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Toolkit 81

Substitution

Example: Philippine Mathematical Olympiad

The next problem is from the 2022 Philippine Mathematical Olympiad — Qualifying Stage.

Problem 20: Let $a,b,c$ be real numbers such that

3ab+2=6b,\qquad 3bc+2=5c,\qquad 3ca+2=4a.

Suppose the only possible values for the product $abc$ are $r/s$ and $t/u$ , where $r/s$ and $t/u$ are both fractions in lowest terms. Find

r+s+t+u.

Solution

3ca+2=4a\ \Rightarrow\ 3ca=4a-2\ \Rightarrow\ c=\frac{4a-2}{3a}\hspace{4em}(1)

3ab+2=6b\ \Rightarrow\ 2=b(6-3a)\ \Rightarrow\ b=\frac{2}{6-3a}\hspace{4em}(2)

Substitute (1) and (2) into $3bc+2=5c$ :

3\left(\tfrac{2}{6-3a}\right)\left(\tfrac{4a-2}{3a}\right)+2=5\left(\tfrac{4a-2}{3a}\right)\ \Rightarrow\ \left(\tfrac{2}{2-a}\right)\left(\tfrac{4a-2}{3a}\right)+2=5\left(\tfrac{4a-2}{3a}\right)

\Rightarrow\ 2(4a-2)+2(2-a)(3a)=5(4a-2)(2-a)\ \Rightarrow\ 8a-4+12a-6a^2=40a-20a^2-20+10a

\Rightarrow\ 14a^2-30a+16=0\ \Rightarrow\ 7a^2-15a+8=0\ \Rightarrow\ (7a-8)(a-1)=0\ \Rightarrow\ a=1\ \text{or}\ a=\tfrac{8}{7}

Case 1

a=1\ \Rightarrow\ b=\tfrac{2}{6-3a}=\tfrac{2}{3}\ \Rightarrow\ c=\tfrac{4a-2}{3a}=\tfrac{2}{3}\ \Rightarrow\ abc=\tfrac{4}{9}=\tfrac{r}{s}

Case 2

a=\tfrac{8}{7}\ \Rightarrow\ b=\tfrac{2}{6-3a}=\tfrac{2}{6-\frac{24}{7}}=\tfrac{7}{9}\ \Rightarrow\ c=\tfrac{4a-2}{3a}=\tfrac{\frac{32}{7}-2}{\frac{24}{7}}=\tfrac{3}{4}

\Rightarrow\ abc=\tfrac{8}{7}\cdot\tfrac{7}{9}\cdot\tfrac{3}{4}=\tfrac{2}{3}=\tfrac{t}{u}\ \Rightarrow\ r=4,\ s=9,\ t=2,\ u=3\ \Rightarrow\ r+s+t+u=18

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