1992 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

If $3(4x+5\pi)=P,$ then $6(8x+10\pi)=$

$2P$

$4P$

$6P$

$8P$

$18P$

Concepts:algebraic scaling distributive property

Difficulty rating: 770

Small Hint:

Compare $8x+10\pi$ with $4x+5\pi$

Big Hint:

Account separately for the change inside the parentheses and the change in the outside coefficient

Solution:

We have $8x+10\pi=2(4x+5\pi),$ so $\begin{gathered} 6(8x+10\pi)\\ {}=12(4x+5\pi)\\ {}=4\bigl[3(4x+5\pi)\bigr]\\ {}=4P. \end{gathered}$

Thus the correct answer is B.

2.

An urn is filled with coins and beads, all of which are either silver or gold. Twenty percent of the objects in the urn are beads. Forty percent of the coins in the urn are silver. What percent of the objects in the urn are gold coins?

$40\%$

$48\%$

$52\%$

$60\%$

$80\%$

Concepts:percent complementary percentage

Difficulty rating: 890

Small Hint:

First find the percent of all objects that are coins

Big Hint:

Of those coins, use the complement of the silver percentage

Solution:

Coins make up $80\%$ of the objects, and $60\%$ of the coins are gold. Thus gold coins make up $0.80\cdot0.60=0.48=48\%$ of all the objects.

Thus the correct answer is B.

3.

If $m\gt0$ and the points $(m,3)$ and $(1,m)$ lie on a line with slope $m,$ then $m=$

$1$

$\sqrt2$

$\sqrt3$

$2$

$\sqrt5$

Concepts:slope formula quadratic equation

Difficulty rating: 1420

Small Hint:

Write the slope between the two given points

Big Hint:

Set $\frac{m-3}{1-m}=m$ and use the positivity condition

Solution:

The slope condition gives $\frac{m-3}{1-m}=m.$ Therefore $m-3=m-m^2,$ so $m^2=3.$ Since $m\gt0,$ we obtain $m=\sqrt3.$

Thus the correct answer is C.

4.

If $a,$ $b,$ and $c$ are positive integers and $a$ and $b$ are odd, then $3^a+(b-1)^2c$ is

odd for all choices of $c$

even for all choices of $c$

odd if $c$ is even; even if $c$ is odd

odd if $c$ is odd; even if $c$ is even

odd if $c$ is not a multiple of $3$ ; even if $c$ is a multiple of $3$

Concepts:parity powers

Difficulty rating: 960

Small Hint:

Determine the parity of each of the two terms

Big Hint:

Because $b$ is odd, $b-1$ is even

Solution:

The power $3^a$ is odd. Since $b-1$ is even, $(b-1)^2c$ is even for every positive integer $c.$ The sum of an odd integer and an even integer is always odd.

Thus the correct answer is A.

5.

$6^6+6^6+6^6+6^6+6^6+6^6=$

$6^6$

$6^7$

$36^6$

$6^{36}$

$36^{36}$

Concepts:repeated addition exponent rule

Difficulty rating: 890

Small Hint:

Factor the repeated summand $6^6$

Big Hint:

There are exactly six identical terms

Solution:

Factoring gives $6\cdot6^6=6^{1+6}=6^7.$

Thus the correct answer is B.

6.

If $x\gt y\gt0,$ then $\frac{x^y y^x}{y^y x^x}=$

$(x-y)^{\frac{y}{x}}$

$\left(\frac{x}{y}\right)^{x-y}$

$1$

$\left(\frac{x}{y}\right)^{y-x}$

$(x-y)^{\frac{x}{y}}$

Concepts:laws of exponents simplifying quotients

Difficulty rating: 1570

Small Hint:

Collect the powers of $x$ and the powers of $y$ separately

Big Hint:

Rewrite $x^{y-x}y^{x-y}$ using a single ratio

Solution:

Combining like bases, $\frac{x^y y^x}{y^y x^x} =x^{y-x}y^{x-y} =\left(\frac{x}{y}\right)^{y-x}.$

Thus the correct answer is D.

7.

