1975 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

The value of $\frac{1}{2-\dfrac{1}{2-\dfrac{1}{2-\frac12}}}$ is

$\frac34$

$\frac45$

$\frac56$

$\frac67$

$\frac65$

Concepts:fraction order of operations

Difficulty rating: 1280

Small Hint:

Evaluate the nested denominator from the inside outward

Big Hint:

The three successive inner values are $\frac32,\frac43,$ and $\frac54$

Solution:

Working outward, $\begin{aligned} 2-\frac12&=\frac32,\\ 2-\frac1{\frac{3}{2}}&=\frac43,\\ 2-\frac1{\frac{4}{3}}&=\frac54. \end{aligned}$ The given value is therefore $\frac{1}{\frac{5}{4}}=\frac{4}{5}.$

Therefore, the correct answer is B.

2.

For which real values of $m$ are the simultaneous equations $\begin{aligned} y&=mx+3,\\ y&=(2m-1)x+4 \end{aligned}$ satisfied by at least one pair of real numbers $(x,y)?$

all $m$

all $m\ne0$

all $m\ne\frac12$

all $m\ne1$

no values of $m$

Concepts:linear equation system of equations slope

Difficulty rating: 1380

Small Hint:

Two nonvertical lines fail to meet only when they are parallel and distinct

Big Hint:

Set the slopes $m$ and $2m-1$ equal

Solution:

The lines fail to intersect only if $m=2m-1,$ which gives $m=1.$ Their $y$ -intercepts are $3$ and $4,$ so at that value they are distinct parallel lines. Every other $m$ gives one intersection.

Therefore, the correct answer is D.

3.

Which of the following inequalities are satisfied for all real numbers $a,$ $b,$ $c,$ $x,$ $y,$ $z$ which satisfy $x\lt a,$ $y\lt b,$ and $z\lt c?$

I. $xy+yz+zx\lt ab+bc+ca$

II. $x^2+y^2+z^2\lt a^2+b^2+c^2$

III. $xyz\lt abc$

None are satisfied.

$\mathrm{I}$ only

$\mathrm{II}$ only

$\mathrm{III}$ only

All are satisfied.

Concepts:inequality counterexample

Difficulty rating: 1710

Small Hint:

The variables are real, so increasing a number need not increase its square or its products

Big Hint:

Try $a=b=1,\ c=-1,$ $x=y=0,\ z=-10$

Solution:

Take $a=b=1,\ c=-1,$ and $x=y=0,\ z=-10.$ The required three coordinate inequalities hold. But I becomes $0\lt-1,$ II becomes $100\lt3,$ and III becomes $0\lt-1,$ all false. Thus none is universally true.

Therefore, the correct answer is A.

4.

If the side of one square is the diagonal of a second square, what is the ratio of the area of the first square to the area of the second?

$2$

$\sqrt2$

$\frac12$

$2\sqrt2$

$4$

Concepts:square (geometry)diagonal area ratio

Difficulty rating: 1210

Small Hint:

If the first side is $s,$ then the second square has diagonal $s$

Big Hint:

A square with diagonal $s$ has area $\frac{s^2}{2}$

Solution:

Let the first square have side $s.$ The second has diagonal $s,$ so its side is $\frac{s}{\sqrt2}$ and its area is $\frac{s^2}{2}.$ The ratio is $\frac{s^2}{\frac{s^2}{2}}=2.$

Therefore, the correct answer is A.

5.

The polynomial $(x+y)^9$ is expanded in decreasing powers of $x.$ The second and third terms have equal values when evaluated at $x=p$ and $y=q,$ where $p$ and $q$ are positive numbers whose sum is one. What is the value of $p?$

$\frac15$

$\frac45$

$\frac14$

$\frac34$

$\frac89$

Concepts:binomial theorem linear equation

Difficulty rating: 1590

Small Hint:

Write the second and third binomial terms explicitly

Big Hint:

After canceling common positive factors, combine $p=4q$ with $p+q=1$

Solution:

The terms are $9p^8q$ and $36p^7q^2.$ Their equality, with $p,q\gt0,$ gives $p=4q.$ Since $p+q=1,$ we get $5q=1$ and $p=\frac{4}{5}.$

Therefore, the correct answer is B.

6.

The sum of the first eighty positive odd integers subtracted from the sum of the first eighty positive even integers is

$0$

$20$

$40$

$60$

$80$

Concepts:arithmetic sequence pairing and grouping

Difficulty rating: 1030

Small Hint:

Pair the $k$ th even number with the $k$ th odd number

Big Hint:

Each difference $2k-(2k-1)$ is the same

Solution:

The desired difference is $\begin{aligned} &(2-1)+(4-3)+\cdots\\ &\qquad +(160-159). \end{aligned}$ It is the sum of $80$ ones, hence $80.$

Therefore, the correct answer is E.

7.

For which nonzero real numbers $x$ is $\frac{|x-|x||}{x}$ a positive integer?

for negative $x$ only

for positive $x$ only

only for $x$ an even integer

for all nonzero real numbers $x$

for no nonzero real numbers $x$

Concepts:absolute value casework

Difficulty rating: 1210

Small Hint:

Evaluate the quotient separately for $x\gt0$ and $x\lt0$

Big Hint:

For $x\lt0,$ replace $|x|$ by $-x$ before evaluating the outer absolute value

Solution:

If $x\gt0,$ the quotient is $0.$ If $x\lt0,$ then $x-|x|=2x\lt0,$ so $|x-|x||=-2x$ and the quotient is $-2.$ Neither value is a positive integer.

Therefore, the correct answer is E.

8.

If the statement “All shirts in this store are on sale.” is false, then which of the following statements must be true?

I. All shirts in this store are at non-sale prices.

II. There is some shirt in this store not on sale.

III. No shirt in this store is on sale.

IV. Not all shirts in this store are on sale.

Note: Originally, statement I read: All shirts in this store are not on sale.

$\mathrm{II}$ only

$\mathrm{IV}$ only

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

$\mathrm{II}$ and $\mathrm{IV}$ only

$\mathrm{I},$ $\mathrm{II}$ and $\mathrm{IV}$ only

Concepts:logical deduction

Difficulty rating: 1380

Small Hint:

Negating “all” asserts the existence of a counterexample

Big Hint:

The false statement does not say that every shirt fails to be on sale

Solution:

The negation of “Every shirt is on sale” is “Some shirt is not on sale,” which is II and is equivalently phrased by IV. Statements I and III make the stronger claim that no shirt is on sale, which need not follow.

Therefore, the correct answer is D.

9.

Let $a_1,$ $a_2,$ $\ldots$ and $b_1,$ $b_2,$ $\ldots$ be arithmetic progressions such that $a_1=25,$ $b_1=75,$ and $a_{100}+b_{100}=100.$ Find the sum of the first one hundred terms of the progression $a_1+b_1,$ $a_2+b_2,$ $\ldots.$

$0$

$100$

$10{,}000$

$505{,}000$

not enough information given to solve the problem

Concepts:arithmetic sequence summation

Difficulty rating: 1430

Small Hint:

The termwise sum of two arithmetic progressions is arithmetic

Big Hint:

Its first and hundredth terms are both $100$

Solution:

Let $c_i=a_i+b_i.$ This is an arithmetic progression with $c_1=25+75=100$ and $c_{100}=100.$ Equal endpoint terms force its common difference to be $0,$ so all $100$ terms equal $100.$ Their sum is $10{,}000.$

Therefore, the correct answer is C.

10.

The sum of the digits in base ten of $\left(10^{4n^2+8}+1\right)^2,$ where $n$ is a positive integer, is

$4$

$4n$

$2+2n$

$4n^2$

$n^2+n+2$

Concepts:place value exponent binomial theorem

Difficulty rating: 1440

Small Hint:

Set $k=4n^2+8$ and square $10^k+1$

Big Hint:

The three nonzero digits occur in distinct places, so there are no carries

Solution:

For $k=4n^2+8,$ $(10^k+1)^2=10^{2k}+2\cdot10^k+1.$ Its only nonzero digits are $1,2,1,$ so their sum is $4.$

Therefore, the correct answer is A.

11.

Let $P$ be an interior point of circle $K$ other than the center of $K.$ Form all chords of $K$ which pass through $P,$ and determine their midpoints. The locus of these midpoints is

a circle with one point deleted

a circle if the distance from $P$ to the center of $K$ is less than one half the radius of $K;$ otherwise a circular arc of less than $360^\circ$

a semicircle with one point deleted

a semicircle

a circle

Concepts:circle chord midpoint

Difficulty rating: 1850

Small Hint:

Join a chord midpoint $M$ to the circle’s center $O$

Big Hint:

Since $OM$ is perpendicular to the chord through $P,$ $\angle OMP=90^\circ$

Solution:

If $M$ is a chord midpoint and $O$ is the center, then $OM$ is perpendicular to the chord, so $\angle OMP=90^\circ$ unless $M=O.$ Thus $M$ lies on the circle with diameter $OP.$ Conversely, each nonendpoint $M$ on that circle gives $PM\perp OM,$ so the line $PM$ cuts a chord of $K$ whose midpoint is $M.$ The endpoint $M=P$ gives the chord through $P$ perpendicular to $OP,$ while $M=O$ gives the diameter through $O$ and $P.$

Therefore, the correct answer is E.

12.

If $a\ne b,$ $a^3-b^3=19x^3,$ and $a-b=x,$ which of the following conclusions is correct?

$a=3x$

$a=3x$ or $a=-2x$

$a=-3x$ or $a=2x$

$a=3x$ or $a=2x$

$a=2x$

Concepts:sum and difference of cubes factoring quadratic

Difficulty rating: 1620

Small Hint:

Factor $a^3-b^3$ and use $x=a-b\ne0$

Big Hint:

Substitute $b=a-x$ after dividing by $x$

Solution:

Factoring and dividing by $x\ne0$ gives $a^2+ab+b^2=19x^2.$ With $b=a-x,$ $3a^2-3ax+x^2=19x^2,$ so $\begin{aligned} a^2-ax-6x^2&=0,\\ (a-3x)(a+2x)&=0. \end{aligned}$ Hence $a=3x$ or $a=-2x.$

Therefore, the correct answer is B.

13.

The equation $x^6-3x^5-6x^3-x+8=0$ has

no real roots

exactly two distinct negative roots

exactly one negative root

no negative roots, but at least one positive root

none of these

Concepts:polynomial inequality

Difficulty rating: 1560

Small Hint:

Determine the sign of every term when $x\lt0$

Big Hint:

Compare the polynomial’s values at $0$ and $1$

Solution:

For $x\lt0,$ each of $x^6,-3x^5,-6x^3,-x,$ and $8$ is positive, so there is no negative root. At $x=0$ the polynomial is $8,$ while at $x=1$ it is $-1.$ Continuity therefore gives a positive root between $0$ and $1.$

Therefore, the correct answer is D.

14.

If the whatsis is so when the whosis is is and the so and so is is $\cdot$ so, what is the whosis $\cdot$ whatsis when the whosis is so, the so and so is so $\cdot$ so, and the is is two (whatsis, whosis, is and so are variables taking positive values)?

whosis $\cdot$ is $\cdot$ so

whosis

is

so

so and so

Concepts:substitution algebraic manipulation logical deduction

Difficulty rating: 1810

Small Hint:

Replace whatsis, whosis, is, and so by $W,H,I,S$

Big Hint:

The condition $S+S=IS$ and positivity force $I=2$

Solution:

The first condition says that $H=I$ and $2S=IS$ imply $W=S.$ Since $S\gt0,$ the latter equation gives $I=2.$ In the requested case, $H=S,\ 2S=S^2,$ and $I=2,$ so positivity gives $H=I=S=2.$ Hence $W=S=2$ and $HW=4=S+S.$

Therefore, the correct answer is E.

15.

In the sequence of numbers $1,$ $3,$ $2,$ $\ldots,$ each term after the first two is equal to the term preceding it minus the term preceding that. The sum of the first one hundred terms of the sequence is

$5$

$4$

$2$

$1$

$-1$

Concepts:recursion pattern recognition summation

Difficulty rating: 1570

Small Hint:

Generate terms until the initial pair $1,3$ returns

Big Hint:

The six-term period has sum $0$

Solution:

The sequence begins $1,3,2,-1,-3,-2,1,3,\ldots$ and repeats every $6$ terms, with period sum $0.$ The first $96$ terms sum to $0,$ and the last four sum to $1+3+2-1=5.$

Therefore, the correct answer is A.

