SAT Prep – Math: Algebra

Example

If $m$ is a positive integer, which of the following COULD represent the lengths of the 3 sides of a triangle?

1. $m, (m+2)\ \&\ (m+4)$
2. $\frac{4m-1}{2}, \frac{3m-1}{3}\ \&\ \frac{2m-1}{2}$
3. $\frac{m}{3},\frac{m}{5}\ \&\ \frac{m}{7}$

1. I only
2. II only
3. III only
4. I and III only
5. II and III only

Solution

Note the Triangle Inequality Theorem for any triangle $\Delta ABC,$ whose three sides are $a, b\ \&\ c$.

The sum of any two sides of a triangle is greater than the third side.

• $a+b>c$;
• $c+b>a$;
• $a+c>b$

Alternatively:

The difference of any two sides of a triangle is less than the third side.

• $|a-b|<c$;
• $|c-b|<a$;
• $|a-c|<b$

We can apply any one of the two rules to test the three roman options.

To save time, we must add the smaller two sides given in the roman options and check the sum against the largest side. If that works, the rule would also work for other combinations of the sides.

1. $m, (m+2)\ \&\ (m+4)$

   We see that the smaller two sides are $m\ \&\ (m+2)$

   The sum of the smaller two sides $=m+(m+2)=2m+2$

   Let's check the sum against the largest side, $(m+4)$.

   Is $2m+2>m+4?$

   Above inequality is true if $m>2$. So, if $m: {3, 4, 5, \dots }$, the statement is correct. Since it is given that $m$ is a positive integer and it is a COULD be true kind of question, Statement I is correct.

   Had this been a MUST be true question, Statement I would be incorrect since for $m=1\ \&\ 2,$ the rule does not hold true.

2. $\frac{4m-1}{2}, \frac{3m-1}{3}\ \&\ \frac{2m-1}{2}$

   Let's simplify the given values of the three sides so that they are easier to compare.

   • $\frac{4m-1}{2}=2m-\frac{1}{2}$;
   • $\frac{3m-1}{3}=m-\frac{1}{3}$;
   • $\frac{2m-1}{2}=m-\frac{1}{2}$

   We see that the smaller two sides are $(m-\frac{1}{3})\ \&\ (m-\frac{1}{2})$

   The sum of the smaller two sides $=m-\frac{1}{3} + m-\frac{1}{2}=2m-\frac{5}{6}$

   Let's check the sum against the largest side, $2m-\frac{1}{2}$.

   Is $2m-\frac{5}{6}>2m-\frac{1}{2}?$

   We have $2m-\frac{5}{6}>2m-\frac{1}{2}$

   $=>-\frac{5}{6}>-\frac{1}{2}$

   $=>\frac{5}{6}<\frac{1}{2}$; multiplying the inequality by $-1$ and reversing the sign of the inequality.

   This is incorrect since 5/6 is greater than 1/2; thus, Statement II is incorrect.

3. $\frac{m}{3},\frac{m}{5}\ \&\ \frac{m}{7}$

   We see that the smaller two sides are $\frac{m}{5}\ \&\ \frac{m}{7}$.

   The sum of the smaller two sides $=\frac{m}{5} + \frac{m}{7}=\frac{12m}{35}$

   Let's check the sum against the largest side, $\frac{m}{3}$.

   Is $\frac{12m}{35}>\frac{m}{3}?$

   We have $\frac{12m}{35}>\frac{m}{3}$

   $=>\frac{1}{\frac{35}{12}}>\frac{1}{3}$

   $=>\frac{1}{2.9}>\frac{1}{3}$

   Or, $2.9<3$

   This is correct; thus, Statement III is also correct.

The correct answer is option D.