<Question>
Let $P(x)$ be a quadratic polynomial with real coefficients satisfying $x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3$ for all real  numbers $x$, and suppose $P(11) = 181$. Find $P(16)$.
</Question>

<think>
<Plan_1>
Okay, let's try to figure out this problem. So, we have a quadratic polynomial P(x) that satisfies two inequalities for all real numbers x: the lower bound is x² - 2x + 2 and the upper bound is 2x² - 4x + 3. Also, we know that P(11) = 181, and we need to find P(16). Hmm, quadratic polynomials... they have the form ax² + bx + c. Since the inequalities hold for all real x, maybe we can compare coefficients or something? But wait, the inequalities are given for all x, so maybe P(x) is somehow sandwiched between these two quadratics. Let me think.

First, let's write down the two quadratics:

<prompt_1.1> Identify the form of the quadratic polynomial P(x) and the given quadratic inequalities. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
Lower bound: x² - 2x + 2

Upper bound: 2x² - 4x + 3

So, for all x, x² - 2x + 2 ≤ P(x) ≤ 2x² - 4x + 3.

Since P(x) is a quadratic, let's denote P(x) = ax² + bx + c. Our goal is to find a, b, c such that the inequalities hold for all x, and also satisfy P(11) = 181. Then use that to find P(16).
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
But how do we relate these inequalities to the coefficients of P(x)? Maybe by considering the difference between P(x) and the lower bound, and between the upper bound and P(x). Let me try that.

<prompt_2.1> Consider the inequalities between P(x) and the given quadratic bounds. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
Let’s define two new quadratics:

1. Q(x) = P(x) - (x² - 2x + 2) = (a - 1)x² + (b + 2)x + (c - 2)

2. R(x) = (2x² - 4x + 3) - P(x) = (2 - a)x² + (-4 - b)x + (3 - c)

Since P(x) is always greater than or equal to the lower bound, Q(x) ≥ 0 for all real x. Similarly, since P(x) is always less than or equal to the upper bound, R(x) ≥ 0 for all real x. Therefore, both Q(x) and R(x) are non-negative quadratic polynomials.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Now, for a quadratic to be non-negative for all real x, two conditions must be satisfied:

<prompt_3.1> Determine the conditions for Q(x) and R(x) to be non-negative for all real x. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
1. The leading coefficient must be non-negative.

2. The discriminant must be less than or equal to zero.

So, applying this to Q(x):

1. a - 1 ≥ 0 ⇒ a ≥ 1

2. The discriminant of Q(x) is [(b + 2)]² - 4*(a - 1)*(c - 2) ≤ 0

Similarly, for R(x):

1. 2 - a ≥ 0 ⇒ a ≤ 2

2. The discriminant of R(x) is [(-4 - b)]² - 4*(2 - a)*(3 - c) ≤ 0

Since a is between 1 and 2, inclusive.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
But we also have the condition that P(11) = 181. Let's write that down:

<prompt_4.1> Use the condition P(11) = 181 to form an equation involving a, b, and c. </prompt_4.1>

Hmm, so we have three variables a, b, c and two inequalities (from Q(x) and R(x) discriminants). This might be tricky. Maybe there's a way to express P(x) as a combination of the two bounding quadratics? Let me think. Alternatively, perhaps Q(x) and R(x) are both perfect squares or have some specific form.

Wait, if both Q(x) and R(x) are non-negative for all x, then maybe they can be written as squares of linear functions.

<prompt_4.2> Explore the possibility of Q(x) and R(x) being perfect squares. </prompt_4.2>

But maybe this approach is too restrictive. Alternatively, perhaps Q(x) and R(x) can be expressed as sums of squares, but not necessarily perfect squares themselves.

Alternatively, maybe the only way both Q(x) and R(x) are non-negative for all x is if they are both perfect squares. Let me check.

Suppose Q(x) is a perfect square.

