Maths Olympiad Prep

Library / HMMT February /854 of 1302

HMMT February, 2015

Algebra Difficulty 5.4 AIME, harder Prove it United States

Problem:

Let a,b,ca, b, c be positive real numbers such that a+b+c=10a + b + c = 10 and ab+bc+ca=25ab + bc + ca = 25. Let m=min⁡{ab,bc,ca}m = \min \{ ab, bc, ca \}. Find the largest possible value of mm.

Solution

Solution:

Answer: 259\sqrt{\frac{25}{9}}

Without loss of generality, we assume that c≥b≥ac \geq b \geq a. We see that 3c≥a+b+c=103c \geq a + b + c = 10. Therefore, c≥103c \geq \frac{10}{3}.

Since
0≤(a−b)2=(a+b)2−4ab=(10−c)2−4(25−c(a+b))=(10−c)2−4(25−c(10−c))=c(20−3c) \begin{aligned} 0 &\leq (a-b)^2 \\ &= (a+b)^2 - 4ab \\ &= (10-c)^2 - 4(25 - c(a+b)) \\ &= (10-c)^2 - 4(25 - c(10-c)) \\ &= c(20 - 3c) \end{aligned}
we obtain c≤203c \leq \frac{20}{3}.

Consider m=min⁡{ab,bc,ca}=abm = \min \{ ab, bc, ca \} = ab, as bc≥ca≥abbc \geq ca \geq ab. We compute
ab=25−c(a+b)=25−c(10−c)=(c−5)2. ab = 25 - c(a+b) = 25 - c(10 - c) = (c - 5)^2.
Since 103≤c≤203\frac{10}{3} \leq c \leq \frac{20}{3}, we get that ab≤259ab \leq \frac{25}{9}. Therefore, m≤259m \leq \frac{25}{9} in all cases and the equality can be obtained when (a,b,c)=(53,53,203)(a, b, c) = \left(\frac{5}{3}, \frac{5}{3}, \frac{20}{3}\right).

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.