The ratio of $w$ to $x$ is $4:3,$ of $y$ to $z$ is $3:2,$ and of $z$ to $x$ is $1:6.$ What is the ratio of $w$ to $y?$

$1:3$

$16:3$

$20:3$

$27:4$

$12:1$

Concepts:ratios proportional reasoning

Difficulty rating: 1260

Small Hint:

Express both $w$ and $y$ as multiples of $x$

Big Hint:

Use $\frac{w}{y}=\frac{w}{x}\cdot\frac{x}{z}\cdot\frac{z}{y}$

Solution:

Multiplying the given ratios, $\begin{aligned} \frac wy &=\frac wx\cdot\frac xz\cdot\frac zy\\ &=\frac43\cdot6\cdot\frac23=\frac{16}{3}. \end{aligned}$ Thus $w:y=16:3.$

Thus the correct answer is B.

8.

A square floor is tiled with congruent square tiles. The tiles on the two diagonals of the floor are black. The rest of the tiles are white. If there are $101$ black tiles, then the total number of tiles is

$121$

$625$

$676$

$2500$

$2601$

Concepts:square grid diagonals counting overlap

Difficulty rating: 1260

Small Hint:

If the floor has $n$ tiles on a side, count the tiles on both diagonals

Big Hint:

Because the total number of black tiles is odd, the two diagonals share one center tile

Solution:

For an odd $n,$ each diagonal contains $n$ tiles and the two diagonals share only the center tile. Hence $2n-1=101,$ so $n=51.$ The floor therefore contains $51^2=2601$ tiles.

Thus the correct answer is E.

9.

Five equilateral triangles, each with side $2\sqrt3,$ are arranged so they are all on the same side of a line containing one side of each. Along this line, the midpoint of the base of one triangle is a vertex of the next. The area of the region of the plane that is covered by the union of the five triangular regions is

$10$

$12$

$15$

$10\sqrt3$

$12\sqrt3$

Concepts:equilateral triangle area inclusion-exclusion overlapping regions

Difficulty rating: 1780

Small Hint:

Add the areas of the five large triangles, then subtract their overlaps

Big Hint:

Each adjacent overlap is an equilateral triangle with half the original side length

Solution:

Each large triangle has area $\frac{\sqrt3}{4}(2\sqrt3)^2=3\sqrt3.$ Each of the four overlaps is an equilateral triangle of side $\sqrt3,$ hence area $\frac{3\sqrt3}{4}.$ There are no triple overlaps, so the union has area $5(3\sqrt3)-4\left(\frac{3\sqrt3}{4}\right)=12\sqrt3.$

Thus the correct answer is E.

10.

The number of positive integers $k$ for which the equation $kx-12=3k$ has an integer solution for $x$ is

$3$

$4$

$5$

$6$

$7$

Concepts:linear equation divisibility positive divisors

Difficulty rating: 1420

Small Hint:

Solve the equation for $x$ in terms of $k$

Big Hint:

The expression $3+\frac{12}{k}$ is integral exactly when $k$ is a positive divisor of $12$

Solution:

Solving gives $x=3+\frac{12}{k}.$ Thus $x$ is an integer precisely when $k$ is a positive divisor of $12.$ The possibilities are $1,$ $2,$ $3,$ $4,$ $6,$ and $12,$ for a total of $6.$

Thus the correct answer is D.

11.

The ratio of the radii of two concentric circles is $1:3.$ If $\overline{AC}$ is a diameter of the larger circle, $\overline{BC}$ is a chord of the larger circle that is tangent to the smaller circle, and $AB=12,$ then the radius of the larger circle is

$13$

$18$

$21$

$24$

$26$

Concepts:circle tangent right triangle similar triangles

Difficulty rating: 1810

Small Hint:

Draw the radius to the point where $\overline{BC}$ touches the smaller circle

Big Hint:

Use the similar right triangles formed by the tangent radius and the diameter

Solution:

Let $O$ be the common center and $T$ the tangency point on $\overline{BC}.$ Since $OT\perp BC$ and $\angle ABC=90^\circ,$ triangles $OTC$ and $ABC$ are similar. If the larger radius is $R,$ then $OT=\frac{R}{3},$ $OC=R,$ and $AC=2R.$ Thus $\frac{OT}{AB}=\frac{OC}{AC}=\frac12.$ Hence $OT=6$ and $R=3OT=18.$

Thus the correct answer is B.

12.

