The3Mathketeers

Below is a diagram of the problem presented with all points labeled appropriately.

There are probably many methods that work, but I decided to use the Law of Cosines to find an angle, which will allow me to find other information about the triangle.

\({\rm BC}^2 = {\rm AC}^2 + {\rm AB}^2 - 2({\rm AC})({\rm AB})\cos m \angle \rm A \\  12^2 = 8^2 + 17^2 - 2 * 8 * 17 * \cos m \angle {\rm A} \\  144 = 64 + 289 - 272 \cos m \angle {\rm A} \\  -272\cos m \angle {\rm A} = -209 \\  \cos m \angle {\rm A} = \frac{209}{272}\)

\(\overline{AD}\) is the base and \(\overline{CD}\) is the height of \(\triangle ACD\). Since \(\triangle ACD\) is a right triangle, finding the remaining sides is somewhat easier although the numbers become ugly.

\(\cos m \angle {\rm A} = \frac{\rm AC}{\rm AD} \\ \frac{209}{272} = \frac{8}{\rm AD} \\ {\rm AD} = \frac{209}{34}\)

Now, we can use Pythagorean's Theorem to find CD. The numbers only get worse from here.

\({\rm AC}^2 = {\rm AD}^2 + {\rm CD}^2 \\ 8^2 = \left(\frac{209}{34}\right)^2 + {\rm CD}^2 \\  {\rm CD}^2 = \frac{73984}{34^2} - \frac{43681}{34^2} \\  {\rm CD}^2 = \frac{30303}{34^2} \\  {\rm CD} = \frac{\sqrt{30303}}{34}\)

Now, we have the base and the height, we can calculate the area of the desired triangle.

\(A_{\triangle \text{ACD}} = \frac{1}{2}bh \\  A_{\triangle \text{ACD}} = \frac{1}{2} * \text{AD} * \text{CD} \\  A_{\triangle \text{ACD}} = \frac{1}{2} * \frac{209}{34} * \frac{\sqrt{30303}}{34} \\ A_{\triangle \text{ACD}} = \frac{209\sqrt{30303}}{2314} \approx 15.7363\)

The given quadratic equation simplifies to \(x^2+(b - 8)x + 26 = 0\). To determine when the quadratic equation has at least one root, we can use the information obtained from calculating the discriminant, which provides insight into the number of real roots a particular quadratic equation can have.

\(\Delta = b^2 - 4ac \\ \Delta = (b - 8)^2 - 4 * 1 * 26 \\ \Delta = b^2 - 16b + 64 - 104 \\ \Delta = b^2 - 16b - 40\)

When \(b^2-16b-40\geq 0\) , then the quadratic equation has at least one real root. I will use the corresponding equation \(b^2-16b-40= 0\) to find the relevant values of b with the quadratic formula.

\(b^2 - 16b - 40 = 0 \\ b_{1, 2} = \frac{16 \pm \sqrt{256 - 4 * 1 * -40}}{2} \\ b_{1, 2} = \frac{16 \pm \sqrt{416}}{2} \\ b_1 = 8 - 2\sqrt{26}, b_2 = 8 + 2\sqrt{26} \)

Since \(b^2-16b-40\) is an upward-facing parabola because the leading coefficient, 1, is positive, this inequality is true b < b_1 or b > b_2, which is when the original quadratic has at least one real root.

Written interval notation, \((-\infty,8-2\sqrt{26}) \cup (8 + 2\sqrt{26}, \infty)\) would describe the range of values for b to guarantee at least one solution.

1) If each leg of 3-4-5 right triangle is the length of the diameter, we can convert that to the radius of each circle, which is used more traditionally in area calculations for circles.

\(r_{1} = \frac{3}{2}, r_{2} = \frac{4}{2}, r_{3} = \frac{5}{2}\)

Now, find the area of each circle with the formula \(A = \pi r^2\). I am not simplifying 4/2 to ease the addition process later.

The sum of these areas is \(A_{\text{total}} = A_1 + A_2 + A_3 = \frac{9 \pi}{4} + \frac{16 \pi}{4} + \frac{25 \pi}{4} = \frac{25 \pi}{2}\).

2) I think this problem requires more information to give a definitive answer. Do we know if the 3 smaller internally tangent circles have equal radii or something like that? I feel like not all the information is given in this problem, so I will not speculate and instead ask for clarification and see if you missed some details.

An infinite geometric series has a defined sum \(S = \frac{a_1}{1 - r}\) where a_1 is the first term and r is the common ratio as long as |r| < 1.

To obtain the next term in the series \(x^2 - 2x^4 + 4x^6 - \cdots \), multiply the previous term by -2x^2, which is the common ratio. We know that the series equals 1, so we can put this information together to find possible values of x.

\(S = \frac{a_1}{1 - r} \\ 1 = \frac{x^2}{1 - (-2x^2)} \\ 2x^2 + 1 = x^2 \\ x^2 + 1 = 0 \\ x^2 = -1\)

At this point, we need not go further as there are no real-valued solutions to x^2 = -1. No real values of x make the series equal 1. I suppose this means that the sum of all values of x is 0?

1) If \(x_1,x_2, \cdots x_6\) represents a sequence of 6 digits where leading zeroes are allowed, then the number of sequences can be calculated as follows:

\(x_1\) is a free choice; it can be any of the digits from 0 to 9 inclusive for a total of 10 options.

\(x_2\) can be any of the digits between 0 to 9 inclusive for a total of 10 options, but \(x_2 \neq x_1\). This excludes one option. Therefore, there are 9 options for \(x_2\).

\(x_3, \cdots ,x_6\) are all the same case as \(x_2\). There are 10 possible digits for each, but \(x_{i + 1} \neq x_i\), so there are 9 options again for each digit.

The number of sequences with no two adjacent digits is \(10 * 9 * 9 * 9 * 9 * 9 = 10 * 9^5 = 590490\)

2)

Assuming we are talking about a standard 6-sided die, the first die is a free choice from 1 to 6 inclusive. Therefore, there are 6 options for the first roll. For the second roll, there will always be 3 options. Consider the two different cases. If the first roll is even, then, to maintain parity, the second roll must be 2, 4, or 6 for a total of 3 options. If the first roll is odd, then, to maintain parity, the second roll must be 1, 3, or 5 for a total of 3 options.

The number of ways of rolling two dices is \(6 * 3 = 18\).

If we graph both equations, it is elementary to count the lattice points. Below is a graph of the bounded region:

There are only two points on or within the boundary. Those points are (0, 0) and (0, 1). Therefore, the number of lattice points is 2.