1967 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

The three-digit number $2a3$ is added to the number $326$ to give the three-digit number $5b9.$ If $5b9$ is divisible by $9,$ then $a+b$ equals:

$2$

$4$

$6$

$8$

$9$

Concepts:place value divisibility

Difficulty rating: 1260

Small Hint:

Use the digit sum of $5b9$

Big Hint:

After finding $b,$ compare the tens digits in the addition

Solution:

Divisibility by $9$ requires $5+b+9$ to be a multiple of $9.$ Since $b$ is a digit, this gives $b=4.$ The addition is then $2a3+326=549,$ so $a=2.$ Hence $a+b=6.$

Therefore, the correct answer is C.

2.

An equivalent of the expression

$\begin{aligned} &\left(\frac{x^2+1}{x}\right) \left(\frac{y^2+1}{y}\right)\\ &\quad+ \left(\frac{x^2-1}{y}\right) \left(\frac{y^2-1}{x}\right),\\ &\qquad xy\ne0, \end{aligned}$

is:

$1$

$2xy$

$2x^2y^2+2$

$2xy+\dfrac{2}{xy}$

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

Concepts:algebraic manipulation fraction

Difficulty rating: 1450

Small Hint:

Expand each product into four terms

Big Hint:

Look for cancellation of the $\frac{x}{y}$ and $\frac{y}{x}$ terms

Solution:

The first product is $xy+\frac{x}{y}+\frac{y}{x}+\frac1{xy},$ while the second is $xy-\frac{x}{y}-\frac{y}{x}+\frac1{xy}.$ Adding gives $2xy+\frac{2}{xy}.$

Therefore, the correct answer is D.

3.

The side of an equilateral triangle is $s.$ A circle is inscribed in the triangle and a square is inscribed in the circle. The area of the square is:

$\dfrac{s^2}{24}$

$\dfrac{s^2}{6}$

$\dfrac{s^2\sqrt2}{6}$

$\dfrac{s^2\sqrt3}{6}$

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

Concepts:equilateral triangle incircle, incenter, and inradius square (geometry)area

Difficulty rating: 1430

Small Hint:

The inradius of an equilateral triangle of side $s$ is $\frac{s\sqrt3}{6}$

Big Hint:

The square’s diagonal is the circle’s diameter

Solution:

The inradius is $r=\frac{s\sqrt3}{6}.$ The inscribed square has diagonal $2r,$ so its area is half the square of that diagonal: $\begin{aligned} \frac{(2r)^2}{2} &=2r^2\\ &=2\left(\frac{s\sqrt3}{6}\right)^2\\ &=\frac{s^2}{6}. \end{aligned}$

Therefore, the correct answer is B.

4.

Given $\dfrac{\log a}{p}=\dfrac{\log b}{q}=\dfrac{\log c}{r}=\log x,$ all logarithms to the same base and $x\ne1.$ If $\dfrac{b^2}{ac}=x^y,$ then $y$ is:

$\dfrac{q^2}{p+r}$

$\dfrac{p+r}{2q}$

$2q-p-r$

$2q-pr$

$q^2-pr$

Concepts:logarithm exponent substitution

Difficulty rating: 1210

Small Hint:

Rewrite the logarithmic equalities as $a=x^p,$ $b=x^q,$ and $c=x^r$

Big Hint:

Substitute those powers into $\frac{b^2}{ac}$

Solution:

The given equalities imply $a=x^p,$ $b=x^q,$ and $c=x^r.$ Therefore $\frac{b^2}{ac} =x^{2q-p-r}.$ Since this is $x^y$ and $x\ne1,$ we have $y=2q-p-r.$

Therefore, the correct answer is C.

5.

A triangle is circumscribed about a circle of radius $r$ inches. If the perimeter of the triangle is $P$ inches and the area is $K$ square inches, then $\frac{P}{K}$ is:

independent of the value of $r$

$\dfrac{\sqrt2}{r}$

$\dfrac{2}{\sqrt r}$

$\dfrac2r$

$\dfrac r2$

Concepts:incircle, incenter, and inradius triangle area perimeter

Difficulty rating: 1440

Small Hint:

Split the triangle into three smaller triangles with altitude $r$

Big Hint:

Express $K$ using the semiperimeter $\frac{P}{2}$

Solution:

The three radii to the points of tangency split the triangle into smaller triangles whose total area is $K=\frac12r(a+b+c)=\frac{Pr}{2}.$ Hence $\frac{P}{K}=\frac{2}{r}.$

Therefore, the correct answer is D.

6.

If $f(x)=4^x,$ then $f(x+1)-f(x)$ equals:

$4$

$f(x)$

$2f(x)$

$3f(x)$

$4f(x)$

Concepts:function exponent

Difficulty rating: 960

Small Hint:

Write $4^{x+1}$ as $4\cdot4^x$

Big Hint:

Factor $4^x$ from the difference

Solution:

We have $f(x+1)=4^{x+1}=4f(x).$ Therefore $f(x+1)-f(x)=3f(x).$

Thus, the correct answer is D.

