2022 JM02

1. The isosceles triangle CDE gives DM⊥CE and DM=1.

   $$CM=ME=1,\quad DM=\sqrt{DE^2-ME^2}=\sqrt{2-1}=1$$

2. MB=2 and CB=3; the similar triangles have the same angle at B.

   $$\frac{AC}{CB}=\frac{DM}{MB}=\frac12\implies AC=3/2$$

AC=3/2.

Checks and common pitfalls: MB includes ME and EB, so MB=2.

1. Compare the squared lengths.

   $$CD^2+DE^2=2+2=4=CE^2$$

2. Therefore triangle CDE is right at D.

   $$\angle CDE=\pi/2$$

∠CDE=π/2.

Checks and common pitfalls: The longest side CE is the hypotenuse.

1. Find the three side lengths.

   $$DB=\sqrt{1+4}=\sqrt5,\quad CD=\sqrt2,\quad CB=3$$

2. Evaluate the cosine opposite CB.

   $$\cos\angle CDB=\frac{2+5-9}{2\sqrt2\sqrt5}=-\frac1{\sqrt{10}}$$

∠CDB=arccos(−1/√10).

Checks and common pitfalls: The negative cosine means the angle is obtuse.

1. Compute the base hypotenuse and altitude by equal area expressions.

   $$AB=\sqrt{(3/2)^2+3^2}=3\sqrt5/2,\quad CX=\frac{AC\,BC}{AB}=3/\sqrt5$$

2. AB is perpendicular to PC and CX, hence to plane PCX and PX. In right triangle PCX, use the tangent ratio.

   $$\tan\angle PXC=\frac{PC}{CX}=\sqrt5\implies\angle PXC=\arctan\sqrt5$$

Dihedral angle arctan√5.

Checks and common pitfalls: Use a plane angle formed by perpendiculars to the common edge AB.

1. Factor the cubic.

   $$f(x)=(x-1)(x^2-2x-2)$$

2. Apply the quadratic formula.

   $$x=1,\quad x=1\pm\sqrt3$$

x=1, 1−√3, 1+√3.

Checks and common pitfalls: f(1)=0 identifies one root, not all roots.

1. First derivative.

   $$f'(x)=3x^2-6x$$

2. Second derivative.

   $$f''(x)=6x-6$$

f′=3x²−6x; f″=6x−6.

Checks and common pitfalls: The constant 2 contributes zero to the derivative.

1. Derivative signs across 0,2 are +,−,+.

   $$f'(x)=3x(x-2)$$

2. Evaluate the corresponding function values.

   $$f(0)=2\text{ (local maximum)},\quad f(2)=-2\text{ (local minimum)}$$

Local maximum 2 at x=0; local minimum −2 at x=2.

Checks and common pitfalls: These are local, not global, extrema of the cubic.

1. The second derivative changes from negative to positive at x=1.

   $$f''(x)=6(x-1)$$

2. The given zero supplies its ordinate.

   $$f(1)=0\implies(1,0)$$

Inflection point (1,0).

Checks and common pitfalls: The concavity check distinguishes an inflection from a mere zero of f″.

1. The curve rises to (0,2), falls to (2,−2), then rises again.

   $$x\text{-intercepts}:1-\sqrt3,1,1+\sqrt3$$

2. It is concave down for x<1 and up for x>1, with inflection (1,0).

   $$x\to\pm\infty\implies f(x)\to\pm\infty$$

The plot contains all requested features and continues beyond the window.

Checks and common pitfalls: The central zero is also the inflection, not an extremum.

1. Find x=2 and x=6.

   $$2x^2-16x+24=2(x-2)(x-6)=0$$

2. Integrate upper minus lower.

   $$A=\int_2^6(-2x^2+16x-24)\,dx=\left[-\frac23x^3+8x^2-24x\right]_2^6=\frac{64}3$$

Area 64/3.

Checks and common pitfalls: Signed integrals must be arranged with a nonnegative height for geometric area.

1. Write the dot-product equation in coordinates.

   $$2(x+1)=2\sqrt{(x-1)^2+y^2}\implies x+1\ge0$$

2. Square and simplify.

   $$(x+1)^2=(x-1)^2+y^2\iff y^2=4x$$

3. Conversely y²=4x forces x≥0, so x+1>0 and reversing the square is valid.

Exactly the parabola y²=4x.

Checks and common pitfalls: Squaring a locus equation requires a converse/sign check.

1. Differentiate implicitly and use the tangent point.

   $$2y\,y'=4\implies a=2/t,\quad t\ne0$$

2. The tangent intercept is t−a·t²/4=t/2.

   $$b=t/2\implies ab=(2/t)(t/2)=1$$

ab=1.

Checks and common pitfalls: The vertex tangent x=0 is vertical and is outside the given line form.

1. Factor the intersection equation.

   $$m^2x^2=4x\iff x(m^2x-4)=0$$

2. Use the nonzero root and recover y.

   $$P=(4/m^2,4/m)$$

P=(4/m²,4/m).

Checks and common pitfalls: m>0 ensures the expressions are defined and P is not the origin.

