AMC 10 · 2013 · #20
Grade 11 geometry-2dPick an answer.
Four messy trig points look like four separate objects. Tool #15 (Organize Information in More Ways): every vertex has the shape (t,t²), so all four live on the single parabola y=t² — that one re-reading is the whole problem. On a parabola the chord through parameters a and b has slope a+b, so 'parallel' turns into 'equal sums.' Tool #1 (Draw a Diagram): plot the four parameters on a number line to see who is where. Tool #14 (Extreme Principle): the smallest and the largest parameter control which pairing can possibly balance. Tool #3 (Eliminate Possibilities): only one of the three ways to split four points into two chords survives. Tool #4 (Introduce a Variable): set p=sin xcos x and s=sin x+cos x; the Pythagorean identity ties them together and produces a quadratic.
All four vertices lie on one parabola
All four points sit on one parabola.
Difference of squares turns the slope of a parabola chord into just the sum of its two parameters, so 'parallel' becomes plain arithmetic.
On this curve the slope of a chord is just the sum of its two parameters, so parallel becomes plain arithmetic.
▸ Why?
A difference of two squares is the two values added multiplied by the two subtracted, and the subtraction cancels.
▸ Why?
Parallel lines are exactly the lines with the same slope, so equal sums is the whole condition.
Locate the four numbers on the line
The angle's range orders the four numbers.
Past 135° the cosine has outgrown the sine in size, which flips cot x below -1 and squeezes tan x between -1 and 0.
11.F-TF.A.2Draw A DiagramOnly the extremes can pair off
Only the extremes can pair off.
If you pair small with small, the other pair is bigger in both slots, so the sums cannot tie — only pairing the extreme low with the extreme high can balance the scale.
6.NS.C.7Extreme PrincipleCollapse the equation with a common factor
A common factor collapses the equation.
The same factor sin x-cos x shows up on both sides, and it is safely nonzero here, so dividing it out is an exact move, not a lossy one.
9.A-SSE.A.2Introduce A VariableTie product and sum with the identity
The Pythagorean identity ties product to sum.
Sum and product of sin x and cos x are never independent — the Pythagorean identity locks them together as s²=1+2p.
11.F-TF.C.8Introduce A VariableSolve and reject the impossible root
Size kills the impossible root.
A product of a sine and a cosine can never exceed 1/2 in size, so one of the two algebraic roots was never a real angle.
9.A-REI.B.4Eliminate PossibilitiesConvert the product to a double angle
The product is a double angle in disguise.
sin(2x) is just twice the product we already solved for, so the quadratic answers the original question directly.
11.F-TF.C.9Introduce A VariableShow the trapezoid really exists
A real angle exists, so choice (A) stands.
Forcing a value is not the same as producing a figure, so we pin down the actual angle and watch the four points really close up into a trapezoid.
10.G-GPE.B.4Draw A DiagramEvery vertex has the shape (t,t²), so all four points sit on one parabola — and on a parabola the chord through parameters a and b has slope a+b, so 'parallel' just means 'equal sums.' Past 135°, cot x is the smallest of the four numbers and sin x the largest, and the only way two sums can tie is smallest-with-largest: sin x+cot x=cos x+tan x. That collapses to sin xcos x=sin x+cos x, and the Pythagorean identity finishes it: sin(2x)=2-2√(2), choice (A).
- All four vertices lie on one parabola
- Locate the four numbers on the line
- Only the extremes can pair off
- Collapse the equation with a common factor
- Tie product and sum with the identity
- Solve and reject the impossible root
- Convert the product to a double angle
- Show the trapezoid really exists