Medians and centroid · 5 marks

You saw how to use the Midpoint Theorem to divide a given triangle into 4 congruent triangles. Can we divide a triangle into 3 congruent triangles? This exercise shows us how to do that and more, provided we are allowed to cut and reassemble.

(i) Draw medians AP, BQ and CR of ∆ABC, meeting in a common point M. Cut along each median to get 6 triangles. Show with justification how to assemble the 6 pieces into 3 congruent triangles! (Hint: Align ∆MPB and ∆MPC along equal sides PB and PC, matching P with itself and B with C.) Prove that you get a triangle. What are its side lengths?
(ii) Repeat the procedure with each of the 3 assembled triangles you got in (i). Show that each of the resulting 9 triangles has sides AB3, BC3 and AC3. The procedure and the result above were unnoticed till 2014, when they were discovered by Lee Sallows, an amateur mathematician!
(iii) Naturally, we can next ask about the possibility of cutting a triangle and reassembling it into 2 congruent triangles. You were asked this question earlier, when studying perimeter and area. Can you solve this now? (Hint: Use median AP in ∆ABC and then a median of ∆APB.)

Answer: (i) Turn ∆MPB through 180° about P (B goes to C) and join it to ∆MPC: since ∠MPB + ∠MPC = 180°, the result is a triangle MCM′ with sides ⅔AP, ⅔BQ, ⅔CR. Pairing the pieces at Q and at R the same way gives two more triangles with the same sides, so the 3 are congruent (SSS). (ii) Each assembled triangle has medians ½BC, ½CA, ½AB, so repeating (i) gives triangles with sides ⅔ × ½ of these: AB3, BC3, AC3. (iii) Cut along median AP, then cut ∆APB along its median PX (X the midpoint of AB); turn ∆PXB about X onto A: it forms a triangle with sides AP, PC, CA, congruent to ∆APC.

Step-by-step solution

Idea: A half-turn about the midpoint of a side swaps its ends. Pieces that share a midpoint can be swung round it, and the straight angle at that midpoint makes the joined pieces a single triangle. The centroid’s 2 : 1 rule gives the lengths.

(i) Draw medians AP, BQ and CR of ∆ABC, meeting in a common point M. Cut along each median to get 6 triangles. Show with justification how to assemble the 6 pieces into 3 congruent triangles! (Hint: Align ∆MPB and ∆MPC along equal sides PB and PC, matching P with itself and B with C.) Prove that you get a triangle. What are its side lengths?

Turn ∆MPB through 180° about P. Since PB = PC, B lands on C, and M lands on M′ on line MP with PM′ = PM. The turned piece ∆M′PC sits next to ∆MPC along PC.½ mark
It is a triangle: ∠MPC + ∠CPM′ = ∠MPC + ∠MPB = 180° (B, P, C collinear), so M, P, M′ are collinear and the two pieces form ∆MCM′.½ mark
Sides (centroid: AM = ⅔AP, MP = ⅓AP, etc.): MM′ = 2MP = ⅔AP; MC = ⅔CR; M′C = MB = ⅔BQ.½ mark
In the same way, ∆MQA turned about Q joins ∆MQC, and ∆MRB turned about R joins ∆MRA. Each new triangle also has sides ⅔AP, ⅔BQ, ⅔CR, so the three triangles are congruent (SSS).½ mark

Three congruent triangles, each with sides ⅔AP, ⅔BQ and ⅔CR (two-thirds of the three medians).

(ii) Repeat the procedure with each of the 3 assembled triangles you got in (i). Show that each of the resulting 9 triangles has sides AB3, BC3 and AC3. The procedure and the result above were unnoticed till 2014, when they were discovered by Lee Sallows, an amateur mathematician!

In ∆MCM′ from (i), P is the midpoint of side MM′ (length ⅔AP), so CP is a median of it: CP = ½BC. In the triangle built at Q, Q is the midpoint of the side of length ⅔BQ and the median to it is CQ = ½CA. In the triangle built at R, the median to the side of length ⅔CR is AR = ½AB.1 mark
The three triangles are congruent, so corresponding medians are equal: each has medians ½BC, ½CA, ½AB. Applying (i) to any of them gives triangles with sides ⅔ × ½BC = BC3, AC3 and AB3. So all 9 triangles have sides AB3, BC3, AC3.1 mark

Each assembled triangle has medians ½AB, ½BC, ½AC, so the next step gives 9 triangles with sides ⅓AB, ⅓BC, ⅓AC.

