If (a+b)/(b+c)=(c+d)/(d+a), then:

(C) either a=c or a+b+c+d=0, or both

1964 AMC 12 — Problems & Online Timed Exam

Timed

1:15:00

1.

What is the value of $\left[\log_{10}\left(5\log_{10}100\right)\right]^2?$

$\log_{10}50$

$25$

$10$

$2$

$1$

Answer: E

Concepts:logarithm order of operations exponent

Difficulty rating: 1030

Small Hint:

Evaluate the inner logarithm first

Big Hint:

After multiplying by $5,$ take the base- $10$ logarithm and then square

Solution:

Since $\log_{10}100=2,$ the expression inside the square brackets is $\log_{10}(5\cdot2)=\log_{10}10=1.$ Its square is $1.$

Therefore, the correct answer is E.

2.

The graph of $x^2-4y^2=0$ is:

a parabola

an ellipse

a pair of straight lines

a point

none of these

Answer: C

Concepts:factoring difference of squares coordinate geometry

Difficulty rating: 940

Small Hint:

Factor the left side as a difference of squares

Big Hint:

Set each linear factor equal to zero

Solution:

Factoring gives $x^2-4y^2=(x-2y)(x+2y).$ Thus every point lies on $x=2y$ or $x=-2y,$ a pair of straight lines.

Therefore, the correct answer is C.

3.

When a positive integer $x$ is divided by a positive integer $y,$ the quotient is $u$ and the remainder is $v,$ where $u$ and $v$ are integers. What is the remainder when $x+2uy$ is divided by $y?$

$0$

$2u$

$3u$

$v$

$2v$

Answer: D

Concepts:modular arithmetic whole number operations substitution

Difficulty rating: 960

Small Hint:

Write $x=uy+v$

Big Hint:

Adding a multiple of $y$ does not change the remainder upon division by $y$

Solution:

The division algorithm gives $x=uy+v.$ Hence $x+2uy=3uy+v.$ The first term is divisible by $y,$ so the remainder is $v.$

Therefore, the correct answer is D.

4.

The expression

$\frac{P+Q}{P-Q}-\frac{P-Q}{P+Q},$

where $P=x+y$ and $Q=x-y,$ is equivalent to:

$\dfrac{x^2-y^2}{xy}$

$\dfrac{x^2-y^2}{2xy}$

$1$

$\dfrac{x^2+y^2}{xy}$

$\dfrac{x^2+y^2}{2xy}$

Answer: A

Concepts:algebraic manipulation substitution difference of squares

Difficulty rating: 1210

Small Hint:

First substitute $P+Q=2x$ and $P-Q=2y$

Big Hint:

Simplify $\frac{x}{y}-\frac{y}{x}$ using a common denominator

Solution:

Since $P+Q=2x$ and $P-Q=2y,$ the expression becomes $\frac{x}{y}-\frac{y}{x}=\frac{x^2-y^2}{xy}.$

Therefore, the correct answer is A.

5.

If $y$ varies directly as $x,$ and if $y=8$ when $x=4,$ the value of $y$ when $x=-8$ is:

$-16$

$-4$

$-2$

$4k,$ $k=\pm1,$ $\pm2,$ $\ldots$

$16k,$ $k=\pm1,$ $\pm2,$ $\ldots$

Answer: A

Concepts:ratio and proportion rate substitution

Difficulty rating: 890

Small Hint:

Write direct variation as $y=kx$

Big Hint:

Use the first pair to find $k$

Solution:

Direct variation gives $y=kx.$ From $8=4k,$ $k=2.$ Therefore, when $x=-8,$ $y=2(-8)=-16.$

Thus, the correct answer is A.

6.

If $x,$ $2x+2,$ $3x+3,$ $\ldots$ are in geometric progression, the fourth term is:

$-27$

$-13\dfrac12$

$12$

$13\dfrac12$

$27$

Answer: B

Concepts:geometric sequence quadratic ratio and proportion

Difficulty rating: 1430

Small Hint:

For three consecutive nonzero terms, the middle term squared equals the product of its neighbors

Big Hint:

After finding $x,$ multiply the third term by the common ratio

Solution:

The geometric-mean relation gives $(2x+2)^2=x(3x+3).$ Under the examination’s ratio convention for a geometric progression, the consecutive ratios must be defined, so $x\ne-1.$ Dividing by $x+1$ yields $4(x+1)=3x$ and $x=-4.$ The first three terms are $-4,-6,-9,$ with ratio $\frac{3}{2}.$ The fourth is $-9(\frac{3}{2})=-\frac{27}{2}.$

Therefore, the correct answer is B.

7.