<prompt_4.3> Determine if Q(x) and R(x) must be perfect squares. </prompt_4.3>
</Plan_4>

<Execution_4>
<execution_4.1>
181 = a*(11)^2 + b*(11) + c = 121a + 11b + c

So, equation (1): 121a + 11b + c = 181
</execution_4.1>

<execution_4.2>
Let me try that.

Suppose Q(x) = (mx + n)^2, then expanding gives m²x² + 2mnx + n². Comparing with Q(x) = (a - 1)x² + (b + 2)x + (c - 2), we have:

a - 1 = m²

b + 2 = 2mn

c - 2 = n²

Similarly, if R(x) = (px + q)^2, then:

2 - a = p²

-4 - b = 2pq

3 - c = q²
</execution_4.2>

<execution_4.3>
Then Q(x) = (mx + n)^2 = m²x² + 2mnx + n²

Comparing coefficients:

a - 1 = m²

b + 2 = 2mn

c - 2 = n²

Similarly, for R(x) to be a perfect square:

R(x) = (px + q)^2 = p²x² + 2pqx + q²

Thus:

2 - a = p²

-4 - b = 2pq

3 - c = q²

But this seems like a strong assumption, but maybe the problem is constructed such that Q(x) and R(x) are perfect squares. Let's see if this leads to a solution.
</execution_4.3>
</Execution_4>

<Plan_5>
Based on execution_4.3:
So, if we make these substitutions:

<prompt_5.1> Formulate equations based on the assumption that Q(x) and R(x) are perfect squares. </prompt_5.1>

Then equation (1): 121a + 11b + c = 181

Substitute a, b, c:

<prompt_5.2> Substitute a, b, c in terms of m and n into equation (1) and solve for m and n. </prompt_5.2>
</Plan_5>

<Execution_5>
<execution_5.1>
From Q(x):

a = 1 + m²

b = 2mn - 2

c = n² + 2

From R(x):

a = 2 - p²

b = -4 - 2pq

c = q² + c

Wait, hold on, from R(x):

c = q² + 3? Wait, no:

Wait, R(x) = (px + q)^2 = p²x² + 2pqx + q². But R(x) is defined as (2 - a)x² + (-4 - b)x + (3 - c). Therefore:

p² = 2 - a

2pq = -4 - b

q² = 3 - c

So, from R(x):

a = 2 - p²

b = -4 - 2pq

c = 3 - q²
</execution_5.1>

<execution_5.2>
From Q(x):

a = m² + 1

b = 2mn - 2

c = n² + 2

Then substitute into equation (1):

121(m² + 1) + 11(2mn - 2) + (n² + 2) = 181

Let's compute that:

121m² + 121 + 22mn - 22 + n² + 2 = 181

Combine constants: 121 - 22 + 2 = 101

So:

121m² + 22mn + n² + 101 = 181

Subtract 101:

121m² + 22mn + n² = 80

Hmm, equation (2): 121m² + 22mn + n² = 80

Now, let's look at R(x):

From R(x) = (px + q)^2:

2 - a = p² ⇒ 2 - (m² + 1) = p² ⇒ 1 - m² = p²

Similarly, -4 - b = 2pq ⇒ -4 - (2mn - 2) = 2pq ⇒ -4 -2mn + 2 = 2pq ⇒ -2 -2mn = 2pq ⇒ -1 - mn = pq

Also, 3 - c = q² ⇒ 3 - (n² + 2) = q² ⇒ 1 - n² = q²

So, we have:

p² = 1 - m²

pq = -1 - mn

q² = 1 - n²

So, we need to solve these equations.

Let me note that p² = 1 - m² and q² = 1 - n². Therefore, 1 - m² and 1 - n² must be non-negative, so |m| ≤ 1 and |n| ≤ 1.

But also, from equation (2): 121m² + 22mn + n² = 80. Hmm, 121m² + 22mn + n² is a quadratic in m and n. Let's see if we can find m and n such that this equation holds, given that |m| ≤ 1 and |n| ≤ 1.