7.

If $\dfrac ab\lt-\dfrac cd,$ where $a,$ $b,$ $c,$ $d$ are real numbers and $bd\ne0,$ then:

$a$ must be negative

$a$ must be positive

$a$ must not be zero

$a$ can be negative or zero, but not positive

$a$ can be positive, negative, or zero

Concepts:inequality counterexample

Difficulty rating: 1430

Small Hint:

The signs of $b,c,d$ are unrestricted except that $b,d\ne0$

Big Hint:

Try $b=d=1$ and choose $c$ so that the right side is positive

Solution:

Take $b=d=1$ and $c=-2.$ The inequality becomes $a\lt2,$ which is satisfied by positive, zero, and negative values of $a.$ Thus none of the three possible signs is forced.

Therefore, the correct answer is E.

8.

To $m$ ounces of an $m\%$ solution of acid, $x$ ounces of water are added to yield an $(m-10)\%$ solution. If $m\gt25,$ then $x$ is:

$\dfrac{10m}{m-10}$

$\dfrac{5m}{m-10}$

$\dfrac{m}{m-10}$

$\dfrac{5m}{m-20}$

not determined by the given information

Concepts:mixture percentage rational equation

Difficulty rating: 1500

Small Hint:

The amount of acid remains $\frac{m^2}{100}$ ounces

Big Hint:

Set that equal to $\frac{m-10}{100}$ times the final volume $m+x$

Solution:

Conservation of acid gives $\frac{m^2}{100} =\frac{m-10}{100}(m+x).$ Hence $m^2=(m-10)(m+x),$ so $10m=(m-10)x$ and $x=\frac{10m}{m-10}.$

Therefore, the correct answer is A.

9.

Let $K,$ in square units, be the area of a trapezoid such that the shorter base, the altitude, and the longer base, in that order, are in arithmetic progression. Then:

$K$ must be an integer

$K$ must be a rational fraction

$K$ must be an irrational number

$K$ must be an integer or a rational fraction

taken alone neither $(A)$ nor $(B)$ nor $(C)$ nor $(D)$ is true

Concepts:trapezoid arithmetic sequence area

Difficulty rating: 1650

Small Hint:

Write the three lengths as $a-d,$ $a,$ and $a+d$

Big Hint:

The resulting area is $a^2,$ but the problem does not restrict the kind of number $a$ is

Solution:

Let the shorter base, altitude, and longer base be $a-d,$ $a,$ and $a+d.$ Then $\begin{aligned} K&=\frac12a\bigl((a-d)+(a+d)\bigr)\\ &=a^2. \end{aligned}$ Depending on $a,$ this can be an integer, a nonintegral rational number, or an irrational number. No one of choices (A) through (D) must hold.

Therefore, the correct answer is E.

10.

If

$\begin{aligned} &\frac{a}{10^x-1}+\frac{b}{10^x+2}\\ &\qquad=\frac{2\cdot10^x+3} {(10^x-1)(10^x+2)} \end{aligned}$

is an identity for positive rational values of $x,$ then the value of $a-b$ is:

$\dfrac43$

$\dfrac53$

$2$

$\dfrac{11}{4}$

$3$

Concepts:rational equation system of equations

Difficulty rating: 1500

Small Hint:

Multiply by $(10^x-1)(10^x+2)$

Big Hint:

Compare the coefficient of $10^x$ and the constant term

Solution:

Clearing denominators gives $\begin{aligned} &a(10^x+2)+b(10^x-1)\\ &\qquad=2\cdot10^x+3. \end{aligned}$ Therefore $a+b=2$ and $2a-b=3.$ Solving gives $a=\frac{5}{3},\ b=\frac{1}{3},$ so $a-b=\frac{4}{3}.$

Therefore, the correct answer is A.

11.

If the perimeter of rectangle $ABCD$ is $20$ inches, the least value of diagonal $AC,$ in inches, is:

$0$

$\sqrt{50}$

$10$

$\sqrt{200}$

none of these

Concepts:rectangle Pythagorean Theorem optimization

Difficulty rating: 1440

Small Hint:

If adjacent sides are $x$ and $10-x,$ square the diagonal

Big Hint:

Complete the square in $x^2+(10-x)^2$

Solution:

Let adjacent sides be $x$ and $10-x.$ Then $\begin{aligned} AC^2&=x^2+(10-x)^2\\ &=2(x-5)^2+50. \end{aligned}$ This is minimized at $x=5,$ giving $AC=\sqrt{50}.$

Therefore, the correct answer is B.

12.

If the (convex) area bounded by the $x$ -axis and the lines $y=mx+4,$ $x=1,$ and $x=4$ is $7,$ then $m$ equals:

$-\dfrac12$

$-\dfrac23$

$-\dfrac32$

$-2$

none of these

Concepts:coordinate geometry trapezoid linear equation

Difficulty rating: 1480

Small Hint:

The parallel vertical sides have lengths $m+4$ and $4m+4$

Big Hint:

Their separation is $3$ , so use the trapezoid-area formula

Solution:

The bounded region is a trapezoid with parallel sides $m+4$ and $4m+4,$ separated by $3.$ Thus $\frac32\bigl((m+4)+(4m+4)\bigr)=7.$ Hence $15m+24=14,$ so $m=-\frac{2}{3}.$ The endpoint heights are positive, as required for the stated convex region.

Therefore, the correct answer is B.

13.

A triangle $ABC$ is to be constructed given side $a$ (opposite angle $A$ ), angle $B,$ and $h_c,$ the altitude from $C.$ If $N$ is the number of noncongruent solutions, then $N$

is $1$

is $2$

must be zero

must be infinite

must be zero or infinite

Concepts:altitude trigonometry logical deduction

Difficulty rating: 2030

Small Hint:

Fix $BC=a$ and draw the ray from $B$ making the given angle

Big Hint:

The distance from $C$ to that ray is fixed, regardless of where $A$ lies on it

Solution:

Fix $BC=a$ and draw the ray from $B$ that makes the prescribed angle $B.$ Its distance from $C$ is fixed. If that distance is not $h_c,$ there is no solution. If it is $h_c,$ then $A$ may be chosen at infinitely many positions on the ray, producing infinitely many noncongruent triangles. Thus $N$ must be zero or infinite.

Therefore, the correct answer is E.

14.

Let $f(t)=\dfrac{t}{1-t},$ $t\ne1.$ If $y=f(x),$ then $x$ can be expressed as:

$f\left(\dfrac1y\right)$

$-f(y)$

$-f(-y)$

$f(-y)$

$f(y)$

Concepts:function rational equation algebraic manipulation

Difficulty rating: 1180

Small Hint:

Solve $y=\frac{x}{1-x}$ for $x$

Big Hint:

Compare $\frac{y}{1+y}$ with $f(-y)$

Solution:

From $y=\frac{x}{1-x}$ we obtain $y-yx=x,$ so $x=\frac{y}{1+y}.$ Since $f(-y)=-\frac{y}{1+y},$ this is $-f(-y).$

Therefore, the correct answer is C.

15.

The difference in the areas of two similar triangles is $18$ square feet, and the ratio of the larger area to the smaller is the square of an integer. The area of the smaller triangle, in square feet, is an integer, and one of its sides is $3$ feet. The corresponding side of the larger triangle, in feet, is:

$12$

$9$

$6\sqrt2$

$6$

$3\sqrt2$

Difficulty rating: 1850

Small Hint:

Let the integer side-scale factor be $k$

Big Hint:

If the smaller area is $T,$ then $T(k^2-1)=18$

Solution:

Let the side-scale factor be the integer $k\gt1$ and the smaller area be the integer $T.$ Then $T(k^2-1)=18.$ Thus $k^2-1$ is a positive divisor of $18.$ The only resulting square $k^2$ is $4,$ so $k=2.$ The corresponding side is $3k=6.$

Therefore, the correct answer is D.

16.

Let the product $(12)(15)(16),$ each factor written in base $b,$ equal $3146$ in base $b.$ Let $s=12+15+16,$ each term expressed in base $b.$ Then $s,$ in base $b,$ is:

$43$

$44$

$45$

$46$

$47$

Concepts:number base polynomial place value

Difficulty rating: 2130

Small Hint:

Translate the three factors and $3146_b$ into polynomials in $b$

Big Hint:

The resulting cubic has the valid base $b=9$

Solution:

The product equation is $\begin{aligned} &(b+2)(b+5)(b+6)\\ &\qquad=3b^3+b^2+4b+6. \end{aligned}$ It simplifies to $\begin{aligned} b^3-6b^2-24b-27&=0,\\ (b-9)(b^2+3b+3)&=0. \end{aligned}$ Thus the valid base is $b=9.$ The required sum is $3b+13=4b+4,$ which is $44_b.$

Therefore, the correct answer is B.

17.

If $r_1$ and $r_2$ are the distinct real roots of $x^2+px+8=0,$ then it must follow that:

$\lvert r_1+r_2\rvert\gt4\sqrt2$

$\lvert r_1\rvert\gt3$ or $\lvert r_2\rvert\gt3$

$\lvert r_1\rvert\gt2$ and $\lvert r_2\rvert\gt2$

$r_1\lt0$ and $r_2\lt0$

$\lvert r_1+r_2\rvert\lt4\sqrt2$

Concepts:quadratic Vieta’s Formulas inequality

Difficulty rating: 1180

Small Hint:

Distinct real roots require the discriminant to be positive

Big Hint:

Use $r_1+r_2=-p$

Solution:

The discriminant condition is $p^2-32\gt0,$ so $\lvert p\rvert\gt4\sqrt2.$ By Vieta’s formulas, $r_1+r_2=-p,$ and therefore $\lvert r_1+r_2\rvert=\lvert p\rvert\gt4\sqrt2.$

Therefore, the correct answer is A.

18.

If $x^2-5x+6\lt0$ and $P=x^2+5x+6,$ then

$P$ can take any real value

$20\lt P\lt30$

$0\lt P\lt20$

$P\lt0$

$P\gt30$

Concepts:quadratic inequality range

Difficulty rating: 1280

Small Hint:

Factor $x^2-5x+6$ to locate $x$

Big Hint:

Check how $x^2+5x+6$ varies on that interval

Solution:

The inequality $(x-2)(x-3)\lt0$ gives $2\lt x\lt3.$ On this interval $P=x^2+5x+6$ is increasing, with endpoint values $P(2)=20$ and $P(3)=30.$ The endpoints are excluded, so $20\lt P\lt30.$

Therefore, the correct answer is B.

19.

The area of a rectangle remains unchanged when it is made $2\dfrac12$ inches longer and $\dfrac23$ inch narrower, or when it is made $2\dfrac12$ inches shorter and $\dfrac43$ inch wider. Its area, in square inches, is:

$30$

$\dfrac{80}{3}$

$24$

$\dfrac{45}{2}$

$20$

Concepts:rectangle system of equations area

Difficulty rating: 1920

Small Hint:

Let the original length and width be $l$ and $w$

Big Hint:

Expand both equal-area equations; the $lw$ terms cancel

Solution:

The two conditions are $\begin{aligned} lw&=\left(l+\frac52\right)\left(w-\frac23\right),\\ lw&=\left(l-\frac52\right)\left(w+\frac43\right). \end{aligned}$ After expanding, these give $-4l+15w=10$ and $8l-15w=20.$ Hence $l=\frac{15}{2},$ $w=\frac{8}{3},$ and $lw=20.$

Therefore, the correct answer is E.

20.

A circle is inscribed in a square of side $m,$ then a square is inscribed in that circle, then a circle is inscribed in the latter square, and so on. If $S_n$ is the sum of the areas of the first $n$ circles so inscribed, then, as $n$ grows beyond all bounds, $S_n$ approaches:

$\dfrac{\pi m^2}{2}$

$\dfrac{3\pi m^2}{8}$

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

$\dfrac{\pi m^2}{4}$

$\dfrac{\pi m^2}{8}$

Concepts:circle area geometric sequence summation

Difficulty rating: 1480

Small Hint:

The first circle has area $\frac{\pi m^2}{4}$

Big Hint:

Each successive circle has half the preceding circle’s area

Solution:

The first circle has area $\pi(\frac{m}{2})^2=\frac{\pi m^2}{4}.$ Each inscribed square reduces the next circle’s squared radius, and hence its area, by a factor of $\frac{1}{2}.$ Thus the limiting sum is $\begin{aligned} &\frac{\pi m^2}{4} \left(1+\frac12+\frac14+\cdots\right)\\ &\qquad=\frac{\pi m^2}{2}. \end{aligned}$

Therefore, the correct answer is A.

21.

In right triangle $ABC$ the hypotenuse $AB=5$ and leg $AC=3.$ The bisector of angle $A$ meets the opposite side in $A_1.$ A second right triangle $PQR$ is then constructed with hypotenuse $PQ=A_1B$ and leg $PR=A_1C.$ If the bisector of angle $P$ meets the opposite side in $P_1,$ the length of $PP_1$ is:

$\dfrac{3\sqrt6}{4}$

$\dfrac{3\sqrt5}{4}$

$\dfrac{3\sqrt3}{4}$

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

$\dfrac{15\sqrt2}{16}$

Concepts:right triangle angle bisector theorem similarity

Difficulty rating: 2150

Small Hint:

Use the angle-bisector theorem in the $3$ - $4$ - $5$ triangle

Big Hint:

The second triangle is a half-scale copy of the first

Solution:

Since $BC=4,$ the angle-bisector theorem gives $A_1B:A_1C=5:3.$ Hence $A_1B=\frac{5}{2}$ and $A_1C=\frac{3}{2},$ so triangle $PQR$ is a half-scale copy of $ABC.$

For the angle bisector from $A$ in the original triangle, $\begin{aligned} AA_1^2 &=5\cdot3\left(1-\frac{4^2}{8^2}\right)\\ &=\frac{45}{4}, \end{aligned}$ so $AA_1=\frac{3\sqrt5}{2}.$ Therefore $PP_1$ is half of this, or $\frac{3\sqrt5}{4}.$

Thus, the correct answer is B.

22.

For natural numbers, when $P$ is divided by $D,$ the quotient is $Q$ and the remainder is $R.$ When $Q$ is divided by $D',$ the quotient is $Q'$ and the remainder is $R'.$ Then, when $P$ is divided by $DD',$ the remainder is:

$R+R'D$

$R'+RD$

$RR'$

$R$

$R'$

Concepts:divisibility modular arithmetic substitution

Difficulty rating: 1260

Small Hint:

Write $P=QD+R$ and $Q=Q'D'+R'$

Big Hint:

Substitute the second equation into the first and isolate the multiple of $DD'$

Solution:

Substitution gives $\begin{aligned} P&=(Q'D'+R')D+R\\ &=Q'(DD')+(R'D+R). \end{aligned}$ The remainder bounds imply $R'D+R\lt DD',$ so this is indeed the remainder.

Therefore, the correct answer is A.

23.

If $x$ is real and positive and grows beyond all bounds, then $\log_3(6x-5)-\log_3(2x+1)$ approaches:

$0$

$1$

$3$

$4$

no finite number

Concepts:logarithm bounding to limit cases

Difficulty rating: 1400

Small Hint:

Combine the logarithms into one logarithm

Big Hint:

Find the limit of $\frac{6x-5}{2x+1}$

Solution:

The difference is $\log_3\left(\frac{6x-5}{2x+1}\right).$ The fraction approaches $3,$ so the expression approaches $\log_3 3=1.$

Therefore, the correct answer is B.

24.

The number of solution-pairs in positive integers of the equation $3x+5y=501$ is:

$33$

$34$

$35$

$100$

none of these

Concepts:Diophantine Equation modular arithmetic counting integers in a range

Difficulty rating: 1500

Small Hint:

Reduce the equation modulo $5$ to find the form of $x$

Big Hint:

Count the positive values $x=2+5k$ that leave $y\gt0$

Solution:

Modulo $5,$ the equation gives $3x\equiv1,$ so $x\equiv2\pmod5.$ Write $x=2+5k.$ Then $y=\frac{501-3x}{5}=99-3k.$ Positivity requires $k=0,1,\ldots,32,$ giving $33$ pairs.

Therefore, the correct answer is A.

25.

For every odd number $p\gt1$ we have:

$(p-1)^{\frac{p-1}{2}}-1$ is divisible by $p-2$

$(p-1)^{\frac{p-1}{2}}+1$ is divisible by $p$

$(p-1)^{\frac{p-1}{2}}$ is divisible by $p$

$(p-1)^{\frac{p-1}{2}}+1$ is divisible by $p+1$

$(p-1)^{\frac{p-1}{2}}-1$ is divisible by $p-1$

Concepts:divisibility modular arithmetic exponent

Difficulty rating: 1500

Small Hint:

Work modulo $p-2$ for choice (A)

Big Hint:

Since $p-1\equiv1\pmod{p-2},$ every positive power has the same residue

Solution:

Because $p\gt1$ is odd, $n=\frac{p-1}{2}$ is a positive integer. Modulo $p-2,$ we have $p-1\equiv1.$ Therefore $\begin{aligned} (p-1)^n-1 &\equiv1^n-1\\ &\equiv0\pmod{p-2}. \end{aligned}$ Thus choice (A) always holds.

Therefore, the correct answer is A.

26.

If one uses only the tabular information $10^3=1000,$ $10^4=10{,}000,$ $2^{10}=1024,$ $2^{11}=2048,$ $2^{12}=4096,$ $2^{13}=8192,$ then the strongest statement one can make for $\log_{10}2$ is that it lies between:

$\dfrac3{10}$ and $\dfrac4{11}$

$\dfrac3{10}$ and $\dfrac4{12}$

$\dfrac3{10}$ and $\dfrac4{13}$

$\dfrac3{10}$ and $\dfrac{40}{132}$

$\dfrac3{11}$ and $\dfrac{40}{132}$

Concepts:logarithm inequality bounding to limit cases

Difficulty rating: 1710

Small Hint:

Compare $10^3$ with $2^{10}$ for a lower bound

Big Hint:

Compare $2^{13}$ with $10^4$ for an upper bound

Solution:

From $10^3\lt2^{10}$ we get $3\lt10\log_{10}2,$ so $\log_{10}2\gt\frac{3}{10}.$ From $2^{13}\lt10^4$ we get $13\log_{10}2\lt4,$ so $\log_{10}2\lt\frac{4}{13}.$ This is the narrowest listed interval justified by the table.

Therefore, the correct answer is C.

27.

Two candles of the same length are made of different materials so that one burns out completely at a uniform rate in $3$ hours and the other in $4$ hours. At what time P.M. should the candles be lighted so that, at $4$ P.M., one stub is twice the length of the other?

$1{:}24$

$1{:}28$

$1{:}36$

$1{:}40$

$1{:}48$

Concepts:rate linear equation date and time

Difficulty rating: 1500

Small Hint:

Let $t$ be the number of hours the candles burn before $4$ P.M.

Big Hint:

Their remaining fractions are $1-\frac{t}{3}$ and $1-\frac{t}{4}$

Solution:

After $t$ hours, the faster and slower candles have fractions $1-\frac{t}{3}$ and $1-\frac{t}{4}$ remaining. The slower stub must be twice the faster: $1-\frac t4=2\left(1-\frac t3\right).$ Thus $t=\frac{12}{5}=2$ hours $24$ minutes. Counting back from $4$ P.M. gives $1{:}36$ P.M.

Therefore, the correct answer is C.