Let $n$ be the number of real values of $p$ for which the roots of

$x^2-px+p=0$

are equal. Then $n$ equals:

$0$

$1$

$2$

a finite number greater than $2$

an infinitely large number

Answer: C

Concepts:quadratic zero product property

Difficulty rating: 1140

Small Hint:

Equal roots make the discriminant zero

Big Hint:

Solve $p^2-4p=0$

Solution:

Equal roots require $(-p)^2-4p=p(p-4)=0.$ Thus $p=0$ or $p=4,$ giving two real values.

Therefore, the correct answer is C.

8.

The smaller root of the equation

$\begin{aligned} &\left(x-\frac34\right)\left(x-\frac34\right)\\ &\quad+\left(x-\frac34\right) \left(x-\frac12\right)=0 \end{aligned}$

is:

$-\dfrac34$

$\dfrac12$

$\dfrac58$

$\dfrac34$

$1$

Answer: C

Concepts:factoring linear equation zero product property

Difficulty rating: 1110

Small Hint:

Factor out the common expression $x-\frac34$

Big Hint:

The remaining linear factor is $2x-\frac54$

Solution:

Combining the two terms and factoring gives $\left(x-\frac34\right)\left(2x-\frac54\right)=0.$ Hence the roots are $\frac{3}{4}$ and $\frac{5}{8},$ of which $\frac{5}{8}$ is smaller.

Therefore, the correct answer is C.

9.

A jobber buys an article at $\$24$ less $12\dfrac12\%.$ He then wishes to sell the article at a gain of $33\dfrac13\%$ of his cost after allowing a $20\%$ discount on his marked price. At what price, in dollars, should the article be marked?

$25.20$

$30.00$

$33.60$

$40.00$

none of these

Answer: E

Concepts:percentage money ratio and proportion

Difficulty rating: 1450

Small Hint:

Compute the discounted cost first, then increase it by one third

Big Hint:

The selling price is $80\%$ of the marked price

Solution:

The cost is $24(1-\frac{1}{8})=21$ dollars. A gain of one third makes the desired selling price $21(\frac{4}{3})=28$ dollars. If $M$ is the marked price, then $0.8M=28,$ so $M=35.$ This is not listed.

Thus, the correct answer is E.

10.

Given a square with side of length $s.$ On a diagonal as base a triangle with three unequal sides is constructed so that its area equals that of the square. The length of the altitude drawn to the base is:

$s\sqrt2$

$\dfrac{s}{\sqrt2}$

$2s$

$2\sqrt{s}$

$\dfrac2{\sqrt{s}}$

Answer: A

Concepts:square (geometry)diagonal triangle area radical

Difficulty rating: 1110

Small Hint:

The square’s diagonal has length $s\sqrt2$

Big Hint:

Set $\frac12(s\sqrt2)h=s^2$

Solution:

The triangle’s base is the square’s diagonal, $s\sqrt2.$ If its altitude is $h,$ equality of areas gives $\frac12(s\sqrt2)h=s^2,$ so $h=s\sqrt2.$

Therefore, the correct answer is A.

11.

Given $2^x=8^{y+1}$ and $9^y=3^{x-9},$ find the value of $x+y.$

$18$

$21$

$24$

$27$

$30$

Answer: D

Concepts:exponent system of equations substitution

Difficulty rating: 1430

Small Hint:

Rewrite $8$ as $2^3$ and $9$ as $3^2$

Big Hint:

Equate exponents to obtain two linear equations in $x$ and $y$

Solution:

Writing both equations with common bases gives $\begin{aligned} x&=3y+3,\\ 2y&=x-9. \end{aligned}$ Substitution yields $y=6$ and $x=21,$ so $x+y=27.$

Therefore, the correct answer is D.

12.

Which of the following is the negation of the statement: For all $x$ of a certain set, $x^2\gt0?$

For all $x,$ $x^2\lt0$

For all $x,$ $x^2\leq0$

For no $x,$ $x^2\gt0$

For some $x,$ $x^2\gt0$

For some $x,$ $x^2\leq0$

Answer: E

Concepts:logical deduction inequality counterexample

Difficulty rating: 1140

Small Hint:

The negation of “for all” begins with “for some”

Big Hint:

Negating $x^2\gt0$ gives $x^2\leq0$

Solution:

The negation of a universal statement is an existential statement, and the negation of $x^2\gt0$ is $x^2\leq0.$ Thus the negation is: for some $x,$ $x^2\leq0.$

Therefore, the correct answer is E.

13.

