1977 AMC 12 Problems

Express every term in terms of $x$

Use $y=2x$ first to obtain $z=4x$

Since $y=2x,$ we have $z=2y=4x.$ Therefore $x+y+z=x+2x+4x=7x.$

Therefore, the correct answer is D.

Separate properties determined by angles from properties determined by size

Compare two equilateral triangles having different side lengths

Every equilateral triangle is equiangular, isosceles, regular, and similar to every other equilateral triangle. Two equilateral triangles with different side lengths are not congruent, however.

Therefore, the correct answer is D.

Find the value of one coin of each of the five types

The five coins in one complete set are worth $91$ cents

One coin of each kind is worth $1+5+10+25+50=91$ cents. Since $\frac{273}{91}=3,$ there are three coins of each of the five kinds, or $5\cdot3=15$ coins.

Therefore, the correct answer is E.

First determine the two base angles of $\triangle ABC$

Use the two given equal-length pairs to find $\angle EDC$ and $\angle BDF$

The base angles of $\triangle ABC$ are each $50^\circ.$ Since $CE=CD,$ triangle $CED$ has vertex angle $50^\circ,$ so $\angle EDC=65^\circ.$ Similarly, $BF=BD$ gives $\angle BDF=65^\circ.$ The three angles above the straight line $CB$ sum to $180^\circ,$ so $$\begin{aligned} \angle EDF&=180^\circ-65^\circ-65^\circ\\ &=50^\circ. \end{aligned}$$

Therefore, the correct answer is C.

Apply the triangle inequality to $AP,$ $PB,$ and $AB$

Recall when equality holds in the triangle inequality

The triangle inequality gives $AP+PB\ge AB.$ Equality holds exactly when $A,$ $P,$ $B$ are collinear with $P$ between $A$ and $B,$ including the endpoints. Thus the locus is the segment from $A$ to $B.$

Therefore, the correct answer is A.

Rewrite the outside reciprocal using a single fraction

Combine the two reciprocals inside the brackets over a common denominator

We have $$\begin{aligned} \left(2x+\frac y2\right)^{-1} &=\frac2{4x+y},\\ (2x)^{-1}+\left(\frac y2\right)^{-1} &=\frac1{2x}+\frac2y\\ &=\frac{y+4x}{2xy}. \end{aligned}$$ Their product is $\frac{1}{xy}=(xy)^{-1}.$

Therefore, the correct answer is D.

Let $u=\sqrt[4]{2}$ and use $u^4=2$

Factor $u^4-1$ into a linear factor, a second linear factor, and a quadratic factor

Let $u=\sqrt[4]{2}.$ Since $$\begin{aligned} &(1-u)\bigl(-(1+u)(1+u^2)\bigr)\\ &\qquad=-(1-u^2)(1+u^2)\\ &\qquad=u^4-1=1, \end{aligned}$$ it follows that $$t=-(1+\sqrt[4]{2})(1+\sqrt2).$$

Therefore, the correct answer is E.

Each of the first three fractions is either $1$ or $-1$

The sign of the final fraction is the product of the first three signs

Let $r,$ $s,$ $t\in\{-1,1\}$ be the signs of $a,$ $b,$ $c.$ The expression is $r+s+t+rst.$ If all three signs are positive it is $4,$ and if all are negative it is $-4.$ If the signs are mixed, direct cancellation gives $0.$ Hence the set is $\{-4,0,4\}.$

Therefore, the correct answer is B.

Assign one variable to each of the three equal arcs and another to arc $AD$

Use both the full-circle arc sum and the external-secant angle theorem at $E$

Let each of arcs $AB,$ $BC,$ $CD$ measure $x^\circ$ and let arc $AD$ measure $y^\circ.$ Then $$\begin{aligned} 3x+y&=360,\\ \frac{x-y}{2}&=40. \end{aligned}$$ Solving gives $y=30.$ The inscribed angle $\angle ACD$ subtends arc $AD,$ so it measures $\frac{y}{2}=15^\circ.$

Therefore, the correct answer is B.

A polynomial’s coefficient sum is obtained by evaluating it at a particular input

Substitute $x=1$ into the given identity

Setting $x=1$ makes the right side the desired coefficient sum. Thus $$\begin{aligned} a_7+a_6+\cdots+a_0 &=(3\cdot1-1)^7\\ &=2^7=128. \end{aligned}$$

Therefore, the correct answer is E.

Prove the translation statement directly from $n\le x\lt n+1$

Test the other two claims using noninteger values between $1$ and $2$

If $n\le x\lt n+1,$ then $n+1\le x+1\lt n+2,$ so $\mathrm{I}$ is true. Taking $x=y=\frac{3}{2}$ gives $[x+y]=3\ne2=[x]+[y]$ and $[xy]=2\ne1=[x][y].$ Thus $\mathrm{II}$ and $\mathrm{III}$ are false.

