one edge of each color.
DPatrick (19:32:52)
And there must be 4 edges of each color.
besttate (19:33:01)
why is the latter true?
DPatrick (19:33:36)
Because edge each connects two vertices. So the red edges must touch 8 total vertices, and each edge hits two, meaning there must be 8/2 = 4 red edges.
DPatrick (19:33:43)
Same for the other two colors (of course).
DPatrick (19:34:09)
There are lots of ways we could proceed from here.
vishalarul (19:34:16)
Try casework.
mustafa (19:34:27)
Start by randomly assigning one line one color
DPatrick (19:34:53)
That's what I would do. I'd start with just one color -- say red -- and see what the different cases are.
DPatrick (19:35:00)
So let's look first at just the red edges, and let's start with the top front edge colored red (if not, rotate the cube so that a red edge is there):
DPatrick (19:35:08)
DPatrick (19:35:10)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/67fdcf1358f2fb2396a0c325181d8aeb.png
DPatrick (19:35:25)
What are the possibilities for placing the other 3 red edges?
roadnottaken (19:35:45)
top back, lower square, back sides.
PI-Dimension (19:35:57)
skew or parallel
DPatrick (19:36:14)
Right...it's kind of hard to describe in text of course.
DPatrick (19:36:26)
There are only 3 ways to finish coloring the red edges:
DPatrick (19:36:31)
DPatrick (19:36:34)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/cf9af4b0a3b327e56eeca9ffbafbc7fb.png
DPatrick (19:36:46)
Are these 3 all different?
besttate (19:36:56)
1st and 3rd are the same
oppenhejo (19:37:02)
1 and 3 are indist
DPatrick (19:37:23)
Right. The 1st and 3rd are the same (the first can be rotated 90 degrees towards you to get the third).
DPatrick (19:37:45)
How do we know that the 1st and 2nd are indeed different?
EunuchOmerta (19:37:51)
parallel vs. skew
robertnishihara (19:37:59)
the lines are all parallel in 2
perfectnumber628 (19:38:05)
in the second, the red edges are all parallel
DPatrick (19:38:41)
Right. In the 2nd cube above, all 4 red edges are parallel. They're not in the 1st one above: they come in two parallel pairs, but the lines between the two pairs are skew.
DPatrick (19:38:51)
So there are essentially only two ways to color the red edges: all parallel, or in two parallel pairs that are skew.
DPatrick (19:39:09)
Let's look at the first (skew) case first. How do we fill in the yellow and blue edges?
besttate (19:39:34)
the 4 verticles must be all monochromatic
vishalarul (19:39:40)
A yellow or blue has to go "between" the reds.
DPatrick (19:40:08)
I think that the easiest way to look at it is to observe that the 8 black (uncolored) edges form a loop.
DPatrick (19:40:16)
This is maybe easier to see if I remove the red edges:
DPatrick (19:40:18)
DPatrick (19:40:21)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/70bc5cc2fed1392722ac6a7fd5598a02.png
DPatrick (19:40:31)
These edges have to be alternately colored blue and yellow, so the only choice is whether to start (say, on the left-most edge) with blue or with yellow; this gives two possibilities:
DPatrick (19:40:42)
DPatrick (19:40:47)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/3a03941f5a98e3d46c6799686e5de9da.png
DPatrick (19:40:52)
Are these two colorings indeed different? How do we know?
EunuchOmerta (19:41:06)
the yellow are parallel in one and skew in the other
robertnishihara (19:41:08)
1 blue all parallel red yellow not 2 yellow parallel others not
DPatrick (19:41:26)
Notice the left one has all blue edges parallel, whereas the right one has all yellow edges parallel. So they're different.
DPatrick (19:41:39)
Now let's go back to the case where all the red edges are parallel:
DPatrick (19:41:46)
DPatrick (19:41:49)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/7c14b6ba12194665d70018dc79ec5d61.png
DPatrick (19:42:00)
What choices do we have to finish coloring this?
vishalarul (19:42:00)
Now the yellow and blue alternate on the opposite faces.
