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 #1
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Analyze the Cone Slices:

 

Imagine the cone is cut into four slices with equal heights. Since the cuts are parallel to the base, each slice is a smaller similar cone.

Relate Lateral Surface Area to Slant Height:

 

The lateral surface area (LSA) of a cone is directly proportional to the slant height (l) of the cone. This means that the ratio of the lateral surface areas of two similar cones is equal to the ratio of their slant heights. Identify Slant Heights:

 

Let L be the slant height of the entire cone (the largest piece). The second-largest piece will have a slant height that is some fraction of L.

 

Analyze Proportion Based on Similar Triangles:

 

Since each slice is a similar cone, the ratio between the heights of the entire cone and the second-largest piece is the same as the ratio between their slant heights.

 

Looking at the slices, we can see that the height of the second-largest piece is 3/4 of the height of the entire cone. Therefore, the slant height of the second-largest piece (l') is also 3/4 of the slant height of the entire cone (L).

 

Calculate the Ratio of Lateral Surface Areas:

 

As mentioned earlier, the ratio of the lateral surface areas (LSA) is equal to the ratio of the slant heights:

 

LSA (second-largest piece) / LSA (largest piece) = l' / L

 

Substituting the values we found:

 

LSA (second-largest piece) / LSA (largest piece) = (3/4)L / L

 

Simplifying:

 

LSA (second-largest piece) / LSA (largest piece) = 3/4

 

Therefore, the ratio of the lateral surface area of the second-largest piece to the lateral surface area of the largest piece is 3:4, expressed as a common fraction.

Apr 17, 2024
 #1
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Since ∠APB=∠PBA, triangles ABP and PBA are isosceles. This means that ∠ABP=∠BAP=21​(180∘−∠APB).

 

Similarly, since ∠BPC=∠PBC, triangles BPC and PBC are isosceles. This means that ∠BPC=∠PBC=21​(180∘−∠BPC).

Summing the angle measures in △ABC gives $ \angle ABC + \angle ACB + \angle BAC = 180^\circ$. Since ∠APB=∠PBA and ∠BPC=∠PBC, we can rewrite this as [ \angle ABP + \angle BAP + \angle BPC + \angle PBC + \angle BAC = 180^\circ. ] Substituting what we found for each angle measure, we get [ \frac{1}{2} (180^\circ - \angle APB) + \frac{1}{2} (180^\circ - \angle APB) + \frac{1}{2} (180^\circ - \angle BPC) + \frac{1}{2} (180^\circ - \angle BPC) + \angle BAC = 180^\circ. ] Simplifying the left side gives [ 2 \cdot 180^\circ - (\angle APB + \angle BPC) + \angle BAC = 180^\circ. ] We are given that ∠ABC−∠ACB=24∘, which is the same as ∠BAP−∠BPC=24∘ since ∠APB=∠BAP and ∠BPC=∠PBC. Substituting again, we get [ 2 \cdot 180^\circ - (\angle BAP + \angle BPC) + \angle BAC = 180^\circ \Rightarrow 2 \cdot 180^\circ - (180^\circ + 24^\circ) + \angle BAC = 180^\circ. ] Solving for ∠BAC gives ∠BAC=36∘.

Since the angles in a triangle sum to 180∘, we can find ∠PBC as follows: [ \angle PBC = 180^\circ - \angle BPC - \angle BAC = 180^\circ - \frac{1}{2} (180^\circ - \angle BPC) - 36^\circ. ] Substituting ∠BAP−∠BPC=24∘ again, we get [ \angle PBC = 180^\circ - \frac{1}{2} (180^\circ - (\angle BAP + 24^\circ)) - 36^\circ = 180^\circ - 90^\circ - 12^\circ = \boxed{78^\circ}. ]

Apr 16, 2024
 #1
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We can solve this problem by considering different cases for how the siblings are arranged across the two rows.

Case 1: Splitting Each Pair

In this case, we have one child from each pair in each row. There are 3!=6 ways to arrange the children in each row (since sibling order within a row doesn't matter). However, we've overcounted since we haven't considered which sibling from each pair sits in which row. So, for each arrangement, we need to multiply by 23 to account for the two choices (sibling 1 or sibling 2) for each of the three pairs. This gives us 6⋅23=48 arrangements for this case.

Case 2: Keeping Pairs Together

Here, both siblings from one or two pairs sit in the same row. There are two sub-cases to consider:

Two Pairs Together: There are (23​)=3 ways to choose which two pairs will sit together (the third pair will be split). Once chosen, there are 2!=2 ways to arrange the two siblings within each pair that sits together, and again 2 ways to decide which row the combined pair sits in. Finally, there are 2!=2 ways to arrange the remaining split pair in the other row. This gives a total of 3⋅2⋅2⋅2=24 arrangements for this sub-case.

One Pair Together: There's only one way to choose which pair will sit together. Within that pair, there are 2! ways to arrange them. There are then 2 ways to decide which row the pair sits in, and 3!=6 ways to arrange the remaining two pairs (who must be split) in the other row. This gives a total of 1⋅2⋅2⋅6=24 arrangements for this sub-case.

Adding the sub-cases of Case 2, we get a total of 24+24=48 arrangements.

Total Arrangements

Summing the arrangements from both cases, we get a total of 48(Case1)+48(Case2)=96​ ways to seat the siblings.

Apr 16, 2024
 #1
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Analyzing the function:

 

This function defines f(x) as the square root of x minus a nested sequence of square roots that keep getting smaller. Intuitively, as x decreases, the nested square roots will also decrease, bringing f(x) closer to an integer.

 

Finding a bound:

 

For f(x) to be an integer, x - sqrt(x - sqrt(x - ...)) must be a perfect square. Let the innermost square root be y. We can rewrite the function as:

f(x) = sqrt(x - y)

 

Squaring both sides:

 

x - y = f(x)^2

 

Since we want the largest possible three-digit x, let's consider the smallest possible value f(x) can take. The smallest perfect square greater than 10 (a three-digit number) is 121. So, let's assume f(x) = 11.

 

Substituting:

 

x - y = 11^2 = 121

 

This tells us that x needs to be at least 121 more than the innermost square root (y) to be a perfect square.

 

Finding the largest three-digit x:

 

We know x must be greater than 121 + y. Since y is a square root and squares are always non-negative, the largest possible value of y for a three-digit x would be the square root of the largest three-digit perfect square, which is 9^2 = 81.

 

Therefore, x must be greater than 121 + 81 = 202.

 

Checking values:

 

We can now check the largest three-digit perfect squares less than 202:

 

196 (14^2) - When plugged into the function, it results in a non-integer value.

 

169 (13^2) - This value works! f(169) = 13, which is an integer.

 

Therefore, the largest three-digit value of x such that f(x) is an integer is x = 169.

Apr 16, 2024