1957 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

The number of distinct lines representing the altitudes, medians, and interior angle bisectors of a triangle that is isosceles, but not equilateral, is:

$9$

$7$

$6$

$5$

$3$

Concepts:isosceles triangle median (geometry)altitude angle bisector

Difficulty rating: 1340

Small Hint:

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

Big Hint:

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

Solution:

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.

2.

In the equation $2x^2-hx+2k=0,$ the sum of the roots is $4$ and the product of the roots is $-3.$ Then $h$ and $k$ have the values, respectively:

$8$ and $-6$

$4$ and $-3$

$-3$ and $4$

$-3$ and $8$

$8$ and $-3$

Concepts:quadratic Vieta’s Formulas

Difficulty rating: 1000

Small Hint:

Apply Vieta’s formulas without first solving the quadratic

Big Hint:

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

Solution:

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.

3.

The simplest form of $1-\dfrac{1}{1+\dfrac{a}{1-a}}$ is:

$a$ if $a\ne0$

$1$

$a$ if $a\ne-1$

$1-a$ with no restriction on $a$

$a$ if $a\ne1$

Concepts:rational equation algebraic manipulation

Difficulty rating: 1210

Small Hint:

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

Big Hint:

Retain every restriction imposed by the original denominator

Solution:

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.

4.

The first step in finding the product $(3x+2)(x-5)$ by use of the distributive property in the form $a(b+c)=ab+ac$ is:

$3x^2-13x-10$

$3x(x-5)+2(x-5)$

$(3x+2)x+(3x+2)(-5)$

$3x^2-17x-10$

$3x^2+2x-15x-10$

Concepts:distributive property algebraic manipulation

Difficulty rating: 940

Small Hint:

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

Big Hint:

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

Solution:

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.

5.

Through the use of theorems on logarithms, $\log\dfrac ab+\log\dfrac bc+\log\dfrac cd-\log\dfrac{ay}{dx}$ can be reduced to:

$\log\dfrac yx$

$\log\dfrac xy$

$1$

$0$

$\log\dfrac{a^2y}{d^2x}$

Concepts:logarithm telescoping algebraic manipulation

Difficulty rating: 1280

Small Hint:

Combine the first three logarithms by multiplying their arguments

Big Hint:

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

Solution:

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.

6.

An open box is constructed by starting with a rectangular sheet of metal $10$ in. by $14$ in. and cutting a square of side $x$ inches from each corner. The resulting projections are folded up and the seams welded. The volume of the resulting box is:

$140x-48x^2+4x^3$

$140x+48x^2+4x^3$

$140x+24x^2+x^3$

$140x-24x^2+x^3$

none of these

Concepts:volume rectangular prism polynomial

Difficulty rating: 1490

Small Hint:

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

Big Hint:

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

Solution:

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.

7.

The area of a circle inscribed in an equilateral triangle is $48\pi.$ The perimeter of this triangle is:

$72\sqrt3$

$48\sqrt3$

$36$

$24$

$72$

Concepts:equilateral triangle incircle, incenter, and inradius circle area

Difficulty rating: 1360

Small Hint:

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

Big Hint:

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

Solution:

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.

8.

The numbers $x,$ $y,$ $z$ are proportional to $2,$ $3,$ $5.$ The sum of $x,$ $y,$ and $z$ is $100.$ The number $y$ is given by the equation $y=ax-10.$ Then $a$ is:

$2$

$\dfrac23$

$3$

$\dfrac52$

$4$

Concepts:ratio and proportion linear equation

Difficulty rating: 1180

Small Hint:

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

Big Hint:

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

Solution:

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.

9.

The value of $x-y^{x-y}$ when $x=2$ and $y=-2$ is:

$-18$

$-14$

$14$

$18$

$256$

Concepts:exponent order of operations substitution

Difficulty rating: 1060

Small Hint:

Evaluate the exponent $x-y$ before the power

Big Hint:

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

Solution:

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.

10.

The graph of $y=2x^2+4x+3$ has its:

lowest point at $(-1,9)$

lowest point at $(1,1)$

lowest point at $(-1,1)$

highest point at $(-1,9)$

highest point at $(-1,1)$

Concepts:quadratic parabola completing the square

Difficulty rating: 1390

Small Hint:

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

Big Hint:

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

Solution:

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.

11.

The angle formed by the hands of a clock at $2:15$ is:

$30^\circ$

$27\frac12^\circ$

$157\frac12^\circ$

$172\frac12^\circ$

none of these

Concepts:clock angle chasing

Difficulty rating: 1210

Small Hint:

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

Big Hint:

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

Solution:

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.

12.

Comparing the numbers $10^{-49}$ and $2\cdot10^{-50},$ we may say:

the first exceeds the second by $8\cdot10^{-1}$

the first exceeds the second by $2\cdot10^{-1}$

the first exceeds the second by $8\cdot10^{-50}$

the second is five times the first

the first exceeds the second by $5$

Concepts:exponent decimal whole number operations

Difficulty rating: 1390

Small Hint:

Express both numbers with the same power of ten

Big Hint:

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

Solution:

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.

13.

A rational number between $\sqrt2$ and $\sqrt3$ is:

$\dfrac{\sqrt2+\sqrt3}{2}$

$\dfrac{\sqrt2\cdot\sqrt3}{2}$

$1.5$

$1.8$

$1.4$

Concepts:radical inequality estimation

Difficulty rating: 1260

Small Hint:

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

Big Hint:

The radical expressions in the first two choices are irrational

Solution:

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.

14.

If $y=\sqrt{x^2-2x+1}$ ${}+\sqrt{x^2+2x+1},$ then $y$ is:

$2x$

$2(x+1)$

$0$

$|x-1|+|x+1|$

none of these

Concepts:radical perfect square absolute value

Difficulty rating: 1210

Small Hint:

Factor each radicand as a perfect square

Big Hint:

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

Solution:

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.

15.

The table below shows the distance $s$ in feet a ball rolls down an inclined plane in $t$ seconds.

$t$	$0$	$1$	$2$	$3$	$4$	$5$
$s$	$0$	$10$	$40$	$90$	$160$	$250$

The distance $s$ for $t=2.5$ is:

$45$

$62.5$

$70$

$75$

$82.5$

Concepts:data and graph interpretation pattern recognition quadratic

Difficulty rating: 1410

Small Hint:

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

Big Hint:

The table follows $s=10t^2$

Solution:

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.

16.

Goldfish are sold at $15$ cents each. The rectangular coordinate graph showing the cost of $1$ to $12$ goldfish is:

a straight line segment

a set of horizontal parallel line segments

a set of vertical parallel line segments

a finite set of distinct points

a straight line

Concepts:data and graph interpretation function counting integers in a range

Difficulty rating: 1180

Small Hint:

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

Big Hint:

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

Solution:

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.

17.

A cube is made by soldering twelve $3$ -inch lengths of wire properly at the vertices of the cube. If a fly alights at one of the vertices and then walks along the edges, the greatest distance it could travel before coming to any vertex a second time, without retracing any distance, is:

$24$ in.

$12$ in.

$30$ in.

$18$ in.

$36$ in.

Concepts:cube geometry graph theory extremal argument

Difficulty rating: 1300

Small Hint:

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

Big Hint:

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

Solution:

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.

18.

Circle $O$ has diameters $\overline{AB}$ and $\overline{CD}$ perpendicular to each other. $\overline{AM}$ is any chord intersecting $\overline{CD}$ at $P.$ Then $AP\cdot AM$ is equal to:

$AO\cdot OB$

$AO\cdot AB$

$CP\cdot CD$

$CP\cdot PD$

$CO\cdot OP$

Concepts:circle chord similarity

Difficulty rating: 1630

Small Hint:

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

Big Hint:

Compare right triangles $APO$ and $ABM$

Solution:

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.

19.

The base of the decimal number system is ten, meaning, for example, that $123=1\cdot10^2+2\cdot10+3.$ In the binary system, which has base two, the first five positive integers are $1,$ $10,$ $11,$ $100,$ $101.$ The numeral $10011$ in the binary system would then be written in the decimal system as:

$19$

$40$

$10011$

$11$

$7$

Concepts:number base place value power of 2

Difficulty rating: 1410

Small Hint:

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

Big Hint:

Only the first, fourth, and fifth digits are nonzero

Solution:

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.

20.

A man makes a trip by automobile at an average speed of $50$ mph. He returns over the same route at an average speed of $45$ mph. His average speed for the entire trip is:

$47\frac7{19}$ mph

$47\frac14$ mph

$47\frac12$ mph

$47\frac{11}{19}$ mph

none of these

Concepts:distance rate and time harmonic mean

Difficulty rating: 1240

Small Hint:

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

Big Hint:

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

Solution:

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.

21.

Start with the theorem “If two angles of a triangle are equal, the triangle is isosceles,” and the following four statements:

$1.$ If two angles of a triangle are not equal, the triangle is not isosceles.

$2.$ The base angles of an isosceles triangle are equal.

$3.$ If a triangle is not isosceles, then two of its angles are not equal.

$4.$ A necessary condition that two angles of a triangle be equal is that the triangle be isosceles.

Which combination of statements contains only those which are logically equivalent to the given theorem?

$1,$ $2,$ $3,$ $4$

$1,$ $2,$ $3$

$2,$ $3,$ $4$

$1,$ $2$

$3,$ $4$

Concepts:logical deduction

Difficulty rating: 1590

Small Hint:

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

Big Hint:

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

Solution:

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.

22.

If $\sqrt{x-1}-\sqrt{x+1}+1=0,$ then $4x$ equals:

$5$

$4\sqrt{-1}$

$0$

$1\frac14$

no real value

Concepts:radical rational equation

Difficulty rating: 1630

Small Hint:

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

Big Hint:

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

Solution:

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.

23.

The graph of $x^2+y=10$ and the graph of $x+y=10$ meet in two points. The distance between these two points is:

less than $1$

$1$

$\sqrt2$

$2$

more than $2$

Concepts:coordinate geometry distance formula system of equations

Difficulty rating: 1280

Small Hint:

Set the two expressions for $y$ equal

Big Hint:

The resulting equation is $x^2=x$

Solution:

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.

24.

If the square of a number of two digits is decreased by the square of the number formed by reversing the digits, then the result is not always divisible by:

$9$

the product of the digits

the sum of the digits

the difference of the digits

$11$

Concepts:digits difference of squares divisibility

Difficulty rating: 1590

Small Hint:

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

Big Hint:

Factor the difference of their squares completely

Solution:

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.

25.

The vertices of triangle $PQR$ have coordinates as follows: $P(0,a),$ $Q(b,0),$ $R(c,d),$ where $a,$ $b,$ $c,$ and $d$ are positive. The origin and point $R$ lie on opposite sides of $PQ.$ The area of triangle $PQR$ may be found from the expression:

$\dfrac{ab+ac+bc+cd}{2}$

$\dfrac{ac+bd-ab}{2}$

$\dfrac{ab-ac-bd}{2}$

$\dfrac{ac+bd+ab}{2}$

$\dfrac{ac+bd-ab-cd}{2}$

Concepts:coordinate geometry triangle area determinant

Difficulty rating: 1630

Small Hint:

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

Big Hint:

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

Solution:

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.

26.

From a point within a triangle, line segments are drawn to the vertices. A necessary and sufficient condition that the three triangles thus formed have equal areas is that the point be:

the center of the inscribed circle

the center of the circumscribed circle

such that the three angles formed at the point each be $120^\circ$

the intersection of the altitudes of the triangle

the intersection of the medians of the triangle

Concepts:centroid triangle area median (geometry)

Difficulty rating: 1340

Small Hint:

Recall how the medians partition a triangle’s area

Big Hint:

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

Solution:

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.

27.

The sum of the reciprocals of the roots of the equation $x^2+px+q=0$ is:

$-\dfrac pq$

$\dfrac qp$

$\dfrac pq$

$-\dfrac qp$

$pq$

Concepts:Vieta’s Formulas rational equation

Difficulty rating: 1180

Small Hint:

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

Big Hint:

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

Solution:

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.

28.

If $a$ and $b$ are positive and $a\ne1,$ $b\ne1,$ then the value of $b^{\log_b a}$ is:

dependent upon $b$

dependent upon $a$

dependent upon $a$ and $b$

zero

one

Concepts:logarithm exponent

Difficulty rating: 1110

Small Hint:

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

Big Hint:

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

Solution:

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.

29.

The relation $x^2(x^2-1)\ge0$ is true only for:

Here $x\ge a$ means that $x$ can take on all values greater than $a$ and the value equal to $a,$ while $x\le a$ has a corresponding meaning with “less than.”

$x\ge1$

$-1\le x\le1$

$x=0,$ $x=1,$ $x=-1$

$x=0,$ $x\le-1,$ $x\ge1$

$x\ge0$

Concepts:inequality factoring zero product property

Difficulty rating: 1360

Small Hint:

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

Big Hint:

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

Solution:

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.

30.

The sum of the squares of the first $n$ positive integers is given by the expression $\dfrac{n(n+c)(2n+k)}6,$ if $c$ and $k$ are, respectively:

$1$ and $2$

$3$ and $5$

$2$ and $2$

$1$ and $1$

$2$ and $1$

Concepts:sum of first n squares polynomial

Difficulty rating: 1180

Small Hint:

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

Big Hint:

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

Solution:

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.

31.

A regular octagon is to be formed by cutting equal isosceles right triangles from the corners of a square. If the square has sides of one unit, the leg of each of the triangles has length:

$\dfrac{2+\sqrt2}{3}$

$\dfrac{2-\sqrt2}{2}$

$\dfrac{1+\sqrt2}{2}$

$\dfrac{1+\sqrt2}{3}$

$\dfrac{2-\sqrt2}{3}$

Concepts:regular polygon special right triangle radical

Difficulty rating: 1630

Small Hint:

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

Big Hint:

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

Solution:

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.

32.

The largest of the following integers which divides each of the numbers of the sequence $1^5-1,$ $2^5-2,$ $3^5-3,$ $\ldots$ $n^5-n,$ $\ldots$ is:

$1$

$60$

$15$

$120$

$30$

Concepts:divisibility modular arithmetic Fermat’s Little Theorem

Difficulty rating: 1790

Small Hint:

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

Big Hint:

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

Solution:

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.

33.

If $9^{x+2}=240+9^x,$ then the value of $x$ is:

$0.1$

$0.2$

$0.3$

$0.4$

$0.5$

Concepts:exponent algebraic manipulation

Difficulty rating: 1280

Small Hint:

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

Big Hint:

After collecting like terms, express $9^x$ as a power of $3$

Solution:

Factoring $9^x,$ $81\cdot9^x-9^x=240,$ so $80\cdot9^x=240$ and $9^x=3.$ Since $9^{\frac{1}{2}}=3,$ we have $x=\frac{1}{2}=0.5.$

Therefore, the correct answer is E.

34.

The points that satisfy the system $x+y=1,$ $x^2+y^2<25,$ where the symbol “ $<$ ” means “less than,” constitute the following set:

only two points

an arc of a circle

a straight line segment not including the end-points

a straight line segment including the end-points

a single point

Concepts:coordinate geometry circle inequality

Difficulty rating: 1300

Small Hint:

Interpret $x^2+y^2<25$ as the interior of a circle

Big Hint:

Intersect that open disk with the line $x+y=1$

Solution:

The inequality describes the open disk inside the circle of radius $5$ centered at the origin. The line $x+y=1$ crosses the circle in two points. Its portion inside the disk is the segment between those points, but the strict inequality excludes the endpoints.

Thus, the correct answer is C.

35.

Side $\overline{AC}$ of right triangle $ABC$ is divided into $8$ equal parts. Seven line segments parallel to $\overline{BC}$ are drawn to $\overline{AB}$ from the points of division. If $BC=10,$ then the sum of the lengths of the seven line segments:

cannot be found from the given information

is $33$

is $34$

is $35$

is $45$

Difficulty rating: 1530

Small Hint:

Each small triangle with vertex $A$ is similar to triangle $ABC$

Big Hint:

The seven parallel lengths are $\frac{10}{8},\frac{20}{8},\ldots,\frac{70}{8}$

Solution:

By similarity, the segment at the $k$ th division point has length $\frac{10k}{8}$ for $k=1,\ldots,7.$ Their sum is $\begin{aligned} \frac{10}{8}(1+2+\cdots+7) &=\frac{10}{8}\cdot28\\ &=35. \end{aligned}$

Therefore, the correct answer is D.

36.

If $x+y=1,$ then the largest value of $xy$ is:

$1$

$0.5$

an irrational number about $0.4$

$0.25$

$0$

Concepts:optimization completing the square quadratic

Difficulty rating: 1180

Small Hint:

Substitute $y=1-x$ into the product

Big Hint:

Complete the square in $x(1-x)$

Solution:

We have $\begin{aligned} xy=x(1-x) &=\frac14\\ &\quad-\left(x-\frac12\right)^2. \end{aligned}$ The squared term is nonnegative, so the largest possible product is $\frac{1}{4}=0.25.$

Thus, the correct answer is D.

37.

In right triangle $ABC,$ $BC=5,$ $AC=12,$ and $AM=x;$ $MN\perp AC,$ $NP\perp BC;$ $N$ is on $\overline{AB}.$ If $y=MN+NP,$ one-half the perimeter of rectangle $MCPN,$ then:

$y=\dfrac12(5+12)$

$y=\dfrac{5x}{12}+\dfrac{12}{5}$

$y=\dfrac{144-7x}{12}$

$y=12$

$y=\dfrac{5x}{12}+6$

Difficulty rating: 1590

Small Hint:

Triangles $AMN$ and $ACB$ are similar

Big Hint:

Use $\frac{MN}{x}=\frac{5}{12}$ and $NP=MC=12-x$

Solution:

By similarity, $MN=\frac5{12}AM=\frac{5x}{12}.$ Also $NP=MC=AC-AM$ $=12-x.$ Hence $\begin{aligned} y=MN+NP &=\frac{5x}{12}+12-x\\ &=\frac{144-7x}{12}. \end{aligned}$

Therefore, the correct answer is C.

38.

From a two-digit number $N$ we subtract the number with the digits reversed and find that the result is a positive perfect cube. Then:

$N$ cannot end in $5$

$N$ can end in any digit other than $5$

$N$ does not exist

there are exactly $7$ values for $N$

there are exactly $10$ values for $N$

Concepts:digits perfect power systematic listing

Difficulty rating: 1630

Small Hint:

If the digits are $a>b,$ the difference from the reversal is $9(a-b)$

Big Hint:

The difference is at most $81,$ so test the positive cubes no larger than $81$

Solution:

Writing $N=10a+b,$ the positive difference is $\begin{aligned} &(10a+b)-(10b+a)\\ &\quad=9(a-b). \end{aligned}$ It is at most $81.$ Among $1,$ $8,$ $27,$ and $64,$ only $27$ is divisible by $9,$ so $a-b=3.$ The digit pairs are $\begin{gathered} (3,0),(4,1),(5,2),(6,3),\\ (7,4),(8,5),(9,6), \end{gathered}$ giving exactly seven values of $N.$

Thus, the correct answer is D.

39.

Two men set out at the same time to walk towards each other from $M$ and $N,$ $72$ miles apart. The first man walks at the rate of $4$ mph. The second man walks $2$ miles the first hour, $2\frac12$ miles the second hour, $3$ miles the third hour, and so on in arithmetic progression. Then the men will meet:

in $7$ hours

in $8\frac14$ hours

nearer $M$ than $N$

nearer $N$ than $M$

midway between $M$ and $N$

Concepts:arithmetic sequence distance rate and time

Difficulty rating: 1810

Small Hint:

After $t$ whole hours, sum the second man’s hourly distances as an arithmetic progression

Big Hint:

Set that sum plus $4t$ equal to $72$

Solution:

In $t$ hours, the second man’s hourly distances form an arithmetic progression with first term $2$ and last term $2+\frac{t-1}{2}.$ His distance is $\frac t2\left(4+\frac{t-1}{2}\right) =\frac{t(t+7)}4.$ Together the men cover $4t+\frac{t(t+7)}4=72,$ or $t^2+23t-288=0.$ The positive root is $t=9.$ The first man then walks $4(9)=36$ miles, exactly half the original distance.