Let $y=mx+b$ be the image when the line $x-3y+11=0$ is reflected across the $x$ -axis. The value of $m+b$ is

$-6$

$-5$

$-4$

$-3$

$-2$

Concepts:reflection equation of a line slope-intercept form

Difficulty rating: 1420

Small Hint:

Reflection across the $x$ -axis replaces each $y$ -coordinate by its negative

Big Hint:

Replace $y$ by $-y$ in the original equation, then solve for $y$

Solution:

A point $(x,y)$ is on the reflected line exactly when $(x,-y)$ is on the original line. Hence $x+3y+11=0,$ or $y=-\frac13x-\frac{11}{3}.$ Therefore $m+b=-\frac13-\frac{11}{3}=-4.$

Thus the correct answer is C.

13.

How many pairs of positive integers $(a,b)$ with $a+b\le100$ satisfy the equation $\frac{a+b^{-1}}{a^{-1}+b}=13?$

$1$

$5$

$7$

$9$

$13$

Concepts:rational expression positive integer solutions divisibility

Difficulty rating: 1800

Small Hint:

Multiply the numerator and denominator by $ab$

Big Hint:

After simplifying, write $a$ as a multiple of $b$ and apply the bound on $a+b$

Solution:

Multiplying the numerator and denominator by $ab$ gives $\frac{a^2b+a}{b+ab^2} =\frac{a(ab+1)}{b(ab+1)} =\frac ab.$ Thus $a=13b.$ The condition $a+b\le100$ becomes $14b\le100,$ so $b$ may be any integer from $1$ through $7.$ There are $7$ pairs.

Thus the correct answer is C.

14.

Which of the following equations have the same graph? $\begin{aligned} &\text{I. }y=x-2\\ &\text{II. }y=\dfrac{x^2-4}{x+2}\\ &\text{III. }(x+2)y=x^2-4 \end{aligned}$

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

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

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

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

None. All the equations have different graphs

Concepts:graph domain restriction extraneous component

Difficulty rating: 1960

Small Hint:

Pay special attention to what happens when $x=-2$

Big Hint:

Equation II omits one point, while equation III includes an entire extra vertical line

Solution:

Equation I is the whole line $y=x-2.$ Equation II simplifies to that line only when $x\ne-2,$ so it omits $(-2,-4).$ In equation III, every point with $x=-2$ satisfies $0=0;$ for other $x$ it gives $y=x-2.$ Thus III is the line together with the entire vertical line $x=-2.$ The three graphs are different.

Thus the correct answer is E.

15.

Let $i=\sqrt{-1}.$ For $n\ge1,$ define a sequence of complex numbers by $\begin{aligned} z_1&=0,\\ z_{n+1}&=z_n^2+i. \end{aligned}$ In the complex plane, how far from the origin is $z_{111}?$

$1$

$\sqrt2$

$\sqrt3$

$\sqrt{110}$

$\sqrt{2^{55}}$

Concepts:complex number recurrence periodic sequence modulus

Difficulty rating: 1960

Small Hint:

Compute the first few terms rather than trying to expand $z_{111}$

Big Hint:

Look for a cycle beginning with $z_3$

Solution:

The first terms are $\begin{aligned} z_1&=0,\quad z_2=i,\quad z_3=-1+i,\\ z_4&=-i,\quad z_5=-1+i. \end{aligned}$ Hence the sequence alternates between $-1+i$ at odd indices and $-i$ at even indices from $z_3$ onward. Therefore $z_{111}=-1+i,$ whose distance from the origin is $\sqrt{(-1)^2+1^2}=\sqrt2.$

Thus the correct answer is B.

16.

If $\frac{y}{x-z}=\frac{x+y}{z}=\frac{x}{y}$ for three positive numbers $x,$ $y,$ and $z,$ all different, then $\frac{x}{y}=$

$\frac12$

$\frac35$

$\frac23$

$\frac53$

$2$

Concepts:proportion algebraic system quadratic equation

Difficulty rating: 2000

Small Hint:

Let $r=\frac{x}{y}$ and express $\frac{z}{y}$ using the middle ratio

Big Hint:

Substitute $\frac{z}{y}=\frac{r+1}{r}$ into the first ratio

Solution:

Let $r=\frac{x}{y}\gt0.$ From $\frac{x+y}{z}=r,$ we get $\frac{z}{y}=\frac{r+1}{r}.$ Then $\frac{y}{x-z} =\frac{1}{r-\frac{r+1}{r}} =r.$ Thus $r^2-r-1=1,$ or $r^2-r-2=0.$ Hence $r=2$ or $-1,$ and positivity gives $r=2.$

Thus the correct answer is E.

17.

