AdvancedMathematicsHigh School (11–12)Geometry › TrianglesNumericalConceptualOlympiad

A triangle ABC has sides a = BC, b = CA and c = AB…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A triangle $ABC$ has sides $a = BC$ , $b = CA$ and $c = AB$ . Its semi-perimeter is $s = \dfrac{a + b + c}{2}$ , its area is $\Delta$ , its inradius is $r$ and its circumradius is $R$ . The incircle touches $BC$ , $CA$ and $AB$ at $D$ , $E$ and $F$ , as in the figure. Put $x = s - a$ , $y = s - b$ and $z = s - c$ . This change of letters is a strong tool: it turns a fact about the sides of a triangle into a fact about three positive numbers that need not satisfy any condition.

A triangle ABC with its incircle, centre I. The incircle touches BC at D, CA at E and AB at F. The sides are labelled outside the triangle, each beside its own side and away from the touch points: a on BC, b on CA and c on AB.

Prove that $AE = AF = x$ , $BF = BD = y$ and $CD = CE = z$ . Deduce that $a = y + z$ , $b = z + x$ and $c = x + y$ .
Prove that $(x + y)(y + z)(z + x) \ge 8xyz$ for all positive numbers $x$ , $y$ , $z$ , with equality only when $x = y = z$ .
You may use $R = \dfrac{abc}{4\Delta}$ , $r = \dfrac{\Delta}{s}$ and Heron's formula in the form $\Delta^2 = sxyz$ . Prove that $R \ge 2r$ for every triangle, with equality only when the triangle is equilateral.
For the triangle with sides $7$ , $8$ and $9$ , find $r$ , $R$ and the ratio $R/r$ .
Among all triangles with perimeter $30$ that have one side equal to $12$ , find the greatest possible area, and say which triangle gives it.

Give each answer that is not a whole number to two decimal places. Each part may use the results of the earlier parts, but every claim needs a proof.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Tangent lengths give the substitution. Then AM-GM proves the inequality, and the formulas for $R$ and $r$ turn it into $R \ge 2r$ .

(a) The two tangents from a point to a circle are equal. So $AE = AF = p$ , $BF = BD = q$ and $CD = CE = t$ for some positive $p$ , $q$ , $t$ . Reading the sides, $c = AF + FB = p + q$ , $a = BD + DC = q + t$ and $b = CE + EA = t + p$ . Adding, $2s = 2(p + q + t)$ , so $p + q + t = s$ . Hence $p = s - (q + t) = s - a = x$ , and in the same way $q = s - b = y$ and $t = s - c = z$ . Then $a = q + t = y + z$ , $b = z + x$ and $c = x + y$ .

(b) For positive numbers, $x + y \ge 2\sqrt{xy}$ , because $x + y - 2\sqrt{xy} = \left(\sqrt{x} - \sqrt{y}\right)^2 \ge 0$ . In the same way $y + z \ge 2\sqrt{yz}$ and $z + x \ge 2\sqrt{zx}$ . All six numbers are positive, so we may multiply: $(x + y)(y + z)(z + x) \ge 8\sqrt{x^2y^2z^2} = 8xyz$ . Equality needs equality in all three, that is $x = y$ , $y = z$ and $z = x$ .

(c) From the given formulas, $\dfrac{R}{2r} = \dfrac{abc}{4\Delta} \cdot \dfrac{s}{2\Delta} = \dfrac{abc \, s}{8\Delta^2} = \dfrac{abc \, s}{8sxyz} = \dfrac{(y + z)(z + x)(x + y)}{8xyz}$ . By (b) this is at least $1$ , so $R \ge 2r$ . Equality holds only when $x = y = z$ , that is, when $s - a = s - b = s - c$ , which means $a = b = c$ .

(d) Here $s = 12$ , $x = 5$ , $y = 4$ and $z = 3$ . So $\Delta = \sqrt{12 \times 5 \times 4 \times 3} = \sqrt{720} = 12\sqrt{5} \approx 26.83$ . Then $r = \dfrac{\Delta}{s} = \sqrt{5} \approx 2.24$ and $R = \dfrac{7 \times 8 \times 9}{4 \times 12\sqrt{5}} = \dfrac{21}{2\sqrt{5}} \approx 4.70$ . The ratio is $\dfrac{R}{r} = \dfrac{abc}{4xyz} = \dfrac{504}{240} = 2.10$ , a little above $2$ , as (c) says it must be.

(e) The perimeter is $30$ , so $s = 15$ . Take $a = 12$ . Then $x = s - a = 3$ , and from (a), $y + z = a = 12$ . Heron's formula gives $\Delta^2 = sxyz = 45yz$ . By AM-GM, $yz \le \left(\dfrac{y + z}{2}\right)^2 = 36$ , so $\Delta^2 \le 1620$ and $\Delta \le 18\sqrt{5} \approx 40.25$ . Equality needs $y = z = 6$ , so $b = z + x = 9$ and $c = x + y = 9$ . The best triangle is the isosceles triangle with sides $9$ , $9$ and $12$ . The trap is to take the equilateral triangle of perimeter $30$ : its area $25\sqrt{3} \approx 43.30$ is larger, but it has no side $12$ .

Check. A triangle with sides $7$ , $8$ , $9$ drawn with coordinates gives the same $r$ and $R$ . For (e), a scan of the second side from $3$ to $15$ in steps of $0.001$ gives the greatest area at $9$ , and (b) and (c) hold for twenty thousand random triangles.

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