Ch 1.2 · Constructions and Doubt

Lesson 4 · Constructions: telling time with shadows

Page 24

This session backfills chapter 1's last two lessons, which sit between session 1 (pages 8 to 23) and everything you have done since. The world's largest sundial, built at Jaipur in 1724, has a 90-foot gnomon whose shadow moves two inches a minute; the Egyptians used the same physics to aim their pyramids at true north. Getting north from a shadow turns out to be a geometry problem, and solving it needs the lesson's two new tools.

Since Euclid, geometric drawings have allowed exactly two instruments: the straightedge for lines and the compass for circles and arcs. A circle is all the points in a plane at one fixed distance (the radius) from a center. Drawings made with only these two tools are constructions, a stricter game than ruler-and-protractor drawing, and the strictness is the point: no numbers allowed, only geometry itself.

New here construction · compass · circle · radius · arc

Lesson 4 · Constructions: telling time with shadows

Page 25

Two recipes you will use for the rest of the book. Construction 1 bisects a line segment: arcs of equal radius from each endpoint cross at two points, and the line through those crossings cuts AB at its midpoint. Construction 2 bisects an angle: one arc finds equal footholds A and B on the sides, equal arcs from those footholds cross at D, and ray CD splits the angle exactly in half.

Do both with a real compass until your hands know them. And notice what the book says twice, with a straight face: there is a simple proof that these work, and you will learn it later. That later is real (it arrives with congruent triangles), and meanwhile you are trusting the tools exactly the way Lesson 5 is about to make you uncomfortable about.

New here bisect

Lesson 4 · Exercises, Set I

Page 26

The Egyptian surveyor's trick, run by you: a post at P throws a morning shadow to A and an afternoon shadow to B on the same circle, and the bisector of ∠APB points true north. The reason is symmetry the book will not prove for chapters: the sun's morning and afternoon positions balance across the north-south line, so the shadow angle balances too, and bisecting recovers the axis.

Questions 7 to 9 then bisect the north-south segment itself, and the crossing that appears runs east-west. Check with your protractor and you will find all four angles are 90°, and question 9's sum is 360, the full turn around a point that session 5's flower and session 6's Stonehenge both used. The construction you are grading is the perpendicular bisector, though the book has not said "perpendicular" yet on this page of the story.

Worked example · page 26

Questions 1 and 2: finding north

Draw a circle with center P, mark shadow points A and B on it, draw PA and PB, and bisect ∠P. The bisecting ray points north. What should the protractor say?

Draw the circle big; small figures hide errors (page 12's lesson). Place A and B anywhere on it, not symmetric-looking, and draw the two shadow rays.
Run Construction 2 from page 25: arc from P through both sides, equal arcs from the two crossing points, ray through their intersection. Label the ray's circle-crossing N and extend it backward through P for south.
Question 2's check: measure ∠APB, then ∠APN. Whatever ∠APB reads (say 74°), ∠APN should read half of it (37°). Your own numbers will differ; the halving is what must not.
If the halves disagree by more than a degree, the usual culprit is a changed compass radius mid-construction. Reset and redo; a construction is only as honest as the radius you kept.

Lesson 4 · Exercises, Sets I and II

Page 27

Two experiments on one 3-4-5-flavored right triangle, and both end in a coincidence that refuses to go away, the page 12 feeling again. Bisect the three sides: the three bisecting lines all pass through one point, and that point sits the same distance from all three corners. Bisect the three angles instead: concurrent again, at a different point, and question 16's answer is no, its corner distances are not equal (it is the sides it stands equidistant from).

Neither concurrence gets explained here; both get proved gloriously later, and the two meeting points have names waiting for them (the book will call them circumcenter and incenter). The Kellogg's Raisin Bran sundial then starts Set II with the honest observation that a real sundial's shadow speeds up and slows down through the day.

Worked example · page 27

Questions 10 to 13: the three side-bisectors

Draw the triangle (right angle at A, legs 3 and 4 inches), bisect each side with straightedge and compass, and study the three bisecting lines.

Draw the triangle with ruler and protractor first; the construction rules only govern the bisecting.
Run Construction 1 on each side separately, arcs from both endpoints, full crossing line each time. Keep the arcs; erasing evidence makes errors invisible.
Question 11: the three lines are concurrent, one common point. That is the surprise; two lines must cross somewhere, but the third has no obvious obligation to join them.
Questions 12 and 13: label the meeting point P and measure PA, PB, PC. All three match. On this triangle P lands at the midpoint of the long side; file that oddity away, because it is a right triangle's private signature.
Self-check by circle: put the compass point on P with radius PA and sweep. The circle should thread all three corners. One point, equidistant from three corners, is a circle's center wearing a disguise, and that is why the concurrence had to happen.

