1957 AMC 12 Problems

Separate the line from the apex from the lines drawn from the two base vertices

At the apex, the altitude, median, and angle bisector coincide; at either base vertex they do not

The altitude, median, and angle bisector from the apex are the same line. At each of the two base vertices, those three lines are distinct. Thus there are $$1+3+3=7$$ distinct lines.

Therefore, the correct answer is B.

Apply Vieta’s formulas without first solving the quadratic

The sum is $\frac{h}{2}$ and the product is $k$

By Vieta’s formulas, the sum and product of the roots are $$\begin{aligned} \frac h2&=4,\\ \frac{2k}{2}&=k=-3. \end{aligned}$$ Hence $h=8$ and $k=-3.$

Therefore, the correct answer is E.

Combine $1+\frac{a}{1-a}$ before taking its reciprocal

Retain every restriction imposed by the original denominator

The original expression requires $a\ne1.$ For such $a,$ $$\begin{aligned} 1+\frac{a}{1-a} &=\frac{1-a+a}{1-a}\\ &=\frac1{1-a}. \end{aligned}$$ Therefore the expression is $1-(1-a)=a.$ The restriction $a\ne1$ remains.

Thus, the correct answer is E.

Match the second factor with $b+c$ in the stated form

Take $a=3x+2,$ $b=x,$ and $c=-5$

Using the property in precisely the stated form, set $a=3x+2,$ $b=x,$ and $c=-5.$ Then $$\begin{gathered} (3x+2)(x-5)\\ =(3x+2)x+(3x+2)(-5). \end{gathered}$$

Thus, the correct answer is C.

Combine the first three logarithms by multiplying their arguments

The product $(\frac{a}{b})(\frac{b}{c})(\frac{c}{d})$ telescopes to $\frac{a}{d}$

Combining the logarithms gives $$\begin{aligned} &\log\left(\frac ab\frac bc\frac cd\frac{dx}{ay}\right)\\ &\quad=\log\left(\frac ad\frac{dx}{ay}\right)\\ &\quad=\log\frac xy. \end{aligned}$$

Therefore, the correct answer is B.

After folding, the height is $x$ and each base dimension loses $2x$

Write the volume as $x(10-2x)(14-2x)$

The box has height $x$ and base dimensions $10-2x$ and $14-2x.$ Hence $$\begin{aligned} V&=x(10-2x)(14-2x)\\ &=140x-48x^2+4x^3. \end{aligned}$$

Thus, the correct answer is A.

Use $\pi r^2=48\pi$ to find the inradius

For an equilateral triangle of side $s,$ the inradius is $\frac{s\sqrt3}{6}$

The inradius satisfies $r^2=48,$ so $r=4\sqrt3.$ If $s$ is the side length, then $$r=\frac{s\sqrt3}{6},$$ which gives $s=24.$ The perimeter is $3s=72.$

Thus, the correct answer is E.

Write $x,y,z$ as $2t,3t,5t$

Their sum determines $t$, after which substitute $x$ and $y$ into the given equation

Let $x=2t,$ $y=3t,$ and $z=5t.$ Since $10t=100,$ we get $t=10,$ so $x=20$ and $y=30.$ Thus $$30=20a-10,$$ and $a=2.$

Therefore, the correct answer is A.

Evaluate the exponent $x-y$ before the power

The powered term is $(-2)^4$, not $-2^4$

Here $x-y=2-(-2)=4.$ Therefore $$\begin{aligned} x-y^{x-y} &=2-(-2)^4\\ &=2-16=-14. \end{aligned}$$

Thus, the correct answer is B.

Complete the square in $2x^2+4x+3$

Because the coefficient of the square is positive, the vertex is a minimum

Completing the square, $$y=2(x+1)^2+1.$$ The squared term is nonnegative, so the graph has its lowest point at $(-1,1).$

Thus, the correct answer is C.

At $2:15,$ the minute hand is at $90^\circ$ from twelve

The hour hand has advanced one quarter of the $30^\circ$ interval from two to three

The minute hand is $90^\circ$ clockwise from twelve. The hour hand is $$2(30^\circ)+\frac14(30^\circ)=67.5^\circ$$ from twelve. Their smaller angle is $90^\circ-67.5^\circ=22.5^\circ,$ which is not listed.

Therefore, the correct answer is E.

Express both numbers with the same power of ten

Rewrite $10^{-49}$ as $10\cdot10^{-50}$

Using a common power of ten, $$\begin{gathered} 10^{-49}-2\cdot10^{-50}\\ =(10-2)10^{-50}\\ =8\cdot10^{-50}. \end{gathered}$$

Thus, the correct answer is C.

