Chapter 7: A Tale of Three Intersecting Lines (Triangles)

The three "sides" then lie along one line and enclose no region, so no triangle is formed. (The longest side would be exactly equal to the sum of the other two.)

No triangle is formed: the points just make a line segment.

Steps (ruler and compass):

1. Draw AB = 4 cm with a ruler.
2. With the compass opened to 4 cm, draw an arc with centre A.
3. With the same opening, draw an arc with centre B cutting the first arc at C.
4. Join AC and BC.

With only a marked ruler it is possible, but only by trial and error (marking C again and again until both AC and BC come out 4 cm). The compass makes it exact: every point on the first arc is 4 cm from A and every point on the second arc is 4 cm from B, so the point C lying on both arcs is 4 cm from A and from B.

Draw AB = 4 cm and arcs of radius 4 cm from A and from B; their meeting point C is 4 cm from both, giving the equilateral triangle ABC.

In each case: draw the longest side as the base AB; draw an arc from A with radius equal to the second length and an arc from B with radius equal to the third; join the meeting point C to A and B.

Draw the base, then arcs of the other two radii from its ends; their intersection is the third vertex. (a) and (c) are isosceles, (e) is equilateral, (b) is right-angled.

Join the centre O to any two points P and Q on the circle: OP = OQ (both are radii), so triangle OPQ is isosceles.

Also, for any chord PQ, a point R on the circle that is equally far from P and Q (on the perpendicular bisector of PQ) gives an isosceles triangle PQR with RP = RQ.

Centre O with any two points P, Q on the circle: OP = OQ = radius, so ΔOPQ is isosceles.

(i) Let the circles meet at P and Q.

• Equilateral: ΔABP (and ΔABQ): AP = AB (radius of circle A) and BP = BA (radius of circle B), so all three sides equal the radius.
• Isosceles: for any point X on circle B, ΔABX has BA = BX (both radii of circle B); for any point Y on circle A, ΔABY has AB = AY. Also ΔAPQ has AP = AQ.

(ii) ΔABC itself is equilateral (each side is a radius, since each circle passes through the other two centres). Any centre with two points on its circle gives an isosceles triangle, e.g. A with two points of circle A.

(i) A, B and an intersection point form an equilateral triangle; A, B and any point of one circle form an isosceles triangle. (ii) ΔABC is equilateral; a centre with two points of its circle gives isosceles triangles.

The arcs do not meet: for base 8 cm, arcs of radius 3 cm and 4 cm from the ends cannot reach each other because $3+4=7<8$. Similarly $2+3=5<6$. No triangle is possible.

More such sets: 1, 2, 5; 4, 5, 10; 2, 2, 4 (the arcs just touch on the base, a flat "triangle"). Pattern: the two shorter lengths add up to less than (or equal to) the longest length.

The arcs do not meet because 3 + 4 < 8 and 2 + 3 < 6; whenever the two shorter lengths add up to at most the longest, no triangle exists.

No. The direct path of 7 cm would be longer than the roundabout path $3+3=6$ cm, which is impossible. On construction, arcs of radius 3 cm from the ends of a 7 cm base do not meet.

Rearranging the lengths does not help: whichever side gets the longest length (7 cm or 30 cm in Fig. 7.4), the other two still add up to less than it. The comparisons involve the same three numbers in every order.

No; 7 > 3 + 3. Rearranging cannot help, since the longest length is always compared with the sum of the same two others.

Yes, by comparing each length with the sum of the other two: $8>3+4=7$ and $6>2+3=5$. The direct path would be longer than the roundabout path, so these triangles cannot exist.

Yes: 8 > 3 + 4 and 6 > 2 + 3.

(a) $25>10+10=20$: no triangle.

(b) $20>5+10=15$: no triangle.

(c) $40>12+20=32$: no triangle.

No triangle exists in any of the three cases.

Yes, always. Arrange the lengths in increasing order $a\le b\le c$. Then

• $a<b+c$ (since $a\le c$ and $b>0$), and
• $b<a+c$ (since $b\le c$ and $a>0$),

so the two smaller lengths are always less than the sum of the other two. Examples: 7, 10, 15 and 12, 14, 18.

So we only need one check: is the longest length less than the sum of the two shorter ones?

Yes; the two smaller lengths always pass. To test for a triangle, check only whether the longest length < sum of the other two.

$8<4+5=9$, so the triangle inequality holds. The other two lengths (4 and 5) are the smaller ones and always pass. Visualising the construction: with base AB = 8 cm, the circle of radius 4 cm about A reaches X with BX = 4 cm, which is less than the 5 cm radius about B, so the circles cross at two points: the triangle exists.

Yes: 8 < 4 + 5, and the two circles cross; the shorter sides never need checking.

Check: longest < sum of the other two?

Possible: (b), (d), (e), (g). Not possible: (a), (c), (f).

(a) Touch (sum of the two smaller = longest): 2, 3, 5; 4, 4, 8; 1, 6, 7.

(b) Do not meet (sum of the two smaller < longest): 2, 3, 6; 1, 4, 9; 3, 3, 7.

Procedure:

1. Arrange the three lengths in increasing order.
2. Add the two smaller lengths.
3. If this sum is greater than the longest length, a triangle exists; if it is equal to or less than the longest, no triangle exists.

Touch: 2, 3, 5 / 4, 4, 8 / 1, 6, 7. Do not meet: 2, 3, 6 / 1, 4, 9 / 3, 3, 7. A triangle exists exactly when the sum of the two smaller lengths exceeds the longest.

(a) $100<1+100=101$: exists (a very thin isosceles triangle).

(b) $9=3+6$: does not exist (the circles only touch; the points are in a line).

(c) $5>1+1$: does not exist.

(d) $12<5+10=15$: exists.

(a) yes (b) no (c) no (d) yes

Yes. For sides $s,s,s$: $s<s+s=2s$ is true for every positive $s$. So the triangle inequality always holds, and an equilateral triangle exists for any side length (including 50).

Yes; s < 2s for every s > 0, so equilateral triangles of every size exist.

The third side $x$ must be more than the difference and less than the sum of the given two.

(a) $99<x<101$: e.g. 99.2, 99.5, 100, 100.4, 100.9

(b) $0<x<10$: e.g. 1, 3, 5, 7.5, 9.9

(c) $4<x<10$: e.g. 4.5, 5, 6, 8, 9.5

(a) any length between 99 and 101 (b) between 0 and 10 (c) between 4 and 10.

Steps: draw one side as the base AB; at A draw the given angle with a protractor; mark C on the new arm at the other given length from A; join BC.

Draw one side, make the given angle at its end, mark the other side on the new arm and join the free ends.

No, as long as the included angle is between 0° and 180°. Whatever the lengths, once the angle is drawn and the two lengths marked on its arms, the two free ends can always be joined, so a triangle always forms. A triangle is impossible only if the "angle" is 180° or more (e.g. 3 cm, 180°, 7 cm: the points lie on one line).

No: for any angle less than 180° and any two lengths, the free ends can always be joined.

Steps: draw the side AB; draw the first angle at A and the second at B, on the same side of AB; the arms meet at C.

Draw the side, construct the two angles at its ends, and mark C where the arms cross.

No. If ∠B is large enough, the arm from B leans away and never meets the arm from A. With ∠A = 40°, for ∠B = 150° the lines do not meet.

The smallest such ∠B is when the arm from B is parallel to the arm from A: then ∠A + ∠B = 180° (interior angles on the same side of the transversal AB), so ∠B = 140°. For every ∠B ≥ 140° there is no triangle.

The length AB plays no part: only the two angles decide whether the arms meet.

No; with ∠A = 40° there is no triangle when ∠B ≥ 140° (e.g. 150°); the length of AB does not matter.

A triangle is possible only when the two angles add up to less than 180°.

Possible when the sum is below 180° (e.g. 30° with 50°); not possible when it is 180° or more (e.g. 30° with 150°).

(a) $35+150=185\ge 180$: No (b) $70+30=100$: Yes (c) $90+85=175$: Yes (d) $50+150=200$: No

Rule: two angles can belong to a triangle exactly when their sum is less than 180°.

(b) and (c) can; (a) and (d) cannot. Rule: the two angles must add up to less than 180°.

The third angle is 50° in both cases. The base length does not change it: a longer base gives a bigger triangle of the same shape. The third angle depends only on the two angles ($180°-60°-70°=50°$).

50°, whatever the base length; the third angle depends only on the other two.

Draw XY through A parallel to BC. Then ∠XAB = ∠B and ∠YAC = ∠C (alternate angles), and ∠XAB + ∠BAC + ∠YAC = 180° (straight angle). So the third angle = 180° − (sum of the other two).