Lesson 4 · Exercises, Set II

Page 28

The lettered fan of the Kellogg dial (22 to 28) is bisection vocabulary practice by eye: which ray splits which angle, which angles match. Answer from the figure's symmetry about the vertical noon line, and say "seems to" honestly; nothing here is proved.

Then the page's real gift: construct an evenly spaced sundial with no protractor at all. A circle, a line through the center, and two arcs of the same radius swung from A and B cut the circle into six perfect slices. Question 31's measure is 60° each, and the reason (worth whispering now, proved much later) is that the radius steps around any circle exactly six times. You are watching the hexagon construction that predates Euclid.

Worked example · page 28

Questions 29 to 31: six equal angles from one radius

Draw a 3-inch circle centered at O, a line through O meeting it at A and B, then arcs centered at A and at B, each passing through O. Why do the crossings cut the circle into six equal parts?

Draw the circle and keep the compass locked at 3 inches for the entire construction; the whole trick lives in that discipline.
Arc centered at A through O: it crosses the circle at C and D. Every point of that arc sits 3 inches from A, so C and D are 3 inches from A and 3 inches from O.
Arc centered at B through O gives E and F the same way. Now look at what you own: six points on the circle (A, C, E, B, F, D in order... check your own lettering against the book's), each neighboring pair 3 inches apart, on a circle of radius 3.
Question 30's word: C, O, F collinear, and E, O, D collinear; the lines CF and ED plus AB make three lines through O.
Question 31: each of the six central angles measures 60°, since six equal slices share 360. Your protractor is the check, not the method; the compass never needed it. That independence from measurement is what "construction" means.

Lesson 4 · Exercises, Sets II and III

Page 29

Bisect the six 60s into twelve 30s, bisect again into twenty-four 15s, and the paper in front of you is a working sundial face: twelve daylight hours, 15° per hour, because the shadow of a properly tilted gnomon turns steadily through 180° from 6 A.M. to 6 P.M. Questions 33 to 36 do the arithmetic (180 in twelve hours, 15 an hour, one degree every four minutes), the same earth-clock numbers as page 96's questions 50 to 53.

Set III is a discovery lab: bisect all four angles of a rectangle and the four bisectors trap a small quadrilateral in the middle. Run the sequence of rectangles before opening the overlay, and keep notes; the shape that appears, and when it vanishes, are the discovery.

Worked example · page 29

Set III: bisecting a rectangle's angles

Bisect all four angles of rectangles measuring 1.5 by 2, 2 by 3, 2 by 4, and 1 by 4 inches. What polygon do the bisectors form, and can it ever shrink to a point?

Each bisector leaves a 90° corner at 45°, so every bisector runs diagonally at 45° to the sides. That single fact drives everything you are about to see.
Questions 2 to 5: in every rectangle the four bisectors enclose a small square, tilted 45°. Different rectangles, same shape, only the size changes.
Question 6: compare sizes against the rectangles. The square grows with the gap between length and width; the 1 by 4 gives a big square, the 2 by 3 a small one... check your drawings: the pattern is that the square's diagonal equals the difference of the rectangle's sides.
Question 7: shrink the difference to zero. A rectangle with equal sides is a square, its bisectors are its two diagonals, and they meet in a single point. So yes: the polygon collapses to a point exactly when the rectangle is a square.
Self-check the claim with one more drawing, a 2 by 2 square. Four bisectors, one crossing. A discovery you can force to happen on demand is the best kind to trust.

Lesson 5 · We can't go on like this

Page 30

The strangest lesson title in the book, over an aerial photograph of quadrilateral fields near Alexandria, and it means exactly what it says. Everything so far has been informal: constructions used without knowing why they work, conclusions guessed from how drawings look. That, the book admits, is Egyptian geometry, 4,000 years old, a pile of rules kept because they seemed useful, with nobody asking why.

This lesson is the pivot the whole course turns on. It will show you, concretely, what trusting appearances costs (a tax formula that cheats, a trisection that lies), and the discomfort it builds is the reason chapters 2 and 3 exist. If you worked sessions 2 through 7 first, you have already seen the cure; here is the disease it was invented for.

