A circle with radius 1 is drawn with an equilateral triangle

Question

1. A circle with radius 1 is drawn with an equilateral triangle inside it (the points of the triangle just touching the circle). Another circle is then drawn inside the triangle, just touching the triangle in the middle of each of its three faces. What is the radius of the smaller circle?

2. i) Integrate lnx/(x^2) between the limits 1 and infinity
ii) Integrate ((lnx)^n)/(x^2) between the same limits.

3. F=xÂ³y+yÂ³x-xy. Draw the loci of points on the x-y plane where F=0.
4. I have four cards on a desk. Each card has a letter on one side and a number on the other. Only one side of each card is visible, and the visible sides show 1, 2, A, and B. I claim that cards with an odd number on one side have a vowel on the other side. Which cards do I need to turn over to prove my claim and why.

5. I draw the curve y=ax^2. I then draw a circle that intersects this curve 4 times. Prove that the sum of the y coordinates of the four intersections is 0.

6. Prove that 3,5,7 is the only case of three consectutive odd numbers being prime. 

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GL

laxieqv · Answer

We will prove that for all such series other than 3,5,7 ...there will be one term perfectly divided by 3


let the three consecutive odd numbers be 

2k+1, 2k+3, 2k+5 respectively

1) 2k+1 is perfectly divided by 3 ----> the problem is proved ---> at least one number is not prime
2) 2k+1 is not perfectly divided by 3: the remainder can be either 1 or 2
a. r = 1 ----> (2k+1 ) -1 is perfectly divided by 3 ---> 2k is perfectly divided by 3 ----> 2k+3 is divisible by 3 -----> the problem is proved.
b. r= 2 ------> (2k+1)-2 = 2k-1 is divisible by 3 ----> (2k-1) + 6= 2k+5 is divisible by 3 -----> the problem is proved.

------> It is true that one of the three consecutive odd numbers is divisible by 3 ( thus, not a prime)

-------> 3,5,7 is the only series that satisfy the condition.

chuckle · Answer

Answer to the first question is 1/2 (Radius of inner circle).

laxieqv · Answer

I claim that cards with an odd number on one side have a vowel on the other side   ------> that means if a card has one side of an odd number , it will has the another of a vowel BUT its doesn't mean that a card having vowel on one side has an odd number on the other side!

---> surely we'll have to open the card containing 1 first of all.
+We don't need to open the card containing 2 coz if the other side is either an odd or an even ---> the claim is still true.
+ Similarly, we don't need to open the card containing A coz if the other side is odd ---> the claim is true ; if the card is even ----> the claim  is not wrong !
+We need to open B coz if the card has an odd ----> the claim is wrong!


In short, we need to open 2 cards: one with 1 and one with B.

laxieqv · Answer

Does this question ask us to draw the roots of the equation : x^3*y + y^3*x- xy=0 on the x-y plane? yx^3 + xy^3 -xy = 0 <=> xy( x^2+y^2) -xy= 0 <=> (x^2+y^2 -1 ) xy= 0 -----> x= 0 or y = 0 or x^2+y^2- 1 = 0 Drawn on the x-y plane, all the roots line on x-axis, y-axis and the circle centrered at the origin and having radius of 1

xennie · Answer

Laxie you continue to amaze me

I hope your applications go well, you certainly deserve a great MBA program.

A circle with radius 1 is drawn with an equilateral triangle

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