Compare the decimal choices with $\sqrt2\approx1.414$ and $\sqrt3\approx1.732$

The radical expressions in the first two choices are irrational

Since $$\begin{gathered} \sqrt2\approx1.414<1.5,\\ 1.5<1.732\approx\sqrt3, \end{gathered}$$ the rational number $1.5$ lies between the two radicals.

Therefore, the correct answer is C.

Factor each radicand as a perfect square

For real $u,$ $\sqrt{u^2}=|u|$

The two radicands are $(x-1)^2$ and $(x+1)^2.$ Hence $$\begin{aligned} y&=\sqrt{(x-1)^2}\\ &\quad+\sqrt{(x+1)^2}\\ &=|x-1|+|x+1|. \end{aligned}$$

Thus, the correct answer is D.

Compare each listed $s$-value with the square of its $t$-value

The table follows $s=10t^2$

The entries follow $s=10t^2.$ Thus, at $t=2.5,$ $$s=10(2.5)^2=10(6.25)=62.5.$$

Therefore, the correct answer is B.

Decide whether the number of fish varies continuously or only through whole numbers

There is one cost point for each integer count from $1$ through $12$

The rule $c=15n$ is linear, but the number $n$ of goldfish can only be one of the twelve integers $1,2,\ldots,12.$ The graph therefore consists of twelve distinct points, not an entire line or segment.

Thus, the correct answer is D.

A path that never returns to a vertex can visit at most all eight cube vertices

Exhibit a path through all eight vertices, which uses seven edges before the next step returns to the starting vertex

The cube has eight vertices. A path can visit all eight once, using seven edges, and then traverse the edge from its last vertex back to its starting vertex; the return is the first repeated vertex. Thus the greatest permitted walk uses eight edges. Its length is $8\cdot3=24$ inches.

Therefore, the correct answer is A.

Because $AB$ is a diameter, angle $AMB$ is a right angle

Compare right triangles $APO$ and $ABM$

Angles $APO$ and $ABM$ are both right angles, and the two triangles share angle $A.$ Therefore triangles $APO$ and $ABM$ are similar. Corresponding sides give $$\frac{AP}{AB}=\frac{AO}{AM}.$$ Hence $AP\cdot AM=AO\cdot AB.$

Thus, the correct answer is B.

Assign powers $2^4,2^3,\ldots,2^0$ to the five binary places

Only the first, fourth, and fifth digits are nonzero

Expanding by binary place value, $$\begin{aligned} (10011)_2 &=1\cdot2^4+1\cdot2^1+1\cdot2^0\\ &=16+2+1=19. \end{aligned}$$

Therefore, the correct answer is A.

Use a convenient one-way distance and divide total distance by total time

For equal distances the arithmetic mean of the two speeds is not the average speed

Let each leg have distance $d.$ The average speed, in miles per hour, is $$\begin{aligned} \frac{2d}{\frac{d}{50}+\frac{d}{45}} &=\frac{2}{\frac{1}{50}+\frac{1}{45}}\\ &=\frac{900}{19}\\ &=47\frac7{19}. \end{aligned}$$

Thus, the correct answer is A.

Write the theorem as $P\Rightarrow Q$ and compare each statement with its inverse, converse, or contrapositive

A statement is always equivalent to its contrapositive, but not generally to its converse or inverse

Let $P$ mean that two angles are equal and $Q$ that the triangle is isosceles. Statement $3$ is the contrapositive $\neg Q\Rightarrow\neg P,$ so it is equivalent to $P\Rightarrow Q.$ Statement $4$ says that $Q$ is necessary for $P,$ which is another wording of $P\Rightarrow Q.$ Statements $1$ and $2$ are the inverse and converse.

Thus only statements $3$ and $4$ are equivalent, so the correct answer is E.

Rearrange to $\sqrt{x+1}=\sqrt{x-1}+1$

After squaring once, isolate $\sqrt{x-1}$

Rearrange and square: $$\begin{aligned} \sqrt{x+1}&=\sqrt{x-1}+1,\\ x+1&=x-1+2\sqrt{x-1}+1. \end{aligned}$$ Thus $2\sqrt{x-1}=1,$ so $x-1=\frac{1}{4}$ and $x=\frac{5}{4}.$ This value satisfies the original equation. Therefore $4x=5.$

Thus, the correct answer is A.