(a) $180°-(36°+72°)=$ 72° (b) $180°-165°=$ 15° (c) $180°-120°=$ 60° (d) $180°-120°=$ 60°

(a) 72° (b) 15° (c) 60° (d) 60°

No: $3\times 70°=210°\ne 180°$.

If two angles are 70°, the third is $180°-140°=$ 40°.

If all three are equal, each is $180°÷3=$ 60° (an equilateral triangle).

No; 40°; 60°

$\angle B+\angle C=180°-50°=130°$, and they are equal, so $\angle B=\angle C=$ 65°.

∠B = ∠C = 65°

The exterior angle equals the sum of the two interior opposite angles: $\angle ACD=\angle A+\angle B$.

Reason: $\angle A+\angle B+\angle ACB=180°$ and $\angle ACD+\angle ACB=180°$, so $\angle ACD=\angle A+\angle B$.

Exterior angle = sum of the two opposite interior angles (∠ACD = ∠A + ∠B).

Fold so that the crease passes through the top vertex A and the base BC folds onto itself (the part of BC on one side lies exactly on the part on the other side). Then the two angles that the crease makes with the base cover each other exactly, so they are equal. Together they make a straight angle (180°), so each is 90°: the crease is perpendicular to the base.

The two angles at the base fold onto each other, so they are equal; since they add up to 180°, each is 90°.

Yes, a right-angled triangle. If ∠B = 90°, then side AB is perpendicular to BC, so AB is the altitude from A to BC (and BC is the altitude from C to AB).

Yes, in a right-angled triangle the two sides forming the right angle are altitudes.

No. Every triangle has at least two acute angles (if two angles were 90° or more, their sum would already be at least 180°). Right-angled and obtuse-angled triangles also have acute angles. So an acute-angled triangle must be one in which all three angles are acute.

No: every triangle has at least two acute angles; an acute-angled triangle has all three angles acute.

Draw BC = 5 cm; arcs of radius 6 cm from B and 5 cm from C meet at A. Then place a set square along a ruler on BC and slide it until its edge touches A; draw AD ⊥ BC.

(Measuring gives AD ≈ 4.8 cm, with D about 3.6 cm from B.)

Construct by SSS (BC = 5 cm base), then drop AD ⊥ BC with a set square; AD ≈ 4.8 cm.

Draw RY = 4 cm, make ∠R = 140° and mark T on the arm with RT = 7 cm; join TY. Since ∠R is obtuse, the perpendicular from T falls outside the triangle: extend YR beyond R and drop TD ⊥ (YR extended).

(The altitude TD ≈ 4.5 cm.)

Construct by SAS, extend YR beyond R and draw TD perpendicular to it; TD ≈ 4.5 cm.

Take AC = 5 cm as the base. Any angles ∠A and ∠C with $\angle A+\angle C=90°$ give ∠B = 90°: e.g. 30° and 60°, 45° and 45°, 20° and 70°. So there are infinitely many such triangles. (All possible positions of B lie on the semicircle with diameter AC.)

Infinitely many: any ∠A, ∠C with ∠A + ∠C = 90° (B can be anywhere on the semicircle on AC).

• Equilateral: every angle is 60°, so it can be neither right-angled nor obtuse-angled.
• Isosceles right-angled: yes, angles 90°, 45°, 45° (e.g. equal sides 4 cm with a 90° angle between them).
• Isosceles obtuse-angled: yes, e.g. angles 120°, 30°, 30° (equal sides 4 cm with a 120° angle between them).

Equilateral: neither. Isosceles: both possible (90°, 45°, 45° and e.g. 120°, 30°, 30°).

Open out (unfold) the two faces the spider walks across so that they lie flat. In the flat net, the shortest path is the straight line from the start to the end corner. Folding the box back, this line crosses the common edge of the two faces at one point (for a cube, at the midpoint of that edge).

For a cube of side $a$, the flat path is the diagonal of a $2a\times a$ rectangle, of length $\sqrt{5}a\approx 2.24a$, shorter than going along two edges and a face diagonal or along three edges ($3a$). For a box with sides $l\ge w\ge h$, the shortest path has length $\sqrt{l^2+(w+h{)}^2}$.

Unfold the box: the straight line in the net is the shortest path; it crosses the shared edge once (at its midpoint for a cube), with length √5 × side for a cube.