A Tale of Three Intersecting Lines

Given: Side lengths 4 cm, 4 cm, 6 cm.

Check Triangle Inequality:
- $4 + 4 = 8 > 6$ ✓
- $4 + 6 = 10 > 4$ ✓
- $4 + 6 = 10 > 4$ ✓

All conditions satisfied, so the triangle exists.

Steps of Construction:

Step 1: Draw base $AB = 6$ cm.

Step 2: With A as centre, draw an arc of radius 4 cm.

Step 3: With B as centre, draw an arc of radius 4 cm such that it intersects the first arc at point C.

Step 4: Join AC and BC.

$\triangle ABC$ is the required triangle with $AB = 6$ cm, $AC = 4$ cm, $BC = 4$ cm.

This is an isosceles triangle (two equal sides of 4 cm).

Given: Side lengths 3 cm, 4 cm, 5 cm.

Check Triangle Inequality:
- $3 + 4 = 7 > 5$ ✓
- $3 + 5 = 8 > 4$ ✓
- $4 + 5 = 9 > 3$ ✓

All conditions satisfied, so the triangle exists.

Steps of Construction:

Step 1: Draw base $AB = 5$ cm.

Step 2: With A as centre, draw an arc of radius 3 cm.

Step 3: With B as centre, draw an arc of radius 4 cm such that it intersects the first arc at point C.

Step 4: Join AC and BC.

$\triangle ABC$ is the required triangle with $AB = 5$ cm, $AC = 3$ cm, $BC = 4$ cm.

This is a scalene triangle (all sides different). It is also a right-angled triangle since $3^2 + 4^2 = 9 + 16 = 25 = 5^2$.

Given: Side lengths 1 cm, 5 cm, 5 cm.

Check Triangle Inequality:
- $1 + 5 = 6 > 5$ ✓
- $1 + 5 = 6 > 5$ ✓
- $5 + 5 = 10 > 1$ ✓

All conditions satisfied, so the triangle exists.

Steps of Construction:

Step 1: Draw base $AB = 1$ cm.

Step 2: With A as centre, draw an arc of radius 5 cm.

Step 3: With B as centre, draw an arc of radius 5 cm such that it intersects the first arc at point C.

Step 4: Join AC and BC.

$\triangle ABC$ is the required triangle with $AB = 1$ cm, $AC = 5$ cm, $BC = 5$ cm.

This is an isosceles triangle (two equal sides of 5 cm).

Given: Side lengths 4 cm, 6 cm, 8 cm.

Check Triangle Inequality:
- $4 + 6 = 10 > 8$ ✓
- $4 + 8 = 12 > 6$ ✓
- $6 + 8 = 14 > 4$ ✓

All conditions satisfied, so the triangle exists.

Steps of Construction:

Step 1: Draw base $AB = 8$ cm.

Step 2: With A as centre, draw an arc of radius 4 cm.

Step 3: With B as centre, draw an arc of radius 6 cm such that it intersects the first arc at point C.

Step 4: Join AC and BC.

$\triangle ABC$ is the required triangle with $AB = 8$ cm, $AC = 4$ cm, $BC = 6$ cm.

This is a scalene triangle (all sides different).

Given: Side lengths 3.5 cm, 3.5 cm, 3.5 cm.

Check Triangle Inequality:
- $3.5 + 3.5 = 7 > 3.5$ ✓
- $3.5 + 3.5 = 7 > 3.5$ ✓
- $3.5 + 3.5 = 7 > 3.5$ ✓

All conditions satisfied, so the triangle exists.

Steps of Construction:

Step 1: Draw base $AB = 3.5$ cm.

Step 2: With A as centre, draw an arc of radius 3.5 cm.

Step 3: With B as centre, draw an arc of radius 3.5 cm such that it intersects the first arc at point C.

Step 4: Join AC and BC.

$\triangle ABC$ is the required equilateral triangle with all sides equal to 3.5 cm.

Concept: In a circle, all radii are equal. Any two radii and the chord joining their endpoints form an isosceles triangle.

Method:
- Let O be the centre of the circle and let A, B be any two points on the circle.
- Then $OA = OB =$ radius.
- Triangle OAB has two equal sides (both equal to the radius), so $\triangle OAB$ is an isosceles triangle.

Conclusion: By choosing any two points on the circle and joining them to the centre O, we always get an isosceles triangle. We can form as many isosceles triangles as we wish by choosing different pairs of points on the circle.

Case 1: Two circles of the same size with centres A and B.

Let the two equal circles intersect at points P and Q.

- Since both circles have the same radius $r$, we have $AP = AQ = BP = BQ = r$.
- Triangle APB: $AP = BP = r$ (both radii of their respective circles), so $\triangle APB$ is isosceles.
- Triangle AQB: Similarly, $AQ = BQ = r$, so $\triangle AQB$ is isosceles.
- Triangle APQ: $AP = AQ = r$, so $\triangle APQ$ is isosceles.
- Triangle BPQ: $BP = BQ = r$, so $\triangle BPQ$ is isosceles.
- Triangle APBQ (quadrilateral), but focusing on $\triangle APB$: if the circles are drawn so that each passes through the other's centre (i.e., $AB = r$), then $AP = BP = AB = r$, making $\triangle APB$ equilateral.