A circle is inscribed in a triangle with side lengths $8,$ $13,$ and $17.$ Let the segments of the side of length $8,$ made by a point of tangency, be $r$ and $s,$ with $r\lt s.$ What is the ratio $r:s?$

$1:3$

$2:5$

$1:2$

$2:3$

$3:4$

Answer: A

Concepts:incircle, incenter, and inradius tangent line ratio and proportion

Difficulty rating: 1210

Small Hint:

Compute the semiperimeter of the triangle

Big Hint:

A tangency segment adjacent to a vertex equals the semiperimeter minus the opposite side

Solution:

The semiperimeter is $\frac{8+13+17}{2}=19.$ At the endpoints of the side of length $8,$ the tangent lengths are $19-13=6$ and $19-17=2.$ Thus $r:s=2:6=1:3.$

Therefore, the correct answer is A.

14.

A farmer bought $749$ sheep. He sold $700$ of them for the price paid for the $749$ sheep. The remaining $49$ sheep were sold at the same price per head as the other $700.$ Based on the cost, the percent gain on the entire transaction is:

$6.5$

$6.75$

$7$

$7.5$

$8$

Answer: C

Concepts:percentage rate money

Difficulty rating: 1180

Small Hint:

Let the total purchase cost be $C$

Big Hint:

The selling price per sheep is $\frac{C}{700}$

Solution:

Let the total cost be $C.$ Since the first $700$ sheep sell for $C,$ each sheep sells for $\frac{C}{700}.$ Total revenue is therefore $\frac{749C}{700}=1.07C,$ a $7\%$ gain.

Thus, the correct answer is C.

15.

A line through the point $(-a,0)$ cuts from the second quadrant a triangular region with area $T.$ The equation of the line is:

$2Tx+a^2y+2aT=0$

$2Tx-a^2y+2aT=0$

$2Tx+a^2y-2aT=0$

$2Tx-a^2y-2aT=0$

none of these

Answer: B

Concepts:slope coordinate geometry triangle area

Difficulty rating: 1500

Small Hint:

If the positive $y$ -intercept is $b,$ then $\frac12ab=T$

Big Hint:

Use the intercept form $-\frac{x}{a}+\frac{y}{b}=1$

Solution:

Let the $y$ -intercept be $b.$ The second-quadrant triangle has area $\frac{ab}{2}=T,$ so $b=\frac{2T}{a}.$ The intercept form is $\frac{x}{-a}+\frac{y}{\frac{2T}{a}}=1.$ Clearing denominators gives $2Tx-a^2y+2aT=0.$

Therefore, the correct answer is B.

16.

Let $f(x)=x^2+3x+2$ and let $S$ be the set of integers $\{0,1,2,\ldots,25\}.$ The number of members $s$ of $S$ such that $f(s)$ has remainder zero when divided by $6$ is:

$25$

$22$

$21$

$18$

$17$

Answer: E

Concepts:modular arithmetic divisibility counting integers in a range

Difficulty rating: 1450

Small Hint:

Factor $f(s)$ as $(s+1)(s+2)$

Big Hint:

The product is always even; determine which residue of $s$ modulo $3$ fails

Solution:

We have $f(s)=(s+1)(s+2),$ a product of consecutive integers, so it is always even. It is divisible by $3$ unless $s\equiv0\pmod3.$ Among $0,1,\ldots,25,$ nine values are multiples of $3.$ Thus $26-9=17$ values work.

Therefore, the correct answer is E.

17.

Given the distinct points $P(x_1,y_1),$ $Q(x_2,y_2)$ and $R(x_1+x_2,y_1+y_2).$ Line segments are drawn connecting these points to each other and to the origin $O.$ Of the three possibilities: $(1)$ parallelogram, $(2)$ straight line, $(3)$ trapezoid, figure $OPRQ,$ depending upon the location of the points $P,$ $Q,$ and $R,$ can be:

$(1)$ only

$(2)$ only

$(3)$ only

$(1)$ or $(2)$ only

all three

Answer: D

Concepts:vector parallelogram logical deduction

Difficulty rating: 1450

Small Hint:

Interpret the position vectors as $R=P+Q$

Big Hint:

Separate the cases in which the vectors $P$ and $Q$ are independent or dependent

Solution:

The relation $\overrightarrow{OR}=\overrightarrow{OP}+\overrightarrow{OQ}$ makes opposite sides of $OPRQ$ parallel and equal whenever $P$ and $Q$ are not collinear, so the figure is a parallelogram. If the two vectors are dependent, all four points lie on a straight line. It cannot be a genuine trapezoid.

Thus, the correct answer is D.

18.

Let $n$ be the number of pairs of values of $b$ and $c$ such that $3x+by+c=0$ and $cx-2y+12=0$ have the same graph. Then $n$ is:

$0$

$1$

$2$

finite but more than $2$

greater than any finite number

Answer: C

Concepts:system of equations slope logical deduction

Difficulty rating: 1500

Small Hint:

Coincident line equations have proportional coefficient triples

Big Hint:

Set $(c,-2,12)=k(3,b,c)$ and use the first and third coordinates

Solution:

For the same graph, $(c,-2,12)=k(3,b,c).$ Thus $c=3k$ and $12=kc=3k^2,$ so $k=\pm2.$ Each value determines one pair through $b=-\frac{2}{k}$ and $c=3k.$ Hence there are two pairs.

Therefore, the correct answer is C.

19.

If $2x-3y-z=0$ and $x+3y-14z=0,$ $z\ne0,$ the numerical value of $\dfrac{x^2+3xy}{y^2+z^2}$ is:

$7$

$2$

$0$

$-\dfrac{20}{17}$

$-2$

Answer: A

Concepts:system of equations substitution algebraic manipulation

Difficulty rating: 1210

Small Hint:

Solve both linear equations for $x$ and $y$ in terms of $z$

Big Hint:

The equations give $y=3z$ and $x=5z$

Solution:

Solving the equations gives $y=3z$ and $x=5z.$ Therefore $\frac{x^2+3xy}{y^2+z^2}=\frac{25z^2+45z^2}{9z^2+z^2}=7.$

Thus, the correct answer is A.

20.

The sum of the numerical coefficients of all the terms in the expansion of $(x-2y)^{18}$ is:

$0$

$1$

$19$

$-1$

$-19$

Answer: B

Concepts:polynomial substitution binomial theorem

Difficulty rating: 1030

Small Hint:

A polynomial’s coefficient sum is found by setting every variable equal to $1$

Big Hint:

Evaluate $(1-2)^{18}$

Solution:

Set $x=y=1.$ The sum of all numerical coefficients is then $(1-2)^{18}=(-1)^{18}=1.$

Therefore, the correct answer is B.

21.

If $\log_{b^2}x+\log_{x^2}b=1,$ where $b\gt0,$ $b\ne1,$ and $x\ne1,$ then $x$ equals:

$\dfrac1{b^2}$

$\dfrac1b$

$b^2$

$b$

$\sqrt b$

Answer: D

Concepts:logarithm substitution algebraic manipulation

Difficulty rating: 1450

Small Hint:

Let $t=\log_bx$

Big Hint:

The equation becomes $\frac t2+\frac1{2t}=1$

Solution:

Let $t=\log_bx.$ Then $\log_{b^2}x=\frac{t}{2}$ and $\log_{x^2}b=\frac{1}{2t}.$ Thus $t+\frac1t=2,$ so $(t-1)^2=0$ and $t=1.$ Hence $x=b.$

Therefore, the correct answer is D.

22.

Given parallelogram $ABCD$ with $E$ the midpoint of diagonal $BD.$ Point $E$ is connected to a point $F$ in $DA$ so that $DF=\frac13DA.$ What is the ratio of the area of triangle $DFE$ to the area of quadrilateral $ABEF?$

$1:2$

$1:3$

$1:5$

$1:6$

$1:7$

Answer: C

Concepts:parallelogram area ratio midpoint triangle area

Difficulty rating: 1670

Small Hint:

Use vectors $A=0,$ $B=u,$ and $D=v$

Big Hint:

Then $E=\frac{u+v}{2}$ and $F=\frac{2v}{3}$ ; compute both areas as fractions of the parallelogram

Solution:

Let the parallelogram’s area be $S=|u\times v|,$ with $A=0,$ $B=u,$ and $D=v.$ Then $E=\frac{u+v}{2}$ and $F=\frac{2v}{3}.$ Determinants give $[DFE]=\frac{S}{12}.$ For quadrilateral $ABEF,$ the two determinant contributions give $[ABEF]=\frac S4+\frac S6=\frac{5S}{12}.$ The desired ratio is $1:5.$

Therefore, the correct answer is C.

23.

Two numbers are such that their difference, their sum, and their product are to one another as $1:7:24.$ The product of the two numbers is:

$6$

$12$

$24$

$48$

$96$

Answer: D

Concepts:ratio and proportion system of equations algebraic manipulation

Difficulty rating: 1210

Small Hint:

Represent the difference, sum, and product by $k,$ $7k,$ and $24k$

Big Hint:

The two numbers are $\frac{7k+k}{2}$ and $\frac{7k-k}{2}$

Solution:

Let the difference be $k$ and the sum be $7k.$ The numbers are $4k$ and $3k,$ so their product is $12k^2.$ But the ratio also says the product is $24k.$ For distinct numbers $k\ne0,$ so $k=2$ and the product is $48.$

Therefore, the correct answer is D.

