Prove it for rectangles
An a by 3 rectangle, sides on grid lines
I = (a − 1)(3 − 1), B = 2a + 6
I + B/2 − 1 = 2a − 2 + a + 3 − 1 = 3a
That is its area, for every a
Maths olympiad · 45 minutes
Students count the dots inside and on the edge of shapes on a grid, find the rule that gives the area themselves, prove it for rectangles, and use it to find tilted areas fast. Slides, a vote that shows who's stuck, problems that climb to olympiad level, and short proofs to mark.
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Move n. The diamond has its corners at (n, 0), (0, n), (−n, 0) and (0, −n). For each n, count the dots on its edge (B) and inside it (I), and work out its area A. Can you find a rule that gives A from I and B?
Twice: at x = −2 and x = 2
A polygon whose corners are all grid points has area A = I + B/2 − 1, where I is the number of grid points inside it and B the number on its edge (Pick's theorem). So the area of a tilted shape, hard to find by counting squares, comes from counting dots. On an edge from (0, 0) to (p, q) there are HCF(p, q) + 1 grid points, ends included, which counts B without drawing.
The lesson goes discovery first. Students count dots and areas for a growing diamond, rectangles and a triangle, fill a table, and guess the rule before it is named. They test it on a shape they didn't use, then prove it for every rectangle in two lines of algebra. Only then do they use it: tilted squares, triangles, counting B on slanted edges, and working backwards. The stretch reaches olympiad uses: why every such area is a whole number or a half, and why no equilateral triangle has all its corners on grid points.
Students should find the area of rectangles and triangles, read coordinates, and find the highest common factor of two numbers.
Taught in Counting routes on a grid.
45 minutes in total. The stretch is optional.
5 min 1 · Hook · students think
Students count dots and find areas of a growing diamond before anyone explains.
“The diamond |x| + |y| = 2 has its corners on grid points. Count the dots on its edge and the dots inside it. Work out its area. Then do the same for the next diamond.”
Counting the edge dots side by side and counting each corner twice: 12 for the diamond with n = 2, not 8.
Give three minutes. For the area, most cut the diamond into four right-angled triangles, or draw the square round it and take away the corners.
Start the table: n = 1 has A = 2, B = 4, I = 1; n = 2 has A = 8, B = 8, I = 5. Ask whether the dots could tell you the area without any cutting.
Collect the class's counts for n = 1 and 2 on the board as a table: A, B, I. Don't confirm a rule yet: the next step gives more shapes.
Shorter: One diamond only (n = 2); start the table in the next step (−2 min).
12 min 2 · Explain · you explain
A table of shapes, the rule guessed and tested, a proof for rectangles, dots on slanted edges.
“We have a table of areas, edge dots and inside dots. Find a rule that gives the area from the dots. Then we test it on a shape nobody used.”
Calling the rule proved because it fits the table: it is a guess until it is proved, or used on a new shape at least.
Prove it for rectangles
An a by 3 rectangle, sides on grid lines
I = (a − 1)(3 − 1), B = 2a + 6
I + B/2 − 1 = 2a − 2 + a + 3 − 1 = 3a
That is its area, for every a
Dots on a slanted edge
(0, 0) to (6, 4): HCF(6, 4) = 2
2 equal steps of (3, 2)
Dots, ends included: 2 + 1 = 3
A tilted square
Corners (0, 0), (2, 1), (1, 3), (−1, 2)
B = 4 (no dots in the middle of an edge), I = 4
A = 4 + 4/2 − 1 = 5
Check: each side² is 2² + 1² = 5
First slide: the table. Rectangles 1 by 1, 2 by 1, 2 by 2, 3 by 2 and the diamond with n = 2. Students look at rows with I = 0 first: A goes up 1 when B goes up 2, so A = B/2 − 1. Then each dot inside adds 1. The guess: A = I + B/2 − 1. Test it on a shape nobody used, the triangle (0, 0), (4, 0), (0, 3): B = 8, I = 3, and 3 + 4 − 1 = 6 = 4 × 3 ÷ 2.
Second slide: name it. This is Pick's theorem (Georg Pick, 1899), for any polygon whose corners are grid points and which has no holes. Prove it for every rectangle: an a by b rectangle has I = (a − 1)(b − 1) and B = 2a + 2b, and the algebra gives ab. The full proof builds from rectangles to triangles to any polygon; it is in the stretch.
Third slide: slanted edges. From (0, 0) to (p, q) the edge goes in HCF(p, q) equal steps, so it has HCF(p, q) + 1 dots, ends included.
Fourth slide: use it. A tilted square, whose area would need cutting, comes straight from its dots; check with side².
Let the class guess the rule before you show it. If they are stuck, give them the shapes with no dots inside first (A = B/2 − 1), then ask what one more dot inside adds.
Shorter: The table and the name; the rectangle proof and slanted edges as one slide to read (−4 min).
8 min 3 · Your turn · students write
Each worked example beside a twin that students solve themselves.
