Reasoning · Grade 6-1 Dual and Semiregular Polyhedra
Problem
Descartes' angle-defect theorem
The angles meeting at a vertex of a regular polyhedron make the interior angle; what is left of 360 is the exterior angle. For the tetrahedron: exterior angle 180 degrees, 4 vertices, total 720. Fill in the cube and octahedron, find the rule, count the icosahedron's vertices.
Your answer
Watch outThe total seems to vary by solid, but it is always 720 degrees.
How to solve
StrategyLook for a Pattern — The table is built to make a pattern jump out: three different solids, three different exterior angles, three different vertex counts, and one column of totals. So I fill in every cell honestly first - drawing or holding a paper net for each vertex so I can see which corners meet - and only then look down the last row. Once I trust the pattern (the totals all come out the same), part (3) runs backwards: I know the total, I can work out one exterior angle of an icosahedron from the five triangles at its vertex, and dividing gives the number of vertices without ever having to count them on a picture.
1STEP 1
Read the definition off the tetrahedron's net
The tetrahedron's exterior angle is 180 degrees.
60° × 3 = 180°, 360° - 180° = 180°
Angles that sit side by side around a point just add up, and a whole turn is 360 degrees. So 'the angle left over' is a plain subtraction, which is Grade 4 work once you can see which corners touch the dot.
4.MD.C.7Draw A Diagram
2STEP 2
Cube: find one exterior angle
The cube's is 90 degrees.
90° × 3 = 270°, 360° - 270° = 90°
You can do this with three sticky notes: lay them so their corners touch at one point and you can see the quarter-turn of table still showing. That visible quarter turn is the 90 degrees.
4.MD.C.7Create A Physical Representation
3STEP 3
Cube: count the vertices and multiply
Times 8 vertices gives 720 degrees.
4 + 4 = 8, 90° × 8 = 720°
Counting corners in two organised layers instead of hunting round the picture is what stops you from double-counting; after that it is one multiplication.
4.NBT.B.5Make A Systematic List
4STEP 4
Regular octahedron: one exterior angle, the vertices, and the sum
Seeing the octahedron as two pyramids stuck together turns 'count the corners of a solid' into 'count a tip, a tip, and a square', which a Grade 4 solver can do without losing track.
4.MD.C.7Create A Physical Representation
5STEP 5
Fill in the table and look down the last row
The table's last row is 720 throughout.
180° × 4 = 720°, 90° × 8 = 720°, 120° × 6 = 720°
This is the whole point of the table: two columns that keep changing and one that refuses to. A quantity that stays put while everything around it moves is worth a name, and this one is called Descartes' theorem.
4.OA.C.5Look For A Pattern
6STEP 6
State the rule (part 2)
The rule is exterior angle times vertices is 720.
(one exterior angle) × (number of vertices) = 720°
Turning 'the totals were all 720' into a multiplication sentence with a box in it is what makes a pattern useful instead of just interesting.
4.OA.C.5Look For A Pattern
The shortfalls at all the vertices always add up to the same fixed total, whatever the solid.
Why?
At each vertex the face angles fall short of one full turn, and that shortfall is what makes the solid close up instead of lying flat.
🧱Angles around a point make 360 degreesAngles that fit side by side all the way around one point fill exactly one full turn, 360 degrees.
Why?
The three counts of a solid are already tied together by vertices minus edges plus faces making 2, which is why the total shortfall cannot vary.
🧱Euler's formula for solidsFor any solid with flat faces, vertices minus edges plus faces always comes out to 2, so a count you cannot see can be recovered.
7STEP 7
Regular icosahedron: find one exterior angle
The icosahedron's exterior angle is 60 degrees.
60° × 5 = 300°, 360° - 300° = 60°
The only geometry fact needed is that every angle of an equilateral triangle is 60 degrees, because the three equal angles of a triangle share 180 degrees. After that it is subtraction from a full turn again.
8.G.A.5Draw A Diagram
8STEP 8
Work backwards to the number of vertices (part 3)
Working backwards gives 12 vertices.
60° × □ = 720° → □ = 720 ÷ 60 = 12
A missing factor is found by division. Dividing 720 by 60 is the same as dividing 72 by 6, so the arithmetic stays easy even though the numbers look big.
5.NBT.B.6Work Backwards
9STEP 9
Check the 12 a completely different way
Counting face corners also gives 12.
20 × 3 = 60, 60 ÷ 5 = 12
Counting the same thing twice in two different groupings is the classic way to be sure of a count, and here it needs nothing beyond multiplying and dividing whole numbers.
4.OA.A.2Make A Systematic List
Answer
12 vertices
720 ÷ 60 = 12
Every number in the table is an angle in degrees, and each exterior angle comes out between 0 and 360, which it must - a vertex with no gap left over would lie flat and never fold into a solid. The three exterior angles rank the solids by pointiness in the order you would expect: the tetrahedron's sharp spike leaves the biggest gap at 180 degrees, the octahedron 120 degrees, the cube 90 degrees, and the nearly-round icosahedron only 60 degrees. The vertex counts move the other way (4, 6, 8, 12), which is what keeps every product at 720. As a fifth test the rule also handles the regular dodecahedron: three regular pentagons of 108 degrees meet at each vertex, so the exterior angle is 360 - 324 = 36 degrees and 720 / 36 = 20 vertices, which is right. Finally, the independent face-corner count (20 x 3 = 60 slots, 5 per vertex, 60 / 5 = 12) confirms the icosahedron's 12.
Takeaway
Every corner of a solid leaves a little gap when you flatten it, and no matter which regular solid you pick those gaps always add to 720 degrees - so one gap divides into 720 to tell you how many corners there are.
Read the definition off the tetrahedron's net
Cube: find one exterior angle
Cube: count the vertices and multiply
Regular octahedron: one exterior angle, the vertices, and the sum