DPatrick (19:42:20)
Right -- we've got to put 2 blue and 2 yellow on each black face.
robertnishihara (19:42:11)
They form two seperate loops which each must be alternately colored
mustafa (19:42:18)
Either yellow and blue are both parallel, or they are both skewed
nameless4365 (19:42:26)
there are two loops, in each we have to alternate blues and yellows, in two possible ways
DPatrick (19:42:41)
The only choice we have to make is whether the colors will 'line up', giving 4 parallel edges of each color, or not.
DPatrick (19:42:46)
DPatrick (19:42:49)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/53f0a700253d9442ff6848cccef4192d.png
DPatrick (19:42:59)
Are these different? Are they different from the previous two complete colorings?
jbano (19:43:10)
yes
robertnishihara (19:43:13)
yes
mustafa (19:43:18)
In one, they are all parallel, and in the other, only red is parallel, and in the first two, yellow and blue were parallel
DPatrick (19:43:28)
The left one has all three colors all parallel. The right one has only the red lines all parallel.
DPatrick (19:43:36)
So there are 4 colorings.
In summary, we get one coloring by making all three colors each all be parallel. We get the other 3 by making exactly one color all parallel, and the other 2 colors come in two skew parallel pairs.
DPatrick (19:43:57)
This is not such a hard problem, but what makes it potentially difficult is carefully organizing your work.
DPatrick (19:44:16)
Let's move on to Problem 2:
DPatrick (19:44:21)
DPatrick (19:44:24)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-38049101.gif
besttate (19:44:35)
switch cot and sin to the other sides
robertnishihara (19:44:36)
Try writing just in sin and cos then combining tan and cot in one fraction
vishalarul (19:44:40)
Get the tan's and cot's on one side and the sin's and cos's on the other side.
SamE (19:44:43)
put the tangent and cotangent together
oppenhejo (19:44:44)
rewrite the tan and cot on the same side
DPatrick (19:44:53)
Yes.
DPatrick (19:44:59)
A big mistake would be to try to find 'formulas' to simplify expressions like tan 7x. That's a huge mess.
DPatrick (19:45:07)
It usually helps to get similar terms on the same side of an equation. So let's rewrite what we have to get the tan 7x and cot 7x terms together.
DPatrick (19:45:10)
DPatrick (19:45:24)
Now what?
13375P34K43V312 (19:45:19)
apply am-gm to the LHS but make sure they're positive
lotrgreengrapes7926 (19:45:22)
Combine tan and cot
mustafa (19:45:25)
Then combine the tan and cot
SamE (19:45:35)
use am-gm on their absolute values
robertnishihara (19:45:36)
tan = sin/cos cot=cos/sin
DPatrick (19:45:56)
Often, it pays to examine equations
qualitatively, meaning roughly looking for the broad strokes or most general features, than to dig into messy computations.
DPatrick (19:46:07)
The left side looks nicer, so it's perhaps more promising to look at it qualitatively.
DPatrick (19:46:12)
cot 7x is just the reciprocal of tan 7x. So the left side is something of the form y + 1/y for some y.
DPatrick (19:46:18)
What do we know about such an expression?
SamE (19:46:29)
|y+1/y|>=2
kostya (19:46:38)
between 2 and inf. or between -2 and - inf.
DPatrick (19:46:58)
If y>0, then it must be at least 2.
DPatrick (19:47:06)
This is the AM-GM inequality, or it's easy to prove directly in this case by completing the square:
DPatrick (19:47:11)
DPatrick (19:47:35)
So if tan 7x > 0, then the LHS (left hand side) is at least 2. (Similarly, if tan 7x < 0, then the LHS is at most -2.)
DPatrick (19:47:45)
But what does this say about the RHS?