28.

Given the two hypotheses: I Some Mems are not Ens and II No Ens are Vees. If “some” means “at least one,” we can conclude that:

Some Mems are not Vees

Some Vees are not Mems

No Mem is a Vee

Some Mems are Vees

Neither $(A)$ nor $(B)$ nor $(C)$ nor $(D)$ is deducible from the given statements

Concepts:logical deduction Venn Diagram counterexample

Difficulty rating: 1880

Small Hint:

Translate the three kinds of objects into sets $M,N,V$

Big Hint:

Test both $M=V$ and $M\cap V=\varnothing$ while keeping the two hypotheses true

Solution:

The hypotheses say only that some element of $M$ lies outside $N,$ and that $N\cap V=\varnothing.$ A model with $M=V$ makes (A), (B), and (C) false while satisfying the hypotheses. A model with $M\cap V=\varnothing$ makes (D) false. Hence none of choices (A) through (D) is forced.

Therefore, the correct answer is E.

29.

$AB$ is a diameter of a circle. Tangents $AD$ and $BC$ are drawn so that $AC$ and $BD$ intersect in a point on the circle. If $AD=a$ and $BC=b,$ $a\ne b,$ the diameter of the circle is:

$\lvert a-b\rvert$

$\dfrac12(a+b)$

$\sqrt{ab}$

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

$\dfrac12\cdot\dfrac{ab}{a+b}$

Concepts:circle tangent line similarity inscribed angle

Difficulty rating: 2150

Small Hint:

Let the intersection point on the circle be $P$ ; then $\angle APB=90^\circ$

Big Hint:

Use the parallel tangents to compare right triangles $ADB$ and $BCA$

Solution:

Let $d=AB$ be the diameter. Since the intersection of $AC$ and $BD$ lies on the circle, those two lines are perpendicular by Thales’ theorem. Also the tangents $AD$ and $BC$ are parallel. The resulting right triangles $ADB$ and $BCA$ are similar, so $\frac da=\frac bd.$ Hence $d^2=ab$ and $d=\sqrt{ab}.$

Therefore, the correct answer is C.

30.

A dealer bought $n$ radios for $d$ dollars, $d,$ a positive integer. He contributed two radios to a community bazaar at half their cost. The rest he sold at a profit of $\$8$ on each radio sold. If the overall profit was $\$72,$ then the least possible value of $n$ for the given information is:

$18$

$16$

$15$

$12$

$11$

Concepts:money linear equation divisibility

Difficulty rating: 1880

Small Hint:

Each radio costs $\frac{d}{n}$ dollars

Big Hint:

Write the total intake from $n-2$ profitable sales and two half-cost sales

Solution:

The $n-2$ regular sales bring $(n-2)(\frac{d}{n}+8),$ while the two bazaar radios bring $\frac{d}{n}$ together. Since the total intake is $d+72,$ $\begin{aligned} &(n-2)\left(\frac dn+8\right)+\frac dn\\ &\qquad=d+72. \end{aligned}$ This reduces to $d=8n(n-11).$ Positivity requires $n\gt11,$ and $n=12$ gives the positive integer $d=96.$

Therefore, the correct answer is D.

31.

Let $D=a^2+b^2+c^2,$ where $a,$ $b$ are consecutive integers and $c=ab.$ Then $\sqrt D$ is:

always an even integer

sometimes an odd integer, sometimes not

always an odd integer

sometimes rational, sometimes not

always irrational

Concepts:algebraic manipulation parity perfect square

Difficulty rating: 1500

Small Hint:

Set $b=a+1$ and expand $D$

Big Hint:

Try to recognize $D$ as the square of $a^2+a+1$

Solution:

With $b=a+1$ and $c=a(a+1),$ $\begin{aligned} D&=a^2+(a+1)^2+a^2(a+1)^2\\ &=(a^2+a+1)^2. \end{aligned}$ Thus $\sqrt D=a^2+a+1.$ Since $a(a+1)$ is even, this integer is always odd.

Therefore, the correct answer is C.

32.

In quadrilateral $ABCD$ with diagonals $AC$ and $BD$ intersecting at $O,$ $BO=4,$ $OD=6,$ $AO=8,$ $OC=3,$ and $AB=6.$ The length of $AD$ is:

$9$

$10$

$6\sqrt3$

$8\sqrt2$

$\sqrt{166}$

Concepts:law of cosines angle chasing

Difficulty rating: 1690

Small Hint:

Use triangle $AOB$ to find $\cos\angle AOB$

Big Hint:

Angles $AOB$ and $AOD$ are supplementary

Solution:

In triangle $AOB,$ $\begin{aligned} \cos\angle AOB &=\frac{8^2+4^2-6^2}{2\cdot8\cdot4}\\ &=\frac{11}{16}. \end{aligned}$ Therefore $\cos\angle AOD=-\frac{11}{16}.$ Applying the law of cosines in triangle $AOD$ gives $\begin{aligned} AD^2 &=8^2+6^2\\ &\quad+2(8)(6)\frac{11}{16}\\ &=166. \end{aligned}$ Hence $AD=\sqrt{166}.$

Therefore, the correct answer is E.

