Fragen 193
Antworten 56


Problem 3:


We can solve this problem by strategically manipulating the inequalities and using our understanding of exponents. Here's how to approach it:

Break down the exponentials:


Notice that 2 raised to the power of 200 appears on the left side of the first inequality. We can rewrite it as (2^2)^100 which is equal to 4^100.

Simplify the inequalities:


With the simplification from step 1, the inequalities become: 4^100 < n^100 < (130n)^{50}


Taking the 50th root:


To compare the terms more effectively, let's take the 50th root of all three parts of the inequalities. This is a valid operation because all terms are positive. Remember that taking the nth root preserves the order of the inequality when dealing with positive numbers.


Applying the 50th root: 2 < n < √(130n)


Squaring both sides:

To get rid of the square root, we can square both sides of the inequality. However, squaring introduces a potential issue: squaring flips the direction of the inequality when the term being squared is negative.


Since n is a positive integer by definition, squaring both sides will preserve the direction of the inequality. So we have: 4 < n^2 < 130n


Analyze the terms:


The leftmost inequality (4 < n^2) tells us that n^2 must be greater than 4. This means n itself must be greater than 2 (since the square of a number cannot be less than the number itself).


The rightmost inequality (n^2 < 130n) tells us that n^2 is less than some multiple of n (specifically, 130 times n). This can only be true if n itself is less than 130.


Identify valid n:


So, to satisfy both inequalities, n must be greater than 2 and less than 130.


Now we consider all the positive integers within this range: 3, 4, 5, ..., 128, 129.


Counting valid integers:


There are a total of 129 - 3 + 1 = 127 positive integers that satisfy the conditions.


Answer: There are 127 positive integers n that satisfy the inequalities: 2^{200} < n^{100} < (130n)^{50}.


Problem 2:


The expression we're given involves squares of several terms. Here's how to find the minimum value for all real numbers x:


Completing the Square: Notice that each term in the expression is a squared term of the form (x + a)^2. To minimize the expression, we can try to manipulate it into the form of a perfect square trinomial (a squared term plus a constant term).


Constant Term Issues: Unfortunately, we cannot directly complete the square for each term because there are constant terms (like 12, 7, etc.) within the parentheses. Completing the square typically involves adding and subtracting a constant term that depends on the coefficient of our x term. However, in this case, adding such a constant term inside the parentheses would also affect the squared term itself, making it no longer a perfect square.


Grouping and Common Factors: Instead of completing the square for each term individually, let's look for ways to group the terms and find common factors. We can rewrite the expression as:


(x + 12)^2 + (x + 7)^2 + (x + 3)^2 + (x - 4)^2 + (x - 8)^2 =

(x^2 + 12x + 144) + (x^2 + 14x + 49) + (x^2 + 6x + 9) + (x^2 - 8x + 16) + (x^2 - 16x + 64)


Notice that each term except the first has a common factor of (x^2 + px + q), where p and q are constants depending on the specific term.


Factoring and Simplifying: Now we can factor out the common factors and group the remaining terms:


(x^2 + 12x + 144) + (x^2 + 14x + 49) + (x^2 + 6x + 9) + (x^2 - 8x + 16) + (x^2 - 16x + 64) =


(x^2 + 12x + 144) + (x^2 + 14x + 49) + (x^2 + 6x + 9) + (x^2 - 8x + 16) + (x^2 - 16x + 64)


= (x^2) + (x^2) + (x^2) + (x^2) + (x^2) + (12x + 14x + 6x - 8x - 16x) + (144 + 49 + 9 + 16 + 64)


= 5(x^2) - 14x + 282


Non-Negative Term: The first term, 5(x^2), is always non-negative because any real number squared is non-negative. The minimum value it can take is 0, which occurs when x = 0.


Minimizing the Expression: The remaining term, -14x + 282, is minimized when -14x is maximized (since it's being subtracted). However, -14x is maximized when x is minimized (remember x is a real number).


Since we have no restrictions on the lower bound of x (it can be any real number), -14x can be infinitely negative as x approaches negative infinity.


Minimum Value: Therefore, the minimum value of the entire expression approaches negative infinity as x approaches negative infinity.


However, the question asks for the minimum value for ALL real numbers x. Since 5(x^2) is always non-negative and -14x can be arbitrarily negative, the minimum value of the expression is achieved when 5(x^2) = 0 (which occurs at x = 0) and -14x reaches its maximum negative value.


Answer: The minimum value of the expression is 282.


Relating medians and angle bisectors:


In a triangle, when a median (such as BM) intersects an altitude (such as AH) inside the triangle, the median is also an angle bisector.

This is a useful property to remember for problems like this.


Angle relationships:


Since BM is a median and an angle bisector of angle B, we have:


∠ABM = ∠MBC (angle bisector property)


We are also given that angle ACB = 41 degrees.


Angle sum in triangle ABC:


The angles in any triangle add up to 180 degrees. Therefore, in triangle ABC:


∠ABC + ∠ACB + ∠BAC = 180°


Solving for ∠MBC:


Since BM is a median, it divides triangle ABC into two triangles with equal areas (triangle ABM and triangle MBC). Because they share the same base (BM), the ratio of their areas is equal to the ratio of their altitudes. In this case, both triangles share the same altitude (AH), so their areas must be equal.


Knowing that triangle ABM and triangle MBC have equal areas, we can express this using their angles:


1/2 * base * height (triangle ABM) = 1/2 * base * height (triangle MBC)


Since the base (BM) is the same for both triangles, we can simplify this to:


∠ABM = ∠MBC (angles are proportional to their areas in a triangle)


We established earlier that due to the angle bisector property, ∠ABM = ∠MBC.


Now, let x represent the measure of angle MBC (which is also equal to angle ABM).


From the angle sum property in triangle ABC:


∠ABC + 41° + ∠BAC = 180°


Since ∠ABM (or x) bisects angle B:


∠ABC + 41° + x = 180°


Substitute ∠ABM with x:


x + 41° + x = 180°


Combine like terms:


2x + 41° = 180°


Subtract 41° from both sides:


2x = 139°


Divide both sides by 2:


x = 69.5°