MaxWong

A fun solution:

\(\displaystyle\sum_{k = 1}^n 1 = n\)

Keeping that in mind:

\(\displaystyle \sum_{n = 1}^\infty \frac{n}{3^n} = \sum_{n=1}^\infty \sum_{k=1}^n 3^{-n}\)

By Fubini's theorem, the sums are interchangeable.

\(\displaystyle \sum_{n = 1}^\infty \frac{n}{3^n} = \sum_{k=1}^\infty \sum_{n=k}^\infty 3^{-n} = \sum_{k =1 }^\infty \dfrac{3^{-k}}{1 - \dfrac13} = \dfrac32 \cdot \dfrac{\dfrac13}{1 - \dfrac13} = \dfrac34\)

We need to know what that number is, so we let x be that number.

\(\dfrac6x = \dfrac94\)

Cross-multiplying,

\(9x = 24\)

Dividing by 9 on both sides:

\(x = \dfrac{24}9 = \dfrac83 = 2\dfrac23\)

Let Q(x) be the quotient and R(x) be the remainder:

\(x^{9999} + x^{8888} + x^{7777} + \cdots + x^{1111} + 1 = Q(x) (x^2 - 1) + R(x)\)

If we let x² = 1 on both sides, then it gives \(x^{9999} + x^{8888} + x^{7777} + \cdots + x^{1111} + 1 = R(x)\) when all x²'s are replaced by 1.

We rewrite the expression more explicitly in terms of x².

\(x^{9999} + x^{8888} + x^{7777} + \cdots + x^{1111} + 1 = (x^2)^{4999}\cdot x+ (x^2)^{4444} + (x^2)^{3888}\cdot x + \cdots + (x^2)^{555}\cdot x + 1\)

Replacing all x² by 1:

\(R(x) = x + 1 + x + 1 + x + 1 + x + 1 + x + 1 = \boxed{5x + 5}\)

Obviously. O is the centroid of \(\triangle PQR\).

Because PN = NR, and QN is perpendicular to PR, PQ = QR.

And, 

\(PQ^2 = PN^2 + NQ^2 = \left(\dfrac12 \cdot 14\right)^2 + 12^2 = 193\\ PQ = \sqrt{193} = QR\)

Using Apollonius's theorem, 

\(MR = \dfrac12 \sqrt{2\cdot 14^2 + 2\cdot \sqrt{193}^2 - \sqrt{193}^2} = \dfrac32 \sqrt{65}\)

Because centroid divides each medians into 2 : 1,

\(OR = \dfrac23 MR = \boxed{\sqrt{65}}\)

This is a visual representation of the problem:

Hope this helps :)

What are the first parts and last parts? You didn't show us!

Let r be the radius.

Let O be the midpoint of AD, i.e., the center of the semicircle.

Let T be the point where the line BC touches the semicircle.

As BA and BT are both tangents to the circle, meeting at the same point:

BT = BA = 12

Then, using Pythagorean theorem on \(\triangle CTO\), \(CO^2 = CT^2 + OT^2\)

Notice that CO = CD + DO = 1 + r, and OT is one of the radii of the semicircle.

\((1 + r)^2 = CT^2 + r^2\\ CT = \sqrt{2r + 1}\)

Also, CA = CD + DO + OA = 1 + r + r = 1 + 2r

Using Pythagorean theorem on \(\triangle CAB\),

\((1 + 2r)^2 + 12^2 = \left(\sqrt{2r + 1} + 12\right)^2\)

Let \(t = \sqrt{2r + 1}\).

\(t^4 + 144 = (t + 12)^2\\ t^4 - t^2 - 24t = 0\\ t(t^3 - t -24) = 0\)

Factorising further using Factor Theorem, (t - 3) is a factor of (t³ - t - 24).

By long division, t³ - t - 24 = (t - 3)(t² + 3t + 8).

\(t(t - 3)(t^2 + 3t + 8) = 0\)

Checking the discriminant of (t² + 3t + 8), there are no real t satisfying t² + 3t + 8 = 0.

Also, t = 0 would imply r = -1/2, which isn't possible either.

The only possible scenario is t = 3.

Recall the definition of t (defined above)

\(\sqrt{2r + 1} = 3\\ 2r + 1 = 9\\ \boxed{r= 4}\)

P.S. Considering the difficulty of this problem, I am suspecting this is some kind of math contest problems 

We only need Law of Indices for this one.

\(6^{3t - 1} = 36^{t - 3}\\ 6^{3t - 1} = (6^2)^{t - 3}\\ 6^{3t- 1} = 6^{2(t - 3)}\\ 6^{3t - 1} = 6^{2t - 6}\\ \)

Now, we equate the powers:

\(3t - 1 =2t - 6\)

You can proceed by solving the linear equation.

Step 1: Calculate the area of the large rectangle, expanding all brackets

Step 2: Calculate the area of the hole, expanding all brackets

Step 3: Subtract the area of the hole from that of the large rectangle.

Try to factor by grouping.

\(x^2 + 13x + 30 = (x^2 + 10x) + (3x + 30) = x(x + 10) + 3(x + 10) = (x + 3)(x + 10)\)

\(x^2 + 5x - 50 = x^2 + 10x - (5x + 50) = x(x + 10) - 5(x + 10) = (x - 5)(x + 10)\)

The rest is trivial.

Hints: Find b first.