The home stretch, we're on problem 18.
And this diagram actually looks like something that this
little tool I'm using is well-suited to draw.
Let me try-- I'll draw the one in the back first, so I can do
different colors and everything.
So I'll do the one in the back in yellow, and it looks
something like this.
Goes like this, then it switches over like that.
And then I'll draw the next one, this orange color, and it
looks something-- no, I'll do it in a more drastic color.
So it does something like this.
It goes like this, crosses over, dips down, then it goes
over again.
And then finally, this last one, and it looks
something like this.
Crosses over and it goes like that, very pretty.
Now let me switch to a smaller-- OK.
Why don't we do white, I never write in white.
In the figure above--so they say this is the start.
And this is step 1.
This is step 2.
OK, what is this?
In the figure above, three wires are braided.
That is, by starting in the order A, B, C-- OK, so this is
A, B, C-- and then the order changes to B-- right, because
the yellow line is B-- A, C.
And then we get B, C, A.
Right?
There's just the order of the strings, left to right.
Or the braids, or the wires.
OK.
That is start with the order A, B, C, the outer left wire
is brought over wire B, right?
This is brought over wire B, to the middle position forming
the order shown in step 1.
Than the outer right wire C, this one, is brought to the
new middle position shown in step 2, and so on.
Alternately bringing each new left and each new right wire
to the middle.
At what numbered step does the braid first repeat the
original order A, B, C?
So what you do is, you first bring the left over the
middle, then you bring the right over the middle, then
you bring the left over the middle, then you bring the
right over the middle.
So at step 3, what do we have to do?
So step 1 we brought the left over the middle.
So we went from A, B, C.
You essentially switch the left and the middle.
And then step 2, you switch the right and the middle.
So step 3, we switch B and C again.
Because we're back on the left-hand side.
So you get C, B, A.
And then step 4, you're going to switch to
the right-hand side.
Step 3 we switched these two.
Now step 4, we're going to switch these two.
So you get C, A, B.
Then in step 5, we're back switched on this side.
So then you get A, C, B.
And then step 6, you're going to switch to the right-hand
side again.
So you get A, B, C.
So by step 6, we repeat.
So that's choice D.
The hard part here, as far as I'm concerned, is just
understanding the problem and their diagram.
And then just seeing the pattern that you're just
switching the letters.
First you switch the left and the middle letter.
Then you switch the right and the middle letter.
Then you switch the left and the middle letter.
Then you switch the right and the middle letter.
Until you get back to A, B, C.
Next problem.
White I don't think is colorful enough.
Let me do magenta.
19.
In a set of 11 different numbers-- they're different--
which of the following cannot affect the
value of the median?
So it's a set of 11 different numbers, they're different, so
let's say it's 1 through 11.
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11.
That's 11 different numbers.
And in this case, what's the median?
Well we have 5 on this side, 5-- the median is this, this
is the middle number, right?
6 is the middle number.
Doubling each number-- well, if you double every number,
the median's going to become 12, so that's not right.
Increasing each number by 10-- once again, if you increase
all of these numbers by 10, this number's going to be 16,
so you would've changed the median.
Increasing the smallest number only.
Well, what if you increase the smallest number to,
I don't know, 12?
So then this 12 would go all the way on
the other side, right?
And then the median would move to 7 because you would have to
have 5 below the median and 5 above the median.
So you can increase that smallest number.
But if you increase it by a large enough amount, it kind
of goes into the greater than the median column, and then
the median would have to shift to the right.
Try it out, write out the list from 2 to 12, and then the
median becomes 7.
OK, so we know that choice B is not right.
Choice C, increasing-- oh no, that was choice C-- choice D,
decreasing the largest number only-- well, the same
argument can apply.
We could take the 11 and decrease it down to, I don't
know, decrease it to 0.
You decrease it to 0, then that'll-- and you could list
them all out, 0 through 10-- and then the median number
will shift to 5, so that will change it.
E, increasing the largest number only.
I can tell you that's the answer from deductive
reasoning, but if we just take this 11 and we make it one
billion, does that change the fact that they're five numbers
larger than 6 and five numbers less than 6?
Ignore the 0.
Five numbers less than 6?
No.
I can make it into a trillion.
I can change it to any number, but the fact remains that I
have the same amount of numbers larger than 6
as I do below 6.
So choice E cannot affect the value of the median.
E.
Next problem.
OK, something for me to draw.
Let's see, they have a 1/4 circle, so I'll just draw a
big circle, and I'll just focus on a 1/4 of it.
Close enough.
That looks like a 1/4 of a circle.
They say that-- oh, and there's a rectangle there too,
fascinating.
So this rectangle in this 1/4 circle, so this rectangle
could look like this.
I know theirs looks taller, but I think this will do.
And then I think I'm almost done.
I think I'm there, I think I can.
And then they shade in some stuff, so if they shaded it, I
will shade it.
A suitably tasteful color.
OK, so I have shaded what they have shaded.
And now, what are they saying?
They're saying that this is R, C, T, B, S, A.
And then they tell us that is a 90 degree angle.
And then they tell us that this distance, from here to
here, from all the way to the top-- so it's essentially the
radius of the circle-- is 6.
The radius of the circle is 6.
In the figure above, arc SBT is 1/4 of a circle with center
R and radius 6.
Fair enough.
If the length plus the width of rectangle ABCR is 8-- so
the length plus the width, so ABCR.
So this is the width, this is the length.
So L plus W is equal to 8.
What is the perimeter of the shaded region?
Fascinating.
Let's do it step by step.
First we can just figure out what this part of the
perimeter is.
And that's the easiest part, right?
Because what's the perimeter of the whole circle?
Well, it's the circumference of the whole circle.
Circumference is equal to 2 pi r.
This is the whole circle we're talking about.
So that's equal to 2 pi times 6, so that's equal to 12 pi.
This is the circumference of the whole circle, so the
circumference of this arc, of this piece, is going to be 1/4
of that, because it's 1/4 of the whole circle.
So it's 1/4 of 12 pi, so that's 3 pi.
Now, what we need to be able to figure out is L and W,
because if we can figure out L and W, we can figure out
everything else about this circle.
And actually, I'm going to continue this problem in the
next video, because I think it might get a little involved,
and I don't want you to get too confused.
I'll see you in the next video.