“Watch me do one. Then you do its twin. Count B edge by edge, and take away the corners you counted twice.”
Counting every corner twice when adding up the edges.
A triangle
(0, 0), (4, 0), (0, 3)
Edges: 4 steps, 3 steps, HCF(4, 3) = 1 step: B = 4 + 3 + 1 = 8
I = 3, so A = 3 + 4 − 1 = 6 (and 4 × 3 ÷ 2 = 6)
Then theirs
(0, 0), (6, 0), (0, 4): find B and I, then check Pick
A tilted square
(0, 0), (2, 1), (1, 3), (−1, 2)
B = 4, I = 4: A = 5
Then theirs
(0, 1), (3, 0), (4, 3), (1, 4): find I, B and the area
Backwards
A triangle with no dots except its corners:
I = 0, B = 3, so A = 0 + 3/2 − 1 = ½
Then theirs
A polygon with I = 2 and B = 5: its area?
Work the first example aloud, writing each line. Then students solve its twin, and check the area with the usual formula.
Do the same for each pair. The tilted square's twin is where Pick saves the most work: check it with side², as in the worked example.
Open the answers slide once most have finished their twins. Ask a student to show their count of B for the first twin: it catches the double-counted corners.
Shorter: Do the first two pairs only (−3 min).
15 min 4 · Who's stuck? · students vote
One question shows who's stuck; then problems that climb from fluency to olympiad level.
“Choose an answer on your own. Nobody talks until everyone has chosen. Then work up the problems as far as you can; the last ones are olympiad level.”
Corners counted on both of their edges; only the corners counted; the dots in the middle of a slanted edge left out.
Every polygon here has its corners on grid points and no holes. Show your working on paper. Questions 6 and 9 need a written explanation.
The question that shows who's stuck
Each wrong answer is a different mistake. While they answer, your Classroom shows who chose which, so you know in seconds whom to help with what.
Support · For question 1: put the numbers into A = I + B/2 − 1. For counting B: count the steps along each edge (the HCF of how far across and how far up), and add the steps: every dot on the edge, corners included, is counted once that way. For the tilted squares, check your area with side².
Everyone chooses before anyone explains. Then ask a student for each answer to point to every dot they counted, and the class to say whether any was counted twice or missed. Shared corners, corners alone and the slanted edge all show up here.
Questions 1–2 build fluency; 3–4 give the same idea a new look; 5–6 ask for reasons; 7–8 are competition style; 9 is an olympiad proof.
Students show their working on paper or the board; the quiz keeps their final answers, and the written ones come to you to mark.
If the class is split, go round the triangle's edge together on the slide, one dot at a time, asking for each dot which edges it is on. Long answers (6, 9) are marked by you: one mark for the idea, one for each step with its reason, one for the conclusion. Model answers are in each question's explanation.
Shorter: After the vote, questions 1–5 only (−6 min).
Longer: Question 9 written up in full, then read aloud and improved together (+10 min).
5 min 5 · Exit · students show
One backwards use of the theorem with a reason shows who can rearrange it and knows when it applies.
“Use Pick's theorem backwards. Then say in a sentence what has to be true of a shape for the theorem to work.”
Rearranging with a slip: B/2 = A − I + 1, so B = 2(A − I + 1).
A number, then a sentence.
A polygon with its corners on grid points has area 9 and 5 dots inside. How many dots are on its edge?
Then: what must be true of a shape for Pick's theorem to work?
The sentences show who knows the conditions (corners on grid points, no holes). Start the next lesson with a shape that breaks one of them. Set the homework.
Longer: Draw a shape with these numbers on squared paper (+5 min).
One to one: let the student build the table themselves and guess the rule before you name it; then give a tilted shape and ask them to find its area two ways (the box around it, and the dots) and compare.
Every polygon has its corners on grid points. Use squared paper. Find each area two ways where you can. Question 6 is a written proof: write it as you would in an olympiad.
Print the worksheet with working space and homework. Teacher answers come with your Library copy.
For a polygon whose corners are all grid points and which has no holes, the area is I + B/2 − 1, where I is the number of grid points inside and B the number on its edge. It was published by Georg Pick in 1899. It turns the area of a tilted shape into counting dots.
An edge from (0, 0) to (p, q) is made of HCF(p, q) equal steps, so it has HCF(p, q) + 1 grid points, ends included. Add the steps of all the edges to get B: that counts every corner once, which avoids the most common mistake, counting each corner on two edges.
Because finding it is the lesson. From a table of shapes, students see that with no dots inside the area goes up by 1 when B goes up by 2, and that each dot inside adds 1. A rule they guessed, tested and then proved for rectangles is one they remember and trust, and it shows the olympiad habit: a pattern is a guess until it is proved.
Areas on grids come up in junior and intermediate maths challenges, and Pick's theorem is a known tool in olympiad rounds and national olympiads, for example to show an area is a multiple of a half. This lesson's problems are written for it; it uses no past papers.
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