Ch 4.1 · Coordinates and Polygons

Lesson 1 · Coordinates and distance

Page 132

Chapter 4 opens with an Addams castaway crawling toward a sign that says only "You are here," which is a map with no coordinate system: a location without references is no information at all. The cure starts small: pick an origin, pick an increasing direction, lay a ruler along the line, and every point owns a number. That is a one-dimensional coordinate system, and it is the Ruler Postulate from page 85 wearing a new name.

Nothing on this page is new machinery; it is a deliberate running start. Answer the margin question (A and C sit at −4 and 3) and notice the book's promise: any placement of the ruler, any units, and distances still come out right, because positive differences do the measuring. Tomorrow the line becomes a plane and the ruler needs a partner.

New here origin

Lesson 1 · Coordinates and distance

Page 133

Two perpendicular lines, the x-axis and y-axis, crossing at the origin: a two-dimensional coordinate system. The axes cut the plane into four quadrants, numbered with Roman numerals counterclockwise, and every point gets an address (x, y), x always first. A(3, 4) and B(−5, 2) in the figures show the perpendicular drop to each axis that reads the address off.

Then the page does something lovely: it computes a slanted distance. The grid triangle has legs 4 and 3, and the Pythagorean Theorem (used on faith since page 65) gives the hypotenuse: √(16 + 9) = 5. The bottom figure redoes it with no grid at all, coordinates alone: subtract, square, add, root. That calculation, done once with numbers, becomes a formula overleaf.

New here axis · quadrant

Lesson 1 · The Distance Formula, then exercises

Page 134

The boxed result is the chapter's power tool: the distance between (x₁, y₁) and (x₂, y₂) is √((x₂ − x₁)² + (y₂ − y₁)²). It is nothing but page 133's triangle in general dress: horizontal leg, vertical leg, Pythagorean finish. Say it as a recipe, not an incantation: subtract the x's, subtract the y's, square both, add, take the root.

The exercises open on Saunderson's pin board, a coordinate grid built for the hands of a blind mathematician, which is a quiet argument that coordinates are a language, not a picture. The worked example reads two figures off the board; do questions 9 and 10 in words ("on the y-axis," "on the x-axis"), because those two special cases get used constantly later.

New here the Distance Formula

Khan Academy Distance formula

Worked example · page 134

Questions 7 and 8: reading the pin board

Point B of the green triangle sits at (−4, 7). Find the coordinates of the triangle's other two corners, and of the red square's four corners.

Anchor yourself: from B, count 4 left of the y-axis and 7 up. Every other point is read the same way, x first, sign from the quadrant.
Question 7: trace the green band to its other corners: F just left of the origin at (−2, −1), and C down in the third quadrant at (−3, −5). (Read your own board carefully; the counting, not my numbers, is the skill.)
Question 8: the red square's corners read G(0, 5), D(6, 3), E(4, −3), and the fourth near (−2, −1). Notice it is a tilted square: no side runs parallel to an axis, yet the coordinates pin it exactly.
Self-check with the new formula: opposite corners of a square must be equally far from the other pair. One distance-formula computation on your read-off corners will confirm or catch a miscount, which is precisely what the formula is for.

Worked example · page 135

Questions 16 to 22: reading the bird chart

Which quadrant holds small broad wings? Where are the bats? Which bird groups sit nearest (0, 0), (10, 15), (−10, −15), (0, 25), and where does the bat region cross the x-axis?

Question 16: small wings means positive x (high loading), broad wings means negative y (low aspect ratio): quadrant IV, where the chart parks pheasants, peacocks, and turkeys.
Question 17: the yellow bat blob spans quadrants II and III and pokes past the y-axis into I: three quadrants, never IV.
Questions 18 to 21: nearest (0, 0) are the pigeons and parrots crowd; near (10, 15) the divers and grebes; near (−10, −15) the owls and hawks; straight up at (0, 25) the swifts.
Question 22: the bat region cuts the x-axis at roughly (−25, 0) and (−3, 0); read your own chart to the nearest 5, since "approximate" is in the question.
Notice what you just did: turned zoology into geometry and back four times. Any two measured quantities make a plane, and that idea, not any bird, is the exercise.

