1966 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

Given that the ratio of $3x-4$ to $y+15$ is constant, and $y=3$ when $x=2,$ then, when $y=12,$ $x$ equals:

$\dfrac18$

$\dfrac87$

$\dfrac73$

$\dfrac72$

$8$

Concepts:linear equation ratio and proportion

Difficulty rating: 960

Small Hint:

Use $x=2$ and $y=3$ to find the constant ratio

Big Hint:

Keep $\frac{3x-4}{y+15}$ equal to that ratio when $y=12$

Solution:

The given pair makes the ratio $\frac{3\cdot2-4}{3+15}=\frac{1}{9}.$ Therefore, when $y=12,$ $\frac{3x-4}{27}=\frac19,$ so $3x-4=3$ and $x=\frac{7}{3}.$

Therefore, the correct answer is C.

2.

When the base of a triangle is increased $10\%$ and the altitude to this base is decreased $10\%,$ the change in area is:

$1\%$ increase

$\dfrac12\%$ increase

$0\%$

$\dfrac12\%$ decrease

$1\%$ decrease

Concepts:percentage triangle area

Difficulty rating: 890

Small Hint:

The area is proportional to the product of base and altitude

Big Hint:

Multiply the scale factors $1.1$ and $0.9$

Solution:

The new area is $1.1\cdot0.9=0.99$ times the old area. It is therefore $1\%$ smaller.

Thus, the correct answer is E.

3.

If the arithmetic mean of two numbers is $6$ and their geometric mean is $10,$ then an equation with the given two numbers as roots is:

$x^2+12x+100=0$

$x^2+6x+100=0$

$x^2-12x-10=0$

$x^2-12x+100=0$

$x^2-6x+100=0$

Concepts:mean Vieta’s Formulas

Difficulty rating: 1180

Small Hint:

The two numbers have sum $12$ and product $100$

Big Hint:

Roots with sum $S$ and product $P$ satisfy $x^2-Sx+P=0$

Solution:

If the numbers are $r,$ and $s,$ then $\frac{r+s}{2}=6$ gives $r+s=12,$ while $\sqrt{rs}=10$ gives $rs=100.$ The monic equation with those roots is $x^2-(r+s)x+rs=0,$ or $x^2-12x+100=0.$

Therefore, the correct answer is D.

4.

Circle $\mathrm{I}$ is circumscribed about a given square and circle $\mathrm{II}$ is inscribed in the given square. If $r$ is the ratio of the area of circle $\mathrm{I}$ to that of circle $\mathrm{II},$ then $r$ equals:

$\sqrt2$

$2$

$\sqrt3$

$2\sqrt2$

$2\sqrt3$

Concepts:area ratio circle area square (geometry)

Difficulty rating: 1030

Small Hint:

For square side $s,$ compare radii $\frac{s}{\sqrt2}$ and $\frac{s}{2}$

Big Hint:

Circle areas are proportional to the squares of their radii

Solution:

For square side $s,$ the outer circle has radius $\frac{s}{\sqrt2},$ while the inner circle has radius $\frac{s}{2}.$ Hence $r=\frac{\pi(\frac{s}{\sqrt2})^2}{\pi(\frac{s}{2})^2}=2.$

Therefore, the correct answer is B.

5.

The number of values of $x$ satisfying the equation

$\frac{2x^2-10x}{x^2-5x}=x-3$

is:

zero

one

two

three

an integer greater than $3$

Concepts:factoring rational equation

Difficulty rating: 1320

Small Hint:

Factor numerator and denominator before solving

Big Hint:

Do not admit values that make the original denominator zero

Solution:

The denominator requires $x\ne0$ and $x\ne5.$ On that domain, $\frac{2x(x-5)}{x(x-5)}=2.$ The remaining equation $2=x-3$ gives $x=5,$ which is excluded. Thus there are no solutions.

Therefore, the correct answer is A.

6.

$AB$ is a diameter of a circle centered at $O.$ $C$ is a point on the circle such that angle $BOC$ is $60^\circ.$ If the diameter of the circle is $5$ inches, the length of chord $AC,$ expressed in inches, is:

$3$

$\dfrac{5\sqrt2}{2}$

$\dfrac{5\sqrt3}{2}$

$3\sqrt3$

none of these

Concepts:chord special right triangle

Difficulty rating: 1110

Small Hint:

Because $AB$ is a diameter, $\angle AOC=120^\circ$

Big Hint:

Split isosceles triangle $AOC$ into two $30$ - $60$ - $90$ triangles

Solution:

The radius is $\frac{5}{2},$ and $\angle AOC=180^\circ-60^\circ=120^\circ.$ A chord subtending $120^\circ$ has length $2\left(\frac52\right)\sin60^\circ=\frac{5\sqrt3}{2}.$

Therefore, the correct answer is C.

7.