24.

Let $y=(x-a)^2+(x-b)^2,$ $a,$ $b$ constants. For what value of $x$ is $y$ a minimum?

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

$a+b$

$\sqrt{ab}$

$\sqrt{\dfrac{a^2+b^2}{2}}$

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

Answer: A

Concepts:quadratic completing the square optimization

Difficulty rating: 1140

Small Hint:

Expand and collect the terms in $x$

Big Hint:

Complete the square or use the vertex formula

Solution:

Expanding and completing the square gives $y=2\left(x-\frac{a+b}{2}\right)^2+\frac{(a-b)^2}{2}.$ The squared term is minimized at $x=\frac{a+b}{2}.$

Therefore, the correct answer is A.

25.

The set of values of $m$ for which $x^2+3xy+x+my-m$ has two factors, with integer coefficients, which are linear in $x$ and $y,$ is precisely:

$0,$ $12,$ $-12$

$0,$ $12$

$12,$ $-12$

$12$

$0$

Answer: B

Concepts:factoring polynomial system of equations

Difficulty rating: 1870

Small Hint:

Write the factors as $(x+ay+b)(x+cy+d)$

Big Hint:

The missing $y^2$ -term and the $3xy$ -term force $\{a,c\}=\{0,3\}$

Solution:

Write the factorization as $(x+ay+b)(x+cy+d).$ Comparing the $y^2$ and $xy$ coefficients gives $ac=0$ and $a+c=3,$ so take $a=0,$ $c=3.$ The other coefficients give $\begin{aligned} b+d&=1,\\ bd&=-m,\\ 3b&=m. \end{aligned}$ Hence $b(d+3)=0.$ If $b=0,$ then $m=0;$ otherwise $d=-3,$ $b=4,$ and $m=12.$ Both values produce valid integer factorizations.

Therefore, the correct answer is B.

26.

In a ten-mile race First beats Second by $2$ miles and First beats Third by $4$ miles. If the runners maintain constant speeds throughout the race, by how many miles does Second beat Third?

$2$

$2\dfrac14$

$2\dfrac12$

$2\dfrac34$

$3$

Answer: C

Concepts:rate ratio and proportion distance rate and time relative speed

Difficulty rating: 1210

Small Hint:

When First finishes, Second and Third have run $8$ and $6$ miles

Big Hint:

Compare Third’s distance when Second completes $10$ miles

Solution:

Relative to First’s speed, Second’s speed is $\frac{8}{10}=\frac{4}{5}$ and Third’s is $\frac{6}{10}=\frac{3}{5}.$ Thus Third runs $\frac{3}{4}$ as fast as Second. When Second finishes $10$ miles, Third has run $10(\frac{3}{4})=7\dfrac12$ miles, so Second wins by $2\dfrac12$ miles.

Therefore, the correct answer is C.

27.

If $x$ is a real number and $|x-4|+|x-3|\lt a,$ where $a\gt0,$ then:

$0\lt a\lt0.01$

$0.01\lt a\lt1$

$0\lt a\lt1$

$0\lt a\leq1$

$a\gt1$

Answer: E

Concepts:absolute value triangle inequality inequality

Difficulty rating: 1260

Small Hint:

Interpret the sum as the distances from $x$ to $3$ and $4$

Big Hint:

Find its minimum for $3\leq x\leq4$ , remembering the inequality is strict

Solution:

By the triangle inequality, $|x-4|+|x-3|\geq|4-3|=1.$ Equality holds for every $x$ between $3$ and $4.$ Therefore the strict inequality has a real solution exactly when $a\gt1.$

Thus, the correct answer is E.

28.

The sum of $n$ terms of an arithmetic progression is $153,$ and the common difference is $2.$ If the first term is an integer, and $n\gt1,$ then the number of possible values for $n$ is:

$2$

$3$

$4$

$5$

$6$

Answer: D

Concepts:arithmetic sequence divisibility factor counting

Difficulty rating: 1650

Small Hint:

If the first term is $a,$ the sum is $n(a+n-1)$

Big Hint:

List the divisors of $153=3^2\cdot17$ that exceed $1$

Solution:

The arithmetic-series formula simplifies to $153=n(a+n-1).$ Thus $n$ must divide $153,$ and every such $n$ gives an integer $a.$ The divisors greater than $1$ are $3,$ $9,$ $17,$ $51,$ and $153,$ so there are five possibilities.

Therefore, the correct answer is D.

29.

