AMC 10 · 2020 · #9

Grade 11 algebra
trigonometric-ratiosperiodic-functiongraph-readingdomain-restrictionsign-analysis identify-subproblemscaseworksign-analysis ↑ Prerequisites: trigonometric-ratiosperiodic-function
📏 Long solution 💡 3 insights
Problem
Count how many values of x in the closed interval from zero to two pi satisfy that the tangent of twice x equals the cosine of half x.

Pick an answer.

(A)
1
(B)
2
(C)
3
(D)
4
(E)
5
How to solve
Strategy Draw a Diagram

There is no algebraic trick that turns tan(2x) = cos(x/2) into a solvable equation — the two sides have different periods and different shapes. So Tool #1 (Draw a Diagram): sketch y = tan(2x) and y = cos(x/2) on the same axes and count crossings. Tool #7 (Identify Subproblems): the asymptotes of tan(2x) chop [0, 2π] into separate branches, and each branch is its own small counting problem. Tool #14 (Extreme Principle): on each branch, only the behaviour at the two ends matters — one side runs to -∞ or the other to +∞, which forces a crossing. Tool #3 (Eliminate Possibilities): add the per-branch counts and match the single choice that fits.

1STEP 1

Turn the equation into two graphs

Read it as two graphs meeting.

tan(2x) = cos(x/2) ⟺ graphs of y = tan(2x) and y = cos(x/2) meet
2STEP 2

Locate the asymptotes of tan(2x)

Locate the tangent's asymptotes.

x = π/4, 3π/4, 5π/4, 7π/4
3STEP 3

Track the cosine across the window

The cosine moves gently across the window.

cos(0/2) = 1, cos(2π/2) = cos(π) = -1
4STEP 4

Split the window into five branches

The window splits into five branches.

[0, π/4), (π/4, 3π/4), (3π/4, 5π/4), (5π/4, 7π/4), (7π/4, 2π]
5STEP 5

The three middle branches each cross once

The middle three each cross once.

h(x) → -∞ (left end), h(x) → +∞ (right end) → exactly 1 root per branch → 3
6STEP 6

Check the first branch from the ends

Check the first branch at its ends.

h(0) = tan(0) - cos(0) = -1 < 0, lim_x → π/4^- h(x) = +∞
7STEP 7

Check the last branch from the ends

Check the last branch too.

lim_x → 7π/4^+ h(x) = -∞, h(2π) = tan(4π) - cos(π) = 0 - (-1) = 1 > 0
8STEP 8

Add the branches and pick the choice

Add the branches.

3 + 1 + 1 = 5 → (E)
9STEP 9

Sanity-check with sample values

Sample values confirm 5.

roots bracketed in (0, 0.5), (1, 2), (3, 3.5), (4, 4.5), (5.6, 6)
Answer
5
The largest choice is 5, and 5 is also the structural ceiling here: tan(2x) has exactly four asymptotes on [0, 2π], giving five branches, and each strictly increasing branch can meet a strictly decreasing bounded curve at most once. So the answer had to be at most 5, and the end-behaviour checks showed every one of the five branches really does deliver a crossing. The numeric spot-checks bracket a root inside each of the five branches, which confirms the count independently of the sketch. The two partial branches are the delicate part, and both survive only because the interval is closed at 0 and at 2π — worth noticing, since an open interval (0, 2π) would still give 5 here (the roots are interior), but the endpoint values h(0) = -1 and h(2π) = 1 are what made the sign argument clean.
💡Key takeaway

Do not try to solve tan(2x) = cos(x/2) — draw it. The asymptotes of tan(2x) sit at π/4, 3π/4, 5π/4, 7π/4, cutting [0, 2π] into 5 branches, and on each branch a rising tangent meets the falling, boxed-in cos(x/2) exactly once. Count the branches and you have counted the answer: 5, choice (E).

  • Turn the equation into two graphs
  • Locate the asymptotes of tan(2x)
  • Track the cosine across the window
  • Split the window into five branches
  • The three middle branches each cross once
  • Check the first branch from the ends
  • Check the last branch from the ends
  • Add the branches and pick the choice
  • Sanity-check with sample values