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Can you help me with the steps, for the second one? I don't get how it works. 

2)   Let's place a point P on a circle opposite to point M. Now we have a right triangle CMP

∠ABC = arctan( AC / BC )

∠CMP = ∠ABC - ∠BCM

CM = cos(∠CMP) * MP

What is the circumradius of a triangle ABC with sides of length 6,8,and 10?  

We recognize that a 6, 8, 10 triangle is a right triangle.  6² + 8² = 10²  

Since it's a right triangle, the hypotenuse is the diameter of the circle it's inscribed in. 

The radius is half the diameter, so ....  radius = 5  

_.

Right scalene Pythagorean triangle.
Sides: a = 6 b = 8 c = 10

Area: T = 24
Perimeter: p = 24
Semiperimeter: s = 12
Circumradius(R)
The circumcircle of a triangle is a circle that passes through all of the triangle's vertices, and the circumradius of a triangle is the radius of the triangle's circumcircle. Circumcenter (center of circumcircle) is the point where the perpendicular bisectors of a triangle intersect.

R=(abc) / 4rs =6.8.10 / 4.2.12 =5 

The circumcenter of a right triangle occurs at the midpoint of the hypotenuse.

The radius of the circumcircle will be the distance from the midpoint to either end of the hypotenuse;

in this case, the radius is 5.

If you find the vertex, you will have the highest point.

One way to find the vertex is to complete the square:

         y  =  -12x² + 48x

         y  =  -12(x² - 4x)         

         y  =  -12(x² - 4x + 4) + 48

         y  =  -12(x - 2)² + 48

When  x = 2  the height will be  y = 48.

3ax² + 3bx + 3c

Factor out the 3:           3[ ax² + bx + c ]

Factor out the a:           3[ a( x² + (b/a)x ) + c ]

Complete the square:  3[ a( x² + (b/a)x + b²/(4a)² ) + c - b²/(4a) ]

Factor:                         3[ a( (x + b/(2a) )² ) + ( c - b²/(4a) )]

Multiply back the 3:         3a( (x + b/(2a) )² ) + ( 3c - 3b²/(4a) )

n = 3a         h  =  -b/(2a)     k  =  3c - 3b²/(4a)

First we find the number of terms which is (103-1)/3+1=102/3+1=34+1=35. Now notice that every pair adds up to 103+1=104. So the sum is 104(35)/2=1820.

1 - How many terms divisible by 2 are there between 2 and 8??

You subtract the last from the first: 8 - 2 =6. This is the number of terms.

6 / 2 = 3. But 3 doesn't count both 8 and  2. When you are subtracting from each other, only one of them is

counted, so you must add 1 to count both of 8  and 2. So that you have: 2, 4, 6, 8 for a total of 4.

Exactly the same thing with 1,000,000,001 as the first term and 9,9999,999,999, the last term.

9,999,999,999 - 1,000,000,001 =8,999,999,998 /2 =4,499,999,999 + 1 =4,500,000,000.

Got it?

Thank you so much! I looked at the original post but it was a tinypic photo so it didn't help me at all. Now time to go and finally understand this problem... Thanks again! 

:)

@answer #1 -- Thanks for your answer! However, I'm not sure I understand what you did! 

I get that the smallest and the largest numbers that meet are the ones you listed. However, why are you subtracting them, then dividing that by 2 and adding 1? I'm confused, can you explain? 

FYI for future answerers, I figured out #3, and #4, I don't need help and those. I still need help on #1 and #2. Thanks! 

I think what s/he meant was

If m is an integer is \(m^3\cdot 3^{k+3}-m^2\cdot 3^{2k+3}+m\cdot 3^{k+2}-1\) an integer?

Btw, try to do the writing problems by yourself on AoPS. They are a refelection of how well you learned and you should try to avoid getting help from other websites, as you can always post on the message board.

-C

Sorry for the typo!. The total should be = 12

y  =  -6t² + 51t

y = 108

Combining:                        -6t² + 51t  =  108

                                 -6t² + 51t - 108  =  0

Multiplying by -1:      6t² -  51t + 108  =  0

Dividing by 3:              2t² - 17t + 36  =  0

Factoring:                      (2t - 9)(t - 4)  =  0

Answer:  Either  t = 4.5  or  t = 4

(1,  4,  9,  16,  36,  64,  81, 144,  324, 576, 1296, 5184) >>Total = 13

Ok, thank you! I will try this out. 

Thanks!

Hello fellow AoPSer!

For the first one you should extend QS such that it intersects the circle again at T. And extend OS to intersect the circle at U and V. Try to find ST from there and set variables to solve an equation with power of a point.

For the second one. Let the intersection of AB and MC be D. Drop an altitude from D to AC. Then use the Angle Bisector Theorem, Similar Triangles, and lastly Power of a Point to get your answer.

Hope this helped!

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Can be done as follows: