The two circles always meet

The idea

If three lengths satisfy the triangle inequality, a triangle exists: with the longest as base, circles with the two smaller lengths as radii intersect, since the radii sum exceeds the base (equal: they only touch; smaller: they never meet).

Is passing the test enough?

The lengths 4 cm, 5 cm and 8 cm pass the test: the longest, 8, is smaller than 4 + 5 = 9.

Is that enough to be sure a triangle exists? Will the two arcs always meet?

• Yes, they will always meet
• No, they might miss each other
• We cannot know without drawing it

Pull circle B

The base AB = 8 cm is the longest length. Circle A has radius 5 cm. Pull circle B bigger and watch the two circles.

Radii together against the base

Pulling circle B showed three cases. Everything depends on the two radii together, compared with AB.

Take the longest length as the base AB, and the two smaller lengths as the radii. If the three lengths satisfy the triangle inequality, the radii together are more than AB. So the circles always cross and a triangle always exists.

If three lengths satisfy the triangle inequality, a triangle exists. With the longest as the base, the circles whose radii are the two smaller lengths cross, since the radii together are more than the base. If they are equal the circles only touch; if less, they never meet.

• The radii together are | What happens
• less than AB | The circles do not meet. No triangle.
• equal to AB | The circles only touch, on AB. No triangle.
• more than AB | The circles cross. A triangle!

Notes

If three lengths satisfy the triangle inequality, a triangle exists. With the longest as the base, the circles with the two smaller lengths as radii cross, since the radii together are more than the base. If equal, they only touch; if less, they never meet.

Check yourself

You want a triangle with sides 6 cm, 8 cm and 12 cm. Which is the base, and which are the radii of the two circles?

A base AB is 8 cm long. A circle of radius 3 cm is drawn at A and a circle of radius 5 cm at B. What happens?

Do 5 cm, 6 cm and 10 cm make a triangle? Decide without drawing.

A base AB is 9 cm long and the circle at A has radius 4 cm. What is the smallest whole number of cm for the radius of the circle at B so that the circles cross?

Answer: 6

The radii together must be more than 9. With 5 cm: 4 + 5 = 9, they only touch. With 6 cm: 4 + 6 = 10, which is more than 9, so the circles cross.

• Base 6 cm, radii 8 cm and 12 cm. The base should be the longest length. The two smaller lengths are the radii.
• Base 12 cm, radii 6 cm and 8 cm — correct. Yes! The longest length is the base. The radii together, 6 + 8 = 14, are more than 12, so the circles cross.
• Base 8 cm, radii 6 cm and 12 cm. The base should be the longest length, which is 12 cm here, not 8 cm.
• They cross in two points, so there is a triangle.. The radii together, 3 + 5 = 8, are only equal to AB. They must be more than AB to cross.
• They only touch, at one point on AB. There is no triangle. — correct. Yes! 3 + 5 = 8 is equal to AB, so the circles touch and no third corner exists.
• They do not meet at all.. The circles do meet, but only just: 3 + 5 = 8 is exactly the length of AB, so they touch at one point.
• Yes. 5 + 6 = 11 is more than 10, so the circles cross. — correct. Yes! The triangle inequality holds, so the circles cross and a triangle exists.
• No. 10 cm is longer than both 5 cm and 6 cm.. Being longer than each of the others does not matter. What matters is whether it is shorter than their sum: 10 < 5 + 6 = 11.
• We cannot know until we try to draw it.. We can know! If the longest is shorter than the sum of the other two, the circles always cross.