Therefore, the correct answer is B.

Let $s$ be the sum of Bob’s and Carl’s ages

Factor $a^2-s^2$ and use the known value of $a-s$

Let $a$ be Al’s age and $s$ the sum of the other two ages. Then $a-s=16$ and $$\begin{aligned} 1632&=a^2-s^2\\ &=(a-s)(a+s)\\ &=16(a+s). \end{aligned}$$ Hence the requested total $a+s$ is $\frac{1632}{16}=102.$

Therefore, the correct answer is D.

Write out $a_3,$ $a_4,$ and $a_5$ in terms of $a_1,$ $a_2$

Equate the first three successive ratios and use positivity

The next terms are $a_3=a_1a_2,$ $a_4=a_1a_2^2,$ and $a_5=a_1^2a_2^3.$ If the sequence is geometric, then $$\begin{aligned} \frac{a_2}{a_1}&=\frac{a_3}{a_2}=a_1,\\ \frac{a_4}{a_3}&=a_2=a_1. \end{aligned}$$ Thus $a_2=a_1^2$ and $a_2=a_1.$ Positivity forces $a_1=a_2=1.$ Conversely, these initial values produce the constant geometric sequence $1,1,1,\ldots.$

Therefore, the correct answer is E.

Move all terms to one side and add $1$

Factor the equation into a product of two integers equal to $1$

Rearranging and completing the product gives $$\begin{aligned} mn-m-n&=0,\\ (m-1)(n-1)&=1. \end{aligned}$$ The integer factor pairs of $1$ are $(1,1)$ and $(-1,-1),$ producing $(m,n)=(2,2)$ and $(0,0).$ Thus there are two ordered pairs.

Therefore, the correct answer is B.

The three circle centers form an equilateral triangle of side $6$

View the outer sides as parallel offsets and compare the inradii of two equilateral triangles

The circle centers form an equilateral triangle of side $6,$ whose inradius is $\frac{6\sqrt3}{6}=\sqrt3.$ Each side of the outer triangle is a parallel tangent line three units farther out, so the outer triangle has inradius $3+\sqrt3.$ An equilateral triangle of inradius $r$ has side $2\sqrt3r,$ hence each outer side is $$2\sqrt3(3+\sqrt3)=6+6\sqrt3.$$ The perimeter is $3(6+6\sqrt3)=18+18\sqrt3.$

Therefore, the correct answer is D.

Compare the term indexed by $n+2$ with the term indexed by $n$

Count the even and odd indices from $0$ through $40$

Let $u_n=i^n\cos(45+90n)^\circ.$ Then $$\begin{aligned} u_{n+2} &=i^{n+2}\cos(225+90n)^\circ\\ &=(-i^n)(-\cos(45+90n)^\circ)\\ &=u_n. \end{aligned}$$ Every even-indexed term equals $\frac{\sqrt2}{2},$ and every odd-indexed term equals $-\frac{i\sqrt2}{2}.$ There are $21$ even indices and $20$ odd indices, so the sum is $\frac{\sqrt2}{2}(21-20i).$

Therefore, the correct answer is D.

List the possible three-element sets of consecutive die values

Each qualifying set has six orderings among the $6^3$ ordered outcomes

The possible sets are $\{1,2,3\},$ $\{2,3,4\},$ $\{3,4,5\},$ and $\{4,5,6\}.$ Each has $3!=6$ orderings, so $24$ of the $6^3=216$ outcomes work. The probability is $\frac{24}{216}=\frac{1}{9}.$

Therefore, the correct answer is B.

Rewrite every logarithm using the same base

Adjacent numerators and denominators cancel in the product

By change of base, $$\begin{aligned} y&=\frac{\log 3}{\log2}\cdot \frac{\log4}{\log3}\cdots \frac{\log32}{\log31}\\ &=\frac{\log32}{\log2}\\ &=\log_2 32=5. \end{aligned}$$

Therefore, the correct answer is B.

Two adjacent circumcenters lie on the perpendicular bisector of the side shared by their triangles

Use the fact that $A,$ $E,$ $C$ are collinear and $B,$ $E,$ $D$ are collinear

Both $P$ and $Q$ lie on the perpendicular bisector of $BE,$ while both $R$ and $S$ lie on the perpendicular bisector of $DE.$ Since $B,$ $E,$ $D$ are collinear, these bisectors are parallel, so $PQ\parallel RS.$ Likewise, $Q,$ $R$ lie on the perpendicular bisector of $CE,$ and $S,$ $P$ lie on that of $AE.$ Since $A,$ $E,$ $C$ are collinear, $QR\parallel SP.$ Therefore $PQRS$ is always a parallelogram.

