Boseo

Let $O_1$ and $O_2$ be the centers of circles $P$ and $Q$ respectively. Let $A$ and $B$ be the points where line $PQ$ intersects the circles $P$ and $Q$ respectively.

Let $C$ be the midpoint of $\overline{AB}$. Let $D$ be the foot of the perpendicular from $O_2$ to $\overline{PQ}$, and let $E$ be the foot of the perpendicular from $O_1$ to $\overline{O_2D}$.

Since $\overline{O_1E}$ and $\overline{O_2D}$ are both perpendicular to $\overline{PQ}$, $O_1EDO_2$ is a rectangle. Therefore, $O_1E=O_2D$.

Since $O_1O_2=12+8=20$, we know $O_1E=O_2D=10$. Because $\angle O_1AQ=\angle O_2BQ=90^\circ$, angles $O_1EA$ and $O_2DE$ are both right angles.

Thus, by the Pythagorean Theorem on $\triangle O_1AE$, \[AE=\sqrt{12^2-10^2}=2\sqrt{7}\]

and by the Pythagorean Theorem on $\triangle O_2DE$, \[ED=\sqrt{20^2-10^2}=10\sqrt3.\]

Finally, we find the length of $\overline{QR}$. By the Pythagorean theorem on $\triangle CDE$,

\[QR = CD = \sqrt{AE^2+DE^2}=\sqrt{(2\sqrt7)^2+(10\sqrt3)^2} = \boxed{4\sqrt{10}}.\]

Let O be the center of the circle, let X, Y, and Z be the points of tangency with AB, AC, and BC, respectively, and let D be the midpoint of BC. Let r be the radius of the circle.

Since BC is a diameter of the circle, DO=r. Therefore, by Pythagoras on triangle BOD, BD=BC2−DO2​=36−r2​.

Since △COD is isosceles with CD a leg, CO=DO2+CD2​=r2+9​. Similarly, AO=r2+4​.

By Power of a Point, OX2=AO⋅AX, so r2=(r2+4​)(r+DX), or r3+4r=r2+4r+rDX, which simplifies to rDX=0. Since DX=0, we have r=0, which is impossible.

Therefore, the only solution is r=2​.

 Let there be n + 1 entries in the row (including the 1s at the beginning and end). Since the average is 2048, the sum of all entries must be 2048 * (n + 1).

The entries in the given row are consecutive multiples of n, except for the 1s at the ends. Therefore, the sum can be expressed as 1 + \binom{n + 1}{1} + \binom{n + 1}{2} + ... + \binom{n + 1}{n + 1}.

This is an arithmetic series with n + 1 terms and a common difference of n. The sum of an arithmetic series is given by the formula: Sn = n/2 * (a1 + an), where Sn is the sum, n is the number of terms, a1 is the first term, and an is the last term.

In this case, a1 = 1 and an = \binom{n + 1}{n + 1}. Plugging these into the formula, we get:

Sn = (n + 1) / 2 * (1 + n^2)

Equating the sums: Now, equate the two expressions for the sum:

2048 * (n + 1) = (n + 1) / 2 * n^2

Cancel out the common factor (n + 1) on both sides:

2048 = (1 + n^2) / 2

Multiply both sides by 2:

4096 = 1 + n^2

Subtract 1 from both sides:

4095 = n^2

Take the square root of both sides:

63 = n

Therefore, n = 63 is the value that makes the average of the entries in the row equal to 2048.

Here's how we can show that the side length of the cube is 𝑠 = 𝑎𝑏𝑐/(𝑎𝑏 + 𝑎𝑐 + 𝑏𝑐):

Diagonal of the cube face:

Each face of the cube is an equilateral triangle with side length equal to the cube's side length, say 𝑠. Since the tetrahedron has right angles at O, the face ABC in which the cube is inscribed is also an equilateral triangle. Therefore, the diagonal of this equilateral triangle (connecting A and B) is √3 times the side length:

AB = √3 * s

Relationship between edges and face diagonal:

We can use the Pythagorean theorem in right triangle ABO to relate the side lengths of the cube (s) and the tetrahedron edges (a and b):

a^2 + b^2 = s^2

Solving for s^2:

s^2 = a^2 + b^2

Expressing s using a, b, and c:

Since OB and OC are perpendicular to each other, we can form another right triangle BOC and apply the Pythagorean theorem again:

b^2 + c^2 = s^2

Substituting the expression for s^2 from step 2:

b^2 + c^2 = a^2 + b^2

Rearranging this equation to isolate b^2:

b^2 = a^2 + c^2 - b^2

2b^2 = a^2 + c^2

b^2 = (a^2 + c^2)/2

Substitute and simplify:

Now, substitute this expression for b^2 in the equation for s^2 from step 2:

s^2 = a^2 + (a^2 + c^2)/2

s^2 = (2a^2 + a^2 + c^2)/2

s^2 = 3a^2 + c^2 / 2

Final step:

Take the square root of both sides to solve for s:

s = √((3a^2 + c^2)/2)

Now, multiply both numerator and denominator by 2b:

s = √((6ab + 2bc)/4b)

Simplifying further:

s = √((3ab + bc)/2b)

Finally, we can rewrite this expression using the givens that ab + ac + bc is the sum of all possible pairwise products of edges:

s = abc/(ab + ac + bc)

Therefore, we have shown that the side length of the cube is indeed 𝑠 = 𝑎𝑏𝑐/(𝑎𝑏 + 𝑎𝑐 + 𝑏𝑐).

Let's call the centers of the circles A and B, and let the tangency points be C and D, as shown in the diagram below:

Since the radii of the circles are 2 and 6, AC=2 and BD=6. Also, AB=10, since the centers of the circles are 10 units apart.

Triangle ABD is a right triangle, so by the Pythagorean Theorem, AD=AB2−BD2​=102−62​=64​=8.

Therefore, the distance between the two tangency points C and D is 8​ units.

Let's call the centers of the circles A and B, and let the tangency points be C and D, as shown in the diagram below:

Since the radii of the circles are 2 and 6, AC=2 and BD=6. Also, AB=10, since the centers of the circles are 10 units apart.

Triangle ABD is a right triangle, so by the Pythagorean Theorem, AD=AB2−BD2​=102−62​=64​=8.

Therefore, the distance between the two tangency points C and D is 8​ units.

We can solve this problem similarly to the previous one by analyzing the discriminant of the quadratic:

Discriminant and no real solutions: Recall that for the quadratic ax^2 + bx + c = 0 to have no real solutions, the discriminant b^2 - 4ac must be negative. In this case, a = 1, b = y, and c = 1.

Therefore, the condition for no real solutions becomes:

y^2 - 4 * 1 * 1 < 0

Simplifying the inequality:

y^2 - 4 < 0

Adding 4 to both sides:

y^2 < 4

Taking the square root of both sides (remembering that we need both positive and negative solutions):

-2 < y < 2

Integer solutions: Now, we need to count the number of integer values of y within this range. Since -2 and 2 are not integers, we consider the closed interval [-1, 1]. However, we need to be careful because the original inequality involved x^2 instead of just x.

Special cases: The original inequality x^2 + yx + 1 < 0 becomes equal to zero for x = -1 and x = 1 when y = -1 and y = 1, respectively. These values of y cannot satisfy the condition of no real solutions because they introduce real roots through x = -1 and x = 1.

Therefore, we exclude -1 and 1 from the possible values of y.

Final count: After excluding -1 and 1, we are left with the closed interval [-1, 0) and (0, 1]. Both intervals contain 1 integer each: 0 for the first and 1 for the second.

Therefore, there are 1 + 1 = 2 possible values of y for which the inequality has no real solutions (excluding y = -1 and y = 1).

Note: This case is different from the previous one because the original inequality involves x^2, which can produce double roots at x = 0. This necessitates excluding y = -1 and y = 1, which would otherwise satisfy the condition but introduce double roots that violate the requirement of no real solutions.

A rhombus with one angle of 90∘ is a square, so in this case, the area is 122=144​.