Lesson 5 · Exercises, Set I

Page 31

The Egyptians taxed four-sided fields with a formula inherited from Babylon: average the two pairs of opposite sides and multiply, A = (a + c)(b + d)/4. It is tidy, plausible, and wrong, and the wrongness has a direction: every landowner it misjudges, it overcharges. The tax office never loses. The exercises put you in 2500 B.C. with three plots to assess.

Run the assessor's arithmetic honestly in the worked example, then hold his three answers; page 32 counts the true squares and the comparison is the lesson. Note which plot is a rectangle, because the formula happens to tell the truth exactly there, and "works on the easy case" is how a wrong formula survives for two thousand years.

Worked example · page 31

Questions 1 to 4: the assessor's arithmetic

Apply A = (a + c)(b + d)/4 to Ramses' plot (sides 10, 10, 10, 10), Cheops's (11, 8, 11, 8), and Ptahotep's (15, 13, 9, 5). Who pays the most tax?

Ramses: (10 + 10)(10 + 10)/4 = (20)(20)/4 = 100 square units.
Cheops: opposite sides pair as 11 with 11 and 8 with 8: (22)(16)/4 = 88.
Ptahotep: consecutive sides 15, 13, 9, 5 pair opposites 15 with 9 and 13 with 5: (24)(18)/4 = 108.
Question 4: Ptahotep is assessed highest at 108, Ramses next at 100, Cheops at 88.
Now look at the figures before page 32 makes you count: Ramses' slanted parallelogram is clearly thinner than a 10-by-10 square, yet it was billed as one. Hold that suspicion; the squares are about to testify.

Lesson 5 · Exercises, Sets I and II

Page 32

The audit. Cheops's rectangle really is 88, the one honest bill. Ramses' parallelogram surrenders to the move you learned from Dudeney on page 23: slice off the left triangle, slide it right, and the plot becomes a rectangle of 80 squares, not 100. Ptahotep's crooked plot takes two auxiliary rectangles and halving to count, and comes in far under its 108 assessment. The formula overcharged both slanted plots, and question 12 asks who was robbed worst.

Set II then raises the stakes from area to possibility itself: the Greeks bisected segments and angles, so they asked about cutting in three. For segments the answer is yes, and page 33 walks the lovely construction; for angles, a storm is coming.

Worked example · page 32

Questions 5 and 6: auditing by rearrangement

Check the assessor: does Cheops's 11 by 8 rectangle really hold 88 squares? And how many squares does Ramses' slanted plot actually contain?

Question 5: count Cheops's grid, 11 columns of 8: 88. The formula told the truth, because for a rectangle "average of opposite sides" is just the sides themselves.
Question 6: do not count Ramses' plot slantwise. Cut the triangle off its left end and slide it onto the right, exactly as the book's figure shows.
The moved triangle changes nothing about area (cutting and rearranging never does; that was the Dudeney lesson on page 23), and the plot becomes a 10-by-8 rectangle: 80 square units.
So Ramses was billed 100 for 80, a 25 percent overcharge, and the slant was the entire error: the formula treats a leaning side's full length as if it stood upright.
Carry the method into 7 to 12: box Ptahotep's plot in rectangles, halve where a diagonal cuts one, and compare his true total against 108. The audit's verdict on question 12: the more a plot leans, the worse the formula lies.

Lesson 5 · Exercises, Set II

Page 33

Trisecting a segment, solved with nothing but the bisection construction used three times and a circle: bisect AB at C, circle through A and B, bisect AD to get a line through the center, and the two cross-segments DG and FE land exactly on the third-points of AB. Work it slowly with your ruler as the final judge; it is the prettiest construction in the chapter, and every step is a page 25 move you already own.

Then the page turns to angles, and presents an "early method" for trisecting one: trisect the chord AB instead, and draw rays through the division points. It looks completely reasonable, which is precisely the trap being set. Question 14 has you build an accurate test case; page 34 springs it.

Worked example · page 33

Question 13: trisecting a segment honestly

Draw AB 3 inches long and trisect it with straightedge and compass, following the book's construction. Why does the ruler check land on 1-inch marks?

Bisect AB (Construction 1); call the midpoint C. One bisection, nothing new.
Compass on C, radius CA: draw the full circle through A and B. The bisecting line from step 1 crosses it at D and E. Draw AD and BE.
Bisect AD, and notice the bisecting line m passes through C, the circle's center. (Every bisector of a chord does; the book plants this quietly and proves it in the circle chapter.) Mark F on AD and G on BE where m crosses them.
Draw DG and FE. Where they cut AB is the trisection: your ruler should read 1 inch and 2 inches. If it misses by more than a hair, the usual sin is a drifted compass radius in step 2.
Count what was used: two bisections, one circle, straight lines. Three equal parts from tools that only know how to halve; that is why the Greeks were confident angles would fall the same way. Hold that confidence up against tomorrow's page.

Lesson 5 · Exercises, Sets II and III

Page 34

The trap springs twice. First the angle "trisection": measure your equilateral-triangle test and the three angles come out unequal, the middle one fatter. The method fails, and the failure is total: no straightedge-and-compass method for trisecting every angle exists. People hunted for 2,000 years; in 1837 it was proved impossible (the book says only "no one has ever found a way," which was the polite version). Two drawings can look equally right and one be wrong; hence the lesson's title.

Set III is the same ambush from the pattern side: points on a circle give 1, 2, 4, 8, 16 regions, and five points scream "powers of two." Count six points yourself before believing anything.

YouTube 3Blue1Brown on this exact circle problem

Worked example · page 34

Set III: cutting up the circle

Points on a circle, all pairs joined: 2 points give 2 regions, 3 give 4, 4 give 8. Build the table, predict six and seven points, then draw and count.

Question 5's table so far: 2 points, 2 regions; 3 points, 4; 4 points, 8; and your five-point drawing (question 2) gives 16. Question 6: each new point has doubled the count.
Questions 7 and 8: the pattern predicts 32 for six points and 64 for seven. Write the predictions down; committing is what makes the check honest (the page 16 star-guessing discipline).
Question 9: draw the six-point circle large, space the points unevenly so no three chords meet at one interior point, and count regions by shading as you go.
The count is 31. Not 32. The doubling was a coincidence that ran out of room, and seven points give 57, nowhere near 64.
Question 10's moral is the lesson's: five confirmations are not a proof. A pattern can hold five times for reasons that quit on the sixth, and only deduction, not accumulation, closes the case. This little circle is famous among mathematicians precisely because it teaches that so painfully.

Chapter review · Summary

Page 35

The 1630 engraving shows Dido having an ox hide cut into strips, a princess solving a maximization problem with a knife; her story waits on page 38. Beneath it, chapter 1 compressed to a page: the Introduction's promise about Euclid's Elements, then lesson by lesson the vocabulary of session 1 (point through concurrent, ray and angle, polygon and polyhedron) and of this session (constructions, bisection, and Lesson 5's warning).

Read each lesson's paragraph and reconstruct the missing detail from memory: for Lesson 2, what the vertex letter's position in an angle's name must be; for Lesson 3, the difference between perimeter and area jobs. The summary gives page numbers for everything that resists. This is the same self-test drill as pages 71 and 123, at the spot where the habit was born.

Chapter review · Exercises, Set I

Page 36

A die from Italy, 2,700 years old, stands in for every cube. The review runs the chapter's vocabulary over it: edges are line segments, faces are squares (polygons), the cube itself a polyhedron. Then the counting, which the worked example organizes, and question 8's little trap: A, G, and C are noncollinear, and the other word for any three points is coplanar, because three points always share a plane (that is why the duck pond table on page 62 needed exactly three ducks).

The unfolded cross pattern (question 9, continuing overleaf) is worth building from paper if folding in your head stalls. Its twelve outside edges lie in surprisingly few lines, and the fold-partners questions on page 37 are the reason to keep the model handy.

Worked example · page 36

Questions 4 to 7: counting a cube

How many edges meet at each corner, and how many edges altogether? How many faces meet at each corner, and how many faces altogether?

Corners first, since the counts hang on them: a cube has 8 corners (four on top, four on bottom).
Question 4: stand at any corner of the die: one edge runs away in each of three directions. Three edges per corner.
Question 5: multiply and repair the double-count: 8 corners × 3 edges = 24, but every edge was counted from both its ends, so 24/2 = 12 edges. (Recount the picture to believe it: 4 top, 4 bottom, 4 vertical.)
Questions 6 and 7: three faces meet at each corner; and counting faces by corners, 8 × 3 = 24 with each face claimed by its 4 corners: 24/4 = 6 faces.
The count-then-divide trick is the takeaway; it counted the pyramid's edges on page 22 and will count diagonals and handshakes for chapters. And notice 8 corners − 12 edges + 6 faces = 2; keep that curiosity in a drawer.

Chapter review · Exercises, Sets I and II

Page 37

The fold-together questions (10 to 14) reward a real paper model: edge JW meets its partner only when the cross closes, and corner R turns out to share its point with two other lettered corners. The flat-versus-solid figure (15 to 17) is the Necker cube's cousin from page 119: seen flat it is a hexagon cut into three quadrilaterals, seen solid it is a cube, and question 17's answer (three faces at most) is why dice photographs always show three.