The two-digit integers from $19$ to $92$ are written consecutively to form the large integer $N=19202122\ldots909192.$ If $3^k$ is the highest power of $3$ that is a factor of $N,$ then $k=$

$0$

$1$

$2$

$3$

more than $3$

Concepts:divisibility by 3 and 9 digit sum arithmetic series

Difficulty rating: 1830

Small Hint:

Use the digit-sum tests for divisibility by $3$ and $9$

Big Hint:

Modulo $9,$ the digit sum of the concatenation equals $19+20+\cdots+92$

Solution:

Modulo $9,$ a number is congruent to its digit sum, so $\begin{aligned} N&\equiv19+20+\cdots+92\\ &=\frac{74(19+92)}2\\ &=4107\equiv3\pmod9. \end{aligned}$ Thus $N$ is divisible by $3$ but not by $9,$ so the highest power of $3$ dividing it is $3^1.$

Thus the correct answer is B.

18.

The increasing sequence of positive integers $a_1,$ $a_2,$ $a_3,$ $\ldots$ has the property that, for every $n\ge1,$ $a_{n+2}=a_n+a_{n+1}.$ If $a_7=120,$ then $a_8$ is

$128$

$168$

$193$

$194$

$210$

Concepts:Fibonacci recurrence Diophantine equation increasing sequence

Difficulty rating: 1960

Small Hint:

Write $a_7$ and $a_8$ in terms of $a_1$ and $a_2$

Big Hint:

From $5a_1+8a_2=120,$ use congruences and $a_2\gt a_1\gt0$

Solution:

Let $a_1=a$ and $a_2=b.$ Repeated use of the recurrence gives $\begin{aligned} a_7&=5a+8b=120,\\ a_8&=8a+13b. \end{aligned}$ Modulo $5,$ the first equation shows that $b$ is a multiple of $5.$ The positive possibilities are $b=5,$ $b=10,$ or $b=15.$ The conditions $b\gt a\gt0$ leave only $b=10$ and $a=8.$ Hence $a_8=8(8)+13(10)=194.$

Thus the correct answer is D.

19.

For each vertex of a solid cube, consider the tetrahedron determined by the vertex and the midpoints of the three edges that meet at that vertex. The portion of the cube that remains when these eight tetrahedra are cut away is called a cuboctahedron. The ratio of the volume of the cuboctahedron to the volume of the original cube is closest to which of these?

$75\%$

$78\%$

$81\%$

$84\%$

$87\%$

Concepts:solid geometry tetrahedron volume volume ratio

Difficulty rating: 1960

Small Hint:

Choose a convenient side length for the cube and compute one removed corner tetrahedron

Big Hint:

Each removed tetrahedron has three mutually perpendicular edges equal to half the cube’s side

Solution:

Take the cube’s side length to be $2.$ Its volume is $8.$ Each corner tetrahedron has three perpendicular edges of length $1,$ so its volume is $\frac16.$ The eight removed tetrahedra have total volume $\frac43,$ leaving $\frac{20}{3}.$ The ratio is $\frac{\frac{20}{3}}{8}=\frac56=83\tfrac13\%,$ which is closest to $84\%.$

Thus the correct answer is D.

20.

Part of an “ $n$ -pointed regular star” is shown. It is a simple closed polygon in which all $2n$ edges are congruent, angles $A_1,$ $A_2,$ $\ldots,$ $A_n$ are congruent, and angles $B_1,$ $B_2,$ $\ldots,$ $B_n$ are congruent. If the acute angle at $A_1$ is $10^\circ$ less than the acute angle at $B_1,$ then $n=$

$12$

$18$

$24$

$36$

$60$

Concepts:polygon angle sum concave polygon regular star

Difficulty rating: 2110

Small Hint:

Let the acute angles at the $A$ - and $B$ -vertices be $\alpha$ and $\beta$

Big Hint:

At each inward $B$ -vertex, the polygon’s interior angle is $360^\circ-\beta$

Solution:

Let the acute angles at the tips and notches be $\alpha$ and $\beta,$ respectively, so $\beta-\alpha=10^\circ.$ The $2n$ -gon has $n$ interior angles $\alpha$ and $n$ reflex interior angles $360^\circ-\beta.$ Therefore $\begin{gathered} n\alpha+n(360^\circ-\beta)\\ {}=(2n-2)180^\circ. \end{gathered}$ Substituting $\beta-\alpha=10^\circ$ gives $350n=360n-360,$ so $n=36.$

Thus the correct answer is D.

