Angles at a vertex = 360°.
Interior angle of regular pentagon
= ((5−2) × 180°) ∕ 5
= 108°.
Let k pentagons meet at a point.
k × 108° = 360°.
k = 360° ∕ 108°.
k = 3⅓.
k is not a whole number.
∴ Regular pentagon cannot tile the plane.
Draw non-convex DART.
D has a reflex angle.
All four sides are joined.
Take two copies.
Make them share one edge.
Rotate the second copy by 180°.
Rotate about the midpoint of the edge.
Angles add to 360°.
Four copies fit around every vertex.
Repeat the same rotation.
No gaps are formed.
No overlaps are formed.
∴ The plane is tiled.
Varignon method also works.
I prefer the rotation method.
It is simpler.
Take any △ABC.
All three sides are joined.
Reflect △ABC about the midpoint of BC.
We get △A′CB.
ABA′C is a parallelogram.
Every parallelogram tiles the plane.
Therefore copies of △ABC tile the plane.
Horizontal lines are equally spaced.
Draw any transversal.
Use the converse of the Midpoint Theorem.
The line joining two midpoints is ∥ to the third side.
It is also half as long.
Apply the same idea to the successive strips.
The meeting point on the middle line is the midpoint.
AB, BC, CD and DA are drawn.
AB and CD cross at E.
Consider △ABE and △CDE.
Each triangle has angle sum 180°.
Total = 180° + 180°.
Subtract the two vertical angles at E.
Therefore the total is less than 360°.
Make the crossing almost flat.
The four angles become very small.
∴ The sum can be 2°.
ABCD is a parallelogram.
Diagonals meet at O.
BO = OD.
DP = BQ.
Therefore BP = DQ.
AB = CD.
∠ABP = ∠CDQ.
BP = DQ.
∴ △ABP ≅ △CDQ.
AP = CQ.
AP ∥ CQ.
Equal and parallel sides.
∴ APCQ is a parallelogram.
△ABC is a right triangle.
Fold A onto B.
The crease is the perpendicular bisector of AB.
The crease meets AC at M.
Every point on the crease is equally distant from A and B.
The midpoint of AC is also equally distant.
∴ The crease passes through the midpoint of AC.
(i)
P, Q, R are midpoints.
Draw medians AP, BQ and CR.
They meet at M.
The medians give 6 triangles.
Align △BMP and △CMP.
Use equal sides PB and PC.
Match P with P.
Match B with C.
The pieces form congruent triangles.
Their sides are ⅔ of a median, ⅔ of a median and one side of △ABC.
(ii)
Repeat the same procedure.
Do it for each of the 3 triangles.
We get 9 triangles.
Each has sides AB∕3.
BC∕3.
AC∕3.
(iii)
Draw median AP.
Then draw a median of △APB.
Cut and reassemble the pieces.
We get 2 congruent triangles.
Each has half the area.
(i)
AB ∥ DC.
EF ∥ AB.
Therefore EF ∥ DC.
E is the midpoint of AD.
By Midpoint Theorem:
F is the midpoint of BC.
EF = (AB + CD) ∕ 2.
(ii)
Let M be the midpoint of BD.
By Midpoint Theorem:
EM ∥ AB.
FM ∥ AB.
Therefore EF ∥ AB.
ABCD is a parallelogram.
O is the midpoint of both diagonals.
AO = CO.
∠OAP = ∠OCQ.
∠AOP = ∠COQ.
∴ △AOP ≅ △COQ.
Therefore OP = OQ.
∴ O is the midpoint of PQ.
ABCD is a trapezium.
AD ∥ BC.
AD = 3 cm.
BC = 5 cm.
E and F are midpoints.
EF ∥ bases.
EF = (3 + 5) ∕ 2.
EF = 4 cm.
Both trapeziums have the same height.
Areas are proportional to parallel sides.
AEFD : EBCF
= (3 + 4) : (4 + 5)
= 7 : 9.
Four points give 3 different cyclic orders.
Rotation and reflection are ignored.
∴ 3 different quadrilaterals.
Exactly 1 is self-intersecting.
The crossed order is self-intersecting.
If the convex hull is a quadrilateral:
2 are convex.
If one point lies inside the triangle:
1 is convex.
1 is non-convex.
△ABC is given.
P lies on AB.
PQ ∥ BC.
AP = 1.
AQ = √5.
(i)
PB = 2.
Draw a line through B ∥ AQ.
QC = 2√5.
(ii)
PB = 1.
Use the same construction.
QC = √5.
(iii)
PB = 3∕2.
Divide AP and PB into equal parts.
QC = (3∕2)√5.
(iv)
Let AP∕PB = m∕n.
Using congruent parallelograms:
AQ∕QC = m∕n.
∴ AP∕PB = AQ∕QC.
(i)
ABCD is a parallelogram.
M is the midpoint of AB.
N is the midpoint of CD.
DM meets AC at X.
BN meets AC at Y.
Using similar triangles:
AX = XY.
XY = YC.
∴ AX = XY = YC.
DM and BN trisect AC.
(ii)
Construct a parallelogram on PQ.
Use the midpoints.
This gives three equal parts.
Another method uses Exercise 11.
Draw suitable parallel lines.
A third method uses the Centroid Theorem.
Use a triangle with PQ as a side.
Yes, it remains true.
Convexity is not required.
The proof does not use convexity.
The Varignon parallelogram becomes a segment:
when the diagonals are perpendicular bisectors of each other.
For non-planar points:
the four midpoints are still coplanar.
They form a parallelogram.
No.
The points need not be midpoints.
Take points dividing the sides in ratio 1 : 2.
The four points can still form a parallelogram.
∴ They need not be midpoints.
(i)
Extend PQ beyond Q to S.
Take QS = PQ.
Then △APQ ≅ △CSQ.
By SAS.
Therefore BCSP is a parallelogram.
PQ = ½ BC.
PQ ∥ BC.
(ii)
Extend PQ to S.
Take QS = PQ.
Use congruent triangles.
Corresponding sides are equal.
Corresponding lines are parallel.
∴ The converse is proved.
Start with a parallelogram.
Equal diagonals ⇒ rectangle.
Perpendicular diagonals ⇒ rhombus.
Rhombus with one right angle ⇒ square.
Equal diagonals that bisect each other ⇒ rectangle.
Take diagonal AC as a hinge.
Rotate one triangle about AC.
The four face angles become less than 360°.
Rotate it almost 180°.
The sum becomes very small.
∴ The sum can be 2°.
Take two copies.
Make their diagonals collinear.
Match the corresponding vertices.
The four side-parallelograms match.
Therefore the copies coincide.
Translate the 4-gon along its diagonals.
Repeat in both directions.
No gaps are formed.
No overlaps are formed.
∴ The plane is tiled.