Case 2: Three circles of the same size with centres A, B, C.

If each circle passes through the centres of the other two (i.e., $AB = BC = CA = r$), then:
- $\triangle ABC$ is equilateral (all sides equal to $r$).
- The intersection points of the circles along with the centres can also form isosceles triangles.
- For example, if P is the intersection of circles centred at A and B, then $AP = BP = r = AB$, so $\triangle APB$ is equilateral.

Conclusion: By choosing the two intersection points of any two same-sized circles as two vertices and one of the centres as the third vertex, we get isosceles triangles. When each circle passes through the other's centre, we get equilateral triangles.

Given: Side lengths 3 cm, 4 cm, 8 cm.

Check Triangle Inequality:
- $3 + 4 = 7$, but $7 < 8$. ✗ (The sum of the two smaller sides is less than the largest side.)

Construction attempt:
Draw base $AB = 8$ cm. Draw an arc of radius 3 cm from A and an arc of radius 4 cm from B. The two arcs do not intersect because the maximum distance they can reach toward each other is $3 + 4 = 7$ cm, which is less than 8 cm (the distance between A and B).

Conclusion: The triangle cannot be constructed. The two arcs do not meet, so no third vertex C exists. This is because the lengths 3, 4, 8 do not satisfy the triangle inequality ($3 + 4 < 8$).

Given: Side lengths 2 cm, 3 cm, 6 cm.

Check Triangle Inequality:
- $2 + 3 = 5$, but $5 < 6$. ✗

The sum of the two smaller sides (2 + 3 = 5) is less than the largest side (6).

Conclusion: The triangle is not possible. The lengths 2, 3, 6 do not satisfy the triangle inequality, so no triangle can be constructed with these side lengths.

Examples of sets for which a triangle is impossible:

1. 1, 2, 5: $1 + 2 = 3 < 5$ ✗
2. 2, 3, 7: $2 + 3 = 5 < 7$ ✗
3. 1, 4, 6: $1 + 4 = 5 < 6$ ✗
4. 5, 6, 15: $5 + 6 = 11 < 15$ ✗
5. 10, 15, 30: $10 + 15 = 25 < 30$ ✗

Pattern observed: In each case, the sum of the two smaller (or any two) lengths is less than or equal to the third (largest) length. That is, the triangle inequality is violated: the longest side is greater than or equal to the sum of the other two sides.

Rule: A triangle is impossible when the largest side $\geq$ sum of the other two sides.

Yes, we can determine this without construction by checking the triangle inequality.

For 3, 4, 8:
$$3 + 4 = 7 < 8$$
The sum of the two smaller sides is less than the largest side. Triangle inequality is violated. So no triangle is possible.

For 2, 3, 6:
$$2 + 3 = 5 < 6$$
Again, the sum of the two smaller sides is less than the largest side. Triangle inequality is violated. So no triangle is possible.

Conclusion: We only need to check whether the sum of the two smaller lengths is greater than the largest length. If not, the triangle does not exist — no construction needed.

Concept: Check the triangle inequality — the sum of any two sides must be greater than the third side. It is sufficient to check: sum of the two smaller sides > largest side.

(a) 10 km, 10 km, 25 km:
$$10 + 10 = 20 < 25$$
Triangle inequality is violated. Triangle does not exist.

(b) 5 mm, 10 mm, 20 mm:
$$5 + 10 = 15 < 20$$
Triangle inequality is violated. Triangle does not exist.

(c) 12 cm, 20 cm, 40 cm:
$$12 + 20 = 32 < 40$$
Triangle inequality is violated. Triangle does not exist.

Conclusion: None of the three sets can form a triangle.

Exploration:

Let the three lengths be $a \leq b \leq c$ (arranged in increasing order).

Comparison 1: $a < b + c$? Since $b \geq a > 0$ and $c \geq a > 0$, we have $b + c \geq a + a > a$. So $a < b + c$ is always true.

Comparison 2: $b < a + c$? Since $c \geq b$ and $a > 0$, we have $a + c > c \geq b$. So $b < a + c$ is always true.

Comparison 3: $c < a + b$? This is not always true — this is the critical comparison.

Conclusion:
- The two smaller lengths are always less than the sum of the other two. No calculation needed for them.
- The only comparison that matters is whether the largest side is less than the sum of the other two sides.
- So to check for the existence of a triangle, we only need to verify: $$\text{largest side} < \text{sum of the other two sides}$$

For example, for 10, 15, 30: $30 > 10 + 15 = 25$, so no triangle exists.