Let

$\frac{35x-29}{x^2-3x+2} =\frac{N_1}{x-1}+\frac{N_2}{x-2}$

be an identity in $x.$ The numerical value of $N_1N_2$ is:

$-246$

$-210$

$-29$

$210$

$246$

Concepts:linear equation partial fractions

Difficulty rating: 1210

Small Hint:

Combine the two fractions over $(x-1)(x-2)$

Big Hint:

Compare the coefficient of $x$ and the constant term

Solution:

Combining the right side gives numerator $\begin{aligned} &N_1(x-2)+N_2(x-1)\\ &\quad=(N_1+N_2)x\\ &\qquad-(2N_1+N_2). \end{aligned}$ Thus $N_1+N_2=35$ and $2N_1+N_2=29.$ Hence $N_1=-6,$ $N_2=41,$ and $N_1N_2=-246.$

Therefore, the correct answer is A.

8.

The length of the common chord of two intersecting circles is $16$ feet. If the radii are $10$ feet and $17$ feet, a possible value for the distance between the centers of the circles, expressed in feet, is:

$27$

$21$

$\sqrt{389}$

$15$

undetermined

Concepts:chord circle Pythagorean Theorem

Difficulty rating: 1450

Small Hint:

The line of centers perpendicularly bisects the common chord

Big Hint:

Find each center’s distance from the chord using half-chord $8$

Solution:

The half-chord has length $8.$ The two perpendicular distances from the centers to its line are $\begin{gathered} \sqrt{10^2-8^2}=6,\\ \sqrt{17^2-8^2}=15. \end{gathered}$ If the centers lie on opposite sides of the chord, their distance is $6+15=21,$ which is a possible value.

Therefore, the correct answer is B.

9.

If $x=(\log_8 2)^{(\log_2 8)},$ then $\log_3 x$ equals:

$-3$

$-\dfrac13$

$\dfrac13$

$3$

$9$

Concepts:exponent logarithm

Difficulty rating: 1180

Small Hint:

Evaluate $\log_8 2$ and $\log_2 8$ separately

Big Hint:

Express the resulting power as a power of $3$

Solution:

We have $\log_8 2=\frac{1}{3}$ and $\log_2 8=3.$ Therefore $x=(\frac{1}{3})^3=3^{-3},$ so $\log_3x=-3.$

Thus, the correct answer is A.

10.

If the sum of two numbers is $1$ and their product is $1,$ then the sum of their cubes is, where $i=\sqrt{-1}:$

$2$

$-2-\dfrac{3\sqrt3i}{4}$

$0$

$-\dfrac{3\sqrt3i}{4}$

$-2$

Concepts:complex number sum and difference of cubes symmetry (algebra)

Difficulty rating: 1110

Small Hint:

Use $u^3+v^3=(u+v)^3-3uv(u+v)$

Big Hint:

Substitute the given sum and product directly

Solution:

If the numbers are $u,$ and $v,$ then $\begin{aligned} u^3+v^3 &=(u+v)^3-3uv(u+v)\\ &=1^3-3=-2. \end{aligned}$

Therefore, the correct answer is E.

11.

The sides of triangle $BAC$ are in the ratio $2:3:4.$ $BD$ is the angle-bisector drawn to the shortest side $AC,$ dividing it into segments $AD$ and $CD.$ If the length of $AC$ is $10,$ then the length of the longer segment of $AC$ is:

$3\dfrac12$

$5$

$5\dfrac57$

$6$

$7\dfrac12$

Concepts:angle bisector theorem ratio and proportion

Difficulty rating: 1180

Small Hint:

Since $AC$ is the shortest side, the adjacent sides are in ratio $3:4$

Big Hint:

Apply the angle-bisector theorem to $AD:DC$

Solution:

The sides adjacent to angle $B$ are in ratio $3:4.$ By the angle-bisector theorem, $AD:DC=3:4.$ The longer of the two parts is therefore $\frac4{3+4}\cdot10=\frac{40}{7}=5\frac57.$

Therefore, the correct answer is C.

12.

The number of real values of $x$ that satisfy the equation

$(2^{6x+3})(4^{3x+6})=8^{4x+5}$

is:

$0$

$1$

$2$

$3$

greater than $3$

Concepts:algebraic manipulation exponent

Difficulty rating: 1110

Small Hint:

Rewrite every term with base $2$

Big Hint:

Compare the total exponent on each side before trying to solve for $x$

Solution:

In base $2,$ the left exponent is $\begin{aligned} &(6x+3)+2(3x+6)\\ &\qquad=12x+15, \end{aligned}$ and the right exponent is $3(4x+5)=12x+15.$ The equation is an identity for every real $x,$ so it has more than three real solutions.

Therefore, the correct answer is E.

13.

The number of points with positive rational coordinates selected from the set of points in the $xy$ -plane such that $x+y\leq5,$ is:

$9$

$10$

$14$

$15$

infinite

Concepts:coordinate geometry inequality

Difficulty rating: 1030

Small Hint:

Try fixing one positive rational coordinate

Big Hint:

There are infinitely many positive rational numbers in an interval

Solution:

For example, set $y=1.$ Every positive rational $x\leq4$ then gives an allowed point $(x,1),$ and there are infinitely many such rationals.

Therefore, the correct answer is E.

14.

The length of rectangle $ABCD$ is $5$ inches and its width is $3$ inches. Diagonal $AC$ is divided into three equal segments by points $E$ and $F.$ The area of triangle $BEF,$ expressed in square inches, is:

$\dfrac32$

$\dfrac53$

$\dfrac52$

$\dfrac13\sqrt{34}$

$\dfrac13\sqrt{68}$

Concepts:coordinate geometry rectangle triangle area

Difficulty rating: 1450

Small Hint:

Place the rectangle at $(0,0),$ $(5,0),$ $(5,3),$ and $(0,3)$

Big Hint:

The trisection points of diagonal $AC$ have easy coordinates

Solution:

Take $A=(0,0),$ $B=(5,0),$ and $C=(5,3).$ Then $E=(\frac{5}{3},1)$ and $F=(\frac{10}{3},2).$ The determinant formula gives $\begin{aligned} [BEF] &=\frac12\left| \left(-\frac{10}{3}\right)2 -\left(-\frac53\right)1 \right|\\ &=\frac52. \end{aligned}$

Therefore, the correct answer is C.

15.

If $x-y\gt x$ and $x+y\lt y,$ then:

$y\lt x$

$x\lt y$

$x\lt y\lt0$

$x\lt0,$ and $y\lt0$

$x\lt0,$ and $y\gt0$

Concepts:algebraic manipulation inequality

Difficulty rating: 960

Small Hint:

Subtract $x$ from the first inequality

Big Hint:

Subtract $y$ from the second inequality

Solution:

From $x-y\gt x$ we get $-y\gt0,$ so $y\lt0.$ From $x+y\lt y$ we get $x\lt0.$

Therefore, the correct answer is D.

16.

If

$\begin{gathered} \dfrac{4^x}{2^{x+y}}=8,\\ \dfrac{9^{x+y}}{3^{5y}}=243, \end{gathered}$

$x$ and $y$ are real numbers, then $xy$ equals:

$1\dfrac34$

$4$

$6$

$12$

$-4$

Concepts:exponent system of equations

Difficulty rating: 1430

Small Hint:

Rewrite both equations with bases $2$ and $3$

Big Hint:

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

Solution:

The first equation becomes $2^{x-y}=2^3,$ so $x-y=3.$ The second becomes $3^{2x-3y}=3^5,$ so $2x-3y=5.$ Solving gives $y=1,$ $x=4,$ and hence $xy=4.$

Therefore, the correct answer is B.

17.

The number of distinct points common to the curves $x^2+4y^2=1$ and $4x^2+y^2=4$ is:

$0$

$1$

$2$

$3$

$4$

Concepts:ellipse substitution system of equations

Difficulty rating: 1430

Small Hint:

Use the first equation to substitute for $x^2$ in the second

Big Hint:

After finding $y,$ count both possible signs of $x$

Solution:

From the first equation, $x^2=1-4y^2.$ Substitution into the second gives $4(1-4y^2)+y^2=4,$ so $y=0.$ Then $x^2=1,$ giving the two points $(1,0)$ and $(-1,0).$

Therefore, the correct answer is C.

18.

In a given arithmetic sequence the first term is $2,$ the last term is $29,$ and the sum of all the terms is $155.$ The common difference is:

$3$

$2$

$\dfrac{27}{19}$

$\dfrac{13}{9}$

$\dfrac{23}{38}$

Concepts:arithmetic sequence summation

Difficulty rating: 1180

Small Hint:

Use $155=\frac{n(2+29)}{2}$ to find the number of terms

Big Hint:

Then use $29=2+(n-1)d$

Solution:

The sum formula gives $155=\frac{31n}{2},$ so $n=10.$ Therefore $29=2+9d,$ and $d=3.$

Thus, the correct answer is A.

19.

Let $s_1$ be the sum of the first $n$ terms of the arithmetic sequence $8,$ $12,$ $\ldots$ and let $s_2$ be the sum of the first $n$ terms of the arithmetic sequence $17,$ $19,$ $\ldots.$ Then $s_1=s_2$ for:

no value of $n$

one value of $n$

two values of $n$

four values of $n$

more than four values of $n$

Concepts:arithmetic sequence quadratic summation

Difficulty rating: 1210

Small Hint:

Write a sum formula for each progression

Big Hint:

Cancel the positive factor $n$ after equating the sums

Solution:

The two sums are $\begin{gathered} s_1=2n^2+6n,\\ s_2=n^2+16n. \end{gathered}$ Equality gives $n(n-10)=0.$ Since a number of terms is positive, only $n=10$ works, so there is one value.

Therefore, the correct answer is B.

20.

If the proposition “ $a=0$ ” is true, the negation of the proposition “For real values of $a$ and $b,$ if $a=0,$ then $ab=0$ ” is:

If $a\ne0,$ then $ab\ne0$

If $a\ne0,$ then $ab=0$

If $a=0,$ then $ab\ne0$

If $ab\ne0,$ then $a\ne0$

If $ab=0,$ then $a\ne0$

Concepts:logical deduction

Difficulty rating: 1180

Small Hint:

The negation of a true conclusion is its opposite statement

Big Hint:

Under the stated assumption $a=0,$ negate the conclusion $ab=0$

Solution:

The stated premise $a=0$ is assumed true. Negating the implication therefore negates its conclusion: $ab=0$ becomes $ab\ne0.$ Thus the required statement is “If $a=0,$ then $ab\ne0.$ ”

Therefore, the correct answer is C.

21.

An “ $n$ -pointed star” is formed as follows: the sides of a convex polygon are numbered consecutively $1,$ $2,$ $\ldots,$ $k,$ $\ldots,$ $n,$ $n\geq5;$ for all $n$ values of $k,$ sides $k$ and $k+2$ are non-parallel, sides $n+1$ and $n+2$ being respectively identical with sides $1$ and $2;$ prolong the $n$ pairs of sides numbered $k$ and $k+2$ until they meet. (A figure is shown for the case $n=5.$ )

Let $S$ be the degree-sum of the interior angles at the $n$ points of the star; then $S$ equals:

$180$

$360$

$180(n+2)$

$180(n-2)$

$180(n-4)$

Concepts:angle chasing angle sum

Difficulty rating: 1880

Small Hint:

Relate each star-tip angle to two adjacent interior angles of the original polygon

Big Hint:

The original polygon’s interior-angle sum is $180(n-2)$ degrees

Solution:

Let the original polygon’s interior angles be $a_1,\ldots,a_n.$ The star angle made from sides $k$ and $k+2$ equals $a_k+a_{k+1}-180^\circ.$ Summing cyclically counts each $a_k$ twice. Hence $\begin{aligned} S&=2\sum_{k=1}^n a_k-180n\\ &=2\cdot180(n-2)-180n\\ &=180(n-4). \end{aligned}$

Therefore, the correct answer is E.

22.

Consider the statements:

$\text{(I) }\sqrt{a^2+b^2}=0$ $\text{(II) }\sqrt{a^2+b^2}=ab$ $\text{(III) }\sqrt{a^2+b^2}=a+b$ $\text{(IV) }\sqrt{a^2+b^2}=a-b$

where we allow $a$ and $b$ to be real or complex numbers. Those statements for which there exist solutions other than $a=0$ and $b=0,$ are:

$(\mathrm{I}),$ $(\mathrm{II}),$ $(\mathrm{III}),$ $(\mathrm{IV})$

$(\mathrm{II}),$ $(\mathrm{III}),$ $(\mathrm{IV})$ only

$(\mathrm{I}),$ $(\mathrm{III}),$ $(\mathrm{IV})$ only

$(\mathrm{III}),$ $(\mathrm{IV})$ only

$(\mathrm{I})$ only

Concepts:complex number counterexample radical

Difficulty rating: 1880

Small Hint:

One nonzero example is enough for each statement

Big Hint:

Try $b=i$ for $(\mathrm{I}),$ equal positive values for $(\mathrm{II}),$ and $b=0$ for $(\mathrm{III})$ and $(\mathrm{IV})$

Solution:

Every statement has a nonzero solution. For $(\mathrm{I}),$ use $(a,b)=(1,i).$ For $(\mathrm{II}),$ use $a=b=\sqrt2,$ giving $\sqrt4=2=ab.$ For both $(\mathrm{III})$ and $(\mathrm{IV}),$ use $(a,b)=(1,0).$ Thus all four statements qualify.

Therefore, the correct answer is A.

23.

If $x$ is real and $4y^2+4xy+x+6=0,$ then the complete set of values of $x$ for which $y$ is real, is:

$x\leq-2$ or $x\geq3$

$x\leq2$ or $x\geq3$

$x\leq-3$ or $x\geq2$

$-3\leq x\leq2$

$-2\leq x\leq3$

Concepts:inequality quadratic

Difficulty rating: 1180

Small Hint:

Treat the equation as a quadratic in $y$

Big Hint:

Require its discriminant $16(x-3)(x+2)$ to be nonnegative

Solution:

As a quadratic in $y,$ the equation has discriminant $\begin{aligned} &(4x)^2-16(x+6)\\ &\qquad=16(x-3)(x+2). \end{aligned}$ This is nonnegative exactly when $x\leq-2$ or $x\geq3.$

Therefore, the correct answer is A.

24.

If $\log_MN=\log_NM,$ $M\ne N,$ $MN\gt0,$ and $M\ne1,$ $N\ne1,$ then $MN$ equals:

$\dfrac12$

$1$

$2$

$10$

a number greater than $2$ and less than $10$

Concepts:algebraic manipulation logarithm

Difficulty rating: 1430

Small Hint:

Write both logarithms using natural logs

Big Hint:

The equality gives $(\ln M)^2=(\ln N)^2$ ; use $M\ne N$

Solution:

Change of base gives $\frac{\ln N}{\ln M}=\frac{\ln M}{\ln N},$ so $(\ln M)^2=(\ln N)^2.$ Equality of the logs themselves would give $M=N,$ which is excluded. Hence $\ln M=-\ln N,$ and $\ln(MN)=0.$ Thus $MN=1.$

Therefore, the correct answer is B.

25.

If $F(n+1)=\dfrac{2F(n)+1}{2}$ for $n=1,$ $2,$ $\ldots,$ and $F(1)=2,$ then $F(101)$ equals:

$49$

$50$

$51$

$52$

$53$

Concepts:arithmetic sequence recursion

Difficulty rating: 960

Small Hint:

Simplify the recurrence to $F(n+1)=F(n)+\frac{1}{2}$

Big Hint:

Count the increments from $n=1$ to $n=101$

Solution:

Each step adds $\frac{1}{2}.$ There are $100$ steps from $F(1)$ to $F(101),$ so $F(101)=2+\frac{100}{2}=52.$

Therefore, the correct answer is D.

26.

Let $m$ be a positive integer and let the lines $13x+11y=700$ and $y=mx-1$ intersect in a point whose coordinates are integers. Then $m$ can be:

$4$ only

$5$ only

$6$ only

$7$ only

one of the integers $4,$ $5,$ $6,$ $7$ and one other positive integer

Concepts:Diophantine Equation divisibility linear equation

Difficulty rating: 1650

Small Hint:

Substitute $y=mx-1$ into the first line

Big Hint:

For integer $x,$ the number $13+11m$ must divide $711$

Solution:

Substitution gives $(13+11m)x=711.$ Thus $13+11m$ must be a positive divisor of $711=3^2\cdot79.$ Among divisors greater than $13,$ only $79$ is congruent to $13\pmod{11}.$ Hence $13+11m=79,$ so $m=6.$

Therefore, the correct answer is C.

27.

At his usual rate a man rows $15$ miles downstream in five hours less time than it takes him to return. If he doubles his usual rate, the time downstream is only one hour less than the time upstream. In miles per hour, the rate of the stream’s current is:

$2$

$\dfrac52$

$3$

$\dfrac72$

$4$

Concepts:rational equation relative speed system of equations

Difficulty rating: 1850

Small Hint:

Let $v$ be the rower’s still-water speed and $c$ the current speed

Big Hint:

Translate the two time differences using speeds $v\pm c$ and $2v\pm c$

Solution:

The two time differences give $\begin{aligned} \frac{15}{v-c}-\frac{15}{v+c}&=5,\\ \frac{15}{2v-c}-\frac{15}{2v+c}&=1. \end{aligned}$ These simplify to $v^2-c^2=6c$ and $4v^2-c^2=30c.$ Substituting $v^2=c^2+6c$ into the second gives $3c^2-6c=0.$ Since $c\gt0,$ we have $c=2.$

Therefore, the correct answer is A.

28.

Five points $O,$ $A,$ $B,$ $C,$ $D$ are taken in order on a straight line with distances $OA=a,$ $OB=b,$ $OC=c,$ and $OD=d.$ $P$ is a point on the line between $B$ and $C$ and such that $AP:PD=BP:PC.$ Then $OP$ equals:

$\dfrac{b^2-bc}{a-b+c-d}$

$\dfrac{ac-bd}{a-b+c-d}$

$-\dfrac{bd+ac}{a-b+c-d}$

$\dfrac{bc+ad}{a+b+c+d}$

$\dfrac{ac-bd}{a+b+c+d}$

Concepts:linear equation ratio and proportion

Difficulty rating: 1850

Small Hint:

Write $p=OP.$ Then $AP=p-a,$ $PD=d-p,$ $BP=p-b,$ and $PC=c-p$

Big Hint:

Cross-multiply the two given ratios after making those substitutions

Solution:

Let $p=OP.$ The given ratio becomes $\frac{p-a}{d-p}=\frac{p-b}{c-p}.$ Cross-multiplication cancels the $p^2$ terms and yields $p(a-b+c-d)=ac-bd.$ Therefore $p=\frac{ac-bd}{a-b+c-d}.$

Thus, the correct answer is B.

29.

The number of positive integers less than $1000$ divisible by neither $5$ nor $7$ is:

$688$

$686$

$684$

$658$

$630$

Concepts:counting integers in a range divisibility inclusion-exclusion

Difficulty rating: 1180

Small Hint:

There are $999$ positive integers below $1000$

Big Hint:

Subtract multiples of $5$ and $7$ , then add back multiples of $35$

Solution:

By inclusion-exclusion, the count is $\begin{aligned} &999-\left\lfloor\frac{999}{5}\right\rfloor -\left\lfloor\frac{999}{7}\right\rfloor\\ &\quad+\left\lfloor\frac{999}{35}\right\rfloor\\ &=999-199-142+28\\ &=686. \end{aligned}$

Therefore, the correct answer is B.

30.

If three of the roots of $x^4+ax^2+bx+c=0$ are $1,$ $2,$ and $3,$ then the value of $a+c$ is:

$35$

$24$

$-12$

$-61$

$-63$

Concepts:polynomial Vieta’s Formulas

Difficulty rating: 1210

Small Hint:

The missing $x^3$ coefficient makes the sum of all four roots zero

Big Hint:

Use the fourth root to compute the pairwise sum $a$ and product $c$

Solution:

The fourth root is $-6$ because the root sum is zero. The sum of pairwise products is $\begin{aligned} a&=2+3-6+6\\ &\quad-12-18=-25, \end{aligned}$ while the product is $c=(1)(2)(3)(-6)=-36.$ Thus $a+c=-61.$

Therefore, the correct answer is D.

31.

Triangle $ABC$ is inscribed in a circle with center $O'.$ A circle with center $O$ is inscribed in triangle $ABC.$ $AO$ is drawn, and extended to intersect the larger circle in $D.$ Then we must have:

$CD=BD=O'D$

$AO=CO=OD$

$CD=CO=BD$

$CD=OD=BD$

$O'B=O'C=OD$

Concepts:angle chasing circumcircle, circumcenter, and circumradius incircle, incenter, and inradius

Difficulty rating: 1880

Small Hint:

Line $AD$ bisects $\angle A,$ so $D$ is the midpoint of arc $BC$

Big Hint:

Compare angles in triangle $COD$ to show $CD=OD$

Solution:

Because $AO$ is an angle bisector, the inscribed angles $\angle BAD$ and $\angle CAD$ are equal. Hence arcs $BD$ and $CD$ , and therefore chords $BD$ and $CD,$ are equal.

Let $\angle BAD=\alpha$ and $\angle BCO=\beta.$ Then $\angle OCD=\alpha+\beta.$ In triangle $AOC,$ the exterior angle $\angle COD$ also equals $\alpha+\beta.$ Thus $CD=OD,$ so $CD=OD=BD.$

Therefore, the correct answer is D.

32.

Let $M$ be the midpoint of side $AB$ of triangle $ABC.$ Let $P$ be a point on $AB$ between $A$ and $M,$ and let $MD$ be drawn parallel to $PC$ and intersecting $BC$ at $D.$ If the ratio of the area of triangle $BPD$ to that of triangle $ABC$ is denoted by $r,$ then:

$\dfrac12\lt r\lt1$ depending upon the position of $P$

$r=\dfrac12$ independent of the position of $P$

$\dfrac12\leq r\lt1$ depending upon the position of $P$

$\dfrac13\lt r\lt\dfrac23$ depending upon the position of $P$

$r=\dfrac13$ independent of the position of $P$

Concepts:area ratio median (geometry)parallel lines

Difficulty rating: 1650

Small Hint:

Triangles $MDP$ and $MDC$ have equal bases on parallel lines

Big Hint:

Add their areas to $[BMD]$ and use that $CM$ is a median

Solution:

Since $MD\parallel PC,$ triangles $MDP$ and $MDC$ have the same base $MD$ and equal altitudes, so their areas are equal. Hence $\begin{aligned} [BPD] &=[BMD]+[MDP]\\ &=[BMD]+[MDC]\\ &=[BMC]. \end{aligned}$ Because $CM$ is a median, $[BMC]=\frac{[ABC]}{2}.$ Thus $r=\frac{1}{2}$ for every allowed $P.$

Therefore, the correct answer is B.

33.

If $ab\ne0$ and $|a|\ne|b|,$ the number of distinct values of $x$ satisfying the equation

$\begin{aligned} \frac{x-a}{b}+\frac{x-b}{a} &=\frac{b}{x-a}\\ &\quad+\frac{a}{x-b} \end{aligned}$

is:

$0$

$1$

$2$

$3$

$4$

Concepts:algebraic manipulation factoring rational equation

Difficulty rating: 1710

Small Hint:

Both sides contain the numerator $(a+b)x-(a^2+b^2)$ after combining terms

Big Hint:

Factor the equation into that numerator times a difference of reciprocals

Solution:

Let $N=(a+b)x-a^2-b^2.$ Combining each side gives $N\left(\frac1{ab} -\frac1{(x-a)(x-b)}\right)=0.$ The first factor gives $x=\frac{a^2+b^2}{a+b}.$ The second gives $(x-a)(x-b)=ab,$ or $x(x-a-b)=0,$ giving $x=0$ and $x=a+b.$ The hypotheses ensure that all three are defined and distinct.

Therefore, the correct answer is D.

34.

Let $r$ be the speed in miles per hour at which a wheel, $11$ feet in circumference, travels. If the time for a complete rotation of the wheel is shortened by $\dfrac14$ of a second, the speed $r$ is increased by $5$ miles per hour. Then $r$ is:

$9$

$10$

$10\dfrac12$

$11$

$12$

Concepts:distance rate and time rational equation unit conversion

Difficulty rating: 1650

Small Hint:

At $r$ miles per hour, one rotation takes $\frac{7.5}{r}$ seconds

Big Hint:

Set $\frac{7.5}{r}-\frac{7.5}{r+5}=\frac{1}{4}$

Solution:

Since $11$ feet is $\frac{11}{5280}$ mile, one rotation at $r$ mph takes $\frac{3600\cdot11}{5280r}=\frac{7.5}{r}$ seconds. Thus $\frac{7.5}{r}-\frac{7.5}{r+5}=\frac14.$ This reduces to $r^2+5r-150=0,$ or $(r-10)(r+15)=0.$ The positive speed is $r=10.$

Therefore, the correct answer is B.

35.

Let $O$ be an interior point of triangle $ABC,$ and let $s_1=OA+OB+OC.$ If $s_2=AB+BC+CA,$ then:

for every triangle $s_1\gt\dfrac12s_2,$ and $s_1\leq s_2$

for every triangle $s_1\geq\dfrac12s_2,$ and $s_1\lt s_2$

for every triangle $s_1\gt\dfrac12s_2,$ and $s_1\lt s_2$

for every triangle $s_1\geq\dfrac12s_2,$ and $s_1\leq s_2$

neither $(A)$ nor $(B)$ nor $(C)$ nor $(D)$ applies to every triangle

Concepts:bounding to limit cases triangle inequality

Difficulty rating: 1880

Small Hint:

Add $AB\lt OA+OB,$ $BC\lt OB+OC,$ and $CA\lt OC+OA$

Big Hint:

Extend $AO$ to side $BC$ to prove $OA+OB\lt AC+CB,$ and cycle

Solution:

Adding the three strict triangle inequalities $\begin{gathered} AB\lt OA+OB,\\ BC\lt OB+OC,\\ CA\lt OC+OA \end{gathered}$ gives $s_2\lt2s_1.$

For the upper bound, extend $AO$ to $D$ on $BC.$ Then $AO+OD\lt AC+CD$ and $OB\lt OD+DB,$ so cancellation gives $OA+OB\lt AC+CB.$ Cycling and adding yields $2s_1\lt2s_2.$ Therefore $s_1\gt \frac{s_2}{2}$ and $s_1\lt s_2.$

Thus, the correct answer is C.

36.

Let

$\begin{aligned} (1+x+x^2)^n &=a_0+a_1x\\ &\quad+a_2x^2+\cdots\\ &\quad+a_{2n}x^{2n} \end{aligned}$

be an identity in $x.$ If we let $s=a_0+a_2+a_4+\cdots+a_{2n},$ then $s$ equals:

$2^n$

$2^n+1$

$\dfrac{3^n-1}{2}$

$\dfrac{3^n}{2}$

$\dfrac{3^n+1}{2}$

Concepts:parity polynomial substitution

Difficulty rating: 1430

Small Hint:

Evaluate the polynomial at $x=1$ and $x=-1$

Big Hint:

Adding those two values cancels all odd-degree coefficients

Solution:

Let $P(x)=(1+x+x^2)^n.$ Then $P(1)=3^n$ is the sum of all coefficients, while $P(-1)=1$ is the even-coefficient sum minus the odd-coefficient sum. Therefore $s=\frac{P(1)+P(-1)}2=\frac{3^n+1}{2}.$

Therefore, the correct answer is E.

37.

