parmen

Problem 2

There are two cases to consider.

(i) Triangle ABC is right-angled.

Then, by Pythagoras' theorem, AC2=62+62=72. Hence, AC=6*sqrt(2)​​.

(ii) Triangle ABC is not right-angled.

Then, by the triangle inequality, AC

If AC<10, then dropping a perpendicular from A to BC gives a right-angled triangle with hypotenuse less than 10, and legs greater than 6. However, Pythagoras' theorem forbids this.

Therefore, the longest possible value of AC is 6*sqrt(2).

Since R1, R2, R3, R4 are the four regions into which the unit circle is divided by the lines x=6 and y=4, we see that regions R1 and R3 are congruent, and regions R2 and R4 are congruent. Therefore, R1+R2+R3+R4=2(R1+R2).

To compute the area of R1, we can construct sector OAC, where O is the center of the circle and A is the point at which the circle intersects the line y=4. Sector OAC has central angle 90 degrees, and its radius is 10, so its area is \frac{1}{4}(10^2)\pi = \frac{25}{2}\pi.

Since regions R1 and R3 are congruent, and region R1 is the upper right quadrant of sector OAC, the area of R1 is \boxed{\frac{25\pi}{4}}.

(B)

We can solve the system of equations for x, y, and z by using the following steps:

Multiply the first equation by −6 and the second equation by 5.

Add the two equations together to get −6xy+30yz+30xz=22.

Divide both sides of the equation by −6.

Simplify the equation to get xy−5yz−5xz=−11.

Add the equation xyz=6 to the equation xy−5yz−5xz=−11 to get 6xy−5yz−5xz=−5.

Factor the equation to get (x−1)(6y−5z)=5.

Since x, y, and z are all non-zero, we must have x−1=5 and 6y−5z=−1.

Solve for x and y to get x=6 and y=65z−1​.

Substitute these values into the equation xyz=6 to get 6z=636z−6​.

Solve for z to get z=21​.

Substitute z=21​ into the equation y=65z−1​ to get y=31​.

Therefore, the only possible value of x+y is 6+1/3​ = 19/3.

(A) For any number a, we can substitute a into the cubic equation f(x)=x3+px2+qx+r, and we get: \begin{align*} f(a)&= a^3 + pa^2 + qa + r \ &= a(a^2 + pa + q) + r \end{align*}The term (a2+pa+q) is the quadratic expression that has the roots a and −p. We can also think of it as the sum of the roots of the quadratic expression. So if we want a cubic expression with the roots a, b, and c, then the quadratic expression with the roots a and −p is (a+b+c). We can use the same logic for the other two quadratic expressions that we need to create. We get: \begin{align*} a(a+b+c) + r = a^3 + ab + ac + r\ b(a+b+c) + r = ab^2 + bb + bc + r\ c(a+b+c) + r = ac^2 + bc + cc + r \end{align*}Subtracting the first equation from the second equation, we get: \begin{align*} b(a+b+c) + r - (a(a+b+c) + r) &= 0\ b(a+b+c) - a(a+b+c) &= 0\ (b-a)(a+b+c) &= 0\ (b-a)\left(t+5\right)&=0 \end{align*}We can do the same for the other pair of equations: \begin{align*} (c-a)\left(t+6\right)&=0\ (c-b)\left(t+5\right)&=0 \end{align*}Multiplying these three equations together, we get the monic cubic expression that we want. \begin{align*} &\left(t+5\right)\left(t+6\right)\left(t+5\right) \ &= (t^2+11t+30)(t+5)\ &= t^3+16t^2+65t+150 \end{align*}Therefore, the monic cubic polynomial in terms of a variable t whose roots are t=a, t=b, and t=c is: \boxed{t^3+16t^2+65t+150}

To find the sine of angle PAQ, we need to find the length of PQ and the length of AP.

To find the length of PQ, we can use the Pythagorean Theorem on triangle BPQ.

PQ2=BP2+PC2=42+12=17

PQ=17​

To find the length of AP, we can use the Pythagorean Theorem on triangle APQ.

AP2=AQ2+PQ2=52+17=42

AP=42​=67​

Now that we know the length of PQ and AP, we can find the sine of angle PAQ using the definition of sine:

sin(PAQ) = opposite/hypotenuse​ = PQ/AP = ​sqrt(17)/(6*sqrt(7)) = sqrt(119)/42.

First, we try to "clear fractions" and see what we get:

x2−7x+10Bx−11​=x−2A​+x−53​ Bx−11=A(x−5)+3(x−2) Expanding, we get: Bx−11=Ax−5A+3x−6 Bx−Ax=3x−5A−6 (B−A)x=−5A−6 If B−A=0, then we would need −5A−6=0, which means A=−56​. But in this case, the equation (B−A)x=−5A−6 simplifies to 0=0, which means that the equation is true for any value of x. Since we know that A and B are unique values, we must have B−A=0. Therefore, we can divide both sides of the equation by B−A to get: x=B−A−5A−6​ Setting x=2, we get: 2=B−A−5A−6​ Solving for A, we get: A=52B−2​ Substituting this value of A into the equation Bx−11=A(x−5)+3(x−2), we get: Bx−11=52B−2​(x−5)+3(x−2) 5Bx−55=2Bx−10+3x−6 3Bx−55=3x−16 3Bx=3x−39 2Bx=−36 B=−x18​ Since B is a unique value, we must have x=0. Therefore, B is a non-zero real number.

Setting x=5, we get: 5B−11=A(5−5)+3(5−2) 5B−11=3(5−2) 5B−11=9 5B=20 B=4 Substituting this value of B into the equation A=52B−2​, we get: A=52(4)−2​ A=56​

Therefore, A+B=6/5​+4=26/5.

Since ∠AOB=90∘, we have [\angle BOM = \frac{1}{2}\angle ABC] and [\angle AON = \frac{1}{2}\angle BAC.] Since N is the midpoint of AC, we have [\angle ACN = 90^\circ - \frac{1}{2}\angle BAC.] Similarly, since M is the midpoint of BC, we have [\angle BCM = 90^\circ - \frac{1}{2}\angle ABC.] Therefore, [\angle MON = \angle BCM + \angle ACN = 180^\circ - \frac{1}{2}\angle ABC - \frac{1}{2}\angle BAC = \boxed{90^\circ + \frac{1}{2}\angle A}.]