Rule: A triangle exists if and only if the sum of the two smaller sides is greater than the largest side.

(a) 2, 2, 5:
$$2 + 2 = 4 < 5$$
Triangle inequality violated. Triangle NOT possible.

(b) 3, 4, 6:
$$3 + 4 = 7 > 6$$ ✓
Triangle inequality satisfied. Triangle IS possible.

(c) 2, 4, 8:
$$2 + 4 = 6 < 8$$
Triangle inequality violated. Triangle NOT possible.

(d) 5, 5, 8:
$$5 + 5 = 10 > 8$$ ✓
Triangle inequality satisfied. Triangle IS possible.

(e) 10, 20, 25:
$$10 + 20 = 30 > 25$$ ✓
Triangle inequality satisfied. Triangle IS possible.

(f) 10, 20, 35:
$$10 + 20 = 30 < 35$$
Triangle inequality violated. Triangle NOT possible.

(g) 24, 26, 28:
$$24 + 26 = 50 > 28$$ ✓
Triangle inequality satisfied. Triangle IS possible.

Background: When constructing a triangle with sides $a$, $b$, $c$ (where $c$ is the base), we draw circles of radii $a$ and $b$ from the two ends of the base $c$.

(a) Circles touch each other at exactly one point (degenerate triangle):
This happens when $a + b = c$ exactly — the two arcs just touch.

Examples:
1. $2, 3, 5$: $2 + 3 = 5$ — circles touch at one point.
2. $4, 6, 10$: $4 + 6 = 10$ — circles touch at one point.
3. $1, 4, 5$: $1 + 4 = 5$ — circles touch at one point.

In each case, the three points are collinear — a degenerate (flat) triangle.

(b) Circles do not intersect (no triangle possible):
This happens when $a + b < c$ — the two arcs are too far apart.

Examples:
1. $2, 3, 8$: $2 + 3 = 5 < 8$ — circles do not intersect.
2. $1, 2, 6$: $1 + 2 = 3 < 6$ — circles do not intersect.
3. $5, 6, 15$: $5 + 6 = 11 < 15$ — circles do not intersect.

**Procedure to check existence of a triangle for given lengths $a$, $b$, $c$:

Step 1: Identify the three given lengths.

Step 2:** Find the largest of the three lengths. Let it be $c$.

Step 3: Find the sum of the other two lengths: $a + b$.

Step 4: Compare:
- If $a + b > c$: The triangle inequality is satisfied and the triangle exists.
- If $a + b = c$: The three points are collinear (degenerate case) — a proper triangle does not exist.
- If $a + b < c$: The triangle inequality is violated — the triangle does not exist.

Note: It is sufficient to check only the largest side against the sum of the other two, since the other two comparisons are automatically satisfied.

Rule: Triangle exists if and only if the largest side < sum of the other two sides.

(a) 1, 100, 100:
Largest side = 100.
$$1 + 100 = 101 > 100$$ ✓
Triangle inequality satisfied. Triangle EXISTS.

(b) 3, 6, 9:
Largest side = 9.
$$3 + 6 = 9$$
$9 = 9$, not strictly greater. Triangle inequality is not satisfied (degenerate case).
Triangle does NOT exist (the three points would be collinear).

(c) 1, 1, 5:
Largest side = 5.
$$1 + 1 = 2 < 5$$
Triangle inequality violated. Triangle does NOT exist.

(d) 5, 10, 12:
Largest side = 12.
$$5 + 10 = 15 > 12$$ ✓
Triangle inequality satisfied. Triangle EXISTS.

For sides 50, 50, 50:
Check triangle inequality:
$$50 + 50 = 100 > 50$$ ✓

The triangle inequality is satisfied. Yes, an equilateral triangle with sides 50, 50, 50 exists.

**In general, for any equilateral triangle with side length $a$ (where $a > 0$):**
$$a + a = 2a > a$$ ✓

The triangle inequality is always satisfied for any positive value of $a$.

Conclusion: An equilateral triangle exists for any positive side length. There is no restriction on the side length of an equilateral triangle.

Rule: If two sides are $a$ and $b$ ($a \leq b$), the third side $x$ must satisfy:
$$b - a < x < a + b$$

(a) Sides: 1 and 100:
$$100 - 1 < x < 100 + 1$$
$$99 < x < 101$$

All numbers strictly between 99 and 101 are valid.

5 possible values: $99.5,\ 99.8,\ 100,\ 100.2,\ 100.5$

(b) Sides: 5 and 5:
$$5 - 5 < x < 5 + 5$$
$$0 < x < 10$$

All numbers strictly between 0 and 10 are valid.

5 possible values: $1,\ 3,\ 5,\ 7,\ 9$

(c) Sides: 3 and 7:
$$7 - 3 < x < 7 + 3$$
$$4 < x < 10$$

All numbers strictly between 4 and 10 are valid.

5 possible values: $4.5,\ 5,\ 6,\ 8,\ 9.5$