1972 AMC 12 Problem 1

Question

The lengths in inches of the three sides of each of four triangles

\mathrm{I},

\mathrm{II},

\mathrm{III},

and

\mathrm{IV}

are as follows:

\begin{aligned} \mathrm{I}\quad&3,4,5\\ \mathrm{II}\quad&4,7\frac12,8\frac12\\ \mathrm{III}\quad&7,24,25\\ \mathrm{IV}\quad&3\frac12,4\frac12,5\frac12 \end{aligned}

Of these four given triangles, the only right triangles are:

\mathrm{I}

and

\mathrm{II}

\mathrm{I}

and

\mathrm{III}

\mathrm{I}

and

\mathrm{IV}

\mathrm{I},

\mathrm{II},

and

\mathrm{III}

\mathrm{I},

\mathrm{II},

and

\mathrm{IV}

Compare the square of each longest side with the sum of the squares of the other two sides

Clear the halves in triangles

\mathrm{II}

and

\mathrm{IV}

before comparing

For triangles

\mathrm{I},

\mathrm{II},

and

\mathrm{III},

the relevant equalities are

\begin{aligned} 3^2+4^2&=5^2,\\ 4^2+\left(\frac{15}{2}\right)^2 &=\left(\frac{17}{2}\right)^2,\\ 7^2+24^2&=25^2. \end{aligned}

For

\mathrm{IV},

\begin{aligned} \left(\frac72\right)^2+\left(\frac92\right)^2 &=\frac{130}{4},\\ \left(\frac{11}{2}\right)^2&=\frac{121}{4}. \end{aligned}

These values are unequal. Thus precisely

\mathrm{I},

\mathrm{II},

and

\mathrm{III}

are right triangles.

Therefore, the correct answer is D.

Problem 1 in other years