Therefore, the correct answer is A.

Reverse each path, starting from the central $T$ in the bottom row

Count the left-going and right-going families separately, then correct for their common path

Reverse the paths and spell TSETNOC from the central bottom $T.$ In the family whose horizontal moves go left, each of the six steps has two choices: up or left. This gives $2^6=64$ paths. By symmetry, $64$ paths have horizontal moves going right. The all-vertical central path belongs to both families, so the total is $$64+64-1=127.$$ This number is not among the first four choices.

Therefore, the correct answer is E.

Subtract the two equations to obtain a factored condition

Check separately the cases $a=-1$ and $x=-1$

Subtracting the second equation from the first gives $$\begin{gathered} (a+1)x+(a+1)\\ =(a+1)(x+1)=0. \end{gathered}$$ If $a=-1,$ the common equation is $x^2-x+1=0,$ which has no real roots. Otherwise $x=-1,$ and substitution into $x^2-x-a=0$ gives $a=2.$ This value works, so exactly one value of $a$ qualifies.

Therefore, the correct answer is B.

First substitute $a=b=0$

Next set $a=0$ and $b=x$

Setting $a=b=0$ gives $2f(0)=4f(0),$ so $f(0)=0.$ Now set $a=0$ and $b=x.$ The equation becomes $$\begin{aligned} f(x)+f(-x) &=2f(0)+2f(x)\\ &=2f(x), \end{aligned}$$ and hence $f(-x)=f(x)$ for every real $x.$

Therefore, the correct answer is C.

Call the roots of the second equation $u$ and $v$ and apply Vieta’s formulas

Use $u^3+v^3=(u+v)^3-3uv(u+v)$

Let the roots of $x^2+mx+n$ be $u,$ $v.$ Then $u+v=-m$ and $uv=n,$ while the roots of the first equation are $u^3,$ $v^3.$ Thus $$\begin{aligned} -p&=u^3+v^3\\ &=(u+v)^3-3uv(u+v)\\ &=-m^3+3mn. \end{aligned}$$ Therefore $p=m^3-3mn.$

Therefore, the correct answer is B.

Decompose $\frac1{(2n-1)(2n+1)}$ into two unit fractions

The last denominator corresponds to $n=128$

For $1\le n\le128,$ $$\begin{gathered} \frac1{(2n-1)(2n+1)}\\ =\frac1{2(2n-1)}-\frac1{2(2n+1)}. \end{gathered}$$ Hence the sum telescopes to $$\frac12\left(1-\frac1{257}\right)=\frac{128}{257}.$$

Therefore, the correct answer is D.

Count the factors of $5$ in $1005!$

Include contributions from multiples of $25,$ $125,$ and $625$

There are more factors of $2$ than of $5,$ so the exponent of $10$ is $$\begin{aligned} n={}&\left\lfloor\frac{1005}{5}\right\rfloor+ \left\lfloor\frac{1005}{25}\right\rfloor\\ &+\left\lfloor\frac{1005}{125}\right\rfloor+ \left\lfloor\frac{1005}{625}\right\rfloor\\ ={}&201+40+8+1=250. \end{aligned}$$ Since $250$ is not listed, the correct choice is “none of these.”

Therefore, the correct answer is E.

Split the quadrilateral along each diagonal and bound every sine by $1$

Combine the two resulting area bounds and analyze when every bound is an equality

Splitting along diagonal $MP$ and using $\sin\theta\le1$ gives $A\le\frac{ab+cd}{2}.$ Splitting along $NQ$ similarly gives $A\le\frac{ad+bc}{2}.$ These bounds remain valid for a nonconvex quadrilateral by taking the appropriate difference of triangle areas. Adding them yields $$\begin{aligned} 2A&\le\frac{ab+ad+bc+cd}{2}\\ &=\frac{(a+c)(b+d)}2, \end{aligned}$$ or $A\le\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right).$ Equality requires equality in all four sine bounds, so all four angles are right angles; conversely a rectangle gives equality. Thus the equality holds exactly for rectangles.

Therefore, the correct answer is B.

Place the corner at the origin with the walls and floor as coordinate planes

A sphere of radius $r$ tangent to all three planes has center $(r,r,r)$

For radius $r,$ the center is $(r,r,r)$ and the given point is $(5,5,10).$ Thus $$2(5-r)^2+(10-r)^2=r^2,$$ which simplifies to $r^2-20r+75=0,$ or $(r-5)(r-15)=0.$ The two radii are $5$ and $15,$ so the sum of the diameters is $2(5+15)=40$ inches.

Therefore, the correct answer is C.