33.

In this diagram semi-circles are constructed on diameters $AB,$ $AC,$ and $CB,$ so that they are mutually tangent. If $CD\perp AB,$ then the ratio of the shaded area to the area of a circle with $CD$ as radius is:

$1:2$

$1:3$

$\sqrt3:7$

$1:4$

$\sqrt2:6$

Concepts:circle area right triangle area ratio

Difficulty rating: 1990

Small Hint:

Subtract the two small semicircle areas from the large one

Big Hint:

In right triangle $ADB,$ the altitude theorem gives $CD^2=AC\cdot CB$

Solution:

Let $AC=u$ and $CB=v.$ The shaded area is the large semicircle minus the two smaller ones: $\frac{\pi}{8}\bigl((u+v)^2-u^2-v^2\bigr) =\frac{\pi uv}{4}.$ Since $D$ lies on the semicircle with diameter $AB,$ triangle $ADB$ is right, and its altitude satisfies $CD^2=uv.$ A circle of radius $CD$ therefore has area $\pi uv.$ The required ratio is $1:4.$

Therefore, the correct answer is D.

34.

Points $D,$ $E,$ $F$ are taken respectively on sides $AB,$ $BC,$ and $CA$ of triangle $ABC$ so that $AD:DB=1:n,$ $BE:CE=1:n,$ and $CF:FA=1:n.$ The ratio of the area of triangle $DEF$ to that of triangle $ABC$ is:

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

$\dfrac1{(n+1)^2}$

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

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

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

Concepts:triangle area area decomposition ratio and proportion

Difficulty rating: 1710

Small Hint:

Subtract the three corner triangles from triangle $ABC$

Big Hint:

Each corner triangle has area ratio $\frac{n}{(n+1)^2}$

Solution:

At each vertex, the two adjacent side fractions used by the corner triangle are $\frac{1}{n+1}$ and $\frac{n}{n+1}.$ Thus each corner triangle has area $\frac{n}{(n+1)^2}$ times $[ABC].$ Therefore $\begin{aligned} \frac{[DEF]}{[ABC]} &=1-\frac{3n}{(n+1)^2}\\ &=\frac{n^2-n+1}{(n+1)^2}. \end{aligned}$

Therefore, the correct answer is A.

35.

The roots of $64x^3-144x^2+92x-15=0$ are in arithmetic progression. The difference between the largest and smallest roots is:

$2$

$1$

$\dfrac12$

$\dfrac38$

$\dfrac14$

Concepts:polynomial arithmetic sequence Vieta’s Formulas

Difficulty rating: 1750

Small Hint:

Write the roots as $t-d,$ $t,$ and $t+d$

Big Hint:

Use their sum to find $t,$ then use their product to find $d^2$

Solution:

Let the roots be $t-d,$ $t,$ and $t+d.$ Their sum is $3t=\frac{144}{64}=\frac{9}{4},$ so $t=\frac{3}{4}.$ Their product is $t(t^2-d^2)=\frac{15}{64}.$ Substituting $t=\frac{3}{4}$ gives $d^2=\frac{1}{4},$ so the difference between the extreme roots is $2\lvert d\rvert=1.$

Therefore, the correct answer is B.

36.

Given a geometric progression of five terms, each a positive integer less than $100.$ The sum of the five terms is $211.$ If $S$ is the sum of those terms in the progression which are squares of integers, then $S$ is:

$0$

$91$

$133$

$195$

$211$

Concepts:geometric sequence divisibility perfect square casework

Difficulty rating: 2380

Small Hint:

Write the rational common ratio in lowest terms as $\frac{c}{d}$

Big Hint:

Integrality forces the middle term to be a multiple of $c^2d^2$ ; use that $211$ is prime

Solution:

Let the common ratio be $\frac{c}{d}$ in lowest terms and the middle term be $a.$ Since all five terms are integers, $a$ is divisible by $c^2d^2,$ so write $a=kc^2d^2.$ The sum $211$ is then divisible by $k.$ Thus $k$ is $1$ or $211.$ But the middle term $a$ is less than $100,$ so $k=1.$

Thus $\begin{aligned} d^4+d^3c+d^2c^2&\\ \quad+dc^3+c^4&=211. \end{aligned}$ The bound gives $c,d\lt4.$ If either is $1,$ the possible sums are $5,$ $31,$ or $121,$ not $211.$ Coprimality therefore leaves $c$ and $d$ equal to $2$ and $3$ in either order. The progression is $16,$ $24,$ $36,$ $54,$ $81.$ Its square terms sum to $16+36+81=133.$

Therefore, the correct answer is C.

37.