In this figure $\angle RFS=\angle FDR,$ $FD=4$ inches, $DR=6$ inches, $FR=5$ inches, and $FS=7\dfrac12$ inches. The length of $RS,$ in inches, is:

undetermined

$4$

$5\dfrac12$

$6$

$6\dfrac14$

Answer: E

Difficulty rating: 1710

Small Hint:

Compare triangles $RFS$ and $FDR$ around the marked equal angles

Big Hint:

The adjacent side ratios are $\frac{5}{4}=\frac{15}{12}$

Solution:

At the equal included angles, $\begin{gathered} \frac{FR}{FD}=\frac54,\\ \frac{FS}{DR}=\frac{15}{12}=\frac54. \end{gathered}$ Thus $\triangle RFS\sim\triangle FDR$ by SAS. Side $RS$ corresponds to $FR=5,$ so $RS=(\frac{5}{4})\cdot5=\frac{25}{4}=6\frac14.$

Therefore, the correct answer is E.

30.

If

$\begin{aligned} (7+4\sqrt3)x^2 &+(2+\sqrt3)x\\ &-2=0, \end{aligned}$

the larger root minus the smaller root is:

$-2+3\sqrt3$

$2-\sqrt3$

$6+3\sqrt3$

$6-3\sqrt3$

$3\sqrt3+2$

Answer: D

Concepts:quadratic radical algebraic manipulation

Difficulty rating: 1670

Small Hint:

Notice that $7+4\sqrt3=(2+\sqrt3)^2$

Big Hint:

For $Ax^2+Bx+C=0,$ the root difference is $\frac{\sqrt{B^2-4AC}}{A}$

Solution:

Let $A=7+4\sqrt3=(2+\sqrt3)^2$ and $B=2+\sqrt3.$ The discriminant is $B^2-4A(-2)=A+8A=9A.$ Hence the root difference is $\frac{\sqrt{9A}}A=\frac3{2+\sqrt3}=6-3\sqrt3.$

Therefore, the correct answer is D.

31.

Let

$\begin{aligned} f(n) &=\frac{5+3\sqrt5}{10} \left(\frac{1+\sqrt5}{2}\right)^n\\ &\quad+\frac{5-3\sqrt5}{10} \left(\frac{1-\sqrt5}{2}\right)^n. \end{aligned}$

Then $f(n+1)-f(n-1),$ expressed in terms of $f(n),$ equals:

$\dfrac12f(n)$

$f(n)$

$2f(n)+1$

$f^2(n)$

$\dfrac12\left(f^2(n)-1\right)$

Answer: B

Concepts:recursion Fibonacci exponent

Difficulty rating: 1900

Small Hint:

Call the two exponential bases $\alpha$ and $\beta$

Big Hint:

Both roots of $r^2-r-1=0$ satisfy $r^{n+1}-r^{n-1}=r^n$

Solution:

Each base $r=\frac{1\pm\sqrt5}{2}$ satisfies $r^2=r+1,$ so $r^2-1=r.$ Therefore $r^{n+1}-r^{n-1}=r^{n-1}(r^2-1)=r^n.$ Applying this identity term by term to the defining linear combination gives $f(n+1)-f(n-1)=f(n).$

Thus, the correct answer is B.

32.

If $\dfrac{a+b}{b+c}=\dfrac{c+d}{d+a},$ then:

$a$ must equal $c$

$a+b+c+d$ must equal zero

either $a=c$ or $a+b+c+d=0,$ or both

$a+b+c+d\ne0$ if $a=c$

$a(b+c+d)=c(a+b+d)$

Answer: C

Concepts:rational equation factoring zero product property

Difficulty rating: 1450

Small Hint:

Cross-multiply the two fractions

Big Hint:

Move all terms to one side and factor out $a-c$

Solution:

Cross-multiplication gives $(a+b)(a+d)=(b+c)(c+d).$ Subtracting the right side and factoring yields $(a-c)(a+b+c+d)=0.$ Hence either $a=c$ or the sum is zero, and both may occur.

Therefore, the correct answer is C.

33.

$P$ is a point interior to rectangle $ABCD$ and such that $PA=3$ inches, $PD=4$ inches, and $PC=5$ inches. Then $PB,$ in inches, equals:

$2\sqrt3$

$3\sqrt2$

$3\sqrt3$

$4\sqrt2$

$2$

Answer: B

Concepts:rectangle British Flag Theorem distance formula

Difficulty rating: 1180

Small Hint:

Use the rectangle identity $PA^2+PC^2=PB^2+PD^2$

Big Hint:

Substitute the three known distances and solve for $PB^2$

Solution:

For any point in a rectangle, the British flag theorem gives $PA^2+PC^2=PB^2+PD^2.$ Thus $9+25=PB^2+16,$ so $PB^2=18$ and $PB=3\sqrt2.$

Therefore, the correct answer is B.