1. Obtain the slope formula.

   $$2y\frac{dy}{dx}=4\implies\frac{dy}{dx}=2/y$$

2. Evaluate at P.

   $$\left.\frac{dy}{dx}\right|_P=\frac2{4/m}=m/2$$

Tangent slope m/2.

Checks and common pitfalls: The tangent slope differs from the secant slope m.

1. Express the angle tangent.

   $$\frac{m-m/2}{1+m^2/2}=\frac m{2+m^2}=\frac14$$

2. Solve the quadratic; both roots are positive.

   $$m^2-4m+2=0\implies m=2\pm\sqrt2$$

m=2±√2.

Checks and common pitfalls: Both positive slopes can produce the same angle difference.

1. Equal distances give equal squared distances.

   $$(x-3)^2+(y-5)^2=(x-5)^2+(y-1)^2$$

2. Collect the linear terms.

   $$4x-8y+8=0\iff x-2y+2=0$$

x−2y+2=0.

Checks and common pitfalls: This is the perpendicular bisector of the segment joining z₁ and z₂.

1. Plot the two fixed points.

   $$z_1\leftrightarrow(3,5),\quad z_2\leftrightarrow(5,1)$$

2. Draw the entire line through the midpoint (4,3) with slope 1/2.

   $$y=x/2+1,\quad x\in\mathbb R$$

The marked points and full perpendicular-bisector line are shown.

Checks and common pitfalls: The locus is a line, not merely the segment between the plotted endpoints.

1. Compute the distance between the fixed points.

   $$|z_2-z_1|=|2-4i|=\sqrt{20}=2\sqrt5$$

2. Halve it, attained at z=4+3i.

   $$\min|z-z_1|=\sqrt5$$

Minimum √5 at z=4+3i.

Checks and common pitfalls: The endpoint z₁ itself is not equidistant from the two fixed points.

1. Raise the angle to seven times its value.

   $$\omega^7=\cos2\pi+i\sin2\pi=1$$

2. Since 2π/7 is not a multiple of 2π, ω−1≠0.

   $$1+\omega+\cdots+\omega^6=\frac{\omega^7-1}{\omega-1}=0$$

ω⁷=1; the seven-term sum is zero.

Checks and common pitfalls: Division by ω−1 needs the explicit ω≠1 check.

1. Apply De Moivre to both signs.

   $$\omega^{\pm n}=\cos(2n\pi/7)\pm i\sin(2n\pi/7)$$

2. Add the conjugate expressions.

   $$\omega^n+\omega^{-n}=2\cos(2n\pi/7)$$

The identity holds for every positive integer n.

Checks and common pitfalls: The negative power is a reciprocal; it is a conjugate because the modulus is one.

1. Expand each squared cosine.

   $$S=\frac14\sum_{n=1}^3(\omega^{2n}+2+\omega^{-2n})$$

2. The six nonconstant powers are exactly ω,…,ω⁶ in a different order.

   $$S=\frac14(6+\omega+\omega^2+\cdots+\omega^6)=\frac14(6-1)=\frac54$$

5/4.

Checks and common pitfalls: The square creates a constant 2 in each conjugate product.

1. Let v=(a,b,c)ᵀ and v^[2]=(a²,b²,c²)ᵀ. Add column 1 to column 2 and use multilinearity to remove the repeated constant column, without division.

   $$D=\det[\,v,\ s\mathbf1,\ t\mathbf1-v^{[2]}\,]=-s\det[\,v,\mathbf1,v^{[2]}\,]=s\det[\,\mathbf1,v,v^{[2]}\,]$$

2. Evaluate the Vandermonde determinant and reorder factors.

   $$D=(a+b+c)(b-a)(c-a)(c-b)=(a-b)(b-c)(c-a)(a+b+c)$$

(a−b)(b−c)(c−a)(a+b+c).

Checks and common pitfalls: The derivation does not divide by a+b+c, so it also covers a+b+c=0.

1. Expand along the first row.

   $$\Delta=\begin{vmatrix}1&1&p\\p&1&1\\1&p&1\end{vmatrix}=1-p-(p-1)+p(p^2-1)=p^3-3p+2$$

2. Require a nonzero determinant.

   $$\Delta=(p-1)^2(p+2)\ne0\iff p\ne1,-2$$

p∈ℝ∖{1,−2}, for any q.

Checks and common pitfalls: The right-hand parameter q does not affect invertibility.

1. At p=1 all left sides equal x+y+z; consistency requires q=1.

   $$p=1,q=1:\quad(x,y,z)=(1-s-t,s,t),\quad s,t\in\mathbb R$$

2. At p=−2 the sum of all equations is 0=q−1, again requiring q=1.

   $$x+y-2z=1,\quad-2x+y+z=-2$$

3. Solve these independent equations and verify the third.

   $$p=-2,q=1:\quad(x,y,z)=(1+t,t,t),\quad t\in\mathbb R$$

For (p,q)=(1,1): (1−s−t,s,t). For (−2,1): (1+t,t,t). No other multiple-solution cases.

Checks and common pitfalls: A singular coefficient matrix may still give no solution when q≠1.