Segments $AD=10,$ $BE=6,$ $CF=24$ are drawn from the vertices of triangle $ABC,$ each perpendicular to a straight line $RS,$ not intersecting the triangle. Points $D,$ $E,$ $F$ are the intersection points of $RS$ with the perpendiculars. If $x$ is the length of the perpendicular segment $GH$ drawn to $RS$ from the intersection point $G$ of the medians of the triangle, then $x$ is:

$\dfrac{40}{3}$

$16$

$\dfrac{56}{3}$

$\dfrac{80}{3}$

undetermined

Concepts:centroid vector mean

Difficulty rating: 1480

Small Hint:

Signed distance from a point to a fixed line is an affine function

Big Hint:

The centroid is the average of the three vertices

Solution:

Because $RS$ does not intersect the triangle, the three perpendicular distances have the same sign. Signed distance to a fixed line is affine, and the centroid is the average of the vertices. Therefore its distance is the average $x=\frac{10+6+24}{3}=\frac{40}{3}.$

Therefore, the correct answer is A.

38.

Given a set $S$ consisting of two undefined elements “pib” and “maa,” and the four postulates:

$\mathrm{P}_1:$ Every pib is a collection of maas.

$\mathrm{P}_2:$ Any two distinct pibs have one and only one maa in common.

$\mathrm{P}_3:$ Every maa belongs to two and only two pibs.

$\mathrm{P}_4:$ There are exactly four pibs.

Consider the three theorems:

$\mathrm{T}_1:$ There are exactly six maas.

$\mathrm{T}_2:$ There are exactly three maas in each pib.

$\mathrm{T}_3:$ For each maa there is exactly one other maa not in the same pib with it.

The theorems which are deducible from the postulates are:

$\mathrm{T}_3$ only

$\mathrm{T}_2$ and $\mathrm{T}_3$ only

$\mathrm{T}_1$ and $\mathrm{T}_2$ only

$\mathrm{T}_1$ and $\mathrm{T}_3$ only

all

Concepts:basic counting graph theory logical deduction

Difficulty rating: 2030

Small Hint:

Label the four pibs $1,$ $2,$ $3,$ and $4$

Big Hint:

Each maa corresponds to an unordered pair of pibs

Solution:

Label the four pibs $1,$ $2,$ $3,$ and $4.$ By $\mathrm P_2$ and $\mathrm P_3,$ every maa is exactly the common maa of one unordered pair of pibs. Thus there are $\binom42=6$ maas, proving $\mathrm T_1.$

A fixed pib $i$ contains the three maas $ij$ with $j\ne i,$ proving $\mathrm T_2.$ For maa $ij,$ the unique maa sharing neither pib is the one belonging to the complementary pair of pibs, proving $\mathrm T_3.$ All three follow.

Therefore, the correct answer is E.

39.

Given the sets of consecutive integers $\{1\},$ $\{2,3\},$ $\{4,5,6\},$ $\{7,8,9,10\},$ $\ldots,$ where each set contains one more element than the preceding one, and where the first element of each succeeding set is one more than the last element of the preceding set. Let $S_n$ be the sum of the elements in the $n$ th set. Then $S_{21}$ equals:

$1113$

$4641$

$5082$

$53361$

none of these

Concepts:arithmetic sequence triangular number summation

Difficulty rating: 1500

Small Hint:

The last number in the $n$ th set is $\frac{n(n+1)}{2}$

Big Hint:

Sum the $n$ consecutive integers ending at that number

Solution:

The $n$ th set ends at $\frac{n(n+1)}{2}$ and contains $n$ consecutive integers. Its sum is $\begin{aligned} S_n &=n\left(\frac{n(n+1)}2\right)\\ &\quad-\frac{n(n-1)}2\\ &=\frac{n(n^2+1)}2. \end{aligned}$ Thus $S_{21}=\frac{21(442)}{2}=4641.$

Therefore, the correct answer is B.

40.

Located inside equilateral triangle $ABC$ is a point $P$ such that $PA=8,$ $PB=6,$ and $PC=10.$ To the nearest integer the area of triangle $ABC$ is:

$159$

$131$

$95$

$79$

$50$

Concepts:equilateral triangle transformation Pythagorean Triple law of cosines area

Difficulty rating: 2370

Small Hint:

Rotate $P$ by $60^\circ$ about $A$ so that $C$ maps to $B$

Big Hint:

The resulting $6$ - $8$ - $10$ triangle determines $\angle APB$

Solution:

Rotate $P$ by $60^\circ$ about $A$ to $P'.$ Since this rotation sends $C$ to $B,$ we have $P'B=PC=10,$ $PP'=PA=8,$ and $PB=6.$ Thus triangle $PP'B$ is right. Also triangle $APP'$ is equilateral, so $\angle APB=60^\circ+90^\circ=150^\circ.$

If the equilateral triangle has side $s,$ the law of cosines in triangle $APB$ gives $\begin{aligned} s^2 &=6^2+8^2-2(6)(8)\cos150^\circ\\ &=100+48\sqrt3. \end{aligned}$ Its area is $\frac{\sqrt3}{4}s^2=36+25\sqrt3,$ whose nearest integer is $79.$

Therefore, the correct answer is D.