Boseo

Since T is the point of external tangency, line OT is the radius of the larger circle, and line TC is the radius of the smaller circle.

Since ∠OTP and ∠TCP are right angles, triangle OTP and triangle TCP are right triangles.

We are given that OT=7 and TC=3. Since the altitude from T to line AB bisects the common external tangent, AP=PB=2AB​.

By the Pythagorean Theorem on triangle OTP, we have OP2=OT2−TP2=72−TP2, and by the Pythagorean Theorem on triangle TCP, we have OP2=TC2+CP2=32+(AP−TP)2=9+(2AB​−TP)2.

Setting these two expressions for OP2 equal, we get:

\begin{align*} 7^2 - TP^2 &= 9 + \left( \frac{AB}{2} - TP \right)^2 \ 49 - TP^2 &= 9 + \frac{AB^2}{4} - AB \cdot TP + TP^2 \ 0 &= \frac{AB^2}{4} - AB \cdot TP + 58 \end{align*}

Since we are looking for TP, let's rewrite this equation to isolate TP. We can factor the quadratic:

\begin{align*} 0 &= \frac{AB^2}{4} - AB \cdot TP + 58 \ 0 &= (AB - 14)(AB - 4) \end{align*}

Thus, either AB−14=0 or AB−4=0. Since a negative length for a line segment doesn't make sense, we must throw out the solution AB=14.

Therefore, AB=4, and applying the Pythagorean Theorem on triangle OTP one more time, we find that TP = sqrt(7^2 - (AB/2)^2) = sqrt(41).

Absolutely, let's find the value of k for the quadratic equation where one solution is twice the other solution.

We can approach this problem by letting the solutions be x1​ and x2​ and setting up the relationship between them along with the quadratic equation. Then, we can use the quadratic formula to solve for k.

Steps to solve: 1. Set up the relationships: Let x1​ and x2​ be the solutions of the quadratic equation. We are given that one solution is twice the other, so we can represent this as:

x1​=2x2​

We can substitute this into the quadratic equation:

x2+12x+k=0

This becomes: (2x2​)2+12(2x2​)+k=0

Expanding the equation: 4x22​+24x2​+k=0

2. Solve for k using the quadratic formula:

Since we know the relationship between x1​ and x2​, we can rewrite the equation in terms of x2​ only. Factoring out x2​, we get:

x2​(4x2​+24+k)=0

For the equation to hold true, either x2​ must be zero (which would mean both solutions are zero) or the expression in the parenthesis must equal zero.

Since we're given that there are two distinct solutions, we know x2​ cannot be zero. Therefore:

4x2​+24+k=0

This is a quadratic equation in terms of x2​. We can solve for k using the quadratic formula, which is:

k=(−b±b2−4ac​)/2a

In this case, a = 4, b = 24, and c = k. Substituting these values into the formula, we get:

k=(−24±242−4∗4∗k​)/2∗4

3. Solve for k:

Since we are given that there is only one solution (one root), the discriminant (the expression under the square root) must be zero. Setting the discriminant to zero and solving for k, we get:

242−4∗4∗k=0 576=16k k=16576​ k=−36

Answer: The value of k for the quadratic equation where one solution is twice the other solution is k = -36.

Consider the expression (3​/2−i/2)n. Let's analyze the powers of this complex number according to the four cases for the remainder r of n when divided by 4:

Case r=0: Then n=4k for some integer k, and

[(\sqrt{3}/2 - i/2)^{4k} = \left( (\sqrt{3}/2)^2 + (i/2)^2 \right)^{2k} = \left( \frac{3}{4} + \frac{1}{4} \right)^{2k} = 1^{2k} = 1.]

Since 1 is real, this case produces a real number.