Set II builds toward the review's quiet showstopper: extend a triangle's sides, bisect the two outside angles at A and B, and the bisectors meet at F. Then CF, drawn back through the far corner, bisects ∠DCE. Nothing in chapter 1 explains it; the worked example makes sure you at least see it happen cleanly.

Worked example · page 37

Questions 19 to 26: the outside bisectors

Draw the 70-30 triangle with a 3-inch base, extend CA and CB past A and B to D and E, bisect ∠DAB and ∠ABE, and let the bisectors meet at F. What do the measurements say, and what does CF do?

Questions 20 to 22 first, by protractor: ∠DAB = 110° (it sits beside the 70 on line DC), ∠ABE = 150° (beside the 30), and ∠DCE = ∠ACB = 80° since 180 − 70 − 30 = 80. Three different chapter-3 facts in one warm-up, done here purely by measuring.
Bisect the two outside angles by construction: the halves should measure 55 and 55 at A (question 23), 75 and 75 at B (question 24).
Extend both bisectors until they cross; label it F. It lands outside the triangle, below the base; make paper room.
Draw CF and measure ∠DCF and ∠FCE (question 25). Both read 40°: CF bisects ∠DCE, question 26's answer.
Three bisectors, built from three unrelated-looking angles, concurrent again. You have now seen concurrence happen four times (pages 12, 27 twice, and here) and can prove none of them; that imbalance is the itch the rest of the book scratches. (The point F even has a name in the trade: an excenter.)

Worked example · page 38

Set III: Dido's land

A 900-yard cord fences three sides of a rectangle; the river is the fourth side. Compare the square layout with a 200-yard version, then hunt the best shape.

Questions 1 and 2, the square: three sides of 300 use the cord exactly. Perimeter counting the river bank: 4 × 300 = 1,200 yards. Area: 300 × 300 = 90,000 square yards.
Questions 3 to 5, the squat version: sides of 200 leave 900 − 400 = 500 for the top. Perimeter 200 + 500 + 200 + 500 = 1,400. Area: 500 × 200 = 100,000. More land from the same cord, and a bigger perimeter courtesy of the free river side; page 21's lesson that area and perimeter answer different questions, now with money on it.
Question 6: try more shapes. Depth 150 gives 600 × 150 = 90,000; depth 250 gives 400 × 250 = 100,000; depth 225 gives 450 × 225 = 101,250.
The pattern peaks at depth 225: the best rectangle is twice as wide as it is deep, splitting the cord as 225 + 450 + 225. Every trial to either side comes in lower; test 210 and 240 to feel the hump.
Why width equals twice depth: the river donates a side, so the cord buys two depths but only one width, and the optimum spends equally on each direction: 450 on depth (225 twice) and 450 on width. Virgil says Dido got Carthage out of it; the geometry says she earned it.

Algebra review · Simplifying expressions

Page 39

The book's first algebra review, and reading it after chapter 3 (as this session's ordering has you do) turns it into a reunion: the commutative, associative, identity, and inverse properties are algebra's axioms, the same rank page 78 gave them, and subtraction and division are not primitive operations at all but defined moves, adding the opposite and multiplying by the reciprocal.

Two vocabulary sharpeners matter for the drill overleaf: in 3a the 3 is a coefficient (repeated addition), while in a³ the 3 is an exponent (repeated multiplication). Keeping coefficient and exponent apart is the entire skill page 40 grades, and confusing them is the single most common algebra slip in geometry problems to come.

Worked example · page 40

Questions 35 and 36: sums versus products

Simplify (2x + 2x + 2x)(3x + 3x) and then (2x · 2x · 2x)(3x · 3x).

Question 35: collapse each parenthesis by addition first. 2x + 2x + 2x = 6x (repeated addition builds a coefficient), and 3x + 3x = 6x.
Multiply: (6x)(6x) = 36x². Coefficients multiply, exponents add: 1 + 1 = 2.
Question 36: now the parentheses hold products. 2x · 2x · 2x = 8x³ (repeated multiplication builds an exponent), and 3x · 3x = 9x².
Multiply: (8x³)(9x²) = 72x⁵.
Same ink, wildly different answers: 36x² against 72x⁵. The plus signs fed coefficients, the dots fed exponents, and the check is to substitute x = 1 into your original and your answer (question 35: 6 × 6 = 36 ✓). That one-number check is the same audit you ran on page 130, and it closes chapter 1's story: even simplifying, you verify.