Three men, Alpha, Beta, and Gamma, working together, do a job in $6$ hours less time than Alpha alone, in $1$ hour less time than Beta alone, and in one-half the time needed by Gamma when working alone. Let $h$ be the number of hours needed by Alpha and Beta, working together, to do the job. Then $h$ equals:

$\dfrac52$

$\dfrac32$

$\dfrac43$

$\dfrac54$

$\dfrac34$

Concepts:rate rational equation system of equations

Difficulty rating: 1850

Small Hint:

Let $t$ be the time all three take together

Big Hint:

Their individual times are $t+6,$ $t+1,$ and $2t$ ; add reciprocal rates

Solution:

If all three together take $t$ hours, then $\frac1{t+6}+\frac1{t+1}+\frac1{2t}=\frac1t.$ This simplifies to $3t^2+7t-6=0,$ so $t=\frac{2}{3}.$ Alpha and Beta alone take $\frac{20}{3}$ and $\frac{5}{3}$ hours, so their combined rate is $\frac{3}{20}+\frac{3}{5}=\frac{3}{4}.$ Hence $h=\frac{4}{3}.$

Therefore, the correct answer is C.

38.

In triangle $ABC$ the medians $AM$ and $CN$ to sides $BC$ and $AB,$ respectively, intersect in point $O.$ $P$ is the midpoint of side $AC,$ and $MP$ intersects $CN$ in $Q.$ If the area of triangle $OMQ$ is $n,$ then the area of triangle $ABC$ is:

$16n$

$18n$

$21n$

$24n$

$27n$

Concepts:area ratio centroid coordinate geometry median (geometry)

Difficulty rating: 1650

Small Hint:

Use an affine model $A=(0,0),$ $B=(2,0),$ and $C=(0,2)$

Big Hint:

Find $M,O,P,Q$ and compare the two areas

Solution:

Area ratios are affine-invariant, so take $A=(0,0),$ $B=(2,0),$ and $C=(0,2).$ Then $\begin{gathered} M=(1,1),\\ O=(\frac{2}{3},\frac{2}{3}),\\ P=(0,1). \end{gathered}$ Line $MP$ is $y=1,$ and it meets median $CN$ at $Q=(\frac{1}{2},1).$ Thus $[OMQ]=\frac{1}{12},$ while $[ABC]=2.$ Their ratio is $24,$ so $[ABC]=24n.$

Therefore, the correct answer is D.

39.

In base $R_1$ the expanded fraction $F_1$ becomes $0.373737\ldots$ and the expanded fraction $F_2$ becomes $0.737373\ldots.$ In base $R_2$ fraction $F_1,$ when expanded, becomes $0.252525\ldots,$ while fraction $F_2$ becomes $0.525252\ldots.$ The sum of $R_1$ and $R_2,$ each written in base ten, is:

$24$

$22$

$21$

$20$

$19$

Concepts:number base repeating decimal system of equations

Difficulty rating: 1990

Small Hint:

A repeating pair $ab$ in base $R$ equals $\frac{aR+b}{R^2-1}$

Big Hint:

Add the equations for $F_1$ and $F_2,$ then subtract them

Solution:

The two descriptions give $\begin{aligned} F_1&=\frac{3R_1+7}{R_1^2-1}\\ &=\frac{2R_2+5}{R_2^2-1},\\ F_2&=\frac{7R_1+3}{R_1^2-1}\\ &=\frac{5R_2+2}{R_2^2-1}. \end{aligned}$ Adding yields $\frac{10}{R_1-1}=\frac{7}{R_2-1},$ or $10R_2-7R_1=3.$ Subtracting yields $\frac{4}{R_1+1}=\frac{3}{R_2+1},$ or $4R_2-3R_1=-1.$ Solving gives $R_1=11,$ $R_2=8,$ whose sum is $19.$

Therefore, the correct answer is E.

40.

In this figure $AB$ is a diameter of a circle, centered at $O,$ with radius $a.$ A chord $AD$ is drawn and extended to meet the tangent to the circle at $B,$ in point $C.$ Point $E$ is taken on $AC$ so that $AE=DC.$ If the coordinates of $E$ are $(x,y),$ then:

$y^2=\dfrac{x^3}{2a-x}$

$y^2=\dfrac{x^3}{2a+x}$

$y^4=\dfrac{x^2}{2a-x}$

$x^2=\dfrac{y^2}{2a-x}$

$x^2=\dfrac{y^2}{2a+x}$

Concepts:circle coordinate geometry similarity tangent line

Difficulty rating: 2060

Small Hint:

Drop perpendiculars from $E$ and $D$ to diameter $AB$

Big Hint:

Use $AE=DC$ between the two parallel tangents, then combine the altitude theorem with similar triangles

Solution:

Drop perpendiculars $EM$ and $DN$ to $AB.$ Since $AE=DC$ and the tangents through $A,B$ are parallel, their projections give $NB=x.$ Hence $AN=2a-x.$ In right triangle $ADB,$ the altitude theorem gives $DN^2=x(2a-x).$ Similar triangles $AME$ and $AND$ give $\frac{DN}{2a-x}=\frac{y}{x},$ so $DN=\frac{y(2a-x)}{x}.$ Substitution and cancellation yield $y^2=\frac{x^3}{2a-x}.$

Therefore, the correct answer is A.