Wait, but 121m² + 22mn + n² = 80. Let's compute the maximum possible value of the left-hand side given |m| ≤1 and |n| ≤1.
</execution_5.2>
</Execution_5>

<Plan_6>
Based on execution_5.2:
If m and n are between -1 and 1, then 121m² is at most 121*(1)^2 = 121, but 22mn is at most 22*1*1 =22, and n² is at most 1. So, the maximum possible value would be 121 +22 +1=144, but that's not helpful. Wait, but 121m² +22mn +n². Let me see. If m and n are real numbers with |m| ≤1 and |n| ≤1, then how big can this expression be?

Alternatively, maybe we can factor this expression. Let's see.

<prompt_6.1> Determine the feasibility of equation (2) given the constraints on m and n. </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
121m² +22mn +n². Hmm, this looks like a quadratic form. Let me check if it's a perfect square. Let's see:

(11m + n)^2 = 121m² +22mn +n². Yes! Exactly. So, 121m² +22mn +n² = (11m + n)^2. Therefore, equation (2) becomes:

(11m + n)^2 = 80

Therefore, 11m + n = ±√80 = ±4√5 ≈ ±8.944
</execution_6.1>
</Execution_6>

<Plan_7>
Based on execution_6.1:
But m and n are real numbers with |m| ≤1 and |n| ≤1. Then 11m + n. The maximum possible value of 11m + n when |m| ≤1 and |n| ≤1 is 11*1 +1=12, and the minimum is 11*(-1) + (-1)= -12. So, 11m +n can range between -12 and 12. However, we have 11m +n = ±4√5 ≈ ±8.944. So, possible.

<prompt_7.1> Solve for m and n given the equation (11m + n)^2 = 80 and |m|, |n| ≤ 1. </prompt_7.1>
</Plan_7>

<Execution_7>
<execution_7.1>
So, let's write:

11m + n = 4√5  ... (3)

or

11m + n = -4√5 ... (4)

But m and n are between -1 and 1. Let's see if this is possible.

Suppose equation (3): 11m + n = 4√5 ≈8.944

But since |11m| ≤11*1=11 and |n| ≤1, the maximum 11m +n can be is 11*1 +1=12, which is more than 8.944. So, possible. Similarly for equation (4): 11m +n = -8.944, which is possible as 11m can be as low as -11, and n as low as -1, so total -12, which is less than -8.944.

But let's see if m and n can satisfy these equations with |m| ≤1 and |n| ≤1.

Let's take equation (3): 11m + n = 4√5

We can express n = 4√5 -11m

But since |n| ≤1, then |4√5 -11m| ≤1

So:

-1 ≤4√5 -11m ≤1

Solving for m:

First inequality: 4√5 -11m ≥ -1 ⇒ -11m ≥ -1 -4√5 ⇒ m ≤ (1 +4√5)/11

Second inequality: 4√5 -11m ≤1 ⇒ -11m ≤1 -4√5 ⇒ m ≥ (4√5 -1)/11

Compute numerical values:

4√5 ≈4*2.236≈8.944

So, (1 +8.944)/11≈9.944/11≈0.904

Similarly, (8.944 -1)/11≈7.944/11≈0.722

Therefore, m must be between approximately 0.722 and 0.904. But m is a real number, but also from p² =1 -m². Since |m| ≤1, p is real.

But let's check if m in that interval is possible.

Similarly, equation (4): 11m +n = -4√5

Then n = -4√5 -11m

Again, |n| ≤1:

|-4√5 -11m| ≤1 ⇒ -1 ≤ -4√5 -11m ≤1

Multiply by -1 (reverse inequalities):

1 ≥4√5 +11m ≥ -1

So:

First inequality: 4√5 +11m ≤1 ⇒11m ≤1 -4√5 ⇒m ≤(1 -4√5)/11≈(1 -8.944)/11≈-7.944/11≈-0.722

Second inequality:4√5 +11m ≥ -1 ⇒11m ≥ -1 -4√5 ⇒m ≥(-1 -4√5)/11≈(-1 -8.944)/11≈-9.944/11≈-0.904

So, m must be between approximately -0.904 and -0.722.