21.

For a finite sequence $A=(a_1,a_2,\ldots,a_n)$ of numbers, the Cesàro sum of $A$ is defined to be $\frac{S_1+S_2+\cdots+S_n}{n},$ where $\begin{aligned} S_k&=a_1+a_2+a_3+\cdots+a_k\\ &\qquad(1\le k\le n). \end{aligned}$ If the Cesàro sum of the $99$ -term sequence $(a_1,a_2,\ldots,a_{99})$ is $1000,$ what is the Cesàro sum of the $100$ -term sequence $(1,a_1,a_2,\ldots,a_{99})?$

$991$

$999$

$1000$

$1001$

$1009$

Concepts:partial sums sequence transformation weighted average

Difficulty rating: 2000

Small Hint:

Convert the first Cesàro sum into the value of $S_1+\cdots+S_{99}$

Big Hint:

Each new partial sum is $1$ plus an old partial sum, with one initial partial sum equal to $1$

Solution:

The given condition says $S_1+\cdots+S_{99}=99{,}000.$ For the new sequence, the partial sums are $1,1+S_1,\ldots,1+S_{99}.$ Their sum is $100+(S_1+\cdots+S_{99})=99{,}100.$ Dividing by $100$ gives the new Cesàro sum $991.$

Thus the correct answer is A.

22.

Ten points are selected on the positive $x$ -axis, $\mathrm{X}^+,$ and five points are selected on the positive $y$ -axis, $\mathrm{Y}^+.$ The fifty segments connecting the ten selected points on $\mathrm{X}^+$ to the five selected points on $\mathrm{Y}^+$ are drawn. What is the maximum possible number of points of intersection of these fifty segments that could lie in the interior of the first quadrant?

$250$

$450$

$500$

$1250$

$2500$

Concepts:combinatorial geometry intersections binomial coefficient

Difficulty rating: 2110

Small Hint:

An interior crossing is determined by choosing two points from each axis

Big Hint:

For each selected pair on each axis, exactly one pair of the four connecting segments crosses

Solution:

Choose two of the $10$ points on the $x$ -axis and two of the $5$ points on the $y$ -axis. Among the four connecting segments, the two with reversed endpoint order cross exactly once. The points can be chosen so no three segments meet at one interior point, so the maximum is $\binom{10}{2}\binom{5}{2}=45\cdot10=450.$

Thus the correct answer is B.

23.

What is the size of the largest subset $S$ of $\{1,2,3,\ldots,50\}$ such that no pair of distinct elements of $S$ has a sum divisible by $7?$

$6$

$7$

$14$

$22$

$23$

Concepts:modular arithmetic residue classes extremal set

Difficulty rating: 2170

Small Hint:

Group the integers by their residues modulo $7$

Big Hint:

Pair residue classes $1$ with $6,$ $2$ with $5,$ and $3$ with $4$ ; treat residue $0$ separately

Solution:

The residue classes $1,2,\ldots,6,0$ contain $8,7,7,7,7,7,7$ numbers, respectively. We may take numbers from at most one class in each complementary pair $(1,6),(2,5),(3,4),$ and at most one multiple of $7.$ Thus $\begin{aligned} |S|&\le\max(8,7)+\max(7,7)\\ &\qquad+\max(7,7)+1=23. \end{aligned}$ Taking every number whose residue modulo $7$ is $1,$ $2,$ or $3,$ together with one multiple of $7,$ attains $23.$

Thus the correct answer is E.

24.

Let $ABCD$ be a parallelogram of area $10$ with $AB=3$ and $BC=5.$ Locate $E,$ $F,$ and $G$ on segments $\overline{AB},$ $\overline{BC},$ and $\overline{AD},$ respectively, with $AE=BF=AG=2.$ Let the line through $G$ parallel to $\overline{EF}$ intersect $\overline{CD}$ at $H.$ The area of the quadrilateral $EFHG$ is

$4$

$4.5$

$5$

$5.5$

$6$

Concepts:parallelogram affine coordinates polygon area

Difficulty rating: 2040

Small Hint:

Use $AB$ and $AD$ as an affine coordinate basis

Big Hint:

In that basis, $E=(\frac{2}{3},0),$ $F=(1,\frac{2}{5}),$ and $G=(0,\frac{2}{5})$ ; use the parallel condition to locate $H$

Solution:

Use $AB$ and $AD$ as basis vectors. Then $\begin{aligned} E&=\left(\frac23,0\right),\\ F&=\left(1,\frac25\right),\\ G&=\left(0,\frac25\right). \end{aligned}$ A line from $G$ parallel to $EF$ meets $CD$ at $H=(\frac{1}{2},1).$ The shoelace formula gives the coordinate area of $EFHG$ as $\frac{1}{2}.$ Affine coordinates scale all areas by the area $10$ of $ABCD,$ so $[EFHG]=\frac12\cdot10=5.$

Thus the correct answer is C.