34.

If $n$ is a multiple of $4,$ the sum

$\begin{aligned} s&=1+2i+3i^2+\cdots\\ &\quad+(n+1)i^n, \end{aligned}$

where $i=\sqrt{-1},$ equals:

$1+i$

$\dfrac12(n+2)$

$\dfrac12(n+2-ni)$

$\dfrac12\left[(n+1)(1-i)+2\right]$

$\dfrac18(n^2+8-4ni)$

Answer: C

Concepts:complex number roots of unity summation

Difficulty rating: 1870

Small Hint:

Group the terms in blocks of four powers of $i$

Big Hint:

Each complete block beginning with coefficient $4j+1$ sums to $-2-2i$

Solution:

Write $n=4m.$ For $j=0,\ldots,m-1,$ the four terms with exponents $4j$ through $4j+3$ sum to $\begin{aligned} &(4j+1)+(4j+2)i\\ &\quad-(4j+3)-(4j+4)i\\ &=-2-2i. \end{aligned}$ The final term is $4m+1.$ Therefore $\begin{aligned} s&=m(-2-2i)+(4m+1)\\ &=\frac{n+2-ni}{2}. \end{aligned}$

Thus, the correct answer is C.

35.

The sides of a triangle are of lengths $13,$ $14,$ and $15.$ The altitudes of the triangle meet at point $H.$ If $AD$ is the altitude to the side of length $14,$ what is the ratio $HD:HA?$

$3:11$

$5:11$

$1:2$

$2:3$

$25:33$

Answer: B

Concepts:altitude coordinate geometry Heron’s Formula

Difficulty rating: 1850

Small Hint:

The $13$ - $14$ - $15$ triangle has area $84$ , so the altitude to side $14$ is $12$

Big Hint:

The altitude foot divides the side of length $14$ into segments $5$ and $9$ ; use coordinates to locate $H$

Solution:

Heron’s formula gives area $84,$ so $AD=\frac{2(84)}{14}=12.$ The adjacent $13$ -side has projection $\sqrt{13^2-12^2}=5,$ leaving $9$ on the base. Put $D=(0,0),\ A=(0,12),$ $B=(-5,0),\ C=(9,0).$ Line $AC$ has slope $-\frac{4}{3},$ so the altitude through $B$ has slope $\frac{3}{4}$ and meets $AD$ at height $\frac{15}{4}.$ Thus $\begin{gathered} HD=\frac{15}{4},\\ HA=12-\frac{15}{4}=\frac{33}{4}, \end{gathered}$ giving $HD:HA=5:11.$

Therefore, the correct answer is B.

36.

In this figure the radius of the circle is equal to the altitude of the equilateral triangle $ABC.$ The circle is made to roll along the side $AB,$ remaining tangent to it at a variable point $T$ and intersecting lines $AC$ and $BC$ in variable points $M$ and $N,$ respectively. Let $n$ be the number of degrees in arc $MTN.$ Then $n,$ for all permissible positions of the circle:

varies from $30^\circ$ to $90^\circ$

varies from $30^\circ$ to $60^\circ$

varies from $60^\circ$ to $90^\circ$

remains constant at $30^\circ$

remains constant at $60^\circ$

Answer: E

Concepts:circle coordinate geometry equilateral triangle inscribed angle arc

Difficulty rating: 2130

Small Hint:

The circle’s center $O$ and vertex $C$ are the same distance above $AB$ , so $CO\parallel AB$

Big Hint:

Extend $NC$ through $C$ to meet the circle again at $D$ , then use reflection across line $CO$

Solution:

Let $O$ be the circle’s center. Both $O$ and $C$ are one triangle altitude above $AB,$ so $CO\parallel AB.$ Extend $NC$ through $C$ to meet the circle again at $D.$ Since $CD$ is opposite to $CN,$ $\begin{aligned} \angle MCD &=180^\circ-\angle MCN\\ &=120^\circ. \end{aligned}$ The line $CO,$ parallel to $AB,$ bisects this angle. Reflection across $CO$ fixes the circle and interchanges rays $CM$ and $CD,$ so it interchanges $M$ and $D.$ Hence $CM=CD,$ and isosceles triangle $MCD$ has base angles $30^\circ.$

Because $D,C,N$ are collinear, $\angle MDN=30^\circ.$ This inscribed angle subtends arc $MTN,$ whose measure is therefore $60^\circ$ for every permissible circle position.

Thus, the correct answer is E.

