Boseo

The probability of Raina getting alternating colors depends on whether she picks an orange ball or a purple ball first. Let's analyze both scenarios:

Scenario 1: Orange - Purple - Orange (OPO)

Probability of picking an orange ball first: 3 out of 8 total balls (3 orange + 5 purple).

Probability of picking a purple ball after an orange: 5 out of 7 remaining balls (1 orange left + 5 purple).

Probability of picking another orange ball after a purple: 2 out of 6 remaining balls (2 orange left).

So, the probability of OPO is (3/8) * (5/7) * (2/6) = 5/56.

Scenario 2: Purple - Orange - Purple (PPO)

Probability of picking a purple ball first: 5 out of 8 total balls.

Probability of picking an orange ball after a purple: 3 out of 7 remaining balls (3 orange left + 4 purple).

Probability of picking another purple ball after an orange: 4 out of 6 remaining balls (4 purple left).

So, the probability of PPO is (5/8) * (3/7) * (4/6) = 5/56.

Total Probability

Since these scenarios (OPO and PPO) are mutually exclusive (they can't happen at the same time), to get the overall probability, we simply add the probabilities of each scenario:

Total Probability = Probability(OPO) + Probability(PPO) = (5/56) + (5/56) = 10/56 = 5/28

Therefore, the probability of Raina drawing three balls with alternating colors (orange-purple-orange or purple-orange-purple) is 5/28.

Since TIP and TOP are isosceles triangles, all their corresponding sides must be equal.

Therefore, TP must be equal to PO, the longest side, which is 11.

So the sum of all possible lengths for TP is simply 11.

Analyze the Setup:

We have a circle with points X, Y, and Z on it.

Line XY intersects a tangent line drawn at point Z at point W.

We are given that WY=1 and YZ=1 (both lengths are the same).

We need to find the length XZ.

Utilize Properties of Tangents and Perpendicular Lines:

Since the tangent line is drawn at point Z, it's perpendicular to the radius drawn from the center of the circle (O) to point Z (radius OZ).

We are also given that WY⊥WZ. This implies that ∠WZY is a right angle.

Apply the Pythagorean Theorem:

Since we have a right angle at point Z (proven in step 2), triangle WYZ is a right triangle. We are given that the lengths of two legs, WY=1 and YZ=1.

By the Pythagorean Theorem:

WZ2=WY2+YZ2

WZ2=12+12

WZ2=2

Taking the square root of both sides (positive for lengths), we find:

WZ=2​

Recognize Symmetry and Solve for XZ:

Since the tangent line is perpendicular to the radius at the point of tangency (point Z), radius OZ bisects XW (symmetry). This means WX=WZ=2​.

Now, consider triangle XYZ. We know the lengths of two sides, YZ=1 and WZ=2​. Since these two segments form a right angle (proven in step 2), triangle XYZ is also a right triangle.

We can again apply the Pythagorean Theorem:

XZ2=YZ2+WZ2

XZ2=12+(2​)2

XZ2=1+2

XZ2=3

Taking the square root of both sides (positive for lengths), we find:

XZ=3​

Therefore, the length YZ is 3​​.

Let's denote lengths:

AP=x (since it's a multiple of PB)

PB=x/3

CP=y (since information about this segment isn't given, we denote it with a variable)

PD=y/3 (similar logic as for PB)

AB=a (what we're solving for)

CD=c (not directly needed, but can be helpful for visualization)

Apply the Power of a Point Theorem:

The Power of a Point Theorem states that for any point P inside a circle, the product of the lengths of the two segments created by drawing secants from that point to the circle is equal. In our case, point P is inside the circle (since chords intersect within the circle), and we can apply the theorem to both secants AB and CD.

For secant AB:

AP⋅PB=x⋅3x​=3x2​

For secant CD:

CP⋅PD=y⋅3y​=3y2​

Since both expressions represent the same power of point P, they must be equal:

3x2​=3y2​

Utilize the given information:

We are given that AP=3⋅PB, which translates to x=3⋅3x​ (substituting the values we defined). This simplifies to x=3x​, which implies x=0. However, a chord cannot have zero length. Therefore, our initial assumption (that x represents a positive length) must be incorrect.

Here's the correction: We can rewrite the given information as x=3PB=3⋅3x​. Solving for x, we get x2=9. Taking the square root of both sides (remembering positive for lengths), we have x=3.

Substitute and solve for AB:

Since we found x=3, we can substitute this value back into the equation we obtained from the Power of a Point Theorem:

3x2​=3y2​

332​=3y2​ (substitute x with 3)

3=y2

Taking the square root (positive for lengths), we have y=3​.