16.

If the first term of an infinite geometric series is a positive integer, the common ratio is the reciprocal of a positive integer, and the sum of the series is $3,$ then the sum of the first two terms of the series is

$\frac13$

$\frac23$

$\frac83$

$2$

$\frac92$

Concepts:geometric sequence divisibility summation

Difficulty rating: 1740

Small Hint:

Write the first term as $a$ and the ratio as $\frac{1}{n}$

Big Hint:

From $\frac{a}{1-\frac{1}{n}}=3,$ use that $a,n$ are positive integers and $n\gt1$

Solution:

Let the first term be $a$ and the ratio $\frac{1}{n}.$ Convergence gives $n\gt1,$ and $\frac{a}{1-\frac{1}{n}}=3,\qquad a=3-\frac3n.$ Thus $n$ divides $3,$ so $n=3$ and $a=2.$ The first two terms sum to $2+\frac{2}{3}=\frac{8}{3}.$

Therefore, the correct answer is C.

17.

A man can commute either by train or by bus. If he goes to work on the train in the morning, he comes home on the bus in the afternoon; and if he comes home in the afternoon on the train, he took the bus in the morning. During a total of $x$ working days, the man took the bus to work in the morning $8$ times, came home by bus in the afternoon $15$ times, and commuted by train (either morning or afternoon) $9$ times. Find $x.$

$19$

$18$

$17$

$16$

not enough information given to solve the problem

Concepts:basic counting

Difficulty rating: 1380

Small Hint:

There are exactly two one-way trips on every working day

Big Hint:

Count all bus trips and all train trips together

Solution:

There were $8+15=23$ bus trips and $9$ train trips, hence $32$ one-way trips in all. Since each working day contributes two trips, $2x=32$ and $x=16.$ The conditional statements are consistent but not needed for this count.

Therefore, the correct answer is D.

18.

A positive integer $N$ with three digits in its base ten representation is chosen at random, with each three-digit number having an equal chance of being chosen. The probability that $\log_2N$ is an integer is

$0$

$\frac3{899}$

$\frac1{225}$

$\frac1{300}$

$\frac1{450}$

Concepts:basic probability power of 2 counting integers in a range

Difficulty rating: 1340

Small Hint:

An integral base- $2$ logarithm means $N$ is a power of $2$

Big Hint:

List the powers of $2$ from $100$ through $999$

Solution:

There are $999-100+1=900$ three-digit integers. The three powers of $2$ among them are $128,256,$ and $512.$ Thus the probability is $\frac{3}{900}=\frac{1}{300}.$

Therefore, the correct answer is D.

19.

Which positive numbers $x$ satisfy the equation $(\log_3x)(\log_x5)=\log_35?$

$3$ and $5$ only

$3,$ $5$ and $15$ only

only numbers of the form $5^n\cdot3^m,$ where $n$ and $m$ are positive integers

all positive $x\ne1$

none of these

Concepts:logarithm algebraic manipulation

Difficulty rating: 1450

Small Hint:

Apply the change-of-base formula to both logarithms on the left

Big Hint:

After cancellation, remember which positive base is forbidden

Solution:

For $x\gt0,\ x\ne1,$ change of base makes the left side $\frac{\log x}{\log3}\cdot\frac{\log5}{\log x}.$ Canceling $\log x$ leaves $\frac{\log5}{\log3}=\log_35.$ The expression is undefined at $x=1.$

Therefore, the correct answer is D.

20.

In the adjoining figure triangle $ABC$ is such that $AB=4$ and $AC=8.$ If $M$ is the midpoint of $BC$ and $AM=3,$ what is the length of $BC?$

$2\sqrt{26}$

$2\sqrt{31}$

$9$

$4+2\sqrt{13}$

not enough information given to solve the problem

Concepts:median (geometry)Stewart’s Theorem

Difficulty rating: 1670

Small Hint:

Use the relation between a triangle’s three sides and a median

Big Hint:

Apollonius’ theorem gives $AB^2+AC^2=2(AM^2+BM^2)$

Solution:

Let $BM=CM=x,$ so $BC=2x.$ By Apollonius’ theorem, $4^2+8^2=2(3^2+x^2).$ Hence $80=18+2x^2,\ x^2=31,$ and $BC=2\sqrt{31}.$

Therefore, the correct answer is B.

21.

Suppose $f(x)$ is defined for all real numbers $x;$ $f(x)\gt0$ for all $x;$ and $f(a)f(b)=f(a+b)$ for all $a$ and $b.$ Which of the following statements are true?

I. $f(0)=1$

II. $f(-a)=\frac{1}{f(a)}$ for all $a$

III. $f(a)=\sqrt[3]{f(3a)}$ for all $a$

IV. $f(b)\gt f(a)$ if $b\gt a$

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

$\mathrm{I},$ $\mathrm{III}$ and $\mathrm{IV}$ only

$\mathrm{I},$ $\mathrm{II}$ and $\mathrm{IV}$ only

$\mathrm{I},$ $\mathrm{II}$ and $\mathrm{III}$ only

All are true.

Concepts:functional equation substitution counterexample

Difficulty rating: 1670

Small Hint:

Substitute $a=0,\ b=-a,$ and then add $a$ three times

Big Hint:

Test the monotonicity claim with the constant function $f(x)=1$

Solution:

Taking $a=0$ and using positivity gives $f(0)=1.$ Taking $b=-a$ gives $f(a)f(-a)=1.$ Also $f(3a)=f(a)^3,$ so positivity permits the positive cube root in III. But $f(x)=1$ satisfies the functional equation and is not strictly increasing, so IV need not hold.

Therefore, the correct answer is D.

22.

If $p$ and $q$ are primes and $x^2-px+q=0$ has distinct positive integral roots, then which of the following statements are true?

I. The difference of the roots is odd.

II. At least one root is prime.

III. $p^2-q$ is prime.

IV. $p+q$ is prime.

$\mathrm{I}$ only

$\mathrm{II}$ only

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

$\mathrm{I},$ $\mathrm{II}$ and $\mathrm{IV}$ only

All are true.

Concepts:Vieta’s Formulas prime factor

Difficulty rating: 1830

Small Hint:

The product of the two positive integer roots is the prime $q$

Big Hint:

The roots must be $1,q,$ and their prime sum forces $q=2$

Solution:

The positive integer roots have product $q,$ so they are $1$ and $q.$ Their sum is $p=q+1,$ which is prime only when $q=2,$ giving $p=3$ and roots $1,2.$ Their difference is $1,$ one root is prime, $p^2-q=7,$ and $p+q=5;$ therefore all four statements hold.

Therefore, the correct answer is E.

23.

In the adjoining figure $AB$ and $BC$ are adjacent sides of square $ABCD;$ $M$ is the midpoint of $AB;$ $N$ is the midpoint of $BC;$ and $AN$ and $CM$ intersect at $O.$ The ratio of the area of $AOCD$ to the area of $ABCD$ is

$\frac56$

$\frac34$

$\frac23$

$\frac{\sqrt3}{2}$

$\frac{\sqrt3-1}{2}$

Concepts:centroid square (geometry)area decomposition

Difficulty rating: 1670

Small Hint:

In triangle $ABC,$ both $AN$ and $CM$ are medians

Big Hint:

Use coordinates for the square and locate the centroid $O$

Solution:

Take $A=(0,0),B=(1,0),$ $C=(1,1),D=(0,1).$ Since $O$ is the centroid of triangle $ABC,$ $O=(\frac{2}{3},\frac{1}{3}).$ Triangles $AOB$ and $COB$ each have area $\frac{1}{6}.$ Removing them from the unit square leaves $[AOCD]=1-\frac16-\frac16=\frac23.$

Therefore, the correct answer is C.

24.

In triangle $ABC,$ $\angle C=\theta$ and $\angle B=2\theta,$ where $0^\circ\lt\theta\lt60^\circ.$ The circle with center $A$ and radius $AB$ intersects $AC$ at $D$ and intersects $BC,$ extended if necessary, at $B$ and at $E$ ( $E$ may coincide with $B$ ). Then $EC=AD$

for no values of $\theta$

only if $\theta=45^\circ$

only if $0^\circ\lt\theta\le45^\circ$

only if $45^\circ\le\theta\lt60^\circ$

for all $\theta$ such that $0^\circ\lt\theta\lt60^\circ$

Concepts:circle isosceles triangle angle chasing

Difficulty rating: 2220

Small Hint:

Because $AB=AE=AD,$ first study the isosceles triangle $ABE$

Big Hint:

Whether $E$ lies on $BC$ or its extension, angle chasing gives $\angle EAC=\angle ECA=\theta$

Solution:

Since $AB=AE,$ triangle $ABE$ is isosceles. For $0^\circ\lt\theta\lt45^\circ,$ $E$ lies between $B,C$ ; the exterior-angle theorem in triangle $AEC$ gives $2\theta=\angle EAC+\theta,$ so $\angle EAC=\theta.$ For $45^\circ\lt\theta\lt60^\circ,$ $E$ lies beyond $B$ ; then $\angle AEC=180^\circ-2\theta,$ hence $\begin{aligned} \angle EAC &=180^\circ\\ &\quad-(180^\circ-2\theta)-\theta\\ &=\theta. \end{aligned}$ At $\theta=45^\circ,\ E=B$ and the conclusion is immediate. Thus triangle $AEC$ is always isosceles, so $EC=EA=AD.$

Therefore, the correct answer is E.

25.

A woman, her brother, her son and her daughter are chess players (all relations by birth). The worst player’s twin (who is one of the four players) and the best player are of opposite sex. The worst player and the best player are the same age. Who is the worst player?

the woman

her son

her brother

her daughter

No solution is consistent with the given information.

Concepts:logical deduction casework ages

Difficulty rating: 1740

Small Hint:

List the only possible same-generation twin pairs among the four people

Big Hint:

If the son is worst, his twin is the daughter and the best can be the brother

Solution:

If the son is worst, the daughter can be his twin and the brother can be best; the son and brother may have the same age, so all conditions can hold. If the woman is worst, the brother is her twin and the daughter must be best, but mother and daughter cannot be the same age. If the brother is worst, the woman is his twin and the son must be best, again impossible in age. If the daughter is worst, her twin is the son and the woman must be best, also impossible in age. Thus only the son works.

Therefore, the correct answer is B.

26.

In acute triangle $ABC$ the bisector of $\angle A$ meets side $BC$ at $D.$ The circle with center $B$ and radius $BD$ intersects side $AB$ at $M;$ and the circle with center $C$ and radius $CD$ intersects side $AC$ at $N.$ Then it is always true that

$\begin{aligned}\angle CND+\angle BMD\\{}-\angle DAC=120^\circ\end{aligned}$

$AMDN$ is a trapezoid

$BC$ is parallel to $MN$

$AM-AN=\frac{3(DB-DC)}2$

$AB-AC=\frac{3(DB-DC)}2$

Concepts:angle bisector theorem ratio and proportion parallel lines

Difficulty rating: 1770

Small Hint:

Use $BM=BD,\ CN=CD,$ and the angle bisector theorem

Big Hint:

Show that $\frac{BM}{CN}=\frac{AB}{AC}$ and apply the converse of the side-splitter theorem

Solution:

The angle bisector theorem gives $\frac{BD}{CD}=\frac{AB}{AC}.$ Since $BM=BD$ and $CN=CD,$ we have $\frac{BM}{CN}=\frac{AB}{AC}.$ Therefore $M$ and $N$ divide $AB$ and $AC$ proportionally from $B$ and $C,$ so the converse of the side-splitter theorem gives $MN\parallel BC.$

Therefore, the correct answer is C.

27.

If $p,$ $q,$ and $r$ are distinct roots of $x^3-x^2+x-2=0,$ then $p^3+q^3+r^3$ equals

$-1$

$1$

$3$

$5$

none of these

Concepts:polynomial Vieta’s Formulas Newton’s Sums

Difficulty rating: 1960

Small Hint:

Use Vieta to find $p+q+r$ and $pq+pr+qr$

Big Hint:

First compute $p^2+q^2+r^2,$ then add the three equations satisfied by the roots

Solution:

Vieta gives $p+q+r=1$ and $pq+pr+qr=1.$ Thus $\begin{aligned} p^2+q^2+r^2 &=(p+q+r)^2\\ &\quad-2(pq+pr+qr)\\ &=-1. \end{aligned}$ Each root $t$ satisfies $t^3=t^2-t+2.$ Summing over the three roots yields $p^3+q^3+r^3=-1-1+6=4.$ This is not among choices A–D.

Therefore, the correct answer is E.

28.

In triangle $ABC$ shown in the adjoining figure, $M$ is the midpoint of side $BC,$ $AB=12,$ and $AC=16.$ Points $E$ and $F$ are taken on $AC$ and $AB,$ respectively, and lines $EF$ and $AM$ intersect at $G.$ If $AE=2AF,$ then $\frac{EG}{GF}$ equals

$\frac32$

$\frac43$

$\frac54$

$\frac65$

not enough information given to solve the problem

Concepts:vector median (geometry)ratio and proportion

Difficulty rating: 2180

Small Hint:

Represent $B,C$ by vectors $\mathbf b,\mathbf c$ from $A$

Big Hint:

If $F=t\mathbf b,$ then $E=\frac32t\mathbf c;$ equate the $\mathbf b,\mathbf c$ coefficients of a point on $EF$ and on $AM$

Solution:

Put $A$ at the origin and write the position vectors of $B,C$ as $\mathbf b,\mathbf c.$ If $F=t\mathbf b,$ then $AF=12t$ and $AE=24t,$ so $E=(\frac{3t}{2})\mathbf c.$ Write $G=(1-u)F+uE.$ Because $G$ also lies on the median $AM,$ its $\mathbf b$ and $\mathbf c$ coefficients are equal: $(1-u)t=\frac32ut.$ Hence $u=\frac{2}{5}.$ Along $FE,\frac{FG}{FE}=\frac{2}{5}$ and $\frac{EG}{FE}=\frac{3}{5},$ so $\frac{EG}{GF}=\frac{3}{2}.$

Therefore, the correct answer is A.

29.

What is the smallest integer larger than $(\sqrt3+\sqrt2)^6?$

$972$

$971$

$970$

$969$

$968$

Concepts:radical binomial theorem floor and ceiling functions

Difficulty rating: 1980

Small Hint:

Pair the expression with its conjugate $(\sqrt3-\sqrt2)^6$

Big Hint:

Their sum is an integer, while the conjugate term lies strictly between $0$ and $1$

Solution:

Let $\alpha=\sqrt3+\sqrt2$ and $\beta=\sqrt3-\sqrt2.$ Adding their sixth powers cancels all odd radical terms: $\begin{aligned} \alpha^6+\beta^6 &=2\bigl(3^3+15(3^2)(2)\\ &\qquad+15(3)(2^2)+2^3\bigr)\\ &=970. \end{aligned}$ Since $0\lt\beta\lt1,$ we have $969\lt\alpha^6=970-\beta^6\lt970.$ Thus the smallest larger integer is $970.$

Therefore, the correct answer is C.

30.

Let $x=\cos36^\circ-\cos72^\circ.$ Then $x$ equals

$\frac13$

$\frac12$

$3-\sqrt6$

$2\sqrt3-3$

none of these

Concepts:trigonometric identity algebraic manipulation

Difficulty rating: 1670

Small Hint:

Let $w=\cos36^\circ$ and $y=\cos72^\circ$

Big Hint:

Use $y=2w^2-1$ and $w=1-2y^2$ , then add the equations

Solution:

Set $w=\cos36^\circ$ and $y=\cos72^\circ.$ The double-angle identities give $y=2w^2-1,\qquad w=1-2y^2.$ Adding yields $\begin{aligned} w+y&=2(w^2-y^2)\\ &=2(w-y)(w+y). \end{aligned}$ Since $w+y\ne0,$ division gives $w-y=\frac{1}{2}.$ Thus $x=\frac{1}{2}.$

Therefore, the correct answer is B.