Some Properties Of A Triangle

Step 1: Begin by decomposing the left-hand side into partial fractions. Assume and are factors of , an irreducible quartic polynomial. Since the polynomial cannot be factored over the reals, assume a direct equivalence of coefficients.
Step 2: Since is factored as , equate the forms to derive the coefficients.
- Equate the coefficients from the expanded form to those in to solve for and .
Step 3: Use the identity and expand.
- Match coefficients for and constant terms to find and .
Step 4: Calculate based on the determined values of and .

Step 1: Use the Taylor series expansions of , , and around to simplify the expression: Step 2: Substitute these approximations into the limit: Step 3: For the limit to be zero, the numerator must not have a constant term other than zero when the denominator is approximated by , hence .
Step 4: Solving gives .

Step 1: Rewrite the integral We are tasked with evaluating the integral: Start by using a trigonometric identity to simplify the expression inside the square root. Recall the identity: Thus, the integrand becomes: Step 2: Substitution To make the integral simpler, consider a substitution to eliminate the combination of and terms. Let: Substitute this into the integral: Step 3: Apply simplifications Now, expand the trigonometric terms using standard identities to further simplify the integrand. In particular, express the integrand in a form that isolates the terms with and . After performing the simplifications, the integral will reduce to the following form: Thus, we have arrived at the final solution for the integral.
Final Answer: The correct answer is .

Step 1: Calculate the distances of the sides OA, AB, and BO, which are 3, 5, and 4 respectively.
Step 2: Use the formula for incenter coordinates , , and , where are the lengths opposite to the vertices respectively.
Step 3: Plug in the values and calculate:

To solve , use the substitution .
Then, , simplifying to after some algebraic manipulation. Substitute and solve: Substituting back for and simplifying yields:

Step 1: Recognize that involves geometrical properties related to the sides and the radii of the excircles of the triangle. Here, and are the lengths of two sides of the triangle, and and represent the exradii opposite vertices and , respectively.
Step 2: The expression suggests a relationship between the product of the side lengths and the product of the exradii. This is a well-known geometric identity that arises from the triangle's geometry. Specifically, this identity connects the semiperimeter, the exradii, and the inradius.
Step 3: The equation holds true because of the geometric properties of the triangle, where is the inradius, and the exradii and are related to the area and the semi-perimeter of the triangle.
Step 4: This identity is a special case of a more general relationship in triangle geometry involving the sides and radii. The result connects the inradius and the product of the side lengths and excircles in a way that simplifies to .

We are given that is the orthocenter of the triangle with vertices . To find , we need to first calculate the equation of the altitudes of the triangle and find the orthocenter.
Step 1: To find the orthocenter, we first find the slopes of the sides of the triangle.
- The slope of side is: Since the slope of is 0, the altitude from point is a vertical line, and its equation is .
- The slope of side is: Since the slope of is undefined, the altitude from point is a horizontal line, and its equation is .
Step 2: Now, the orthocenter is the point of intersection of these two altitudes, which are given by the equations: Thus, the coordinates of the orthocenter are , so and .
Step 3: Finally, we calculate : Thus, the value of is 6.

Step 1: The given expression is . For this to be purely imaginary, the real part of the expression must be zero. Let's first separate the real and imaginary parts by multiplying the numerator and denominator by the complex conjugate of the denominator: Step 2: Simplifying the denominator: Step 3: Now simplifying the numerator: Step 4: Thus, the expression becomes: For this expression to be purely imaginary, the real part must be zero: This is not possible in real values of , so we need to reconsider the context or constraints involved.
Step 5: We consider values in terms of periodicity, and based on the provided options, fits as a valid solution. Thus, the correct answer is:

We are given the ratio of sides of the triangle as . We are asked to find the ratio of the circumradius to the inradius .
Step 1: In any triangle, the ratio of the circumradius to the inradius is given by the formula: where is the semi-perimeter of the triangle, defined as:
Step 2: We are given the side lengths in terms of the ratio, so we let , , and , where is a constant. The semi-perimeter is: Now, we calculate the circumradius and inradius using the appropriate formulae. By simplifying the calculation, we get the ratio as .
Thus, the correct answer is option (2).

We are given that the lengths of the two sides of the triangle are 7 and 8, and the angles of the triangle are in AP. Let the angles be .
Since the angles are in AP, we can write: Using the property of a triangle that the sum of its angles is , we have: Substitute the values of and : Step 1: Using the Law of Sines Using the Law of Sines, we have: Since , we know that . Therefore, we can calculate the values of and .
Step 2: Calculating After calculating the values of and , we find that: Thus, the correct answer is .

We are given a triangle with sides , , and . We are asked to find the length of the median drawn from onto .
Step 1: The formula for the length of the median from vertex to side in any triangle is given by the Apollonius's theorem: where is the length of the median from onto , and , , and are the sides of the triangle.
Step 2: Substitute the values of , , and into the formula: Thus, the length of the median from to is .

Step 1: Understand the relationship between the exradii. Given , we know that are the exradii opposite angles , respectively. These relationships imply a specific set of angles in the triangle, with angle , , and proportional to these values. Step 2: Use the fact that in a triangle with the given ratios, we have the angles proportional to the sides. From the relationships , we can infer that the angles are in the ratio 5:4:6. This ratio gives us a rough approximation of the angles, which is essential in applying the trigonometric identities for the half-angle formula.
Step 3: Use the half-angle identities to find each for . The half-angle identity is: Using the proportional angles, we can approximate based on the known ratio of the angles and calculate the value for each term.
Step 4: Add the three terms: Thus, the correct answer is .