But let's see if this is possible. Let's check if with these m and n, the other equations hold.

But this seems complicated. Let's proceed step by step.
</execution_7.1>
</Execution_7>

<Plan_8>
Based on execution_7.1:
Let's first consider equation (3): 11m +n =4√5

Express n =4√5 -11m

Then, from q² =1 -n². Since |n| ≤1, q² =1 -n² ≥0, so q is real.

<prompt_8.1> Analyze the feasibility of equation (3) and consider equation (4). </prompt_8.1>
</Plan_8>

<Execution_8>
<execution_8.1>
Similarly, p² =1 -m², so p is real.

But let's see if we can find m such that m is in [0.722, 0.904] and n =4√5 -11m is in [-1,1].

But m is between ~0.722 and ~0.904. Let's compute n for m=0.722:

n=4√5 -11*0.722≈8.944 -7.942≈1.002, which is just over 1. So, n≈1.002, which is slightly more than 1. Not allowed, since |n| ≤1. Similarly, at m=0.904:

n=4√5 -11*0.904≈8.944 -9.944≈-1.0. So, n≈-1.0. But at m=0.904, n≈-1.0, which is allowed (since |n| ≤1). However, between m=0.722 and m=0.904, n would decrease from ~1.002 to -1.0. But since n must be ≤1, the upper bound is slightly over 1 when m is just above 0.722. Therefore, there is no m in [0.722, 0.904] such that n=4√5 -11m is exactly 1. So, perhaps equation (3) is not possible? Or maybe m and n are not rational?

Alternatively, maybe the initial assumption that Q(x) and R(x) are perfect squares is incorrect. Maybe they are not perfect squares but still non-negative. Then, perhaps we need a different approach.

Wait, but if we suppose that Q(x) and R(x) are squares, then we might get a unique solution, but maybe it's not the only solution. However, given that we have P(11)=181, which is a specific point, perhaps this uniquely determines P(x). Let me check.

Alternatively, maybe P(x) is the average of the two bounding quadratics? Let's see.

Let’s compute the average of x² -2x +2 and 2x² -4x +3:
</execution_8.1>
</Execution_8>

<Plan_9>
Based on execution_8.1:
( x² -2x +2 + 2x² -4x +3 ) /2 = (3x² -6x +5)/2 = 1.5x² -3x +2.5

But is this the case? Not necessarily, unless P(x) is the average. But the problem states that P(x) is between these two for all x, so P(x) could be any quadratic in between. However, given that P(11)=181, which is a specific value, perhaps only one such quadratic passes through (11,181) and stays between the bounds.

Alternatively, perhaps the quadratic P(x) touches the upper bound at x=11.

<prompt_9.1> Consider if P(x) could be the average of the bounding quadratics and if it satisfies the given conditions. </prompt_9.1>
</Plan_9>

<Execution_9>
<execution_9.1>
But not sure.

Alternatively, maybe P(x) is tangent to one of the bounding quadratics at some point. But since we have a specific point x=11, maybe P(11)=181 is the point where P(x) touches the upper bound.

Let’s check the upper bound at x=11:

Upper bound: 2*(11)^2 -4*(11) +3 = 2*121 -44 +3 = 242 -44 +3=201

But P(11)=181, which is less than 201, so P(11) is below the upper bound. So, it's not touching the upper bound there. Similarly, check the lower bound at x=11:

Lower bound: (11)^2 -2*(11) +2=121 -22 +2=101. So, 101 ≤ P(11)=181 ≤201. So, 181 is in between.

Alternatively, maybe P(x) is the lower bound plus some multiple of (upper bound - lower bound). Let's see.

Let’s denote the difference between upper and lower bounds:

Upper - Lower = (2x² -4x +3) - (x² -2x +2) = x² -2x +1 = (x -1)^2

So, the difference is (x -1)^2. Therefore, Upper = Lower + (x -1)^2.

Therefore, if we write P(x) = Lower + k*(x -1)^2, where 0 ≤k ≤1, then P(x) would satisfy Lower ≤ P(x) ≤ Upper.

So, let's try this approach. Let’s suppose:

P(x) = (x² -2x +2) + k*(x -1)^2

Then, expand this:

= x² -2x +2 +k*(x² -2x +1)

= x² -2x +2 +k x² -2k x +k

= (1 +k)x² + (-2 -2k)x + (2 +k)

Therefore, coefficients are:

a =1 +k

b= -2 -2k

c=2 +k

Now, check if this satisfies the upper bound.

Since Upper = Lower + (x -1)^2, then P(x) = Lower +k*(x -1)^2 ≤ Lower + (x -1)^2 = Upper. So, yes, P(x) is between Lower and Upper for all x, as required.

Therefore, P(x) is of this form for some k between 0 and1. Then, given that P(11)=181, we can solve for k.

So, let's compute P(11):
</execution_9.1>
</Execution_9>

<Plan_10>
Based on execution_9.1:
(1 +k)*(11)^2 + (-2 -2k)*(11) + (2 +k) =181

Compute each term:

<prompt_10.1> Solve for k using the condition P(11) = 181. </prompt_10.1>
</Plan_10>

<Execution_10>
<execution_10.1>
(1 +k)*121 + (-2 -2k)*11 + (2 +k) =181

Compute each part:

121*(1 +k) =121 +121k

(-2 -2k)*11 =-22 -22k

(2 +k) =2 +k

Adding them together:

121 +121k -22 -22k +2 +k = (121 -22 +2) + (121k -22k +k) =101 +100k

Therefore:

101 +100k =181

Subtract 101:

100k =80

Therefore, k=80/100=0.8

So, k=0.8. Therefore, P(x) is:

P(x) = (1 +0.8)x² + (-2 -2*0.8)x + (2 +0.8)

Simplify:

1.8x² -3.6x +2.8

So, a=1.8, b=-3.6, c=2.8

Therefore, to find P(16):

P(16)=1.8*(16)^2 -3.6*(16) +2.8

Compute each term:

16²=256

1.8*256= Let's compute 256*1=256, 256*0.8=204.8, so total 256 +204.8=460.8

-3.6*16= -57.6

+2.8

So total:

460.8 -57.6 +2.8 = (460.8 -57.6) +2.8 =403.2 +2.8=406
</execution_10.1>
</Execution_10>

<Plan_11>
Based on execution_10.1:
Therefore, the answer is \boxed{406}.

But let me verify this. Let's check if P(11)=181 with this polynomial:

<prompt_11.1> Verify the solution by checking if P(11) = 181 and ensuring the polynomial satisfies the given inequalities. </prompt_11.1>
</Plan_11>

<Execution_11>
<execution_11.1>
P(11)=1.8*(121) -3.6*11 +2.8= 217.8 -39.6 +2.8= (217.8 -39.6)=178.2 +2.8=181. Correct.

Also, check if P(x) is between the two bounds. Since P(x) = Lower +0.8*(x -1)^2, which is adding 0.8 times a square to the lower bound. Therefore, since (x -1)^2 is always non-negative, P(x) is always greater than or equal to Lower(x). And since 0.8*(x -1)^2 ≤ (x -1)^2, then P(x) = Lower +0.8*(x -1)^2 ≤ Lower + (x -1)^2 = Upper. Therefore, the inequalities are satisfied. So, this seems correct.
</execution_11.1>
</Execution_11>

<Final_answer>
Therefore, $P(16) = \boxed{406}$.
</Final_answer>
</think>