37.

Given two positive numbers $a,$ $b$ such that $a\lt b.$ Let A.M. be their arithmetic mean and let G.M. be their positive geometric mean. Then A.M. minus G.M. is always less than:

$\dfrac{(b+a)^2}{ab}$

$\dfrac{(b+a)^2}{8b}$

$\dfrac{(b-a)^2}{ab}$

$\dfrac{(b-a)^2}{8a}$

$\dfrac{(b-a)^2}{8b}$

Answer: D

Concepts:AM-GM Inequality inequality radical

Difficulty rating: 1670

Small Hint:

Rewrite $\frac{a+b}{2}-\sqrt{ab}$ using $\sqrt b-\sqrt a$

Big Hint:

Factor $b-a=(\sqrt b-\sqrt a)(\sqrt b+\sqrt a)$ to compare with choice D

Solution:

We have $\frac{a+b}{2}-\sqrt{ab} =\frac{(\sqrt b-\sqrt a)^2}{2}.$ After canceling the positive factor $(\sqrt b-\sqrt a)^2,$ comparison with choice D reduces to $\frac12 <\frac{(\sqrt b+\sqrt a)^2}{8a}.$ This is true because $\sqrt b+\sqrt a\gt2\sqrt a.$ Thus A.M. minus G.M. is always less than the expression in choice D.

Therefore, the correct answer is D.

38.

The sides $PQ$ and $PR$ of triangle $PQR$ are respectively of lengths $4$ inches and $7$ inches. The median $PM$ is $3\dfrac12$ inches. Then $QR,$ in inches, is:

$6$

$7$

$8$

$9$

$10$

Answer: D

Concepts:median (geometry)Stewart’s Theorem law of cosines

Difficulty rating: 1210

Small Hint:

Use Apollonius’s theorem for the median to $QR$

Big Hint:

Substitute $PQ=4,\ PR=7,\ PM=\frac{7}{2}$

Solution:

Apollonius’s theorem gives $\begin{aligned} QR^2 &=2(4^2+7^2)-4\left(\frac72\right)^2\\ &=81. \end{aligned}$ Therefore $QR=9.$

Thus, the correct answer is D.

39.

The magnitudes of the sides of triangle $ABC$ are $a,$ $b,$ $c,$ as shown, with $c\leq b\leq a.$ Through interior point $P$ and the vertices $A,$ $B,$ $C,$ lines are drawn meeting the opposite sides in $A',$ $B',$ $C',$ respectively. Let $s=AA'+BB'+CC'.$ Then, for all positions of point $P,$ $s$ is less than:

$2a+b$

$2a+c$

$2b+c$

$a+2b$

$a+b+c$

Answer: A

Concepts:extremal argument bounding to limit cases inequality

Difficulty rating: 1850

Small Hint:

A point on a segment is closer to a fixed vertex than the farther endpoint of that segment

Big Hint:

Bound $AA'$ , $BB'$ , and $CC'$ separately using their adjacent side lengths

Solution:

For a fixed vertex, squared distance is a convex function along the opposite side, so its maximum occurs at an endpoint. Since each cevian endpoint is interior to its side, $\begin{gathered} AA'\lt\max(AB,AC)=b,\\ BB'\lt\max(BA,BC)=a,\\ CC'\lt\max(CA,CB)=a. \end{gathered}$ Adding gives $s\lt2a+b.$

Therefore, the correct answer is A.

40.

A watch loses $2\dfrac12$ minutes per day. It is set right at $1$ P.M. on March $15.$ Let $n$ be the positive correction, in minutes, to be added to the time shown by the watch at a given time. When the watch shows $9$ A.M. on March $21,$ $n$ equals:

$14\dfrac{14}{23}$

$14\dfrac1{14}$

$13\dfrac{101}{115}$

$13\dfrac{83}{115}$

$13\dfrac{13}{23}$

Answer: A

Concepts:clock date and time rate fraction

Difficulty rating: 1670

Small Hint:

The watch advances $\frac{575}{576}$ as fast as real time

Big Hint:

From the displayed $1$ P.M. on March $15$ to $9$ A.M. on March $21$ is $8400$ displayed minutes

Solution:

The watch runs at $\frac{575}{576}$ of the correct rate. The displayed elapsed time is $5$ days $20$ hours, or $8400$ minutes. Thus the real elapsed time is $8400(\frac{576}{575}),$ and the correction is $\begin{aligned} n&=8400\left(\frac{576}{575}-1\right)\\ &=\frac{8400}{575} =\frac{336}{23}\\ &=14\frac{14}{23}. \end{aligned}$

Therefore, the correct answer is A.