DPatrick (19:47:50)
roadnottaken (19:48:02)
It cannot be between -2 and 2.
tjhance_2 (19:48:03)
sin6x+cos4x=-2 or 2, so sin6x=cos4x=1 or -1
EunuchOmerta (19:48:04)
cos4x = sin 6x = 1 or cos4x = sin6x = -1
iostream.h (19:48:07)
cos and sin have range [-1,1], so cos 4x = sin 6x = 1 or -1 for all x in the solution set.
DPatrick (19:48:38)
The RHS is the sum of a sine and a cosine. Each of these terms is
at most 1.
DPatrick (19:49:02)
But if they're positive, they have to sum to
at least 2 (by our analysis of the left side).
DPatrick (19:49:10)
So the only way that they can sum to 2 is if they're each 1!
DPatrick (19:49:18)
DPatrick (19:49:21)
DPatrick (19:49:40)
What's the easiest way to finish the problem?
SamE (19:49:38)
so now we analyze the values that make the cosine and sine what we want
EunuchOmerta (19:49:49)
so cos4x = 1 at values of the form pi k / 2
DPatrick (19:50:09)
Right. We can just check values by hand to finish.
DPatrick (19:50:28)
cos 4x = +/- 1 is the most restrictive, so this gives us the smallest list of values that we'd have to check.
DPatrick (19:50:45)
13375P34K43V312 (19:50:53)
none of these work
vishalarul (19:50:56)
None of these work!
DPatrick (19:51:03)
But all of these have tan 7x either 0 or undefined, so none of these work.
DPatrick (19:51:10)
PI-Dimension (19:51:03)
pi/4 or5pi/4
DPatrick (19:51:29)
DPatrick (19:51:46)
The moral of the story is that often examining equations qualitatively can produce very useful insight, especially in situations where there's no good quantitative approach.
DPatrick (19:52:07)
Problem 3/3/18
DPatrick (19:52:12)
DPatrick (19:52:18)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-37168846.gif
DPatrick (19:52:29)
Obviously we'll want to start by sketching a picture. Let's start with just the circles:
DPatrick (19:52:34)
DPatrick (19:52:37)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/12f01d4b97ce65181331a96a52dc686b.png
13375P34K43V312 (19:52:22)
AC+CB=8
robertnishihara (19:52:24)
ac+cb=8
DPatrick (19:52:56)
Indeed, so we need X on circle C with AX+XB = 8.
SamE (19:52:32)
X lies on an ellipse
tjhance_2 (19:52:33)
X and C are on an ellipse with A and B as foci
roadnottaken (19:52:47)
That last equation makes the term "ellipse" spring to mind.
calc rulz (19:52:47)
AX+XB=AX+CB looks like the condition for an ellipse
robertnishihara (19:52:57)
A, B foci of ellipse
kostya (19:53:00)
X is on an ellipse
DPatrick (19:53:23)
Indeed, that's the key observation!
DPatrick (19:53:47)
The locus of all points P such that AP+PB is a constant (in this case, 8) is an ellipse with foci A and B.
DPatrick (19:53:52)
We can now sketch it:
DPatrick (19:53:55)
DPatrick (19:53:58)
http://www.artofproblemsolving.com/Admin/latexrender/pictures/fde74fdd23c9d12433d541408f6b8bae.png
EunuchOmerta (19:53:29)
cooridnization time!
vishalarul (19:53:47)
Should we go to coordinates after this?
besttate (19:54:03)
coordinates?
DPatrick (19:54:27)
This seems like a job for analytic geometry.
DPatrick (19:54:43)
If we put the origin in the logical place -- the point of tangency between A and B -- then we're just looking for the y-coordinate of X, which is the height of triangle ABX.
DPatrick (19:54:54)
So we need equations for circle C and the ellipse.
besttate (19:55:13)
we need to assign A B and C to coordinates first, right?
DPatrick (19:56:00)
Well, the logical place for the origin is between A and B. Then A = (-2,0), B = (2,0), and C = (0,2*sqrt(3)).