(iii) Naturally, we can next ask about the possibility of cutting a triangle and reassembling it into 2 congruent triangles. You were asked this question earlier, when studying perimeter and area. Can you solve this now? (Hint: Use median AP in ∆ABC and then a median of ∆APB.)

Cut along median AP: pieces ∆APC and ∆APB. Let X be the midpoint of AB and cut ∆APB along its median PX into ∆PXA and ∆PXB.½ mark
Turn ∆PXB through 180° about X: B lands on A and P on P′ with X the midpoint of PP′. ∠AXP + ∠AXP′ = 180°, so with ∆PXA it forms ∆PAP′ with sides AP, AP′ = BP = PC and PP′ = 2PX = AC (Midpoint Theorem: PX = ½AC). So ∆PAP′ ≅ ∆APC (SSS): two congruent triangles.½ mark

Yes: cut along AP and along PX (X the midpoint of AB); turning ∆PXB about X gives a triangle congruent to ∆APC.

(i) Swinging the pieces round the midpoints gives 3 congruent triangles with sides ⅔AP, ⅔BQ, ⅔CR. (ii) These have medians ½BC, ½CA, ½AB, so the next round gives 9 triangles with sides AB/3, BC/3, AC/3. (iii) Cut along median AP and median PX of ∆APB; the turned piece makes a triangle congruent to ∆APC.

Check: A(0, 0), B(6, 0), C(1, 5): medians AP ≈ 4.301, BQ ≈ 6.042, CR ≈ 5.385. The assembled ∆MCM′ has sides ≈ 2.867, 4.028, 3.590 (two-thirds of the medians ✓) and medians ≈ 3.536, 2.550, 3.000, which are ½BC, ½CA, ½AB (BC ≈ 7.071, CA ≈ 5.099, AB = 6) ✓. Two-thirds of these are 2.357, 1.700, 2.000 = BC/3, CA/3, AB/3 ✓.

Answer to write in the exam

(i)

Turn ∆MPB through 180° about P: B → C (PB = PC), M → M′ with PM′ = PM

∠MPC + ∠CPM′ = ∠MPC + ∠MPB = 180° ⇒ M, P, M′ collinear ⇒ ∆MCM′ is a triangle

MM′ = 2MP = ⅔AP, MC = ⅔CR, M′C = MB = ⅔BQ (centroid divides medians 2 : 1)

Similarly pairs at Q and R give triangles with the same sides

∴ 3 congruent triangles (SSS), sides ⅔AP, ⅔BQ, ⅔CR

(ii)

In ∆MCM′: P midpoint of MM′ (= ⅔AP) ⇒ median CP = ½BC

Similarly the median to the side ⅔BQ is CQ = ½AC, and to the side ⅔CR is AR = ½AB

The 3 triangles are congruent ⇒ each has medians ½BC, ½AC, ½AB

Applying (i): sides = ⅔ × ½ of these

∴ Each of the 9 triangles has sides AB/3, BC/3, AC/3

(iii)

Cut along median AP; cut ∆APB along its median PX (X midpoint of AB)

Turn ∆PXB through 180° about X ⇒ B → A, P → P′, P′X = XP

∆PAP′ has sides AP, AP′ = BP = PC, PP′ = 2PX = AC (Midpoint Theorem)

∴ ∆PAP′ ≅ ∆APC (SSS): two congruent triangles

Common mistakes that cost marks

Joining ∆MPB and ∆MPC by sliding instead of turning about P; then B does not land on C.
Claiming the joined pieces form a triangle without showing M, P, M′ are collinear (the straight angle at P).
Writing the sides as the full medians AP, BQ, CR. They are two-thirds of the medians.

How this can come in the exam

MCQ (1 mark)

G is the centroid of ∆ABC and the median from A is 18 cm long. In the reassembly above, the side of the new triangle that comes from this median is

6 cm
9 cm
12 cm
18 cm

Show answer

(C) 12 cm
MM′ = 2 × MP = 2 × ⅓ × 18 = 12 cm (two-thirds of the median).

Try one yourself

The medians of a triangle are 9 cm, 12 cm and 15 cm. Find the sides of each of the three congruent triangles obtained by the cut-and-reassemble method.