Show that the sum of angles of a non-planar quadrilateral is always less than 360°. Can you find a non-planar quadrilateral ABCD for which ∠A + ∠B + ∠C + ∠D = 2°? (Hint: Think of a diagonal, say AC, as a hinge around which triangles ABC and ADC can rotate.) What happens to each angle of ABCD as you do this rotation?
Step-by-step solution
Idea: In a triangle with two fixed sides, the angle between them grows or shrinks with the third side. Folding brings B and D together, so BD shrinks and the angles at A and C shrink, while the angles at B and D (inside the rigid triangles) stay the same.
- Split ABCD by diagonal AC into triangles ABC and ADC; treat AC as a hinge. Turning ∆ADC about AC does not change ∠ABC or ∠ADC (they are angles of rigid triangles).½ mark
- When the figure is flat and convex, ∠A = ∠BAC + ∠CAD and ∠C = ∠BCA + ∠ACD, so ∠A + ∠B + ∠C + ∠D = (angles of ∆ABC) + (angles of ∆ADC) = 360°. In this position BD is as long as it can be.1 mark
- Folding (any non-planar position) makes BD shorter. In ∆BAD the sides AB, AD are fixed, so a shorter BD means a smaller ∠BAD; likewise ∠BCD is smaller. Hence ∠A < ∠BAC + ∠CAD and ∠C < ∠BCA + ∠ACD, and the angle sum is less than 360°. (If the flat figure is non-convex at A, then ∠BAC + ∠CAD > 180° ≥ ∠BAD anyway.)1½ marks
- Sum = 2°: take ∆ABC with AB = CB and ∠ABC = 0.5° (so ∠BAC = ∠BCA = 89.75°), and let ∆ADC be its mirror image across AC. Folding shut, ∠A and ∠C shrink towards 0° while ∠B + ∠D = 1° stays fixed, so the sum falls from 360° towards 1°; it passes 2° on the way (computation: when the two triangles make an angle of 0.5°). So such an ABCD exists.1 mark
Check: Computer check with A(−1, 0, 0), C(1, 0, 0), B(0, 229.18, 0) (∠ABC = 0.5°) and D = B turned 0.5° about AC: ∠A + ∠B + ∠C + ∠D = 2.000° ✓. At a fold of 90° the sum is about 181°, at 180° (flat) it is 360° ✓.
Answer to write in the exam
Hinge ∆ABC and ∆ADC on AC: ∠B, ∠D fixed
Flat convex position: ∠A + ∠B + ∠C + ∠D = 180° + 180° = 360°, BD largest
Folding ⇒ BD shorter ⇒ ∠BAD, ∠BCD smaller (AB, AD and CB, CD fixed)
∴ Non-planar ABCD: ∠A + ∠B + ∠C + ∠D < 360°
AB = CB, ∠ABC = 0.5°, ∆ADC mirror image: sum goes from 360° down towards 1° as it folds ⇒ equals 2° at a fold angle of 0.5°
Common mistakes that cost marks
- Assuming every angle shrinks when folding. ∠B and ∠D stay fixed; only ∠A and ∠C change.
- Thinking the sum can reach 0°: ∠B + ∠D stays fixed, so pick them small (here 1° in total).
- Using the planar proof (two triangles add to 360°) without noticing that ∠A is no longer ∠BAC + ∠CAD.
How this can come in the exam
Which of these could be the angle sum of a non-planar quadrilateral?
- 300°
- 360°
- 400°
- 720°
Show answer
(A) 300°
The angle sum of a non-planar quadrilateral is always less than 360°, so only 300° is possible.
Try one yourself
A non-planar quadrilateral has ∠B = 80° and ∠D = 60°. Can its angle sum be 150°? Can it be 100°?
Show answer
150°: possible (sum must be less than 360° and more than ∠B + ∠D = 140°). 100°: impossible, since ∠A and ∠C are positive and the sum exceeds 140°.
More questions like this
- Let us see a third method to tile the plane using a 4-gon. Focus on only two coloured copies of SOME sharing a vertex. What do you see? It appears that each 4-gon is just a shifted copy of the other. Let us see exactly how. Draw two copies of SOME as shown in the figure so that the diagonals EO and EʹOʹ are collinear with O = Eʹ. Now place a cutout of one copy on top of SOME with diagonal EO drawn on it. Slide this cutout so that segment EO moves along EOʹ until EO matches EʹOʹ. Verify that your cutout exactly matches SʹOʹMʹEʹ.
- Is there a 4-gon with given side lengths?
- Counting diagonals of a polygon.
- Sum of angles of a polygon. What is the sum of angles of a (planar non-self-intersecting) n-gon? We know that the answer is 180° for n = 3 and 360° for n = 4. Find the next few values. Then find a formula in terms of n and prove it.
- Multiple converses to a theorem. Let us see how multiple statements can be considered converses to the Midpoint Theorem and how the converse of the Midpoint Theorem (the line through the midpoint of one side parallel to another side bisects the third side) is one of them. To formulate a converse we should express the original statement in “If … then …” form. For a complex statement, there may be multiple ways to do that. The Midpoint Theorem starts with ∆ABC and points P and Q on sides AB and AC respectively. The theorem has two assumptions and two conclusions, which we have named for further discussion.
Assumptions: (P MID) P is the midpoint of AB, and (Q MID) Q is the midpoint of AC.
Conclusions: (PRLL) PQ ‖ BC, and (HALF) PQ = BC2.
Let us use these four named conditions to discuss various possible statements.
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