EunuchOmerta (19:55:14)
C: x^2 + (y - 2 root 3)^2 = 4
tjhance_2 (19:55:34)
c=2*sqrt(3) so its equation is x^2+(y-2sqrt(3))^2=2^2
DPatrick (19:56:22)
The circle is easier: it's centered at (0,2*sqrt(3)) and has radius 2:
DPatrick (19:56:25)
DPatrick (19:56:41)
The ellipse is a little trickier.
EunuchOmerta (19:55:25)
ellipse: x^2/ 16 + y^2/12 = 1
roadnottaken (19:55:26)
x^2/16+y^2/12=1
vishalarul (19:56:53)
We need the major and minor axes for the ellipse.
DPatrick (19:57:18)
Right. We have to notice that it has x-radius 4 and y-radius 2*sqrt(3), giving:
DPatrick (19:57:20)
vishalarul (19:57:23)
Now we solve for y; we don't care about x.
DPatrick (19:57:50)
Right...given that we want y and don't care about x, what's the easiest thing to do?
besttate (19:57:50)
substitue for x^2 using first equation
vishalarul (19:58:03)
Should we solve for x^2 and substitute?
not_trig (19:57:49)
multiply by 16 for ellipse's equation
not_trig (19:57:55)
and subtract
bpms (19:58:05)
Multiply the second by 16 and subtract
DPatrick (19:58:27)
Either way is good -- the point is to eliminate x and solve for y.
DPatrick (19:58:38)
I would solve for x^2 in the circle equation and plug it into the ellipse:
DPatrick (19:58:43)
vishalarul (19:58:38)
The rest is algebra.
DPatrick (19:59:00)
Yep, we just clean this up to a quadratic:
DPatrick (19:59:04)
DPatrick (19:59:16)
Quadratic formula time!
DPatrick (19:59:17)
knexpert (19:59:26)
y is positive because it is above the x axis
robertnishihara (19:59:28)
but y must be positive
DPatrick (19:59:36)
The negative solution clearly doesn't work (and leads to an imaginary value of x), so we have that:
DPatrick (19:59:40)
PI-Dimension (19:59:48)
multiple by 2!
tjhance_2 (19:59:55)
1/2 * base * height formula
not_trig (19:59:55)
so [AXB] = 1/2(4)(6(\sqrt{5}-\sqrt{3}))
DPatrick (20:00:09)
DPatrick (20:00:48)
On to Problem 4:
DPatrick (20:00:54)
DPatrick (20:00:56)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-22108893.gif
DPatrick (20:01:16)
Obviously, if you can solve (b), then you can just plug in numbers to solve (a), but the point of (a) is to provide a specific example for you to experiment with, and also a value with which to check your answer to part (b).
DPatrick (20:01:42)
So let's experiment with the 12-player tournament.
JerryS (20:01:41)
Suppose that 4 other players at least tie her
knexpert (20:01:42)
maximize the score of 5 of the players in part a and add 1
DPatrick (20:02:08)
One possible "worst-case" scenario for Alice if is she is a member of a group of 5 -- call them the Winners -- and the other 7 people are in a group called the Losers. All the players within each group tie, and each Winner beats each Loser.
DPatrick (20:02:30)
We'll use the notation W-T-L to mean that a player gets W wins, T ties, and L losses. (Note that this is different as to how sports records are usually reported in the newspaper.) Of course, we must have W+T+L = 11, and the number of points earned is 2W+T.
not_trig (20:02:21)
that's 18 points
DPatrick (20:03:01)
Right, in our scenario described above, Alice (and the other 4 Winners) each finish 7-4-0, for 18 points. So 18 is not enough for Alice to have more points than 8 others.
DPatrick (20:03:20)
So we know that answer to part (a) is
at least 19.
DPatrick (20:03:33)
How can we prove that if Alice has >=19, then she must beat at least 8 people?
besttate (20:03:14)
19 will work though, if you try
VDLmath (20:03:24)
So what if we try 19?