Now, consider segment AP : its total length is x+PB=3+33​=4. Since AP=x=3, segment PB must have a length of PB=14−3​=1.

Finally, to find the length of AB, we add the lengths AP and PB:

AB=AP+PB=3+1=4​.

Since PR and QS​ are perpendicular chords, they bisect each other. This means that point O, the center of the circle, lies on the line segment connecting P and S.

We are given that PR=10 and QS=24. Since PR is bisected by O, segments PO and OR each have length 2PR​=5. Similarly, segments QO and OS each have length 2QS​=12.

Now, consider right triangle POS. We know the length of both legs, PO=5 and OS=12, so we can use the Pythagorean Theorem to find the length of the hypotenuse, which is segment RS.

By the Pythagorean Theorem:

RS2=PO2+OS2

RS2=52+122

RS2=25+144

RS2=169

Taking the square root of both sides (remembering that since we're dealing with lengths, we only care about the positive square root), we find:

RS=169​

RS=13​.

We can use the Tangent-Secant Theorem to solve for AB.

The Tangent-Secant Theorem states that the square of the tangent segment (XY in this case) is equal to the product of the whole secant segment (XB) and the external secant segment (XA).

In this case, we are given that XY=8 and XA=10. Let XB=b.

Therefore, applying the Tangent-Secant Theorem:

XY2=XB⋅XA

82=b⋅10

64=10b

b=1064​

b=6.4

Therefore, the length of segment AB is $ \boxed{6.4}$.

Folding the sheet of paper as described brings vertex B to coincide with vertex D.

The resulting figure is a pentagon with side lengths AB, BC, CD, DE, and EA.

We know that AB=8.5 inches and AD=11 inches. Since ABCD is a rectangle, CD=AB=8.5 inches.

Segment DE is the hypotenuse of a right triangle with legs AD=11 inches and BE=2AB​=4.25 inches.

By the Pythagorean Theorem, we can find DE=112+4.252​≈156.25​≈12.5 inches (rounded to the nearest tenth).

The final side length, EA, is the hypotenuse of another right triangle with legs AB=8.5 inches and BE=4.25 inches.

Using the Pythagorean Theorem again, we find that EA=8.52+4.252​≈90.25​≈9.5 inches (rounded to the nearest tenth).

Adding up all the side lengths, we find the perimeter of the pentagon is

AB + BC + CD + DE + EA = 8.5 + 8.5 + 8.5 + 12.5 + 9.5 = 47.5​ inches (to the nearest tenth).

Analyzing the Cube Cuts

Imagine the large cube is 3 units on each side. When we cut it into smaller cubes, each side of the larger cube will be made of 3 smaller cubes.

(a) Cubes with One Black Face

A cube will have exactly one black face only if it's on the edge of the larger cube but not on a corner.

Edges: There are 12 edges on a cube.

Corner exceptions: On each edge, there are 2 smaller cubes that touch a corner.

Since corner cubes will have 3 black faces, we subtract these exceptions from the total edge cubes. There are 8 corners, so there are 8×2=16 corner exception cubes.

Therefore, the number of cubes with one black face is the total number of edge cubes minus the corner exceptions: 12 edges - 16 corner exceptions = -4 cubes

This seems like a negative number of cubes, which doesn't make sense. The mistake lies in assuming all the edge cubes have one black face.

Here's the correction: We only counted the  edges once, but each edge actually has 2 cubes that qualify (one on each side). So, we need to multiply the number of edges by 2:

Total one-black-face cubes = (2 cubes/edge) x (12 edges) - 16 corner exceptions = 24 - 16 = 8 cubes

(b) Cubes with No Black Faces

These cubes must be completely inside the larger cube, not touching any of the faces.

Inner core: Since each side of the larger cube is made of 3 smaller cubes, the inner core will be a cube with sides of length 1 unit less (3 - 2 = 1 unit). The volume of this inner core is therefore 1 x 1 x 1 = 1 cube.

Inner cubes: This inner core cube is itself made of smaller cubes. Each side has 1 cube, so there are a total of 1 x 1 x 1 = 1 smaller cube inside.

Therefore, there is only 1 cube with no black faces.

(c) Probability of Top Face Black

When a small cube is rolled, there are 6 possible faces that could land on top.

Out of these 6 faces, only 1 face is painted black (since we're considering the cubes with at least one black face).

Therefore, the probability of the top face being black is the number of black faces divided by the total number of faces: Probability = 1 black face / 6 total faces = 1/6.