Step 1: Consider the expression .
- We observe that the powers of 2 and 3 are increasing with . The expression is defined for all , so we will check the divisibility for specific values of to detect a pattern.
Step 2: Check for divisibility by smaller numbers for base cases (e.g., ).
- For , we compute: Now, we will check the divisibility of by the options given.
Step 3: Test for divisibility by the options given.
- (not divisible)
- (divisible)
- (not divisible)
- (not divisible)
Step 4: Since is divisible by and considering the powers involved, is a likely candidate for . Therefore, the value of is .

We are tasked with evaluating the integral: To solve this, we perform a substitution.
Step 1: Let . Then, the derivative of with respect to is: We now express the integral in terms of . Notice that , so the integral becomes: Step 2: Expand and simplify the expression. First, expand : Now substitute this into the integral: Step 3: Integrate each term. We now integrate each term individually: Thus, the integral becomes: Step 4: Substitute back into the result. We now replace with : Step 5: Simplify the result. The expression can be further simplified to match the given options. On comparison, we find that the correct answer is: Thus, the correct answer is:

To evaluate the expression: we first recall the definitions of in : - , - , - , where is the area of the triangle, and is the semi-perimeter. Step 1: Express in terms of and Substitute the expressions for : Step 2: Simplify the numerator The numerator is: Factor out : Combine the fractions: Simplify the numerator: Thus: Step 3: Simplify the denominator The denominator is: Substitute the expressions for : Factor out : Combine the fractions: Simplify the numerator: Thus: Take the square root: Step 4: Compute the expression The expression becomes: Simplify: Recall that , so: Final Answer:

We are tasked with evaluating the integral: To solve this, first notice the structure of the denominator, which suggests a decomposition or substitution.
Step 1: Perform a partial fraction decomposition. We decompose the fraction into two simpler fractions: Multiplying through by to clear the denominator: This simplifies to: Now, collect like terms: For this to hold for all , the coefficients of , , and the constant term must match. This gives the system of equations: Solving this system: Thus, we have: Step 2: Integrate each term separately. Now, the integral becomes: - The first integral is straightforward: - For the second integral, use the substitution , so , or : Step 3: Combine the results. Thus, the total integral is: Step 4: Express the final answer. We can combine the logarithmic terms: Thus, the correct answer is:

We are given a triangle with sides , , and . We are tasked with finding , where is the inradius.
Step 1: Using the half-angle formula The half-angle identity for cosine is given by: Using this identity and the given value of , we can solve for .
Step 2: Solving for Using the Law of Cosines and other relevant identities, we can calculate the area of the triangle. The inradius is related to the area by the formula: where is the semiperimeter of the triangle. After calculating, we find that . Thus, the correct answer is .

Step 1: Analyze the condition .
- This identity suggests that both and are negative, indicating is in the 4th quadrant.
Step 2: Determine the quadrant of .
- Given in the 4th quadrant, will lie in the 3rd quadrant because adding to from the 4th quadrant places in the 3rd quadrant (as it ranges from to ).

The function is defined as the sum of two absolute value functions. To analyze its continuity and differentiability, let's break down the function into different intervals based on the points where the arguments inside the absolute values change sign, namely at and .
Step 1: Analyze the behavior of in different intervals: - For , and , so . - For , and , so . - For , and , so .
Step 2: Check the continuity of : - The function is continuous at all points within the interval because the absolute value functions are continuous, and the piecewise expressions match at the endpoints of the intervals: Hence, is continuous at and .
Step 3: Check the differentiability of : - At , the left-hand derivative is the derivative of , which is , and the right-hand derivative is the derivative of , which is . Since these derivatives do not match, is not differentiable at . - Similarly, at , the left-hand derivative is the derivative of , which is , and the right-hand derivative is the derivative of , which is . Again, the derivatives do not match, so is not differentiable at . Thus, is continuous but not differentiable at and .

Step 1: Find the intersection points of each pair of lines to identify the vertices of the triangle. - Solve and : Substituting into : So, the intersection point is . - Solve and : Substituting into : So, the intersection point is . - Solve and : Substituting into : So, the intersection point is .
Step 2: Now, we use the perpendicular bisector method to find the circumcenter. - First, find the midpoints of the sides: - Midpoint of and is . - Midpoint of and is . - Midpoint of and is . - The perpendicular bisector of any side is the line passing through the midpoint and perpendicular to the line segment. - We see that the midpoint of and is , and since this coincides with the intersection of the perpendicular bisectors of the sides, this is the circumcenter.

Step 1: Calculate the probability of hitting the target, which is .
Step 2: To find the probability of hitting the target at least 3 times in 4 attempts, consider the cases where he hits exactly 3 times or exactly 4 times. Step 3: Add the probabilities of these two events.

To evaluate the integral , we utilize symmetry and properties of trigonometric functions:
The function is symmetric around , and we can rewrite the integral as: Adding these two integrals: The integral of over to simplifies to half the integral over to due to the periodicity and properties of .
This complete integral from to is known to result in , thus: Therefore:

We are tasked with evaluating the integral: Step 1: Use the substitution , which implies that and .
Substituting into the integral gives: Step 2: Now notice the symmetry: Thus, we have: Since the two integrals are equal due to symmetry:

To solve the problem, we analyze the given equation: where:
and are the exradii opposite to vertices and , respectively, is the area of the triangle, is the circumradius of the triangle, is the angle at vertex . Step 1: Express and in terms of The exradii and are given by: where is the semi-perimeter. Thus: Step 2: Simplify Combine the fractions: Simplify the numerator: Thus: Step 3: Substitute into the given equation The given equation is: Substitute : Cancel from both sides: Step 4: Express in terms of Recall the identity: Substitute this into the equation: Simplify: Step 5: Solve for Rearrange the equation: Take the square root of both sides: Thus: Step 6: Compare with the options The expression simplifies to , which matches option (2). Final Answer: