1961 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

When simplified, $\left(-\dfrac{1}{125}\right)^{-\frac{2}{3}}$ becomes:

$\dfrac1{25}$

$-\dfrac1{25}$

$25$

$-25$

$25\sqrt{-1}$

Concepts:exponent radical algebraic manipulation

Difficulty rating: 1230

Small Hint:

Apply the cube root before squaring

Big Hint:

The negative exponent then takes the reciprocal

Solution:

Because the cube root is real, $\left(-\frac1{125}\right)^{\frac{2}{3}} =\left(-\frac15\right)^2=\frac1{25}.$ The negative exponent takes the reciprocal, giving $25.$

Therefore, the correct answer is C.

2.

An automobile travels $\frac{a}{6}$ feet in $r$ seconds. If this rate is maintained for $3$ minutes, how many yards does it travel in the $3$ minutes?

$\dfrac{a}{1080r}$

$\dfrac{30r}{a}$

$\dfrac{30a}{r}$

$\dfrac{10r}{a}$

$\dfrac{10a}{r}$

Concepts:distance rate and time unit conversion

Difficulty rating: 1140

Small Hint:

Convert $3$ minutes to seconds

Big Hint:

After finding the distance in feet, divide by $3$ to convert to yards

Solution:

The speed is $\frac{a}{6r}$ feet per second. In $180$ seconds the automobile travels $180\cdot\frac{a}{6r}=\frac{30a}{r}$ feet, or $\frac{10a}{r}$ yards.

Thus, the correct answer is E.

3.

If the graphs of $2y+x+3=0$ and $3y+ax+2=0$ are to meet at right angles, the value of $a$ is:

$\pm\dfrac23$

$-\dfrac23$

$-\dfrac32$

$6$

$-6$

Concepts:slope coordinate geometry linear equation

Difficulty rating: 1320

Small Hint:

Put both equations into slope-intercept form

Big Hint:

The product of the slopes of perpendicular nonvertical lines is $-1$

Solution:

The two slopes are $-\frac{1}{2}$ and $-\frac{a}{3}.$ Perpendicularity requires $\left(-\frac12\right)\left(-\frac a3\right)=-1,$ so $a=-6.$

Therefore, the correct answer is E.

4.

Let the set consisting of the squares of the positive integers be called $u;$ thus $u$ is the set $1,$ $4,$ $9,$ $\ldots.$ If a certain operation on one or more members of the set always yields a member of the set, we say that the set is closed under that operation. Then $u$ is closed under:

addition

multiplication

division

extraction of a positive integral root

none of these

Concepts:perfect square counterexample

Difficulty rating: 1140

Small Hint:

Test each operation on arbitrary squares $m^2$ and $n^2$

Big Hint:

The product of two squares has an immediate square form

Solution:

For positive integers $m,n,$ $m^2n^2=(mn)^2,$ which is again the square of a positive integer. The other operations fail, for example $1+4=5,\frac{1}{4},$ and $\sqrt4=2.$

Thus, the correct answer is B.

5.

Let $S=(x-1)^4+4(x-1)^3$ $+6(x-1)^2+4(x-1)+1.$ Then $S$ equals:

$(x-2)^4$

$(x-1)^4$

$x^4$

$(x+1)^4$

$x^4+1$

Concepts:binomial theorem substitution

Difficulty rating: 1030

Small Hint:

Compare the coefficients with the expansion of $(u+1)^4$

Big Hint:

Use $u=x-1$

Solution:

The expression is the binomial expansion $((x-1)+1)^4=x^4.$

Therefore, the correct answer is C.

6.

When simplified, $\log 8\div\log\dfrac18$ becomes:

$6\log2$

$\log2$

$1$

$0$

$-1$

Concepts:logarithm algebraic manipulation

Difficulty rating: 960

Small Hint:

Write $\frac{1}{8}$ as $8^{-1}$

Big Hint:

Relate $\log(\frac{1}{8})$ to $\log8$

Solution:

Since $\log(\frac{1}{8})=\log(8^{-1})=-\log8,$ $\frac{\log8}{\log(\frac{1}{8})}=-1.$

Thus, the correct answer is E.

7.

When simplified, the third term in the expansion of $\left(\frac{a}{\sqrt{x}}-\frac{\sqrt{x}}{a^2}\right)^6$ is:

$\dfrac{15}{x}$

$-\dfrac{15}{x}$

$-\dfrac{6x^2}{a^9}$

$\dfrac{20}{a^3}$

$-\dfrac{20}{a^3}$

Concepts:binomial theorem exponent algebraic manipulation

Difficulty rating: 1280

Small Hint:

The third term uses two copies of the second binomial term

Big Hint:

Use the coefficient $\binom62$ and simplify the powers of $a$ and $x$

Solution:

The third term is $\begin{aligned} &\binom62\left(\frac a{\sqrt x}\right)^4 \left(-\frac{\sqrt x}{a^2}\right)^2\\ &\qquad=15\cdot\frac{a^4}{x^2}\cdot\frac{x}{a^4} =\frac{15}{x}. \end{aligned}$

Therefore, the correct answer is A.

8.

Let the two base angles of a triangle be $A$ and $B,$ with $B$ larger than $A.$ The altitude to the base divides the vertex angle $C$ into two parts, $C_1$ and $C_2,$ with $C_2$ adjacent to side $a.$ Then:

$C_1+C_2=A+B$

$C_1-C_2=B-A$

$C_1-C_2=A-B$

$C_1+C_2=B-A$

$C_1-C_2=A+B$

Concepts:altitude angle chasing angle sum

Difficulty rating: 1470

Small Hint:

The altitude creates two right triangles

Big Hint:

Express each part of $C$ as a complement of the opposite base angle

Solution:

The two right triangles give $\begin{aligned} A+C_1&=90^\circ,\\ B+C_2&=90^\circ. \end{aligned}$ Subtracting the second relation from the first gives $C_1-C_2=B-A.$

Thus, the correct answer is B.

9.

Let $r$ be the result of doubling both the base and the exponent of $a^b,$ $b\ne0.$ If $r$ equals the product of $a^b$ by $x^b,$ then $x$ equals:

$a$

$2a$

$4a$

$2$

$4$

Concepts:exponent algebraic manipulation

Difficulty rating: 1450

Small Hint:

Write the doubled expression as $(2a)^{2b}$

Big Hint:

Express both sides as a single quantity raised to the $b$ th power

Solution:

We have $r=(2a)^{2b}=(4a^2)^b$ and also $r=a^bx^b=(ax)^b.$ Thus $ax=4a^2,$ so $x=4a.$

Therefore, the correct answer is C.

10.

Each side of triangle $ABC$ is $12$ units. $D$ is the foot of the perpendicular dropped from $A$ on $BC,$ and $E$ is the midpoint of $AD.$ The length of $BE,$ in the same unit, is:

$\sqrt{18}$

$\sqrt{28}$

$6$

$\sqrt{63}$

$\sqrt{98}$

Concepts:equilateral triangle median (geometry)Pythagorean Theorem

Difficulty rating: 1340

Small Hint:

In the equilateral triangle, find $BD$ and $AD$

Big Hint:

Use right triangle $BDE$ after halving $AD$

Solution:

In the equilateral triangle, $BD=6$ and $AD=6\sqrt3.$ Hence $DE=3\sqrt3,$ and right triangle $BDE$ gives $\begin{aligned} BE^2&=6^2+(3\sqrt3)^2\\ &=36+27=63. \end{aligned}$ Therefore $BE=\sqrt{63}.$

Thus, the correct answer is D.

11.

Two tangents are drawn to a circle from an exterior point $A;$ they touch the circle at points $B$ and $C,$ respectively. A third tangent intersects segment $AB$ in $P$ and $AC$ in $R,$ and touches the circle at $Q.$ If $AB=20,$ then the perimeter of triangle $APR$ is:

$42$

$40.5$

$40$

$39\dfrac78$

not determined by the given information

Concepts:tangent line perimeter algebraic manipulation

Difficulty rating: 1440

Small Hint:

Tangent segments from the same exterior point have equal lengths

Big Hint:

Replace $PQ$ by $PB$ and $RQ$ by $RC$ in the perimeter

Solution:

Equal tangent segments give $PQ=PB$ and $RQ=RC.$ Thus $\begin{aligned} AP+PR+RA &=AP+PQ\\ &\quad+RQ+RA\\ &=AP+PB\\ &\quad+RC+RA\\ &=AB+AC. \end{aligned}$ Also $AB=AC=20,$ so the perimeter is $40.$

Therefore, the correct answer is C.

12.

The first three terms of a geometric progression are $\sqrt2,$ $\sqrt[3]2,$ $\sqrt[6]2.$ Find the fourth term.

$1$

$\sqrt[7]2$

$\sqrt[8]2$

$\sqrt[9]2$

$\sqrt[10]2$

Concepts:geometric sequence exponent

Difficulty rating: 1320

Small Hint:

Write each radical as a power of $2$

Big Hint:

Check the differences between the successive exponents

Solution:

The exponents are $\frac{1}{2},$ $\frac{1}{3},$ and $\frac{1}{6}.$ Each term is obtained by multiplying by $2^{-\frac{1}{6}},$ so the next exponent is $\frac{1}{6}-\frac{1}{6}=0.$ The fourth term is $2^0=1.$

Thus, the correct answer is A.

13.

The symbol $|a|$ means $a$ if $a$ is a positive number or zero, and $-a$ if $a$ is a negative number. For all real values of $t$ the expression $\sqrt{t^4+t^2}$ is equal to:

$t^3$

$t^2+t$

$|t^2+t|$

$t\sqrt{t^2+1}$

$|t|\sqrt{1+t^2}$

Concepts:radical absolute value factoring

Difficulty rating: 1320

Small Hint:

Factor $t^2$ from under the radical

Big Hint:

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

Solution:

Factoring inside the radical gives $\begin{aligned} \sqrt{t^2(t^2+1)} &=\sqrt{t^2}\sqrt{t^2+1}\\ &=|t|\sqrt{1+t^2}. \end{aligned}$

Therefore, the correct answer is E.

14.

A rhombus is given with one diagonal twice the length of the other diagonal. Express the side of the rhombus in terms of $K,$ where $K$ is the area of the rhombus in square inches.

$\sqrt K$

$\dfrac12\sqrt{2K}$

$\dfrac13\sqrt{3K}$

$\dfrac14\sqrt{4K}$

none of these are correct

Concepts:rhombus diagonal area Pythagorean Theorem

Difficulty rating: 1510

Small Hint:

Let the diagonals be $d$ and $2d$ , then use the rhombus area formula

Big Hint:

Half-diagonals form the legs of a right triangle whose hypotenuse is a side

Solution:

Let the diagonals be $d$ and $2d.$ Then $K=\frac{d(2d)}{2}=d^2.$ A side has length $\begin{aligned} \sqrt{\left(\frac d2\right)^2+d^2} &=\frac{\sqrt5}{2}d\\ &=\frac12\sqrt{5K}. \end{aligned}$ This is not among the first four choices.

Thus, the correct answer is E.

15.

If $x$ men working $x$ hours a day for $x$ days produce $x$ articles, then the number of articles (not necessarily an integer) produced by $y$ men working $y$ hours a day for $y$ days is:

$\dfrac{x^3}{y^2}$

$\dfrac{y^3}{x^2}$

$\dfrac{x^2}{y^3}$

$\dfrac{y^2}{x^3}$

$y$

Concepts:rate ratio and proportion algebraic manipulation

Difficulty rating: 1450

Small Hint:

Compute the production per man-hour from the first situation

Big Hint:

The second situation uses $y^3$ man-hours

Solution:

The first group uses $x^3$ man-hours to make $x$ articles, so the rate is $\frac{1}{x^2}$ article per man-hour. The second group supplies $y^3$ man-hours and therefore makes $\frac{y^3}{x^2}$ articles.

Thus, the correct answer is B.

16.

An altitude $h$ of a triangle is increased by a length $m.$ How much must be taken from the corresponding base $b$ so that the area of the new triangle is one-half that of the original triangle?

$\dfrac{bm}{h+m}$

$\dfrac{bh}{2h+2m}$

$\dfrac{b(2m+h)}{m+h}$

$\dfrac{b(m+h)}{2m+h}$

$\dfrac{b(2m+h)}{2(h+m)}$

Concepts:triangle area linear equation algebraic manipulation

Difficulty rating: 1510

Small Hint:

Let $t$ be the amount removed, so the new base is $b-t$

Big Hint:

Set $\frac12(b-t)(h+m)$ equal to half the original area

Solution:

If $t$ is removed, the new area condition is $\frac12(b-t)(h+m)=\frac12\left(\frac12bh\right).$ Hence $b-t=\frac{bh}{2(h+m)},$ so $\begin{aligned} t&=b-\frac{bh}{2(h+m)}\\ &=\frac{b(2m+h)}{2(h+m)}. \end{aligned}$

Therefore, the correct answer is E.

17.

In the base ten number system the number $526$ means $5\cdot10^2+2\cdot10+6.$ In the Land of Mathesis, however, numbers are written in the base $r.$ Jones purchases an automobile there for $440$ monetary units (abbreviated m.u.). He gives the salesman a $1000$ m.u. bill, and receives, in change, $340$ m.u. The base $r$ is:

$2$

$5$

$7$

$8$

$12$

Concepts:number base place value quadratic

Difficulty rating: 1490

Small Hint:

Translate $1000_r-440_r=340_r$ into powers of $r$

Big Hint:

Simplify the resulting quadratic and use a base larger than every digit shown

Solution:

In ordinary notation the transaction says $r^3-(4r^2+4r)=3r^2+4r.$ Thus $r(r^2-7r-8)=0,$ so $r=8$ or $r=-1.$ A numeral base must be positive and exceed $4,$ hence $r=8.$

Therefore, the correct answer is D.

18.

The yearly changes in the population census of a town for four consecutive years are, respectively, $25\%$ increase, $25\%$ increase, $25\%$ decrease, $25\%$ decrease. The net change over the four years, to the nearest percent, is:

$-12$

$-1$

$0$

$1$

$12$

Concepts:percentage exponent estimation

Difficulty rating: 1340

Small Hint:

Represent each increase by a factor of $1.25$ and each decrease by $0.75$

Big Hint:

Compare $1.25^2\cdot0.75^2$ with $1$

Solution:

The final population is multiplied by $(1.25)^2(0.75)^2=\frac{225}{256}\approx0.879.$ This is a decrease of about $12.1\%,$ which rounds to $12\%.$

Thus, the correct answer is A.

19.

Consider the graphs of $y=2\log x$ and $y=\log2x.$ We may say that:

they do not intersect

they intersect at $1$ point only

they intersect at $2$ points only

they intersect at a finite number of points but greater than $2$

they coincide

Concepts:logarithm factoring coordinate geometry

Difficulty rating: 1110

Small Hint:

Use $2\log x=\log(x^2)$ for $x>0$

Big Hint:

Equate the logarithm arguments and respect the logarithm’s domain

Solution:

On the domain $x>0,$ equality requires $\log(x^2)=\log(2x),$ so $x^2=2x.$ The only positive solution is $x=2,$ giving exactly one intersection.

Therefore, the correct answer is B.

20.

The set of points satisfying the pair of inequalities $y>2x$ and $y>4-x$ is contained entirely in quadrants:

$\mathrm{I}$ and $\mathrm{II}$

$\mathrm{II}$ and $\mathrm{III}$

$\mathrm{I}$ and $\mathrm{III}$

$\mathrm{III}$ and $\mathrm{IV}$

$\mathrm{I}$ and $\mathrm{IV}$

Concepts:inequality coordinate geometry

Difficulty rating: 1410

Small Hint:

Show that every feasible point has positive $y$

Big Hint:

Check whether feasible points can have either sign of $x$

Solution:

If $x\ge0,$ then $y>2x\ge0.$ If $x<0,$ then $y>4-x>0.$ Thus every feasible point is above the $x$ -axis, while both positive and negative $x$ -values occur. The region lies in quadrants $\mathrm{I}$ and $\mathrm{II}.$

Therefore, the correct answer is A.

21.

Medians $AD$ and $CE$ of triangle $ABC$ intersect in $M.$ The midpoint of $AE$ is $N.$ Let the area of triangle $MNE$ be $k$ times the area of triangle $ABC.$ Then $k$ equals:

$\dfrac16$

$\dfrac18$

$\dfrac19$

$\dfrac1{12}$

$\dfrac1{16}$

Concepts:median (geometry)centroid area ratio determinant

Difficulty rating: 1300

Small Hint:

Use coordinates or vectors with $A$ as the origin

Big Hint:

Points $E,$ $N,$ and $M$ divide the same two median directions by simple fractions

Solution:

Set $A=\mathbf0,$ $B=\mathbf b,$ and $C=\mathbf c.$ Then $\begin{aligned} E&=\frac{\mathbf b}{2},\\ N&=\frac{\mathbf b}{4},\\ M&=\frac{\mathbf b+\mathbf c}{3}. \end{aligned}$ A determinant calculation gives $[MNE]=\frac1{24} \left|\det(\mathbf b,\mathbf c)\right|,$ while $[ABC]=\frac12 \left|\det(\mathbf b,\mathbf c)\right|.$ Their ratio is $\frac{1}{12}.$

Thus, the correct answer is D.

22.

If $3x^3-9x^2+kx-12$ is divisible by $x-3,$ then it is also divisible by:

$3x^2-x+4$

$3x^2-4$

$3x^2+4$

$3x-4$

$3x+4$

Concepts:polynomial factoring substitution

Difficulty rating: 1180

Small Hint:

Apply the factor theorem at $x=3$ to determine $k$

Big Hint:

Then factor out $x-3$

Solution:

Let $P(x)=3x^3-9x^2+kx-12.$ Since $x-3$ is a factor, $P(3)=81-81+3k-12=0,$ so $k=4.$ Direct division gives $P(x)=(x-3)(3x^2+4).$

Therefore, the correct answer is C.

23.

Points $P$ and $Q$ are both in the line segment $AB$ and on the same side of its midpoint. $P$ divides $AB$ in the ratio $2:3,$ and $Q$ divides $AB$ in the ratio $3:4.$ If $PQ=2,$ then the length of $AB$ is:

$60$

$70$

$75$

$80$

$85$

Concepts:ratio and proportion fraction linear equation

Difficulty rating: 1390

Small Hint:

Express $\frac{AP}{AB}$ and $\frac{AQ}{AB}$ from the two division ratios

Big Hint:

Their difference is the fraction of $AB$ represented by $PQ$

Solution:

The division ratios give $\frac{AP}{AB}=\frac25$ and $\frac{AQ}{AB}=\frac37.$ Hence $PQ=(\frac{3}{7}-\frac{2}{5})AB=\frac{AB}{35}.$ Since $PQ=2,$ we obtain $AB=70.$

Thus, the correct answer is B.

24.

Thirty-one books are arranged from left to right in order of increasing prices. The price of each book differs by $\$2$ from that of each adjacent book. For the price of the book at the extreme right a customer can buy the middle book and an adjacent one. Then:

the adjacent book referred to is at the left of the middle book

the middle book sells for $\$36$

the cheapest book sells for $\$4$

the most expensive book sells for $\$64$

none of these are correct

Concepts:arithmetic sequence money casework

Difficulty rating: 1530

Small Hint:

Let the cheapest price be $p$ ; identify the $16$ th and $31$ st prices

Big Hint:

Test the two books adjacent to the middle book and reject any case forcing a nonpositive price

Solution:

The prices are $p,$ $p+2,$ $\ldots,$ $p+60,$ and the middle price is $p+30.$ If the right neighbor is used, then $(p+30)+(p+32)=p+60,$ which gives $p=-2,$ impossible for a selling price. The left neighbor gives $(p+30)+(p+28)=p+60,$ so $p=2.$ Thus the adjacent book must be the one to the left.

Therefore, the correct answer is A.

25.

Triangle $ABC$ is isosceles with base $AC.$ Points $P$ and $Q$ are respectively in $CB$ and $AB$ and such that $AC=AP=PQ=QB.$ The number of degrees in angle $B$ is:

$25\dfrac57$

$26\dfrac13$

$30$

$40$

not determined by the information given

Concepts:isosceles triangle angle chasing angle sum

Difficulty rating: 1780

Small Hint:

Let $\angle B=m$ and use each pair of equal segments to create isosceles triangles

Big Hint:

Chase the angles through triangles $BQP,$ $AQP,$ and $APC$

Solution:

Let $\angle B=m.$ Since $PQ=QB,$ triangle $BQP$ gives $\angle QPB=m$ and $\angle AQP=2m.$ Since $AP=PQ,$ triangle $AQP$ gives $\angle QAP=2m,$ so $\angle QPA=180^\circ-4m.$ As $B,P,C$ are collinear, $\angle APC=3m.$ Now $AP=AC$ gives $\angle ACP=3m.$ Finally $AB=BC$ makes both base angles of $ABC$ equal to $3m,$ and $m+3m+3m=180^\circ.$ Therefore $m=\frac{180^\circ}{7}=25\dfrac57^\circ.$

Thus, the correct answer is A.

26.

For a given arithmetic series the sum of the first $50$ terms is $200,$ and the sum of the next $50$ terms is $2700.$ The first term in the series is:

$-12\dfrac12$

$-21.5$

$-20.5$

$3$

$3.5$

Concepts:arithmetic sequence summation system of equations

Difficulty rating: 1280

Small Hint:

Write the first sum in terms of the first term $a$ and common difference $d$

Big Hint:

The next $50$ terms have first term $a+50d$

Solution:

The two sums give $\begin{aligned} 25(2a+49d)&=200,\\ 25(2a+149d)&=2700. \end{aligned}$ Thus $2a+49d=8$ and $2a+149d=108.$ Subtraction gives $d=1,$ and then $2a=-41,$ so $a=-20.5.$

Therefore, the correct answer is C.

27.

Given two equiangular polygons $P_1$ and $P_2$ with different numbers of sides; each angle of $P_1$ is $x$ degrees and each angle of $P_2$ is $kx$ degrees, where $k$ is an integer greater than $1.$ The number of possibilities for the pair $(x,k)$ is:

infinite

finite, but greater than two

two

one

zero

Concepts:equiangular polygon angle sum inequality

Difficulty rating: 1710

Small Hint:

An equiangular $n$ -gon has angle $180^\circ-\frac{360^\circ}{n}$

Big Hint:

Use $kx<180^\circ$ and $k\ge2$ to constrain the smaller angle $x$

Solution:

Every polygon angle is at least $60^\circ,$ and every convex polygon angle is less than $180^\circ.$ Since $k\ge2$ and $kx<180^\circ,$ we need $x<90^\circ.$ The only possible equiangular polygon angle in $[60^\circ,90^\circ)$ is the triangle angle $x=60^\circ.$ Then $k=2$ gives $kx=120^\circ,$ the angle of a regular hexagon; larger $k$ is impossible. Hence there is one pair.

Thus, the correct answer is D.

28.

If $2137^{753}$ is multiplied out, the units’ digit in the final product is:

$1$

$3$

$5$

$7$

$9$

Concepts:units digit modular exponentiation pattern recognition

Difficulty rating: 1320

Small Hint:

Only the units digit of the base matters

Big Hint:

List the four-term cycle of powers of $7$ modulo $10$

Solution:

Powers of $7$ have units digits $7,$ $9,$ $3,$ $1$ in a cycle of length $4.$ Since $753\equiv1\pmod4,$ the units digit of $2137^{753}$ is $7.$

Therefore, the correct answer is D.

29.

Let the roots of $ax^2+bx+c=0$ be $r$ and $s.$ The equation with roots $ar+b$ and $as+b$ is:

$x^2-bx-ac=0$

$x^2-bx+ac=0$

$x^2+3bx+ca+2b^2=0$

$x^2+3bx-ca+2b^2=0$

$\begin{gathered}x^2+bx(2-a)\\{}+a^2c+b^2(a+1)=0\end{gathered}$

Concepts:Vieta’s Formulas quadratic algebraic manipulation

Difficulty rating: 1500

Small Hint:

Use $r+s=-\frac{b}{a}$ and $rs=\frac{c}{a}$

Big Hint:

Compute the sum and product of $ar+b$ and $as+b$

Solution:

The transformed roots have sum $a(r+s)+2b=-b+2b=b$ and product $\begin{gathered} (ar+b)(as+b)\\ =a^2rs+ab(r+s)+b^2\\ =ac. \end{gathered}$ Therefore their monic equation is $x^2-bx+ac=0.$

Thus, the correct answer is B.

30.

If $\log_{10}2=a$ and $\log_{10}3=b,$ then $\log_5 12$ equals:

$\dfrac{a+b}{a+1}$

$\dfrac{2a+b}{a+1}$

$\dfrac{a+2b}{1+a}$

$\dfrac{2a+b}{1-a}$

$\dfrac{a+2b}{1-a}$

Concepts:logarithm algebraic manipulation

Difficulty rating: 1280

Small Hint:

Use the change-of-base formula with common logarithms

Big Hint:

Write $12=2^2\cdot3$ and $5=\frac{10}{2}$

Solution:

By change of base, $\begin{aligned} \log_5 12 &=\frac{\log12}{\log5}\\ &=\frac{2\log2+\log3} {\log10-\log2}\\ &=\frac{2a+b}{1-a}. \end{aligned}$

Therefore, the correct answer is D.

31.

In triangle $ABC$ the ratio $AC:CB$ is $3:4.$ The bisector of the exterior angle at $C$ intersects $BA$ extended at $P$ ( $A$ is between $P$ and $B$ ). The ratio $PA:AB$ is:

$1:3$

$3:4$

$4:3$

$3:1$

$7:1$

Concepts:angle bisector theorem ratio and proportion

Difficulty rating: 1300

Small Hint:

Apply the exterior angle bisector theorem to the full lengths $PA$ and $PB$

Big Hint:

Use $PB=PA+AB$

Solution:

The exterior angle bisector theorem gives $\frac{PA}{PB}=\frac{AC}{CB}=\frac34.$ Since $PB=PA+AB,$ write $PA=3t,$ $PB=4t.$ Then $AB=t,$ so $PA:AB=3:1.$

Thus, the correct answer is D.

32.

A regular polygon of $n$ sides is inscribed in a circle of radius $R.$ The area of the polygon is $3R^2.$ Then $n$ equals:

$8$

$10$

$12$

$15$

$18$

Concepts:regular polygon trigonometry area decomposition

Difficulty rating: 1300

Small Hint:

Divide the polygon into $n$ congruent triangles with vertex at the center

Big Hint:

Use area $\frac12nR^2\sin(\frac{360^\circ}{n})$ and test the choices

Solution:

The polygon’s area is $\frac n2R^2\sin\frac{360^\circ}{n}.$ For $n=12,$ this becomes $6R^2\sin30^\circ=3R^2,$ as required.

Therefore, the correct answer is C.

33.

The number of solutions of $2^{2x}-3^{2y}=55,$ in which $x$ and $y$ are integers, is:

$0$

$1$

$2$

$3$

more than three, but finite

Concepts:difference of squares Diophantine Equation factor

Difficulty rating: 1550

Small Hint:

Factor the left side as a difference of squares

Big Hint:

The two positive factors multiply to $55$ and have the same parity

Solution:

Factor: $(2^x-3^y)(2^x+3^y)=55.$ First, $x>0.$ If $y\le0,$ then $55<4^x\le56,$ which is impossible for an integral power of $4.$ Hence $y>0,$ so both factors are positive odd integers. The factor pair $5,$ $11$ gives $2^{x+1}=16$ and $2\cdot3^y=6,$ so $(x,y)=(3,1).$ The factor pair $1,$ $55$ would require $2^x=28,$ impossible. Hence there is exactly one solution.

Thus, the correct answer is B.

34.

Let $S$ be the set of values assumed by the fraction $\frac{2x+3}{x+2}$ when $x$ is any member of the interval $x\ge0.$ Let $M$ be the least upper bound of $S,$ and let $m$ be the greatest lower bound of $S.$ We may then say:

$m$ is in $S,$ but $M$ is not in $S$

$M$ is in $S,$ but $m$ is not in $S$

both $m$ and $M$ are in $S$

neither $m$ nor $M$ is in $S$

$M$ does not exist either in or outside $S$

Concepts:function inequality bounding to limit cases

Difficulty rating: 1500

Small Hint:

Rewrite the fraction as $2-\frac1{x+2}$

Big Hint:

Evaluate the lower endpoint and examine the limit as $x$ increases

Solution:

We have $\frac{2x+3}{x+2}=2-\frac1{x+2}.$ For $x\ge0,$ this increases from $\frac{3}{2}$ toward $2$ without reaching $2.$ Thus $S=[\frac32,2),$ so its greatest lower bound $m=\frac{3}{2}$ belongs to $S,$ while its least upper bound $M=2$ does not.

Therefore, the correct answer is A.

35.

The number $695$ is to be written with a factorial base of numeration, that is, $\begin{aligned} 695={}&a_1+a_2\cdot2!+a_3\cdot3!\\ &+\cdots+a_n\cdot n!, \end{aligned}$ where $a_1,$ $a_2,$ $a_3,$ $\ldots,$ $a_n$ are integers such that $0\le a_k\le k,$ and $n!$ means $n(n-1)(n-2)\cdots2\cdot1.$ Find $a_4.$

$0$

$1$

$2$

$3$

$4$

Concepts:number base factorial

Difficulty rating: 1500

Small Hint:

Begin with the largest factorial not exceeding $695$

Big Hint:

After removing the $5!$ contribution, divide the remainder by $4!$

Solution:

Since $5!=120,$ $695=5\cdot120+95.$ Next $4!=24,$ and $95=3\cdot24+23.$ Therefore the coefficient $a_4$ is $3.$

Thus, the correct answer is D.

36.

In triangle $ABC$ the median from $A$ is perpendicular to the median from $B.$ If $BC=7$ and $AC=6,$ find the length of $AB.$

$4$

$\sqrt{17}$

$4.25$

$2\sqrt5$

$4.5$

Concepts:median (geometry)vector coordinate geometry

Difficulty rating: 1710

Small Hint:

Put $A$ at the origin and represent $B$ and $C$ by vectors

Big Hint:

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

Solution:

Let $A=\mathbf0,$ $B=\mathbf b,$ and $C=\mathbf c,$ with $|\mathbf b|=\ell,$ $|\mathbf c|=6,$ and $|\mathbf c-\mathbf b|=7.$ The median directions are $\frac{\mathbf b+\mathbf c}{2}$ from $A$ and $\frac{\mathbf c}{2}-\mathbf b$ from $B.$ Their dot product is zero, so $(\mathbf b+\mathbf c)\mathbin{\cdot}(\mathbf c-2\mathbf b)=0.$ Using $\mathbf b\mathbin{\cdot}\mathbf c=\frac{\ell^2+36-49}{2},$ this simplifies to $\ell^2=17.$ Hence $AB=\sqrt{17}.$

Therefore, the correct answer is B.

37.

In racing over a distance $d$ at uniform speed, $A$ can beat $B$ by $20$ yards, $B$ can beat $C$ by $10$ yards, and $A$ can beat $C$ by $28$ yards. Then $d,$ in yards, equals:

not determined by the given information

$58$

$100$

$116$

$120$

Concepts:distance rate and time ratio and proportion rational equation

Difficulty rating: 1550

Small Hint:

Translate each winning margin into a ratio of speeds

Big Hint:

Multiply the ratios $\frac{v_B}{v_A}$ and $\frac{v_C}{v_B}$ to get $\frac{v_C}{v_A}$

Solution:

The margins give $\begin{aligned} \frac{v_B}{v_A}&=\frac{d-20}{d},\\ \frac{v_C}{v_B}&=\frac{d-10}{d},\\ \frac{v_C}{v_A}&=\frac{d-28}{d}. \end{aligned}$ Therefore $\frac{(d-20)(d-10)}{d^2}=\frac{d-28}{d}.$ Expanding and simplifying yields $2d=200,$ so $d=100.$

Thus, the correct answer is C.

38.

Triangle $ABC$ is inscribed in a semicircle of radius $r$ so that its base $AB$ coincides with diameter $AB.$ Point $C$ does not coincide with either $A$ or $B.$ Let $s=AC+BC.$ Then, for all permissible positions of $C:$

$s^2\le8r^2$

$s^2=8r^2$

$s^2\ge8r^2$

$s^2\le4r^2$

$s^2=4r^2$

Concepts:inscribed angle right triangle Cauchy-Schwarz Inequality

Difficulty rating: 1300

Small Hint:

The angle subtended by the diameter is a right angle

Big Hint:

If the legs are $p$ and $q,$ compare $(p+q)^2$ with $2(p^2+q^2)$

Solution:

By Thales’ theorem, $ABC$ is right at $C.$ Put $p=AC$ and $q=BC.$ Then $p^2+q^2=AB^2=4r^2.$ Since $(p-q)^2\ge0,$ we have $2pq\le p^2+q^2.$ Therefore $s^2=(p+q)^2\le2(p^2+q^2)=8r^2.$

Thus, the correct answer is A.

39.

Any five points are taken inside or on a square with side length $1.$ Let $a$ be the smallest possible number with the property that it is always possible to select one pair of points from these five such that the distance between them is equal to or less than $a.$ Then $a$ is:

$\dfrac{\sqrt3}{3}$

$\dfrac{\sqrt2}{2}$

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

$1$

$\sqrt2$

Concepts:pigeonhole principle extremal argument square (geometry)

Difficulty rating: 1670

Small Hint:

Partition the unit square into four congruent smaller squares

Big Hint:

To prove sharpness, look for five points whose closest-pair distance reaches the bound

Solution:

Divide the square into four squares of side $\frac{1}{2}.$ Two of the five points lie in the same small square, so their distance is at most its diagonal, $\frac{\sqrt2}{2}.$ This bound is attainable by placing points at the four corners and the center: the shortest distance is then $\frac{\sqrt2}{2}.$ Hence the least guaranteed value is $\frac{\sqrt2}{2}.$

Therefore, the correct answer is B.

40.

Find the minimum value of $\sqrt{x^2+y^2}$ if $5x+12y=60.$

$\dfrac{60}{13}$

$\dfrac{13}{5}$

$\dfrac{13}{12}$

$1$

$0$

Concepts:distance formula Cauchy-Schwarz Inequality optimization

Difficulty rating: 1300

Small Hint:

Interpret $\sqrt{x^2+y^2}$ as distance from the origin

Big Hint:

Apply Cauchy-Schwarz to $5x+12y$

Solution:

By Cauchy-Schwarz, $\begin{aligned} 60&=5x+12y\\ &\le\sqrt{5^2+12^2}\\ &\qquad\cdot\sqrt{x^2+y^2}\\ &=13\sqrt{x^2+y^2}. \end{aligned}$ Thus the distance is at least $\frac{60}{13}.$ Equality occurs when $(x,y)$ is proportional to $(5,12),$ so the minimum is $\frac{60}{13}.$

Therefore, the correct answer is A.