25.

In triangle $ABC,$ $\angle ABC=120^\circ,$ $AB=3,$ and $BC=4.$ If perpendiculars constructed to $\overline{AB}$ at $A$ and to $\overline{BC}$ at $C$ meet at $D,$ then $CD=$

$3$

$\frac8{\sqrt3}$

$5$

$\frac{11}{2}$

$\frac{10}{\sqrt3}$

Concepts:coordinate geometry perpendicular lines distance

Difficulty rating: 2110

Small Hint:

Place $B=(0,0)$ and $A=(3,0)$

Big Hint:

Use the $120^\circ$ angle to locate $C,$ then parametrize the line through $C$ perpendicular to $BC$

Solution:

Set $B=(0,0)$ and $A=(3,0).$ Then $\begin{aligned} C&=4(\cos120^\circ,\sin120^\circ)\\ &=(-2,2\sqrt3). \end{aligned}$ The perpendicular to $AB$ at $A$ is $x=3.$ A vector perpendicular to $BC=(-2,2\sqrt3)$ is $(\sqrt3,1),$ so the other perpendicular is $C+t(\sqrt3,1).$ Its $x$ -coordinate is $3$ when $t=\frac{5}{\sqrt3}.$ Since $|(\sqrt3,1)|=2,$ $CD=2t=\frac{10}{\sqrt3}.$

Thus the correct answer is E.

26.

Semicircle $\overset{\frown}{AB}$ has center $C$ and radius $1.$ Point $D$ is on $\overset{\frown}{AB}$ and $CD\perp AB.$ Extend $\overline{BD}$ and $\overline{AD}$ to $E$ and $F,$ respectively, so that circular arcs $\overset{\frown}{AE}$ and $\overset{\frown}{BF}$ have $B$ and $A$ as their respective centers. Circular arc $\overset{\frown}{EF}$ has center $D.$ The area of the shaded “smile,” $AEFBDA,$ is

$(2-\sqrt2)\pi$

$2\pi-\pi\sqrt2-1$

$\left(1-\frac{\sqrt2}{2}\right)\pi$

$\frac{5\pi}{2}-\pi\sqrt2-1$

$(3-2\sqrt2)\pi$

Concepts:sector area semicircle isosceles right triangle composite region

Difficulty rating: 2260

Small Hint:

Use $AC=BC=CD=1$ to identify two isosceles right triangles and determine the relevant radii

Big Hint:

Express the smile as one $90^\circ$ sector plus two $45^\circ$ sectors, then subtract triangle $ABD$ and semicircle $ADB$

Solution:

Triangles $ACD$ and $BCD$ are isosceles right triangles, so $AD=BD=\sqrt2,$ $\angle ADB=90^\circ,$ and $DE=DF=2-\sqrt2.$ The smile is the $90^\circ$ sector $EDF,$ plus the congruent $45^\circ$ sectors $ABE$ and $BAF,$ minus triangle $ABD$ and the original semicircle. Its area is $\begin{aligned} &\frac14\pi(2-\sqrt2)^2 +2\left(\frac18\pi\cdot2^2\right)\\ &\qquad-\frac12(2)(1)-\frac12\pi\\ &=2\pi-\pi\sqrt2-1. \end{aligned}$

Thus the correct answer is B.

27.