Set the two expressions for $y$ equal

The resulting equation is $x^2=x$

At an intersection, $$10-x^2=10-x,$$ so $x=0$ or $x=1.$ The points are $(0,10)$ and $(1,9).$ Their distance is $$\sqrt{(1-0)^2+(9-10)^2}=\sqrt2.$$

Therefore, the correct answer is C.

Represent the number as $10a+b$ and its reversal as $10b+a$

Factor the difference of their squares completely

If the digits are $a,b,$ then $$\begin{aligned} &(10a+b)^2-(10b+a)^2\\ &\quad=[9(a-b)][11(a+b)]\\ &\quad=99(a-b)(a+b). \end{aligned}$$ This is always divisible by $9,$ $11,$ the digit sum, and the digit difference. It need not be divisible by the digit product; for example, $21^2-12^2=297$ is not divisible by $2.$

Thus, the correct answer is B.

Use the determinant formula for the area of a triangle from its coordinates

The opposite-side condition determines the sign inside the absolute value

The signed doubled area is $$\begin{vmatrix} 0&a&1\\ b&0&1\\ c&d&1 \end{vmatrix} =ab-ac-bd.$$ The line $PQ$ has equation $ax+by=ab.$ Since the origin gives a value below $ab$ and $R$ is on the opposite side, $ac+bd>ab.$ Thus the area is $$\frac{ac+bd-ab}{2}.$$

Therefore, the correct answer is B.

Recall how the medians partition a triangle’s area

The intersection of the medians divides the triangle into six small triangles paired into three equal-area regions

The three medians meet at the centroid and divide the triangle into six small triangles of equal area. Each of the three triangles having the centroid and one side of the original triangle consists of two of those small triangles, so their areas are equal. Conversely, equal areas give equal barycentric coordinates, which uniquely locate the centroid.

Thus, the necessary and sufficient point is the intersection of the medians, and the correct answer is E.

Call the roots $r$ and $s$, and combine $\frac{1}{r}+\frac{1}{s}$

Use $r+s=-p$ and $rs=q$

If the roots are $r,s,$ then Vieta’s formulas give $r+s=-p$ and $rs=q.$ Therefore $$\frac1r+\frac1s=\frac{r+s}{rs}=-\frac pq.$$

Thus, the correct answer is A.

Use the inverse relationship between a base-$b$ logarithm and exponentiation by $b$

If $u=\log_b a,$ then $b^u=a$

By the definition of a logarithm, $$b^{\log_b a}=a.$$ Thus the value depends on $a,$ not on the choice of admissible base $b.$

Therefore, the correct answer is B.

The factor $x^2$ is positive except at $x=0$

Away from zero, the sign is controlled by $x^2-1$

At $x=0,$ the product is zero. For $x\ne0,$ the factor $x^2$ is positive, so the product is nonnegative exactly when $$x^2-1\ge0,$$ or $x\le-1$ or $x\ge1.$ Including the isolated value $x=0$ gives the set in choice D.

Thus, the correct answer is D.

Compare the expression with the standard sum-of-squares formula

Alternatively, substitute $n=1$ and $n=2$ to obtain two equations

The standard formula is $$\begin{aligned} 1^2+2^2+\cdots+n^2 &=\frac{n(n+1)}6\\ &\quad\cdot(2n+1). \end{aligned}$$ Therefore $c=1$ and $k=1.$

Thus, the correct answer is D.

If each cut-off leg is $x,$ an uncut horizontal octagon side has length $1-2x$

A slanted octagon side is the hypotenuse $x\sqrt2$, and regularity makes these equal

Let $x$ be a leg of each corner triangle. The horizontal and vertical octagon sides have length $1-2x,$ while the four slanted sides have length $x\sqrt2.$ Thus $$1-2x=x\sqrt2,$$ so $$x=\frac1{2+\sqrt2}=\frac{2-\sqrt2}{2}.$$

Therefore, the correct answer is B.

Show $n^5-n$ is always divisible by $2,$ $3,$ and $5$

Use consecutive factors for $2$ and $3,$ and residues modulo $5$ for the remaining factor

Factor $$\begin{aligned} n^5-n &=n(n-1)(n+1)\\ &\quad\cdot(n^2+1). \end{aligned}$$ Among three consecutive integers, one is divisible by $3$ and at least one is even, so the expression is divisible by $6.$ Also $n^5\equiv n\pmod5$ for every integer $n,$ so it is divisible by $5.$ Hence every term is divisible by $30.$ Taking $n=2$ gives $2^5-2=30,$ so no larger listed integer can divide every term.

Thus, the correct answer is E.

Rewrite $9^{x+2}$ as $81\cdot9^x$