DPatrick (20:04:01)
What do you mean by "try 19"?
DPatrick (20:04:23)
The issue here is that we can't just show that there's some configuration with 19 points that works. We need to show that
all configurations with 19 points work.
robertnishihara (20:04:15)
Suppose the worst case scenario where she has 19
not_trig (20:04:19)
19 cannot be split among 5 winners
vishalarul (20:04:20)
Look at the worst-case scenario if she got 19 points.
roadnottaken (20:04:05)
Contradiction
SamE (20:04:32)
If Alice and 4 others have at least 19 points each...
besttate (20:04:31)
try to create a worst case senario, there are not enough points to go around for the rest of the winners to score 19
DPatrick (20:04:47)
Right.
DPatrick (20:05:01)
Suppose that there are at least 5 players with at least 19 points.
DPatrick (20:05:12)
These players must have at leat 5*19 = 95 points among them.
DPatrick (20:05:40)
The other 7 players, just from playing each other, must have at least 2*C(7,2) = 42 points from their games.
DPatrick (20:06:04)
So how does this lead to a contradiction?
not_trig (20:06:07)
the total points scored are 2xC(12,2) = 132
robertnishihara (20:06:17)
42+95>12*11
VDLmath (20:06:18)
the total points is supposed to be 132
DPatrick (20:06:45)
Right. The tournament has a total of 2*C(12,2) = 132 points (there are C(12,2) = 66 games, and each game awards 2 points).
DPatrick (20:07:02)
But if at least 5 players have at least 19 points, this leads to a total of at least 95+42 = 137 points, which is too many.
DPatrick (20:07:25)
So there can be at most 4 players with at least 19 points, and hence if Alice has 19, she must beat at least 8.
DPatrick (20:07:45)
Experimenting with this specific case should provide the road map for the general solution.
DPatrick (20:07:58)
We now have an n-player tournament.
SamE (20:07:57)
But our proof for part (a) still isn't done.
DPatrick (20:08:36)
What's left to do? We did both parts: we showed 18 doesn't work, and we showed that >=19 has to work. It's important to do both parts, but once we've established both parts, we're done.
SamE (20:08:14)
We need to prove it for ALL numbers less than 19, not just 18.
DPatrick (20:09:09)
Actually, we don't. Read the question carefully and hopefully you'll see why.
DPatrick (20:09:28)
Let's move on to the general case.
DPatrick (20:09:54)
As with the specific case above, we first find a case where Alice beats only k-1 other players. Divide the players in (n-k+1) Winners (including Alice) and (k-1) Losers.
DPatrick (20:10:09)
How much does each Winner score?
robertnishihara (20:10:26)
2(k-1)+n-k
kostya (20:10:54)
(k-1)*2+n-k
DPatrick (20:11:19)
Each Winner has k-1 wins and n-k ties, for a total of 2(k-1)+(n-k) = n + k -2 points.
DPatrick (20:11:30)
So n+k-2 is not enough to guarantee beating k players.
DPatrick (20:12:00)
Now for the other half...we need to show that there can't be n-k+1 players with at least n+k-1 points.
DPatrick (20:12:38)
If at least (n-k+1) players have at least (n+k-1) points, then the remaining (k-1) players must have at least (k-1)(k-2) points among them.
not_trig (20:12:37)
Total points = 2(C(n,2))
DPatrick (20:13:02)
Right, we compare the number of points alloted above to the total number of points in the tournament, which is 2*C(n,2).
DPatrick (20:13:08)
DPatrick (20:13:23)
And there is a total of n^2-n points in the tournament.
Eternica (20:13:33)
but (k-1)<n
VDLmath (20:13:34)
This is a contradiction
DPatrick (20:13:59)
DPatrick (20:14:08)
Of course n-k >= 0, otherwise the problem wouldn't make sense.
DPatrick (20:14:18)
So we have too many points, and have a contradiction.