Worked example · page 136

Questions 26 to 30: three current arrows

Read the endpoints of arrows AB, CD, and EF from the grid and compute each length with the distance formula. Do the results match their look?

Read endpoints first, compute second (mixing the two steps is where errors live). From the grid: A(4, 1) to B(6, 6); C(−4, −1) to D(−2, 4); E(2, −6) to F(4, −1). Check each against your own figure.
AB: √((6 − 4)² + (6 − 1)²) = √(4 + 25) = √29.
CD: √((−2 − (−4))² + (4 − (−1))²) = √(4 + 25) = √29. Mind the signs: subtracting a negative is where question 28 earns its keep.
EF: √((4 − 2)² + (−1 − (−6))²) = √29 again. All three arrows are exactly √29 ≈ 5.4 units.
Question 26 and 30 close the loop: the arrows are parallel and equally long, drawn in different places. Same displacement, different position; hold that idea gently, because "same size and shape, different place" is this chapter's entire subject.

Worked example · page 137

Questions 35 to 38: the window's diagonal

A window's corners A(100, 200) and C(900, 800) are given. Find B and D, then the pixel distance A to C and B to D, and convert at 0.25 mm per pixel.

Questions 35 and 36: a screen window's sides run with the axes, so B shares C's x and A's y: B(900, 200); D shares A's x and C's y: D(100, 800).
Question 37: AC = √((900 − 100)² + (800 − 200)²) = √(800² + 600²) = √(640,000 + 360,000) = √1,000,000 = 1,000 pixels.
Spot the family: 800 and 600 are 4 and 3 scaled by 200, so the diagonal had to be 5 × 200. The 3-4-5 triangle from the Greek stamp keeps paying rent.
Question 38: BD runs (900, 200) to (100, 800): √((−800)² + 600²) = 1,000 pixels too. A rectangle's diagonals match, now proved by arithmetic rather than looks.
Convert: 1,000 pixels × 0.25 mm = 250 mm, a 25-centimeter diagonal. Question 39's click at (410, 670) lands on the QuickTime icon, and you resolved it exactly the way the machine does.

Lesson 1 · Exercises, Sets II and III

Page 138

The dimension ladder (42 to 46) is a beautiful pattern climb: one dimension is √((x₂ − x₁)²) (positive difference in disguise), two is the page 134 formula, three adds a z-term for the cube (diagonal 3√3), and four adds one more for the hypercube, whose corner-to-corner distance comes out exactly 6. You cannot draw four dimensions, but the formula does not care; that indifference is the whole point of coordinates.

The triangle coordinates (47 to 49) are a plant: triangles ABC and CDO have sides 15, 20, 25 each, so question 49 asks whether a tracing would fit. Hold your answer; page 163 makes it a theorem (SSS). Set III's circle challenge is the worked example, and it is the best exercise in the lesson.

Worked example · page 138

Set III: six points, one circle, two liars

A circle is centered at the origin. Points A(63, 16), B(25, 60), C(−36, 54), D(−52, 39), E(−33, −56), F(20, −62) all appear to lie on it, but one is inside and one outside. Which, and how can you tell?

The definition from page 24 is the tool: on the circle means exactly one radius from the center. With the center at O, each point's distance is √(x² + y²), so compare x² + y² and skip the square roots entirely.
A: 63² + 16² = 3969 + 256 = 4225. B: 625 + 3600 = 4225. So the radius squared is 4225 (radius 65), and A and B are honest.
D: 2704 + 1521 = 4225. E: 1089 + 3136 = 4225. Also honest.
C: 1296 + 2916 = 4212, just under 4225: C is inside the circle. F: 400 + 3844 = 4244, just over: F is outside.
The gaps are 13 and 19 parts in 4225, far too small for any eye or printer, which is the book's point in the parentheses: drawn dots have width, geometric points do not, and only computation can referee. Five distance formulas just outperformed a perfectly accurate drawing.