A circle of radius $r$ has chords $\overline{AB}$ of length $10$ and $\overline{CD}$ of length $7.$ When $\overline{AB}$ and $\overline{CD}$ are extended through $B$ and $C,$ respectively, they intersect at $P,$ which is outside the circle. If $\angle APD=60^\circ$ and $BP=8,$ then $r^2=$

$70$

$71$

$72$

$73$

$74$

Concepts:power of a point secants right triangle circle diameter

Difficulty rating: 2330

Small Hint:

Apply the secant-secant theorem to find $PC$ and $PD$

Big Hint:

Once $PA=18,$ $PC=9,$ and $\angle APC=60^\circ,$ identify a $30^\circ$ - $60^\circ$ - $90^\circ$ triangle

Solution:

Power of $P$ gives $PA\cdot PB=PC\cdot PD.$ Here $PA=18,$ $PB=8,$ and $PD=PC+7,$ so $PC(PC+7)=144.$ Thus $PC=9$ and $PD=16.$ Since $PA=2PC$ and $\angle APC=60^\circ,$ triangle $APC$ is right at $C,$ with $AC=9\sqrt3.$ Therefore $AD$ is a diameter, and $\begin{aligned} (2r)^2&=AD^2=AC^2+CD^2\\ &=243+49=292. \end{aligned}$ Hence $r^2=73.$

Thus the correct answer is D.

28.

Let $i=\sqrt{-1}.$ The product of the real parts of the roots of $z^2-z=5-5i$ is

$-25$

$-6$

$-5$

$\frac14$

$25$

Concepts:complex quadratic square root of a complex number Vieta

Difficulty rating: 2310

Small Hint:

Apply the quadratic formula and write $\sqrt{21-20i}=a+bi$

Big Hint:

Solve $a^2-b^2=21$ and $2ab=-20$ to determine the real parts of the two roots

Solution:

The quadratic formula gives $z=\frac{1\pm\sqrt{21-20i}}2.$ Since $(5-2i)^2=21-20i,$ the roots are $\begin{aligned} \frac{1+(5-2i)}2&=3-i,\\ \frac{1-(5-2i)}2&=-2+i. \end{aligned}$ Their real parts have product $3(-2)=-6.$

Thus the correct answer is B.

29.

An “unfair” coin has a $\frac23$ probability of turning up heads. If this coin is tossed $50$ times, what is the probability that the total number of heads is even?

$2^5\left(\frac23\right)^{50}$

$\frac12\left(1-\frac1{3^{50}}\right)$

$\frac12$

$\frac12\left(1+\frac1{3^{50}}\right)$

$\frac23$

Concepts:binomial parity probability recurrence complementary states

Difficulty rating: 2310

Small Hint:

Track the difference between the probabilities of an even and an odd number of heads

Big Hint:

One toss multiplies that difference by $P(T)-P(H)$

Solution:

Let $E_n$ and $O_n$ be the probabilities of an even and odd number of heads after $n$ tosses. Then $E_n+O_n=1,$ while $\begin{aligned} E_{n+1}-O_{n+1} &=\left(\frac13-\frac23\right)\\ &\qquad\cdot(E_n-O_n). \end{aligned}$ Since $E_0-O_0=1,$ we have $E_{50}-O_{50}=(-\frac{1}{3})^{50}=3^{-50}.$ Solving the sum and difference equations gives $E_{50}=\frac12\left(1+\frac1{3^{50}}\right).$

Thus the correct answer is D.

30.

Let $ABCD$ be an isosceles trapezoid with bases $AB=92$ and $CD=19.$ Suppose $AD=BC=x$ and a circle with center on $\overline{AB}$ is tangent to segments $\overline{AD}$ and $\overline{BC}.$ If $m$ is the smallest possible value of $x,$ then $m^2=$

$1369$

$1679$

$1748$

$2109$

$8825$

Concepts:isosceles trapezoid circle tangent to lines optimization coordinate geometry

Difficulty rating: 2400

Small Hint:

By symmetry, place the trapezoid with its bases centered on the same vertical axis and put the circle’s center at the midpoint of $AB$

Big Hint:

Write the distance from the center to a leg and require the tangency point to lie on the leg segment; the minimum occurs at an endpoint case

Solution:

Place $A=(-46,0),$ $B=(46,0),$ $D=(-\frac{19}{2},h),$ and $C=(\frac{19}{2},h).$ Symmetry forces the circle’s center to be $O=(0,0).$ The horizontal offset along each leg is $\frac{73}{2},$ so $x^2=h^2+\left(\frac{73}{2}\right)^2.$ The perpendicular from $O$ to a leg first has its foot on the segment when the foot reaches the upper endpoint; this boundary condition is $OD\perp AD.$ Thus $\begin{aligned} (-\frac{19}{2},h)\cdot(\frac{73}{2},h)&=0,\\ h^2&=\frac{1387}{4}. \end{aligned}$ Therefore $m^2=\frac{1387+5329}{4}=1679.$

Thus the correct answer is B.