DPatrick (20:14:25)
Thus the answer is n+k-1.
DPatrick (20:14:40)
We check that this matches part (a): plug in n=12 and k=8, and we get 19.
DPatrick (20:14:56)
We also check that this matches the extreme cases. If k=n-1 (meaning Alice beats everybody), then the only way that we can guarantee that this happens is if Alice scores the max 2n-2 points. (If she scores 2n-3 points by going (n-2)-1-0, then the person she ties might also go (n-2)-1-0, so she wouldn't beat everybody.)
EunuchOmerta (20:14:52)
wait, what if k = 0
DPatrick (20:15:07)
Aha! Our analysis doesn't work for k=0.
DPatrick (20:15:15)
This is a special case.
DPatrick (20:15:46)
The answer for k=0 is m=0. (The condition is trivial if we don't require Alice to beat anybody.)
calc rulz (20:15:32)
Will we lose points if we don't mention that?
13375P34K43V312 (20:15:57)
did we have to include this?
SplashD (20:16:02)
Did we have to note that?
DPatrick (20:16:17)
I'm not sure.
DPatrick (20:16:31)
Let's go on to Problem 5.
DPatrick (20:16:39)
DPatrick (20:16:43)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-237663838.gif
besttate (20:17:00)
cal f(1) some number z, for cleaner notation
DPatrick (20:17:39)
Sure, we want f(1), so let's give it a name. Let's call f(1) = c (so it matches my notes).
vishalarul (20:17:06)
Plug in 1 for x.
VDLmath (20:17:25)
plug in 1
DPatrick (20:18:00)
With functional equations, often the best strategy is the naive strategy: plug in numbers and see what happens.
DPatrick (20:18:10)
In this case, it seems most obvious to plug in x=1:
besttate (20:17:49)
c(f(c+1))=1
Goistein_2 (20:18:08)
f(1)f(f(1)+1/1)=1, when f(1)=c, so f(c+1)=1/c
vishalarul (20:18:10)
DPatrick (20:18:22)
DPatrick (20:18:27)
So f(c+1) = 1/c.
besttate (20:18:37)
c+1 is in the domain
DPatrick (20:19:06)
Right, so why not plug it in too?
DPatrick (20:19:12)
DPatrick (20:19:24)
So f(1/c + 1/(c+1)) = c.
DPatrick (20:19:33)
What do we conclude?
besttate (20:19:27)
1/c+1/(c+1)=1 then!
iostream.h (20:19:35)
note that f is injective, so 1/c + 1/(c+1) = 1
vishalarul (20:19:37)
But f(1)=c.
DPatrick (20:20:00)
Right! f(1) = c, and we know that the function is strictly increasing.
DPatrick (20:20:26)
So if f( 1/c + 1/(c+1)) = c, and f(1) = c too, we must have 1 = 1/c + 1(c+1).
besttate (20:20:02)
c^2-c-1=0 (simple algebra)
vishalarul (20:20:34)
Solve for c!
DPatrick (20:20:49)
DPatrick (20:20:53)
Which is it, plus or minus?
not_trig (20:20:56)
minus!
EunuchOmerta (20:20:58)
minus
Goistein_2 (20:21:01)
minus
DPatrick (20:21:13)
How come?
.cpp (20:21:30)
Because it's a strictly increasing function.
calc rulz (20:21:38)
If we use the plus then the fcn is not increasing!
not_trig (20:21:42)
f(1) cannot be greater than 1
DPatrick (20:21:54)
We know that f(1) = c and f(c+1) = 1/c. If c is the positive root, then c+1 > 1, but 1/c < c, so that can't work since f is supposed to be an increasing function.
DPatrick (20:22:06)
DPatrick (20:22:17)
With a little persistence, one can write down the explicit formula for f(x); I'll leave that as an exercise.
DPatrick (20:22:35)
That's it for Round 3.
DPatrick (20:22:39)
Round 4 Problems will be posted by the end of the month, possibly sooner.