Lesson 2 · Polygons and congruence

Page 139

A Palm Beach subdivision from the air: identical lots, identical houses, identical pools, and a roofline that hands the book its subject. Page 19 called a polygon "a figure bounded by line segments"; that was description. Now comes a real definition: a polygon is a connected set of at least three segments in one plane, each segment meeting exactly two others, one at each endpoint.

Read the definition the lawyerly way you read the horse definitions on page 46, clause by clause: "at least three," "same plane," "exactly two," "one at each endpoint." Every clause exists to exclude something, and page 140's rogues' gallery shows exactly what. Sides equal vertices always, and an n-gon is named by that shared count; the roof outline is a 5-gon.

New here polygon

Lesson 2 · Polygons and congruence

Page 140

First the failures: two figures that pass the casual eye but flunk the definition, one because segments dangle, one because a segment meets three others. Then the page builds this chapter's central word three stories tall. Congruent figures have the same size and shape (a tracing of one fits exactly on the other). The fitting itself pairs up vertices, a correspondence, and a correspondence that makes everything coincide is a congruence.

For triangles the definition sharpens: congruent iff some correspondence makes all corresponding sides and angles equal. And the notation carries the correspondence silently: △ABC ≅ △EFD says A pairs with E, B with F, C with D, so six equalities can be read off without a figure. Write the triangles' names in matching order, always; the habit is about to become the engine of every proof in the chapter.

New here correspondence · congruence

Worked example · page 141

Questions 3 to 5: the dot-to-dot polygon

The finished puzzle is a polygon. How many sides does it have, what is it called, and would it still be a polygon if the last dot were not joined back to the first?

Question 3: the dots number 1 through 31, and closing the figure joins 31 back to 1, so there are exactly 31 segments: 31 sides. No recount needed, and that shortcut is the definition talking: sides equal vertices.
Question 4: page 139's naming rule makes it a 31-gon. Not every polygon has a Greek name, and the n-gon convention exists for exactly this reason.
Question 5: leave the last segment out and vertex 1 and vertex 31 each touch only one segment. The definition demands every segment meet exactly two others, one at each endpoint, so the open chain is not a polygon.
Say the failure precisely, citing the clause, the way page 140's non-polygons are dismissed. Definitions with teeth are what let the rest of this chapter say "polygon" and mean something checkable.

Lesson 2 · Exercises, Set I

Page 142

Three figures, three jobs. The patented paper clip (a real 1990 patent) is chapter 3 in miniature: two perpendicularity statements make right angles at C and D, the crossing at the middle makes vertical angles, and the whole diagram fails the polygon definition because its segments cross rather than meeting at endpoints. The origami duck comes back from page 101 carrying congruent triangles this time; fold logic finds them, and the worked example organizes the hunt.

The art gallery's four-winged floor plan has 20 walls (count them in pairs as you go around), making it a 20-gon, and question 24's guard puzzle has a famous professional cousin: for this star-shaped gallery one guard, standing central, sees every wall.

Worked example · page 142

Questions 12 to 16: congruent triangles in the duck

In the folded square (diagonal AC, creases from the folds, E the midpoint region on AB), name the triangles that appear congruent to △AFG, △ACD, △CDF, △ACE, and one triangle congruent to nothing else.

Use the folds, not your eye: every fold in page 101's construction laid one region exactly onto another, and regions that coincided are congruent by definition.
Question 12: the fold along AC carried △AFG onto △AEG... read your figure's letters: the small triangle above the diagonal pairs with its mirror below. Match your lettering by the fold that swaps B and D.
Questions 13 to 15 follow the same rule: △ACD pairs with △ACB (the big halves of the square); △CDF with △CBE; △ACE with △ACF.
Question 16: hunt for a triangle no fold ever moved onto another; the odd-sized one (the book intends △GCE or its cousin in your lettering) has no partner.
Check any claimed pair by the tracing test: could you cut one out and lay it on the other? For fold-partners the paper already did it, which is why origami is a congruence machine.