Case r=1: Then n=4k+1 for some integer k, and

\begin{align*} (\sqrt{3}/2 - i/2)^{4k + 1} &= (\sqrt{3}/2 - i/2) \cdot (\sqrt{3}/2 - i/2)^{4k} \ &= (\sqrt{3}/2 - i/2) \cdot 1 \ &= \sqrt{3}/2 - i/2. \end{align*}

Since this expression has a nonzero imaginary part, it is not real.

Case r=2: Then n=4k+2 for some integer k, and

\begin{align*} (\sqrt{3}/2 - i/2)^{4k + 2} &= (\sqrt{3}/2 - i/2)^2 \cdot (\sqrt{3}/2 - i/2)^{4k} \ &= \left( \frac{3}{4} - \frac{2 \sqrt{3}}{4} i - \frac{1}{4} \right) \cdot 1 \ &= \frac{3}{4} - \frac{2 \sqrt{3}}{4} i. \end{align*}

Since this expression has a nonzero imaginary part, it is not real.

Case r=3: Then n=4k+3 for some integer k, and

\begin{align*} (\sqrt{3}/2 - i/2)^{4k + 3} &= (\sqrt{3}/2 - i/2) \cdot (\sqrt{3}/2 - i/2)^{4k + 2} \ &= (\sqrt{3}/2 - i/2) \cdot \left( \frac{3}{4} - \frac{2 \sqrt{3}}{4} i \right) \ &= \frac{3 \sqrt{3}}{4} + \frac{1}{4} - \frac{3}{4} i. \end{align*}

Since this expression has a nonzero imaginary part, it is not real.

In conclusion, the smallest positive integer value of n such that (3​/2−i/2)^n is real is 4​.

Solving for the Area of Triangle ABC

We are given the following information:

Triangle ABC with point X on side BC such that AX = 13, BX = 10, and CX = 10.

The circumcircles of triangles ABX and ACX have the same radius (denoted by r).

1. Using Similar Triangles:

Since the circumcircles of triangles ABX and ACX have the same radius, they are similar. Furthermore, triangles ABX and ACX share angle A. Therefore, these triangles are also similar by AA Similarity.

This implies that the ratio between corresponding side lengths in these triangles is constant. In particular:

AX / AB = BX / AX (ratio of corresponding heights from X to the base)

2. Finding AB:

We are given that AX = 13 and BX = 10. Substituting these values into the equation above:

13 / AB = 10 / 13

Cross-multiplying:

AB * 10 = 13 * 13

Solving for AB:

AB = (13 * 13) / 10 = 16.9

3. Area of Triangle ABC:

Since we have the base (BC = BX + CX = 10 + 10 = 20) and the height (corresponding to base BC, which is the altitude drawn from A to BC), we can find the area of triangle ABC using the area formula:

Area of Triangle ABC = ½ * Base * Height

Height of Triangle ABC (altitude from A to BC) = AX (since triangles ABX and ACX are similar, the height from X in each triangle is proportional to the corresponding side length opposite X).

Area of Triangle ABC = ½ * 20 * 13 = 130

Therefore, the area of triangle ABC is 130 square units.

For f(x)=f(k−x) to hold for all real numbers x, the function f must be symmetric about the line x=k. This means that for any input value x, the output f(x) must be the same as the output f(k−x).

Let's analyze the terms of f(x):

2x2 and x3 are even functions, meaning their output doesn't change sign when the input is negated (i.e., f(−x)=f(x) for even functions).

7x and −5x are odd functions, meaning their output flips sign when the input is negated (i.e., f(−x)=−f(x) for odd functions).

18x is a simple linear term that changes sign when the input is negated.

For f to be symmetric about x=k, the sum of the odd terms (linear terms) must be zero at x=k. In other words, when x=k, the contribution from the linear terms (−7x−5x+18x) must cancel out.

Setting x=k in these terms, we get:

−7k−5k+18k=0

Simplifying the equation:

6k=0

Therefore, k=0.

Since f(x)=f(k−x) holds for all real numbers x when k=0, our solution is k=0​.