Therefore, the correct answer is E.

40.

If the parabola $y=-x^2+bx-8$ has its vertex on the $x$ -axis, then $b$ must be:

a positive integer

a positive or a negative rational number

a positive rational number

a positive or a negative irrational number

a negative irrational number

Concepts:quadratic parabola radical

Difficulty rating: 1360

Small Hint:

A parabola whose vertex lies on the $x$ -axis has a repeated root

Big Hint:

Set the discriminant of $-x^2+bx-8=0$ equal to zero

Solution:

The vertex lies on the $x$ -axis exactly when the quadratic has one repeated root. Its discriminant must satisfy $b^2-4(-1)(-8)=b^2-32=0.$ Hence $b=\pm\sqrt{32}=\pm4\sqrt2,$ a positive or negative irrational number.

Thus, the correct answer is D.

41.

Given the system of equations

$\begin{aligned} ax+(a-1)y&=1,\\ (a+1)x-ay&=1. \end{aligned}$

For which one of the following values of $a$ is there no solution for $x$ and $y?$

$1$

$0$

$-1$

$\pm\dfrac{\sqrt2}{2}$

$\pm\sqrt2$

Concepts:system of equations determinant

Difficulty rating: 1590

Small Hint:

A $2\times2$ system can fail to have a unique solution when its coefficient determinant is zero

Big Hint:

Compute $a(-a)-(a-1)(a+1)$

Solution:

The coefficient determinant is $\begin{aligned} &a(-a)-(a-1)(a+1)\\ &\quad=1-2a^2. \end{aligned}$ It vanishes when $a=\pm\frac{\sqrt2}{2}.$ For either value, the two coefficient rows are proportional but the two right sides are not in the same ratio, so the system is inconsistent.

Thus, the correct answer is D.

42.

If $S=i^n+i^{-n},$ where $i=\sqrt{-1}$ and $n$ is an integer, then the total number of possible distinct values for $S$ is:

$1$

$2$

$3$

$4$

more than $4$

Concepts:complex number roots of unity modular arithmetic

Difficulty rating: 1530

Small Hint:

Powers of $i$ repeat with period four

Big Hint:

Check one representative of each residue class of $n$ modulo $4$

Solution:

If $n$ is even, $i^n$ is $1$ or $-1$ and equals its reciprocal, so $S=2$ or $-2.$ If $n$ is odd, the two terms are $i$ and $-i,$ in some order, so $S=0.$ Thus the distinct values are $-2,\ 0,\ 2,$ three values.

Therefore, the correct answer is C.

43.

We define a lattice point as a point whose coordinates are integers, zero admitted. Then the number of lattice points on the boundary and inside the region bounded by the $x$ -axis, the line $x=4,$ and the parabola $y=x^2$ is:

$24$

$35$

$34$

$30$

not finite

Concepts:lattice point parabola counting integers in a range

Difficulty rating: 1550

Small Hint:

Only the integer $x$ -coordinates $0,$ $1,$ $2,$ $3,$ and $4$ occur

Big Hint:

For a fixed integer $x,$ count the integer $y$ -values from $0$ through $x^2$

Solution:

For each $x$ among $0,$ $1,$ $2,$ $3,$ and $4,$ there are $x^2+1$ integer values from $y=0$ through $y=x^2.$ Therefore the number of lattice points is $\begin{aligned} &(0^2+1)+(1^2+1)+(2^2+1)\\ &\quad +(3^2+1)+(4^2+1)\\ &=1+2+5+10+17\\ &=35. \end{aligned}$

Thus, the correct answer is B.

44.

In triangle $ABC,$ $AC=CD$ and $\angle CAB-\angle ABC=30^\circ.$ Then $\angle BAD$ is:

$30^\circ$

$20^\circ$

$22\frac12^\circ$

$10^\circ$

$15^\circ$

Concepts:isosceles triangle angle chasing

Difficulty rating: 1630

Small Hint:

Because $AC=CD,$ triangle $ACD$ is isosceles

Big Hint:

Express $\angle CDA$ using $\angle DAB$ and $\angle ABC$

Solution:

Let $\alpha=\angle CAB,$ $\beta=\angle ABC,$ and $x=\angle BAD.$ Since $D$ lies on $BC,$ $\angle ADC=x+\beta$ as an exterior angle of triangle $ABD.$ Because $AC=CD,$ triangle $ACD$ is isosceles, so $\angle CAD=\angle ADC.$ But $\angle CAD=\alpha-x.$ Hence $\alpha-x=x+\beta,$ so $2x=\alpha-\beta=30^\circ$ and $x=15^\circ.$

Thus, the correct answer is E.

45.

If two real numbers $x$ and $y$ satisfy the equation $\dfrac xy=x-y,$ then:

$x\ge4$ or $x\le0,$ where $x\ge a$ means that $x$ can take any value greater than $a$ or equal to $a$

$y$ can equal $1$

both $x$ and $y$ must be irrational

$x$ and $y$ cannot both be integers

both $x$ and $y$ must be rational

Concepts:rational equation quadratic inequality

Difficulty rating: 1830

Small Hint:

Clear the denominator and regard the result as a quadratic equation in $y$

Big Hint:

Real $y$ requires the discriminant $x^2-4x$ to be nonnegative

Solution:

Since $y\ne0,$ multiplying by $y$ gives $x=xy-y^2,$ or $y^2-xy+x=0.$ For a real value of $y,$ its discriminant must satisfy $x^2-4x=x(x-4)\ge0.$ Thus $x\le0$ or $x\ge4.$

Therefore, the correct answer is A.

46.

Two perpendicular chords intersect in a circle. The segments of one chord are $3$ and $4;$ the segments of the other are $6$ and $2.$ Then the diameter of the circle is:

$\sqrt{89}$

$\sqrt{56}$

$\sqrt{61}$

$\sqrt{75}$

$\sqrt{65}$

Concepts:chord coordinate geometry perpendicular bisector

Difficulty rating: 1870

Small Hint:

Place the chord intersection at the origin and the perpendicular chords on the coordinate axes

Big Hint:

The circle’s center lies on both chord perpendicular bisectors

Solution:

Place the intersection at $(0,0)$ with endpoints $(-3,0),$ $(4,0),$ $(0,-2),$ and $(0,6).$ The perpendicular bisectors of the chords are $x=\frac{1}{2}$ and $y=2,$ so the center is $(\frac{1}{2},2).$ Using the endpoint $(4,0),$ $\begin{aligned} r^2 &=\left(4-\frac12\right)^2+(0-2)^2\\ &=\frac{65}{4}. \end{aligned}$ Therefore the diameter is $2r=\sqrt{65}.$

Thus, the correct answer is E.

47.

In circle $O,$ the midpoint of radius $\overline{OX}$ is $Q;$ at $Q,$ $\overline{AB}\perp\overline{XY}.$ The semicircle with $\overline{AB}$ as diameter intersects $\overline{XY}$ in $M.$ Line $AM$ intersects circle $O$ in $C,$ and line $BM$ intersects circle $O$ in $D.$ Line $AD$ is drawn. Then, if the radius of circle $O$ is $r,$ $AD$ is:

$r\sqrt2$

$r$

not a side of an inscribed regular polygon

$\dfrac{r\sqrt3}{2}$

$r\sqrt3$

Concepts:circle inscribed angle special right triangle

Difficulty rating: 2070

Small Hint:

Since $\overline{XY}$ perpendicularly bisects $\overline{AB},$ compare $MA$ and $MB$

Big Hint:

Use the semicircle to find $\angle AMB,$ then use that $B,$ $M,$ and $D$ are collinear

Solution:

Line $XY$ is the perpendicular bisector of $AB,$ so $MA=MB.$ Since $M$ lies on the semicircle with diameter $AB,$ $\angle AMB=90^\circ.$ Thus triangle $AMB$ is an isosceles right triangle and $\angle ABM=45^\circ.$ Because $B,M,D$ are collinear, the inscribed angle $\angle ABD=45^\circ,$ so its intercepted arc $AD$ measures $90^\circ.$ Therefore chord $AD$ is a side of an inscribed square and has length $AD=r\sqrt2.$

Thus, the correct answer is A.

48.

Let $ABC$ be an equilateral triangle inscribed in circle $O.$ $M$ is a point on arc $BC.$ Lines $AM,$ $BM,$ and $CM$ are drawn. Then $AM$ is:

equal to $BM+CM$

less than $BM+CM$

greater than $BM+CM$

equal to, less than, or greater than $BM+CM,$ depending upon the position of $M$

none of these

Concepts:cyclic quadrilateral Ptolemy’s Theorem equilateral triangle

Difficulty rating: 1990

Small Hint:

Apply Ptolemy’s theorem to cyclic quadrilateral $ABMC$

Big Hint:

Use $AB=BC=CA$ to cancel the common side length

Solution:

In cyclic quadrilateral $ABMC,$ Ptolemy’s theorem gives $\begin{aligned} AM\cdot BC &=AB\cdot CM\\ &\quad+AC\cdot BM. \end{aligned}$ Since $ABC$ is equilateral, $AB=BC=AC.$ Dividing by this common length yields $AM=CM+BM.$

Therefore, the correct answer is A.

49.

The parallel sides of a trapezoid are $3$ and $9.$ The non-parallel sides are $4$ and $6.$ A line parallel to the bases divides the trapezoid into two trapezoids of equal perimeters. The ratio in which each of the non-parallel sides is divided is:

$4:3$

$3:2$

$4:1$

$3:1$

$6:1$

Concepts:trapezoid ratio and proportion perimeter

Difficulty rating: 1830

Small Hint:

A segment parallel to the bases divides both legs in the same fraction

Big Hint:

Let the upper leg segments be $4t$ and $6t$ ; the common dividing segment cancels when the two perimeters are equated

Solution:

Let the upper pieces of the legs of lengths $4$ and $6$ be $4t$ and $6t,$ respectively. The dividing segment appears once in each perimeter and cancels. Equal perimeters therefore give $\begin{aligned} 3+4t+6t &=9+4(1-t)\\ &\quad+6(1-t). \end{aligned}$ Hence $20t=16,$ so $t=\frac{4}{5}.$ Each leg is divided in the ratio $t:(1-t)=\frac45:\frac15=4:1.$

Thus, the correct answer is C.

50.

In circle $O,$ $G$ is a moving point on diameter $\overline{AB}.$ $\overline{AA'}$ is drawn perpendicular to $\overline{AB}$ and equal to $AG.$ $\overline{BB'}$ is drawn perpendicular to $\overline{AB},$ on the same side of diameter $\overline{AB}$ as $\overline{AA'},$ and equal to $BG.$ Let $O'$ be the midpoint of $\overline{A'B'}.$ Then, as $G$ moves from $A$ to $B,$ point $O':$

moves on a straight line parallel to $AB$

remains stationary

moves on a straight line perpendicular to $AB$

moves in a small circle intersecting the given circle

follows a path which is neither a circle nor a straight line

Concepts:coordinate geometry midpoint invariant

Difficulty rating: 1790

Small Hint:

Place $A=(0,0),$ $B=(L,0),$ and $G=(g,0)$

Big Hint:

Write coordinates for $A'$ and $B'$ , then average them

Solution:

Let $A=(0,0),$ $B=(L,0),$ and $G=(g,0).$ The perpendicular constructions on the same side give $A'=(0,g),\qquad B'=(L,L-g).$ Their midpoint is $\begin{aligned} O' &=\left(\frac L2,\frac{g+L-g}{2}\right)\\ &=\left(\frac L2,\frac L2\right), \end{aligned}$ independent of $g.$ Thus $O'$ remains stationary.

Therefore, the correct answer is B.