Lesson 2 · Exercises, Sets I and II

Page 143

The magic-magazine figure drills correspondence notation the honest way: four congruent zigzag polygons in four colors, and you write the vertex pairings tracing would produce. Go one vertex at a time, matching corners by their look (the notch, the point, the square corner); rushing the middle letters is the standard error, and page 140's warning about order is the fix.

Set II introduces a pair of words with a mechanical test each: a polygon is convex when no side's line enters it, concave when one does, and question 40's rubber band snapping around nails is the same test physicalized. The tessellation's blue pieces fail, the yellow hexagons pass, and question 38 plants a flag the chapter will salute later: equal sides alone do not force congruence.

New here convex · concave

Worked example · page 143

Questions 39 to 43: convex or concave

State the line test and the rubber-band test, then settle question 43: can a triangle be concave?

Question 39: extend each side into a full line. If some side's line passes through the polygon's interior, the polygon is concave; if every side's line keeps the whole polygon on one side, it is convex. A dent is exactly a side whose line cuts back in.
Question 40: nails at the vertices, rubber band around all of them. The band takes the polygon's own shape only when the polygon is convex; around a concave one, the band bridges the dent.
Draw the four examples (41 and 42) with the tests in mind: to make a concave pentagon or quadrilateral, push one vertex inward until a side's line slices the interior.
Question 43: try the same trick on a triangle. Three vertices leave nothing to push between; each side's line is fenced off by the opposite vertex alone, and the band always fits. A triangle cannot be concave, which is one more reason triangles are the chapter's chosen atoms.

Worked example · page 144

Question 55: two triangles congruent to a third

Given △ABC ≅ △XYZ and △DEF ≅ △XYZ, prove △ABC ≅ △DEF. Supply the four reasons.

Statement 1 is the hypothesis: Given.
Statement 2 unpacks both congruences into their twelve equalities (six parts each). Reason: the definition of congruent triangles, which is exactly the license to trade the symbol ≅ for a list of equal parts.
Statement 3 splices the lists: ∠A = ∠X and ∠D = ∠X force ∠A = ∠D, and likewise five more times. Reason: substitution (two things equal to the same thing), page 79's property earning its keep.
Statement 4 repacks: six matching equalities under the correspondence ABC ↔ DEF is precisely what △ABC ≅ △DEF means. Reason: the definition of congruent triangles again, now run in reverse.
Definition open, algebra, definition shut: that rhythm is most of congruence proofs, and the corollary now joins your equipment list, no figure ever drawn.

Worked example · page 145

Set III: counting Sierpinski's sides

The first polygon has some number of sides; each later polygon replaces every piece with four smaller ones. How many sides do the third, fourth, tenth, and nth have?

Questions 1 and 2: count the first figure honestly: 12 sides. Count the second by groups, not one by one: each of the 12 became 4, so 48.
Question 3: the rule, once seen, does the counting: third generation 48 × 4 = 192, fourth 192 × 4 = 768. Verify a corner of the printed third figure against the rule; countable chaos is still countable.
Question 4: the tenth continues the quadrupling nine times from 12: 12 × 4⁹ = 12 × 262,144 = 3,145,728 sides.
Question 5: the nth polygon has 12 × 4ⁿ⁻¹ sides. Show the reasoning as the pattern, stated once: start at 12, multiply by 4 per generation, n − 1 generations after the first.
Contrast with page 34's circle regions, where the obvious pattern lied. Here the rule is not a guess: the construction itself replaces each side with four, so the multiplication is built in, not observed. Knowing why a pattern holds is what makes it safe to ride to three million.