1959 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

Each edge of a cube is increased by $50\%.$ The percent of increase of the surface area of the cube is:

$50$

$125$

$150$

$300$

$750$

Concepts:percentage surface area power scaling of length, area, and volume

Difficulty rating: 890

Small Hint:

Compare the new edge length with the old edge length

Big Hint:

Surface area changes by the square of the linear scale factor

Solution:

If the original edge length is $s,$ the new edge length is $1.5s.$ Thus the surface area is multiplied by $(1.5)^2=2.25.$ The increase is $1.25$ times the original area, or $125\%.$

Thus, the correct answer is B.

2.

Through a point $P$ inside triangle $ABC$ a line is drawn parallel to the base $\overline{AB},$ dividing the triangle into two equal areas. If the altitude to $\overline{AB}$ has length $1,$ then the distance from $P$ to $\overline{AB}$ is:

$\dfrac12$

$\dfrac14$

$2-\sqrt2$

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

$\dfrac{2+\sqrt2}{8}$

Concepts:similarity area ratio altitude

Difficulty rating: 1360

Small Hint:

Let $x$ be the distance from $P$ to the base

Big Hint:

The small triangle above the parallel line has altitude $1-x$ and half the original area

Solution:

Let $x$ be the requested distance. The triangle above the parallel line is similar to the original triangle, with linear ratio $1-x.$ Its area is half the original area, so $(1-x)^2=\frac12.$ Since $0\lt x\lt1,$ we have $1-x=\frac{1}{\sqrt2},$ and therefore $x=1-\frac1{\sqrt2}=\frac{2-\sqrt2}{2}.$

Therefore, the correct answer is D.

3.

If the diagonals of a quadrilateral are perpendicular to each other, the figure would always be included under the general classification:

rhombus

rectangle

square

isosceles trapezoid

none of these

Concepts:kite diagonal counterexample

Difficulty rating: 960

Small Hint:

Look for a quadrilateral with perpendicular diagonals that is not one of the four named special types

Big Hint:

A general kite supplies a useful counterexample

Solution:

A kite can have perpendicular diagonals without being a rhombus, rectangle, square, or isosceles trapezoid. Therefore perpendicular diagonals alone do not force any of the first four classifications.

Thus, the correct answer is E.

4.

If $78$ is divided into three parts which are proportional to $1,$ $\dfrac13,$ $\dfrac16,$ the middle part is:

$9\dfrac13$

$13$

$17\dfrac13$

$18\dfrac13$

$26$

Concepts:ratio and proportion fraction

Difficulty rating: 1320

Small Hint:

Write the three parts as $x,\frac{x}{3},\frac{x}{6}$

Big Hint:

Their sum is $78$

Solution:

Let the parts be $x,\frac{x}{3},$ and $\frac{x}{6}.$ Then $x+\frac x3+\frac x6=\frac32x=78,$ so $x=52.$ The middle part is $\frac{x}{3}=\frac{52}{3}=17\frac13.$

Thus, the correct answer is C.

5.

The value of $(256)^{0.16}\cdot(256)^{0.09}$ is:

$4$

$16$

$64$

$256.25$

$-16$

Concepts:exponent radical

Difficulty rating: 1110

Small Hint:

Add the exponents because the bases are equal

Big Hint:

The sum of the decimal exponents is $\frac{1}{4}$

Solution:

Using the product rule for powers, $\begin{aligned} 256^{0.16}\cdot256^{0.09} &=256^{0.25}\\ &=256^{\frac{1}{4}}=4. \end{aligned}$

Therefore, the correct answer is A.

6.

Given the true statement: If a quadrilateral is a square, then it is a rectangle. It follows that, of the converse and the inverse of this true statement:

only the converse is true

only the inverse is true

both are true

neither is true

the inverse is true, but the converse is sometimes true

Concepts:logical deduction counterexample

Difficulty rating: 1140

Small Hint:

Write the converse and inverse explicitly

Big Hint:

Use a nonsquare rectangle to test the converse and a nonrectangular nonsquare to test the inverse

Solution:

The converse says that every rectangle is a square, which is false. The inverse says that every quadrilateral that is not a square is not a rectangle, which is also false because a nonsquare rectangle is a counterexample.

Thus neither statement is true, and the correct answer is D.

7.

The sides of a right triangle are $a,$ $a+d,$ and $a+2d,$ with $a$ and $d$ both positive. The ratio of $a$ to $d$ is:

$1{:}3$

$1{:}4$

$2{:}1$

$3{:}1$

$3{:}4$

Concepts:Pythagorean Theorem factoring ratio and proportion

Difficulty rating: 1210

Small Hint:

The longest side $a+2d$ must be the hypotenuse

Big Hint:

Apply the Pythagorean theorem and factor the resulting quadratic in $\frac{a}{d}$

Solution:

The Pythagorean theorem gives $a^2+(a+d)^2=(a+2d)^2.$ Simplifying, $\begin{aligned} a^2-2ad-3d^2 &=0,\\ (a-3d)(a+d) &=0. \end{aligned}$ Positivity excludes $a=-d,$ so $a=3d.$

Thus the ratio is $3:1,$ and the correct answer is D.

8.

The value of $x^2-6x+13$ can never be less than:

$4$

$4.5$

$5$

$7$

$13$

Concepts:completing the square quadratic optimization

Difficulty rating: 890

Small Hint:

Complete the square in $x^2-6x+13$

Big Hint:

A square is always nonnegative

Solution:

Completing the square, $x^2-6x+13=(x-3)^2+4.$ This expression is at least $4,$ with equality at $x=3.$

Thus, the correct answer is A.

9.

A farmer divides his herd of $n$ cows among his four sons so that one son gets one-half the herd, a second son one-fourth, a third son one-fifth, and the fourth son $7$ cows. Then $n$ is:

$80$

$100$

$140$

$180$

$240$

Concepts:fraction linear equation

Difficulty rating: 1140

Small Hint:

Add the three fractional shares of the herd

Big Hint:

The fraction left over corresponds to $7$ cows

Solution:

The first three sons receive $\frac12+\frac14+\frac15=\frac{19}{20}$ of the herd. Thus the remaining $\frac{1}{20}$ of the herd is $7,$ so $n=140.$

Therefore, the correct answer is C.

10.

In triangle $ABC,$ with $\overline{AB}=\overline{AC}=3.6,$ a point $D$ is taken on $\overline{AB}$ at a distance $1.2$ from $A.$ Point $D$ is joined to point $E$ in the prolongation of $\overline{AC}$ so that triangle $AED$ is equal in area to triangle $ABC.$ Then $\overline{AE}$ equals:

$4.8$

$5.4$

$7.2$

$10.8$

$12.6$

Concepts:triangle area area ratio

Difficulty rating: 1280

Small Hint:

Compare the altitudes from $B$ and $D$ to the line $AC$

Big Hint:

Because $\frac{AD}{AB}=\frac{1}{3},$ the altitude from $D$ is one-third the altitude from $B$

Solution:

Triangles $AED$ and $ABC$ use bases on the same line $AC.$ Since $\frac{AD}{AB}=\frac{1.2}{3.6}=\frac13,$ the perpendicular distance from $D$ to line $AC$ is one-third the corresponding distance from $B.$ Equality of areas therefore requires $AE\left(\frac h3\right)=AC(h),$ so $AE=3AC=10.8.$

Thus, the correct answer is D.

11.

The logarithm of $0.0625$ to the base $2$ is:

$0.025$

$0.25$

$5$

$-4$

$-2$

Concepts:logarithm power of 2 decimal

Difficulty rating: 1030

Small Hint:

Rewrite $0.0625$ as a fraction

Big Hint:

Express $\frac{1}{16}$ as a power of $2$

Solution:

Since $0.0625=\frac1{16}=2^{-4},$ we have $\log_2(0.0625)=-4.$

Therefore, the correct answer is D.

12.

By adding the same constant to each of $20,$ $50,$ $100$ a geometric progression results. The common ratio is:

$\dfrac53$

$\dfrac43$

$\dfrac32$

$\dfrac12$

$\dfrac13$

Concepts:geometric sequence algebraic manipulation

Difficulty rating: 1210

Small Hint:

Let $c$ be the added constant

Big Hint:

For three consecutive geometric terms, the square of the middle term equals the product of the outer terms

Solution:

The geometric-progression condition is $(50+c)^2=(20+c)(100+c).$ Expanding and cancelling $c^2$ gives $500=20c,$ so $c=25.$ The common ratio is $\frac{50+25}{20+25}=\frac{75}{45}=\frac53.$

Thus, the correct answer is A.

13.

The arithmetic mean (average) of a set of $50$ numbers is $38.$ If two numbers, namely, $45$ and $55,$ are discarded, the mean of the remaining set of numbers is:

$36.5$

$37$

$37.2$

$37.5$

$37.52$

Concepts:mean

Difficulty rating: 960

Small Hint:

Recover the original sum from the mean and number of entries

Big Hint:

Subtract $45+55$ and divide by the $48$ remaining entries

Solution:

The original sum is $50\cdot38=1900.$ After removing $45$ and $55,$ the remaining sum is $1800,$ so the new mean is $\frac{1800}{48}=37.5.$

Therefore, the correct answer is D.

14.

Given the set $S$ whose elements are zero and the even integers, positive and negative. Of the five operations applied to any pair of elements: $(1)$ addition, $(2)$ subtraction, $(3)$ multiplication, $(4)$ division, $(5)$ finding the arithmetic mean (average), those operations that yield only elements of $S$ are:

all

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

$1,$ $2,$ $3,$ $5$

$1,$ $2,$ $3$

$1,$ $3,$ $5$

Concepts:parity counterexample whole number operations

Difficulty rating: 1060

Small Hint:

Test whether each operation always takes two even integers to an even integer

Big Hint:

Use small counterexamples for division and averaging

Solution:

Even integers are closed under addition, subtraction, and multiplication. Division fails, since $\frac{2}{4}=\frac{1}{2}\notin S.$ Averaging fails, since the mean of $0$ and $2$ is $1\notin S.$

Thus only operations $1,$ $2,$ and $3$ always work, and the correct answer is D.

15.

In a right triangle the square of the hypotenuse is equal to twice the product of the legs. One of the acute angles of the triangle is:

$15^\circ$

$30^\circ$

$45^\circ$

$60^\circ$

$75^\circ$

Concepts:right triangle Pythagorean Theorem isosceles triangle

Difficulty rating: 1140

Small Hint:

Combine the given relation with the Pythagorean theorem

Big Hint:

Compare $a^2+b^2$ with $2ab$

Solution:

If the legs are $a,b$ and the hypotenuse is $c,$ then $a^2+b^2=c^2=2ab.$ Hence $(a-b)^2=0,$ so the legs are equal. The right triangle is isosceles and its acute angles are $45^\circ.$

Thus, the correct answer is C.

16.

The expression $\frac{x^2-3x+2}{x^2-5x+6}\div \frac{x^2-5x+4}{x^2-7x+12},$ when simplified, is:

$\dfrac{(x-1)(x-6)}{(x-3)(x-4)}$

$\dfrac{x+3}{x-3}$

$\dfrac{x+1}{x-1}$

$1$

$2$

Concepts:rational equation factoring algebraic manipulation

Difficulty rating: 1210

Small Hint:

Factor all four quadratics completely

Big Hint:

Replace division by multiplication by the reciprocal

Solution:

Factoring and multiplying by the reciprocal gives $\begin{aligned} &\frac{(x-1)(x-2)}{(x-2)(x-3)}\\ &\quad{}\cdot \frac{(x-3)(x-4)}{(x-1)(x-4)}=1 \end{aligned}$ for every value in the domain of the original expression.

Therefore, the correct answer is D.

17.

If $y=a+\dfrac bx,$ where $a$ and $b$ are constants, and if $y=1$ when $x=-1,$ and $y=5$ when $x=-5,$ then $a+b$ equals:

$-1$

$0$

$1$

$10$

$11$

Concepts:system of equations function substitution

Difficulty rating: 1180

Small Hint:

Substitute the two given input-output pairs

Big Hint:

Solve $a-b=1$ and $a-\frac{b}{5}=5$

Solution:

The two conditions give $a-b=1,\qquad a-\frac b5=5.$ Subtracting yields $\frac{4b}{5}=4,$ so $b=5$ and $a=6.$ Therefore $a+b=11.$

Thus, the correct answer is E.

18.

The arithmetic mean (average) of the first $n$ positive integers is:

$\dfrac n2$

$\dfrac{n^2}{2}$

$n$

$\dfrac{n-1}{2}$

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

Concepts:arithmetic sequence mean

Difficulty rating: 840

Small Hint:

Use the sum $1+2+\cdots+n$

Big Hint:

Divide the sum by the number $n$ of terms

Solution:

The sum of the first $n$ positive integers is $\frac{n(n+1)}{2}.$ Dividing by $n$ gives the mean $\frac{n+1}{2}.$

Therefore, the correct answer is E.

19.

With the use of three different weights, namely, $1$ lb., $3$ lb., and $9$ lb., how many objects of different weights can be weighed, if the objects to be weighed and the given weights may be placed in either pan of the scale?

$15$

$13$

$11$

$9$

$7$

Concepts:number base systematic listing

Difficulty rating: 1360

Small Hint:

Each weight may go with the object, against the object, or remain unused

Big Hint:

The weights $1,3,9$ can represent every integer from $1$ through their sum

Solution:

Using coefficients $-1,0,1$ for the weights $1,3,9,$ balanced ternary represents every integer weight from $1$ through $1+3+9=13.$ Thus there are $13$ different positive object weights that can be measured.

Therefore, the correct answer is B.

20.

It is given that $x$ varies directly as $y$ and inversely as the square of $z,$ and that $x=10$ when $y=4$ and $z=14.$ Then, when $y=16$ and $z=7,$ $x$ equals:

$180$

$160$

$154$

$140$

$120$

Concepts:ratio and proportion exponent

Difficulty rating: 1140

Small Hint:

Write $x=\frac{ky}{z^2}$

Big Hint:

Compare the new and old values by ratios so the constant cancels

Solution:

Because $x=\frac{ky}{z^2},$ $\frac{x_{\rm new}}{10} =\frac{16}{4}\left(\frac{14}{7}\right)^2 =4\cdot4=16.$ Hence $x_{\rm new}=160.$

Thus, the correct answer is B.

21.

If $p$ is the perimeter of an equilateral triangle inscribed in a circle, the area of the circle is:

$\dfrac{\pi p^2}{3}$

$\dfrac{\pi p^2}{9}$

$\dfrac{\pi p^2}{27}$

$\dfrac{\pi p^2}{81}$

$\dfrac{\pi p^2\sqrt3}{27}$

Concepts:equilateral triangle circumcircle, circumcenter, and circumradius circle area

Difficulty rating: 1210

Small Hint:

The triangle side length is $\frac{p}{3}$

Big Hint:

An equilateral triangle with side $s$ has circumradius $\frac{s}{\sqrt3}$

Solution:

The side length is $\frac{p}{3},$ so the circumradius is $R=\frac{\frac{p}{3}}{\sqrt3}=\frac{p}{3\sqrt3}.$ Therefore the circle’s area is $\pi R^2=\frac{\pi p^2}{27}.$

Thus, the correct answer is C.

22.

The line joining the midpoints of the diagonals of a trapezoid has length $3.$ If the longer base is $97,$ then the shorter base is:

$94$

$92$

$91$

$90$

$89$

Concepts:trapezoid midpoint

Difficulty rating: 1150

Small Hint:

Recall the length of the segment joining the diagonal midpoints of a trapezoid

Big Hint:

It equals half the difference of the base lengths

Solution:

If the shorter base is $b,$ the diagonal-midpoint segment has length half the difference of the bases: $\frac{97-b}{2}=3.$ Thus $97-b=6$ and $b=91.$

Therefore, the correct answer is C.

23.

The set of solutions for the equation $\log_{10}(a^2-15a)=2$ consists of:

two integers

one integer and one fraction

two irrational numbers

two non-real numbers

no numbers, that is, the set is empty

Concepts:logarithm quadratic

Difficulty rating: 1110

Small Hint:

Convert the logarithmic equation to exponential form

Big Hint:

Solve $a^2-15a=100$ and check that the logarithm’s argument is positive

Solution:

The equation is equivalent to $a^2-15a=100,$ or $(a-20)(a+5)=0.$ Thus $a=20$ or $a=-5.$ In both cases the logarithm’s argument is $100,$ so both solutions are valid integers.

Therefore, the correct answer is A.

24.

A chemist has $m$ ounces of salt water that is $m\%$ salt. How many ounces of salt must he add to make a solution that is $2m\%$ salt?

$\dfrac{m}{100+m}$

$\dfrac{2m}{100-2m}$

$\dfrac{m^2}{100-2m}$

$\dfrac{m^2}{100+2m}$

$\dfrac{2m}{100+m}$

Concepts:mixture percentage rational equation

Difficulty rating: 1300

Small Hint:

The original amount of salt is $\frac{m^2}{100}$ ounces

Big Hint:

If $x$ ounces of pure salt are added, equate $\frac{\frac{m^2}{100}+x}{m+x}$ to $\frac{2m}{100}$

Solution:

Let $x$ ounces of salt be added. The concentration equation is $\frac{\frac{m^2}{100}+x}{m+x}=\frac{2m}{100}.$ Multiplying through and collecting the $x$ -terms gives $x(100-2m)=m^2,$ so $x=\frac{m^2}{100-2m}.$

Therefore, the correct answer is C.

25.

The symbol $|a|$ means $+a$ if $a$ is greater than or equal to zero, and $-a$ if $a$ is less than or equal to zero; the symbol $\lt$ means “less than”; the symbol $\gt$ means “greater than.” The set of values $x$ satisfying the inequality $|3-x|\lt4$ consists of all $x$ such that:

$x^2\lt49$

$x^2\gt1$

$1\lt x^2\lt49$

$-1\lt x\lt7$

$-7\lt x\lt1$

Concepts:absolute value inequality

Difficulty rating: 1060

Small Hint:

Rewrite $|3-x|\lt4$ as a compound inequality

Big Hint:

Solve $-4\lt3-x\lt4$ for $x$

Solution:

The absolute-value inequality is equivalent to $-4\lt3-x\lt4.$ Subtracting $3$ and multiplying by $-1$ reverses the inequalities, giving $-1\lt x\lt7.$

Thus, the correct answer is D.

26.

The base of an isosceles triangle is $\sqrt2.$ The medians to the legs intersect each other at right angles. The area of the triangle is:

$1.5$

$2$

$2.5$

$3.5$

$4$

Concepts:isosceles triangle median (geometry)coordinate geometry area

Difficulty rating: 1660

Small Hint:

Place the base endpoints symmetrically about the origin and the third vertex on the perpendicular bisector

Big Hint:

Write direction vectors for the two medians and set their dot product equal to zero

Solution:

Put the base endpoints at $A=(-\frac{\sqrt2}{2},0)$ and $B=(\frac{\sqrt2}{2},0),$ and let the third vertex be $C=(0,h).$ The median from $A$ has direction $\left(\frac{3\sqrt2}{4},\frac h2\right),$ while the median from $B$ has direction $\left(-\frac{3\sqrt2}{4},\frac h2\right).$ Their dot product is zero, so $-\frac98+\frac{h^2}{4}=0,$ and $h=\frac{3\sqrt2}{2}.$ Thus the area is $\frac12\cdot\sqrt2\cdot\frac{3\sqrt2}{2}=\frac32.$

Therefore, the correct answer is A.

27.

Which one of the following statements is not true for the equation

$ix^2-x+2i=0,$

where $i=\sqrt{-1}?$

The sum of the roots is $2$

The discriminant is $9$

The roots are imaginary

The roots can be found by using the quadratic formula

The roots can be found by factoring, using imaginary numbers

Concepts:complex number quadratic Vieta’s Formulas

Difficulty rating: 1360

Small Hint:

Use Vieta’s formula for the sum of the roots before solving the equation

Big Hint:

The sum is the negative of the $x$ -coefficient divided by the $x^2$ -coefficient

Solution:

By Vieta’s formula, the sum of the roots is $-\frac{-1}{i}=\frac1i=-i,$ not $2.$ Also the discriminant is $(-1)^2-4(i)(2i)=1-8i^2=9.$ The quadratic formula gives the imaginary roots $i$ and $-2i,$ so the remaining statements are true.

Thus, the correct answer is A.

28.

In triangle $ABC,$ $\overline{AL}$ bisects angle $A$ and $\overline{CM}$ bisects angle $C.$ Points $L$ and $M$ are on $\overline{BC}$ and $\overline{AB},$ respectively. The sides of triangle $ABC$ are $a,$ $b,$ and $c.$ Then $\dfrac{AM}{MB}=k\dfrac{CL}{LB},$ where $k$ is:

$1$

$\dfrac{bc}{a^2}$

$\dfrac{a^2}{bc}$

$\dfrac cb$

$\dfrac ca$

Concepts:angle bisector theorem ratio and proportion

Difficulty rating: 1300

Small Hint:

Apply the angle bisector theorem separately to $CM$ and $AL$

Big Hint:

Use the standard notation $a=BC,\ b=CA,\ c=AB$

Solution:

By the angle bisector theorem, $\begin{aligned} \frac{AM}{MB}&=\frac{AC}{CB}=\frac ba,\\ \frac{CL}{LB}&=\frac{AC}{AB}=\frac bc. \end{aligned}$ Therefore $k=\frac{\frac{b}{a}}{\frac{b}{c}}=\frac ca.$

Thus, the correct answer is E.

29.

On an examination of $n$ questions a student answers correctly $15$ of the first $20.$ Of the remaining questions he answers one third correctly. All the questions have the same credit. If the student’s mark is $50\%,$ how many different values of $n$ can there be?

$4$

$3$

$2$

$1$

the problem cannot be solved

Concepts:percentage linear equation divisibility

Difficulty rating: 1280

Small Hint:

Express the total number correct in terms of $n$

Big Hint:

Set $15+\frac{n-20}{3}$ equal to $\frac{n}{2}$

Solution:

The score condition gives $15+\frac{n-20}{3}=\frac n2.$ Multiplying by $6$ yields $90+2n-40=3n,$ so $n=50.$ This value also makes the number of remaining correct answers an integer. Hence there is exactly one possible value of $n.$

Therefore, the correct answer is D.

30.

$A$ can run around a circular track in $40$ seconds. $B,$ running in the opposite direction, meets $A$ every $15$ seconds. What is $B$ ’s time to run around the track, expressed in seconds?

$12\dfrac12$

$24$

$25$

$27\dfrac12$

$55$

Concepts:relative speed rate

Difficulty rating: 1140

Small Hint:

Measure each runner’s speed in laps per second

Big Hint:

Because they run in opposite directions, their speeds add to $\frac{1}{15}$ lap per second

Solution:

If $B$ ’s lap time is $t,$ their relative speed is $\frac1{40}+\frac1t=\frac1{15}.$ Thus $\frac{1}{t}=\frac{1}{24},$ so $t=24$ seconds.

Therefore, the correct answer is B.

31.

A square, with an area of $40,$ is inscribed in a semicircle. The area of a square that could be inscribed in the entire circle with the same radius is:

$80$

$100$

$120$

$160$

$200$

Concepts:circle square (geometry)Pythagorean Theorem

Difficulty rating: 1360

Small Hint:

Let the first square have side $s$ and place its lower side on the semicircle’s diameter

Big Hint:

A top vertex has horizontal coordinate $\frac{s}{2}$ and vertical coordinate $s$ relative to the center

Solution:

Let the semicircle have radius $r$ and the square have side $s,$ where $s^2=40.$ A top vertex of the square is $\frac{s}{2}$ horizontally and $s$ vertically from the center, so $r^2=s^2+\left(\frac s2\right)^2=\frac54s^2=50.$ A square inscribed in the full circle has diagonal $2r,$ hence area $2r^2=100.$

Thus, the correct answer is B.

32.

The length $l$ of a tangent, drawn from a point $A$ to a circle, is $\dfrac43$ of the radius $r.$ The (shortest) distance from $A$ to the circle is:

$\dfrac12r$

$r$

$\dfrac12l$

$\dfrac23l$

a value between $r$ and $l$

Concepts:tangent line right triangle circle

Difficulty rating: 1280

Small Hint:

Join $A$ to the center and to the point of tangency

Big Hint:

Use the right triangle with legs $r$ and $l=\frac{4r}{3}$

Solution:

If $O$ is the center and $T$ the point of tangency, then $OT\perp AT.$ Hence $\begin{aligned} AO &=\sqrt{r^2+l^2}\\ &=\sqrt{r^2+\frac{16r^2}{9}}\\ &=\frac{5r}{3}. \end{aligned}$ The shortest distance from $A$ to the circle is $AO-r=\frac{2r}{3}.$ Since $l=\frac{4r}{3},$ this distance equals $\frac{l}{2}.$

Therefore, the correct answer is C.

33.

A harmonic progression is a sequence of numbers such that their reciprocals are in arithmetic progression. Let $S_n$ represent the sum of the first $n$ terms of the harmonic progression; for example, $S_3$ represents the sum of the first three terms. If the first three terms of a harmonic progression are $3,$ $4,$ $6,$ then:

$S_4=20$

$S_4=25$

$S_5=49$

$S_6=49$

$S_2=\dfrac12S_4$

Concepts:arithmetic sequence pattern recognition

Difficulty rating: 1280

Small Hint:

Write the reciprocals $\frac{1}{3},\frac{1}{4},\frac{1}{6}$ and identify their common difference

Big Hint:

Continue the arithmetic progression of reciprocals one more term

Solution:

The reciprocals begin $\frac13,\quad\frac14,\quad\frac16,$ with common difference $-\frac{1}{12}.$ The next reciprocal is $\frac{1}{12},$ so the fourth term is $12.$ Therefore $S_4=3+4+6+12=25.$

Thus, the correct answer is B.

34.

Let the roots of $x^2-3x+1=0$ be $r$ and $s.$ Then the expression $r^2+s^2$ is:

a positive integer

a positive fraction greater than $1$

a positive fraction less than $1$

an irrational number

an imaginary number

Concepts:Vieta’s Formulas quadratic algebraic manipulation

Difficulty rating: 1110

Small Hint:

Use $r+s=3$ and $rs=1$

Big Hint:

Rewrite $r^2+s^2$ as $(r+s)^2-2rs$

Solution:

Vieta’s formulas give $r+s=3$ and $rs=1.$ Therefore $\begin{aligned} r^2+s^2 &=(r+s)^2-2rs\\ &=9-2=7, \end{aligned}$ which is a positive integer.

Thus, the correct answer is A.

35.

The symbol $\ge$ means “greater than or equal to”; the symbol $\le$ means “less than or equal to.” In the equation $(x-m)^2-(x-n)^2=(m-n)^2,$ $m$ is a fixed positive number, and $n$ is a fixed negative number. The set of values $x$ satisfying the equation is:

$x\ge0$

$x\le n$

$x=0$

the set of all real numbers

none of these

Concepts:difference of squares linear equation inequality

Difficulty rating: 1280

Small Hint:

Factor the difference of squares on the left

Big Hint:

Because $m\ne n,$ divide by $m-n$ and solve for $x$

Solution:

Factoring the left side gives $\begin{aligned} &(n-m)(2x-m-n)\\ &\qquad=(m-n)^2. \end{aligned}$ Since $m\ne n,$ division by $n-m$ yields $2x-m-n=n-m,$ so $x=n.$ This single negative value is not any of choices A through D.

Therefore, the correct answer is E.

36.

The base of a triangle is $80,$ and one of the base angles is $60^\circ.$ The sum of the lengths of the other two sides is $90.$ The shortest side is:

$45$

$40$

$36$

$17$

$12$

Concepts:law of cosines algebraic manipulation

Difficulty rating: 1550

Small Hint:

Let the side adjacent to the $60^\circ$ angle be $x$ , so the third side is $90-x$

Big Hint:

Apply the law of cosines with included sides $80$ and $x$

Solution:

Let the side adjacent to the $60^\circ$ base angle be $x,$ and let the opposite side be $90-x.$ The law of cosines gives $\begin{aligned} (90-x)^2 &=80^2+x^2\\ &\quad{}-2(80)(x)\cos60^\circ. \end{aligned}$ Simplifying yields $8100-180x=6400-80x,$ so $x=17.$ The side lengths are $17,\ 73,$ and $80,$ and the shortest is $17.$

Thus, the correct answer is D.

37.

When simplified the product $\begin{aligned} &\left(1-\dfrac13\right)\left(1-\dfrac14\right)\\ &\quad{}\cdot\left(1-\dfrac15\right)\cdots\left(1-\dfrac1n\right) \end{aligned}$ becomes:

$\dfrac1n$

$\dfrac2n$

$\dfrac{2(n-1)}n$

$\dfrac{2}{n(n+1)}$

$\dfrac{3}{n(n+1)}$

Concepts:telescoping fraction

Difficulty rating: 1140

Small Hint:

Rewrite each factor $1-\frac{1}{k}$ as $\frac{k-1}{k}$

Big Hint:

Write out the first few factors and cancel adjacent numerators and denominators

Solution:

The product telescopes: $\frac23\cdot\frac34\cdot\frac45\cdots\frac{n-1}{n} =\frac2n.$

Therefore, the correct answer is B.

38.

If $4x+\sqrt{2x}=1,$ then $x:$

is an integer

is fractional

is irrational

is imaginary

may have two different values

Concepts:radical substitution quadratic

Difficulty rating: 1360

Small Hint:

Set $u=\sqrt{2x},$ so $x=\frac{u^2}{2}$ and $u\ge0$

Big Hint:

Solve the resulting quadratic $2u^2+u-1=0$

Solution:

Let $u=\sqrt{2x}\ge0.$ Since $x=\frac{u^2}{2},$ the equation becomes $2u^2+u=1,$ or $(2u-1)(u+1)=0.$ Thus $u=\frac{1}{2}$ and $x=\frac{u^2}{2}=\frac{1}{8}.$ The other quadratic root is inadmissible because $u\ge0.$ Hence $x$ is fractional.

Therefore, the correct answer is B.

39.

Let $S$ be the sum of the first nine terms of the sequence

$\begin{gathered} x+a,\quad x^2+2a,\\ x^3+3a,\quad\ldots. \end{gathered}$

Then $S$ equals:

$\dfrac{50a+x+x^8}{x+1}$

$50a-\dfrac{x+x^{10}}{x-1}$

$\dfrac{x^9-1}{x+1}+45a$

$\dfrac{x^{10}-x}{x-1}+45a$

$\dfrac{x^{11}-x}{x-1}+45a$

Concepts:geometric sequence arithmetic sequence summation

Difficulty rating: 1280

Small Hint:

Separate the powers of $x$ from the multiples of $a$

Big Hint:

Use the geometric-series sum for $x+x^2+\cdots+x^9$

Solution:

Adding the first nine terms gives $\begin{aligned} S={}&(x+x^2+\cdots+x^9)\\ &{}+(1+2+\cdots+9)a. \end{aligned}$ Therefore $S=\frac{x^{10}-x}{x-1}+45a.$ The expression is understood by continuity at $x=1,$ where both forms equal $9+45a.$

Thus, the correct answer is D.

40.

In triangle $ABC,$ $\overline{BD}$ is a median. $\overline{CF}$ intersects $\overline{BD}$ at $E$ so that $BE=ED.$ Point $F$ is on $\overline{AB}.$ Then, if $BF=5,$ $BA$ equals:

$10$

$12$

$15$

$20$

none of these

Concepts:median (geometry)vector mass points

Difficulty rating: 1590

Small Hint:

Use coordinates or masses, noting that $D$ is the midpoint of $AC$ and $E$ is the midpoint of $BD$

Big Hint:

Write $E$ both as $\frac{B+D}{2}$ and as a point on line $CF$

Solution:

Use vectors with $A=\mathbf0,\ B=\mathbf b,$ and $C=\mathbf c.$ Since $D=\frac{\mathbf c}{2}$ and $E$ is the midpoint of $BD,$ $E=\frac12\mathbf b+\frac14\mathbf c.$ A point $F$ on $AB$ has the form $F=t\mathbf b.$ Since $E$ lies on $CF,$ comparison of the $\mathbf c$ -coefficient shows that $E=\tfrac14C+\tfrac34F.$ Hence $\frac34t=\frac12,$ so $t=\frac{2}{3}.$ Thus $\frac{AF}{AB}=\frac{2}{3},$ and $\frac{BF}{BA}=\frac{1}{3}.$ Since $BF=5,$ $BA=15.$

Therefore, the correct answer is C.

41.

On the same side of a straight line three circles are drawn as follows: a circle with a radius of $4$ inches is tangent to the line, the other two circles are equal, and each is tangent to the line and to the other two circles. The radius of the equal circles is:

$24$

$20$

$18$

$16$

$12$

Concepts:tangent circles distance formula

Difficulty rating: 1590

Small Hint:

The small circle lies symmetrically between the two equal circles

Big Hint:

If an equal circle has radius $R,$ compare the center distance $\sqrt{R^2+(R-4)^2}$ with $R+4$

Solution:

Let the equal circles have radius $R.$ Their centers are $2R$ apart, so the center of the radius- $4$ circle lies midway between them. The horizontal and vertical separations between its center and either large center are $R$ and $R-4.$ Tangency gives $R^2+(R-4)^2=(R+4)^2.$ Simplifying yields $R^2-16R=0,$ and positivity gives $R=16.$

Thus, the correct answer is D.

42.

Given three positive integers $a,$ $b,$ and $c.$ Their greatest common divisor is $D;$ their least common multiple is $M.$ Then, which two of the following statements are true?

$(1)$ The product $MD$ cannot be less than $abc.$

$(2)$ The product $MD$ cannot be greater than $abc.$

$(3)$ $MD$ equals $abc$ if and only if $a,$ $b,$ $c$ are each prime.

$(4)$ $MD$ equals $abc$ if and only if $a,$ $b,$ $c$ are relatively prime in pairs. (This means: no two have a common factor greater than $1.$ )

$1,$ $2$

$1,$ $3$

$1,$ $4$

$2,$ $3$

$2,$ $4$

Concepts:greatest common divisor least common multiple prime factorization

Difficulty rating: 1790

Small Hint:

For one prime, order its exponents in $a,b,c$ as $u\le v\le w$

Big Hint:

Compare the exponent $u+w$ in $MD$ with the exponent $u+v+w$ in $abc$

Solution:

For any prime, let its exponents in $a,b,c$ be $u\le v\le w.$ Its exponent in $MD$ is $u+w,$ while its exponent in $abc$ is $u+v+w.$ Thus $MD\le abc,$ proving statement $(2).$ Equality holds exactly when $v=0$ for every prime, meaning no prime divides two of $a,b,c.$ That is precisely pairwise relative primality, proving statement $(4).$

Therefore, the correct answer is E.

43.

The sides of a triangle are $25,$ $39,$ and $40.$ The diameter of the circumscribed circle is:

$\dfrac{133}{3}$

$\dfrac{125}{3}$

$42$

$41$

$40$

Concepts:Heron’s Formula circumcircle, circumcenter, and circumradius triangle area

Difficulty rating: 1750

Small Hint:

Use Heron’s formula with semiperimeter $52$

Big Hint:

After finding the area $K,$ use $abc=4KR$ and double the circumradius

Solution:

The semiperimeter is $52,$ so Heron’s formula gives $K=\sqrt{52\cdot27\cdot13\cdot12}=468.$ If $R$ is the circumradius, then $R=\frac{25\cdot39\cdot40}{4\cdot468}=\frac{125}{6}.$ Hence the diameter is $2R=\frac{125}{3}.$

Thus, the correct answer is B.

44.

The roots of $x^2+bx+c=0$ are both real and greater than $1.$ Let $s=b+c+1.$ Then $s:$

may be less than zero

may be equal to zero

must be greater than zero

must be less than zero

must be between $-1$ and $1$

Concepts:Vieta’s Formulas inequality quadratic

Difficulty rating: 1360

Small Hint:

Call the two roots $u$ and $v$ and express $b,c$ using Vieta’s formulas

Big Hint:

Factor $1-(u+v)+uv$

Solution:

If the roots are $u,v>1,$ then $b=-(u+v)$ and $c=uv.$ Therefore $\begin{aligned} s&=1-u-v+uv\\ &=(u-1)(v-1)>0. \end{aligned}$

Thus $s$ must be greater than zero, and the correct answer is C.

45.

If $\begin{aligned} &(\log_3x)(\log_x2x)(\log_{2x}y)\\ &\qquad=\log_xx^2, \end{aligned}$ then $y$ equals:

$\dfrac92$

$9$

$18$

$27$

$81$

Concepts:logarithm telescoping

Difficulty rating: 1280

Small Hint:

Use $(\log_a b)(\log_b c)=\log_a c$ twice

Big Hint:

Simplify the entire left side to $\log_3y$

Solution:

The logarithms telescope: $\begin{aligned} &(\log_3x)(\log_x2x)(\log_{2x}y)\\ &\qquad=\log_3y. \end{aligned}$ Also $\log_xx^2=2,$ so $\log_3y=2$ and $y=3^2=9.$

Therefore, the correct answer is B.

46.

A student on vacation for $d$ days observed that

$(1)$ it rained $7$ times, morning or afternoon;

$(2)$ when it rained in the afternoon, it was clear in the morning;

$(3)$ there were five clear afternoons;

$(4)$ there were six clear mornings.

Then $d$ equals:

$7$

$9$

$10$

$11$

$12$

Concepts:basic counting complementary counting

Difficulty rating: 1590

Small Hint:

Count rainy mornings and rainy afternoons from the numbers of clear half-days

Big Hint:

Condition $(2)$ ensures that a rainy afternoon and rainy morning never occur on the same day

Solution:

There are $d-6$ rainy mornings and $d-5$ rainy afternoons. By condition $(2),$ these are distinct rainy occasions, and their total is $7.$ Hence $(d-6)+(d-5)=7,$ so $2d=18$ and $d=9.$

Thus, the correct answer is B.

47.

Assume that the following three statements are true:

I. All freshmen are human.

II. All students are human.

III. Some students think.

Given the following four statements:

$(1)$ All freshmen are students.

$(2)$ Some humans think.

$(3)$ No freshmen think.

$(4)$ Some humans who think are not students.

Those which are logical consequences of I, II, and III are:

$2$

$4$

$2,$ $3$

$2,$ $4$

$1,$ $2$

Concepts:logical deduction counterexample

Difficulty rating: 1320

Small Hint:

Use the person whose existence is asserted in statement III

Big Hint:

Combine “some students think” with “all students are human”

Solution:

Statement III supplies at least one student who thinks. By statement II, that student is human, so some human thinks and statement $(2)$ follows. Nothing relates freshmen to students, says whether freshmen think, or guarantees a thinking human outside the students. Thus none of $(1),$ $(3),$ or $(4)$ follows.

Therefore, the correct answer is A.

48.

Given the polynomial $\begin{aligned} &a_0x^n+a_1x^{n-1}+\cdots\\ &\qquad{}+a_{n-1}x+a_n, \end{aligned}$ where $n$ is a positive integer or zero, and $a_0$ is a positive integer. The remaining $a$ ’s are integers or zero. Set $\begin{aligned} h={}&n+a_0+|a_1|+|a_2|\\ &{}+\cdots+|a_n|. \end{aligned}$ [See example $25$ for the meaning of $|x|.$ ] The number of polynomials with $h=3$ is:

$3$

$5$

$6$

$7$

$9$

Concepts:polynomial casework

Difficulty rating: 1730

Small Hint:

Since $n+a_0\le3$ and $a_0\ge1,$ consider only $n=0,1,2$

Big Hint:

For each degree, count the integer coefficient tuples with the required sum of absolute values

Solution:

We count by degree. If $n=0,$ then $a_0=3,$ giving one polynomial. If $n=1,$ then $a_0+|a_1|=2.$ This gives $a_0=2,a_1=0$ or $a_0=1,a_1=\pm1,$ for three polynomials. If $n=2,$ then $a_0=1$ and $a_1=a_2=0,$ giving one more. No higher degree is possible. The total is $1+3+1=5.$

Thus, the correct answer is B.

49.

For the infinite series $\begin{aligned} &1-\dfrac12-\dfrac14+\dfrac18\\ &{}-\dfrac1{16}-\dfrac1{32}+\dfrac1{64}\\ &{}-\dfrac1{128}-\cdots, \end{aligned}$ let $S$ be the (limiting) sum. Then $S$ equals:

$0$

$\dfrac27$

$\dfrac67$

$\dfrac9{32}$

$\dfrac{27}{32}$

Concepts:geometric sequence pairing and grouping summation

Difficulty rating: 1590

Small Hint:

Group the terms in consecutive blocks of three

Big Hint:

Each block is $\frac{1}{8}$ times the preceding block

Solution:

Group the series as $\begin{aligned} S={}&\left(1-\frac12-\frac14\right)\\ &+\left(\frac18-\frac1{16}-\frac1{32}\right) +\cdots. \end{aligned}$ The first block is $\frac{1}{4},$ and successive blocks form a geometric series with ratio $\frac{1}{8}.$ Hence $S=\frac{\frac{1}{4}}{1-\frac{1}{8}}=\frac27.$

Therefore, the correct answer is B.

50.

A club with $x$ members is organized into four committees in accordance with these two rules:

$(1)$ Each member belongs to two and only two committees.

$(2)$ Each pair of committees has one and only one member in common.

Then $x:$

cannot be determined

has a single value between $8$ and $16$

has two values between $8$ and $16$

has a single value between $4$ and $8$

has two values between $4$ and $8$

Concepts:double counting combinations bijection

Difficulty rating: 1360

Small Hint:

Associate each member with the pair of committees to which that member belongs

Big Hint:

Rule $(2)$ says every pair of the four committees occurs exactly once

Solution:

Each member belongs to exactly one unordered pair of committees. Conversely, each pair of committees has exactly one common member. Thus the members are in one-to-one correspondence with the pairs of four committees, so $x=\binom42=6.$ This is a single value between $4$ and $8.$

Therefore, the correct answer is D.