the power of a point powerpoint (with some other kinds of problems too) april 1, 2009
TRANSCRIPT
![Page 1: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/1.jpg)
The
Power of a Point PowerPoint
(with some other kinds of problems too)
April 1, 2009
![Page 2: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/2.jpg)
![Page 3: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/3.jpg)
Quadrilateral ABCD is circumscribed about circle O. Find the perimeter of ABCD.
![Page 4: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/4.jpg)
![Page 5: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/5.jpg)
Answer to Quadrilateral ABCD:28
![Page 6: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/6.jpg)
![Page 7: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/7.jpg)
Solve the equation log2x216 =x, where x is real.
![Page 8: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/8.jpg)
![Page 9: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/9.jpg)
Answer to log2x216: x = 3
![Page 10: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/10.jpg)
![Page 11: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/11.jpg)
If the sum of two complex numbers is 1 and their product
is 1, what is the sum of their squares?
![Page 12: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/12.jpg)
![Page 13: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/13.jpg)
Answer to sum of squares:-1
![Page 14: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/14.jpg)
![Page 15: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/15.jpg)
Hexagon ABCDEF is circumscribed about a circle.
AB = 8, CD = 9, EF = 10, and BC = 7.
Find the value of DE + FA.
![Page 16: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/16.jpg)
![Page 17: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/17.jpg)
Answer to Hexagon ABCDEF:20
![Page 18: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/18.jpg)
![Page 19: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/19.jpg)
Find the shaded area.
![Page 20: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/20.jpg)
![Page 21: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/21.jpg)
Answer to shaded area:30
![Page 22: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/22.jpg)
![Page 23: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/23.jpg)
If 19C8 (“19 choose 8”) is 75,582,
what is 19C9?
![Page 24: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/24.jpg)
![Page 25: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/25.jpg)
Answer to 19C9: 92,378
![Page 26: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/26.jpg)
![Page 27: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/27.jpg)
Find the center of the circle with equation
x2 + y2 - 6x + 4√2 y = 64.
![Page 28: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/28.jpg)
![Page 29: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/29.jpg)
Answer to the center of the circle: (3, -2√2)
![Page 30: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/30.jpg)
![Page 31: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/31.jpg)
What base 8 number does 1100102 represent?
![Page 32: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/32.jpg)
![Page 33: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/33.jpg)
Answer to base 8 number:628
![Page 34: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/34.jpg)
![Page 35: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/35.jpg)
AC = 4, CD = 5, radius OD = 18. Find AB.
![Page 36: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/36.jpg)
![Page 37: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/37.jpg)
Answer to AB: 171
4
![Page 38: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/38.jpg)
![Page 39: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/39.jpg)
Find the product of the roots of 6x2 + 17x – 42 = 0
![Page 40: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/40.jpg)
![Page 41: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/41.jpg)
Answer to product of the roots:-7
![Page 42: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/42.jpg)
![Page 43: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/43.jpg)
Find the constant term of the expansion of (x2 – 2/x)6
![Page 44: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/44.jpg)
![Page 45: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/45.jpg)
Answer to the constant term: 240
![Page 46: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/46.jpg)
![Page 47: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/47.jpg)
A circle of radius 2 rolls around the outside of a square of side 4. Find the length of the path
made by the center of the circle.
![Page 48: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/48.jpg)
![Page 49: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/49.jpg)
Answer to the circle rolling around the square:
16 + 4π
![Page 50: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/50.jpg)
![Page 51: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/51.jpg)
In how many distinct ways can the letters in LJUBLJANA be
arranged?
Bonus for double points: Of what country is Ljubljana the
capital?
![Page 52: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/52.jpg)
![Page 53: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/53.jpg)
Answer to the number of ways 45,360
Bonus answer: Slovenia
![Page 54: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/54.jpg)
![Page 55: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/55.jpg)
The End of Ciphering
Find the fallacy in each of the following April Fools problems.
![Page 56: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/56.jpg)
Three traveling salesmen stop at an inn. There is only one small room left. They are tired and take it anyway. The room is $30, so each of the 3 contributes $10. In the morning the manager arrives and decides to give them a partial refund. He gives the bellboy $5 to give to the salesmen. The bellboy realizes the men don’t expect a refund, so he gives them back only $3 and keeps $2 for himself. The men split the refund, taking $1 each. As each man had originally paid $10, but received $1 back, it ended up costing each man $9. They are happy with this and the bellboy is happy as he has $2 in his pocket.Question: each of the 3 men ended up paying $9. 3X9=27+2 (money in bellboy’s pocket) = 29. We started with $30. What happened to the extra $1?
![Page 57: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/57.jpg)
Step 1: Let a=b. Step 2: Then a2 = ab, Step 3: a2 + a2 = a2 + ab , Step 4: 2a2 = a2 + ab , Step 5: 2a2 - 2ab = a2 + ab -2ab, Step 6: 2a2 - 2ab = a2 - ab Step 7: This can be written as 2(a2 - ab) = 1(a2 – ab) Step 8: and canceling the (a2 – ab) from both sides gives 1=2.
![Page 58: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/58.jpg)
1: -1/1 = 1/-1 2: Taking the square root of both sides: √(-1/1) = √(1/-1) 3: Simplifying: √(-1) / √(1) = √(1) / √(-1) 4: In other words, i/1 = 1/i. 5: Therefore, i / 2 = 1 / (2i), 6: i/2 + 3/(2i) = 1/(2i) + 3/(2i), 7: i (i/2 + 3/(2i) ) = i ( 1/(2i) + 3/(2i) ), 8: (i2)/2 + (3i)/2i = i/(2i) + (3i)/(2i), 9: (-1)/2 + 3/2 = 1/2 + 3/2, 10: and this shows that 1=2.
![Page 59: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/59.jpg)
Step 3 is wrong. The problem is that there is no rule that guarantees √(a/b) = √(a) / √(b), except in the case in which a and b are both positive.
If this surprises you, think about the questionWhy should √(a/b) equal √(a)/√(b) ?
![Page 60: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/60.jpg)
If you were to try to convince someone of this, you'd have to start with the
definition of what a "square root" is: it's a number whose square is the number you started with. So all that has to be
true is that √(a) squared is a, √(b) squared is b, and √(a/b) squared is a/b.
So, when you square √(a/b), you will get a/b, and when you square √(a)/√(b), you will also get a/b. That's all that the
definition of square root tells you.
![Page 61: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/61.jpg)
Now, the only way two numbers x and y can have the same square is if x = ±y.
So, what is true is that √(a/b) = ± √(a)/√(b), but in general
there's no reason it has to be √(a/b) = +√(a)/√(b),
rather than √(a/b) = -√(a)/√(b), unless a and b are both positive, for
then (because by convention we take the positive square root) everything in
the above equation is positive.
![Page 62: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/62.jpg)
In our case, it is true that √(-1/1) = √(-1) / √(1), but √(-1/1) (that is, i)
is -√(1)/√(-1) (that is, -1/i) not +√(1)/√(-1) (that is, 1/i)
The fallacy comes from using the latter instead of the former.
![Page 63: The Power of a Point PowerPoint (with some other kinds of problems too) April 1, 2009](https://reader036.vdocument.in/reader036/viewer/2022062423/56649ec85503460f94bd59ef/html5/thumbnails/63.jpg)
In fact, the whole proof really boils down to the fact that
(-1)(-1) = 1, so √ (-1 * -1) = 1,
but √(-1) * √(-1) = i2 = -1 (not 1). The proof tried to claim that these two were equal (but in a more disguised way where it was harder to spot the
mistake).