|By Aim78 (Aim78) on Thursday, September 30, 2004 - 08:43 pm: Edit|
This one is on page 608 of 10 Reals, #25.
I got it right, but I'm not confident about how I got the answer. I feel like I assumed a few things that I shouldn't have. Anyone want to use geometry to explain why the intersection is 90 degrees?
|By Legendofmax (Legendofmax) on Thursday, September 30, 2004 - 09:18 pm: Edit|
Well it says CD bisects EF, and since EF divides the square (all sides equal length, all angles equal) into two triangles, that means angle C-intersection-E and E-intersection-D have to be 90 degrees.
|By Aim78 (Aim78) on Thursday, September 30, 2004 - 11:55 pm: Edit|
We must be looking at different problems - this is the second edition red book.
|By Legendofmax (Legendofmax) on Thursday, September 30, 2004 - 11:58 pm: Edit|
What test date/section/question number?
|By Aim78 (Aim78) on Friday, October 01, 2004 - 01:24 am: Edit|
Sat 2000 section 1 #25
|By Bakk (Bakk) on Friday, October 01, 2004 - 04:38 pm: Edit|
Actually the 90% angles are not relevant to finding the SAT solution.
Take a close look at what ETS did give you, and focus on the fact that you have 2 isosceles (and congruent) triangles PQS and SQR, and also that triangle PQR is isosceles. If angle QSR is equal to 70 degrees, so is angle QRS. This means that angle SQR is 180 - (2*70) or equal to 40 degrees. Because of congruency, angle PQR is equal to 80 degrees. Since PQR is isosceles, x is 1/2 of (180 minus 80), or 50.
It takes a lot longer to explain than to solve this problem!
As always, quickly redrawing the figure and focusing on EVERY part of the problem statement helps solve this type of problem.
|By Aim78 (Aim78) on Friday, October 01, 2004 - 08:41 pm: Edit|
Ah, thank you. I didn't notice all of the isosceles triangles.
Here's another problem from Sunday May 2000 test.
25. Circle C has radius SQRT(2). Squares with sides of length 1 are to be drawn so that, for each square, one vertex is on circle C and the rest of the square is inside circle C. What is the greatest number of such squares that can be drawn if the squares donot have overlapping areas?
If you just take 2pi/1 = 2pi = ~4, how are you guaranteed that all of the squares fit inside? There must be a better way.
|By Aim78 (Aim78) on Friday, October 01, 2004 - 10:05 pm: Edit|
New problem, folks.
|By Bakk (Bakk) on Saturday, October 02, 2004 - 01:11 am: Edit|
Aim, draw your circle, and then divide it in four equal parts (diameters at 90 degree angles).
In one of the four quadrants, draw another radius that bisects the 90 degree angle. Consider this newly drawn radius the diagonal of one your squares. Guess what the diagonal of a square with lengths of 1 is equal to?
Since the highest number of the solution is 4, it's pretty easy to see that you could repeat that in each quadrant or 4 times.
The reality is that all you need to know is the value of the diagonal of a 1 unit squared square, but a question 25 deserves a small diagram.
|By Mr_Sanguine (Mr_Sanguine) on Saturday, October 02, 2004 - 12:56 pm: Edit|
Can anyone answer this?
How many positive integers less than 1001 are divisible by either 2, or 5, or both?
|By Adidasty (Adidasty) on Saturday, October 02, 2004 - 05:16 pm: Edit|
|By Aim78 (Aim78) on Saturday, October 02, 2004 - 05:19 pm: Edit|
Thanks Bakk, I didn't even think about the relationship between 1 and root 2.
Sanguine, I would say 600. 500 are divisible by 2 and 200 are divisible by 5. However, there's an overlap because half of the numbers divisible by 5 are also divisible by 2. So only 100 are only divisible by 5. 100+500 = 600.
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