BuilderBoi

We can rewrite the equation as: \(x^2-2x+9=0\)

Using Vieta's, the sum of the roots (a+b) is \(\large{-{b \over a}} = 2\), and the sum of the roots (ab) is \(\large {c \over a} = 9\)

We can also rewrite \(ab + a+b\) as \(ab + (a+b)\). 

Can you take it from here?

We can write this as: \((x-2) \times (x+2) \times (x-2) \times (x+2) \times (x+2)\)

Recall the identity: \((a-b)(a+b) = a^2-b^2\)

This means we can rewrite the equation as: \((x^2-4) \times (x^2-4) \times (x-2)\)

We know that \((x^2-4)(x^2-4) = x^2 \times x^2 -4 \times x^2 - 4\times x^2 + 16 = x^4-8x^2+16\)

Now, we have: \((x^4-8x^2+16)(x+2)\). 

Can you expand this?

The only numbers it can be are: 16 and 61, 25 and 52, and 34 and 43. 

Now, can you tell me which pair of numbers has a difference of 27?

Hint: The correct answer is the smaller number of the correct pair

What do you think it is?

Subsitute x = 0, x = 1, x = 2, and x = 3.

See what you get, and determine which graph makes more sense for the blue line. 

To find the circumcenter, we first need to know the radius of the circle. 

We can rewrite the circle equation to: \(x^2-4x+y^2+8y=-11\)

We can now complete the square, and we get: \((x-2)^2+(y+4)^2=4+16-11\)

The left hand is equal to 9, meaning the radius is 3. 

We now know the circumcenter is 3. 

The equation for the circumcenter is: \(\large{{abc} \over {4[ABC]}}\)

But, we have an equilateral triangle, so \(a = b= c\). This means we can substitute both b and c for a. 

This gives us the equation: \(\large{{a^3} \over 4[ABC] } = 3\)

The formula for the area of an equilateral triangle is \({\sqrt 3 \over 4} s^2\)

We can substitute this in, and we get: \({{a^3} \over {{4 \sqrt3 \over 4} a^2}} = 3\)

We can simplify and divide by \(a^2\) to get: \(\large{a \over \sqrt3} = 3\)

This means that the side of the triangle is \(3 \sqrt 3 \). 

Can you find the area from here?

Hint: Use the formula for the area of the triangle

We can rewrite \(\large{{u \over v} + {v \over u} }\) as \(\large{{u^2 \over xy} + {v^2 \over xy}}\). Because these have common denominators, we can simplify furthur: \(\large{{x^2+y^2} \over {xy}}\). Now recall the identity: \(a^2+b^2=(a+b)^2-2ab\). We can then apply this here, to get: \(\large{{(u+v)^2-2uv} \over uv}\).

Using Vieta's, we know that \(u+v= -{ b\over a} = -{5 \over 3}\) and \(uv = {c \over a} = {11 \over 3} \)

Now, we have to substitute these values and simplify. 

Can you take it from here?

By the intersecting chord theorem, we have this equation: \(AQ \times BQ = CQ \times DQ\)

Substituting what we know, we have: \(6 \times 12 = DQ \times CQ\)

Let \(CQ = x\) and \(DQ = y\). 

We can form the system: \(xy = 72\) and \(x+y = 30\)

Now, we have to solve for and take the smaller option for x. 

Can you do it from here?

Hint: You will substitute into a quadratic, and you will have to use the formula: \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

For the equation to have a double root, the discriminant (\(b^2-4ac\)) must equal 0. Substituting what we know, we have: \(b^2-4 \times 16 = 0 \)

Now, we have to solve for b. 

Can you take it from here?

Here is the link to the original problem, if anyone is interested in other solutions: https://gogeometry.blogspot.com/2016/03/geometry-problem-1194-equilateral.html

Notice that 414 is a multiple of 9. This means that the remainder isn't dependent on the value of n. 

The only other thing that can change the remainder is 12. 

This means that the remainder of \(414n+12\) will always be the same as 12. 

Can you take it from here?