1986 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

$[x-(y-x)]-[(x-y)-x]=$

$2y$

$2x$

$-2y$

$-2x$

$0$

Concepts:simplifying expressions distributive property

Difficulty rating: 840

Small Hint:

Remove the innermost parentheses before subtracting the second bracket

Big Hint:

Track the minus sign in front of each bracket carefully

Solution:

Expanding the two bracketed expressions gives $\begin{aligned} &[x-(y-x)]-[(x-y)-x]\\ &\qquad=(2x-y)-(-y)=2x. \end{aligned}$

Thus the correct answer is B.

2.

If the line $L$ in the $xy$ -plane has half the slope and twice the $y$ -intercept of the line $y=\frac23x+4,$ then an equation for $L$ is

$y=\frac13x+8$

$y=\frac43x+2$

$y=\frac13x+4$

$y=\frac43x+4$

$y=\frac13x+2$

Concepts:slope-intercept form linear equation

Difficulty rating: 890

Small Hint:

Read the slope and intercept from $y=mx+b$

Big Hint:

Apply the two requested changes to the coefficient and constant separately

Solution:

The original line has slope $\frac{2}{3}$ and $y$ -intercept $4.$ Therefore $L$ has slope $\frac{1}{3}$ and $y$ -intercept $8,$ so its equation is $y=\frac13x+8.$

Thus the correct answer is A.

3.

In the figure, $\triangle ABC$ has a right angle at $C$ and $\angle A=20^\circ.$ If $BD$ is the bisector of $\angle ABC,$ then $\angle BDC=$

$40^\circ$

$45^\circ$

$50^\circ$

$55^\circ$

$60^\circ$

Concepts:angle sum angle bisector right triangle

Difficulty rating: 1230

Small Hint:

First find $\angle ABC$ from the angle sum of $\triangle ABC$

Big Hint:

Use the bisector and then the angle sum of right triangle $BDC$

Solution:

Since $\angle C=90^\circ$ and $\angle A=20^\circ,$ we have $\angle ABC=70^\circ.$ The bisector gives $\angle DBC=35^\circ.$ Therefore, in right triangle $BDC,$ $\angle BDC=90^\circ-35^\circ=55^\circ.$

Thus the correct answer is D.

4.

Let $S$ be the statement

“If the sum of the digits of the whole number $n$ is divisible by $6,$ then $n$ is divisible by $6.$ ”

A value of $n$ which shows $S$ to be false is

$30$

$33$

$40$

$42$

none of these

Concepts:counterexample divisibility digit sum

Difficulty rating: 870

Small Hint:

A counterexample must satisfy the hypothesis but not the conclusion

Big Hint:

Divisibility by $6$ requires both divisibility by $2$ and by $3$

Solution:

The digit sum of $33$ is $6,$ which is divisible by $6.$ However, $33$ is odd, so it is not divisible by $6.$ Thus $33$ makes the implication false.

Therefore the correct answer is B.

5.

Simplify $\left(\sqrt[6]{27}-\sqrt{6\frac34}\right)^2.$

$\frac34$

$\frac{\sqrt3}{2}$

$\frac{3\sqrt3}{4}$

$\frac32$

$\frac{3\sqrt3}{2}$

Concepts:radical simplification exponent mixed number

Difficulty rating: 1420

Small Hint:

Rewrite $27$ as a power of $3$ and $6\frac34$ as an improper fraction

Big Hint:

Simplify both radicals before squaring their difference

Solution:

We have $\sqrt[6]{27}=(3^3)^{\frac{1}{6}}=\sqrt3$ and $\sqrt{6\frac34}=\sqrt{\frac{27}{4}}=\frac{3\sqrt3}{2}.$ Hence the expression is $\begin{aligned} \left(\sqrt3-\frac{3\sqrt3}{2}\right)^2 &=\left(-\frac{\sqrt3}{2}\right)^2\\ &=\frac34. \end{aligned}$

Thus the correct answer is A.

6.

Using a table of a certain height, two identical blocks of wood are placed as shown in Figure $1.$ Length $r$ is found to be $32$ inches. After rearranging the blocks as in Figure $2,$ length $s$ is found to be $28$ inches. How high is the table?

$28$ inches

$29$ inches

$30$ inches

$31$ inches

$32$ inches

Concepts:system of equations length cancellation

Difficulty rating: 1260

Small Hint:

Let $h$ be the table height and let the block dimensions be $\ell$ and $w$

Big Hint:

Write one vertical-distance equation for each figure and add them

Solution:

Let $h$ be the table height and let $\ell,w$ be the long and short block dimensions. Figure $1$ gives $h+\ell-w=32,$ while Figure $2$ gives $h+w-\ell=28.$ Adding cancels the block dimensions: $2h=60,$ so $h=30$ inches.

Thus the correct answer is C.

7.

The sum of the greatest integer less than or equal to $x$ and the least integer greater than or equal to $x$ is $5.$ The solution set for $x$ is

$\frac52$

$\{x\mid 2\le x\le3\}$

$\{x\mid 2\le x\lt3\}$

$\{x\mid 2\lt x\le3\}$

$\{x\mid 2\lt x\lt3\}$

Concepts:floor function ceiling function interval

Difficulty rating: 1280

Small Hint:

Treat integer and noninteger values of $x$ separately

Big Hint:

For noninteger $x,$ the ceiling is one more than the floor

Solution:

If $x$ is an integer, then $\lfloor x\rfloor+\lceil x\rceil=2x,$ which cannot equal $5.$ If $x$ is not an integer, then $\lfloor x\rfloor+\lceil x\rceil=2\lfloor x\rfloor+1.$ This equals $5$ exactly when $\lfloor x\rfloor=2,$ so $2\lt x\lt3.$

Thus the correct answer is E.

8.

The population of the United States in $1980$ was $226{,}504{,}825.$ The area of the country is $3{,}615{,}122$ square miles. There are $(5280)^2$ square feet in one square mile. Which number below best approximates the average number of square feet per person?

$5{,}000$

$10{,}000$

$50{,}000$

$100{,}000$

$500{,}000$

Concepts:estimation unit conversion scientific notation

Difficulty rating: 1130

Small Hint:

Divide total square feet by population

Big Hint:

Round the given values to two significant digits before multiplying

Solution:

The average is approximately $\begin{aligned} &\frac{(3.615\times10^6)(5280)^2} {2.265\times10^8}\\ &\qquad\approx 4.45\times10^5 \end{aligned}$ square feet per person. Of the choices, this is closest to $500{,}000.$

Thus the correct answer is E.

9.

The product $\begin{aligned} &\left(1-\frac1{2^2}\right) \left(1-\frac1{3^2}\right)\\ &\quad{}\cdots \left(1-\frac1{9^2}\right) \left(1-\frac1{10^2}\right) \end{aligned}$ equals

$\frac5{12}$

$\frac12$

$\frac{11}{20}$

$\frac23$

$\frac7{10}$

Concepts:telescoping product difference of squares

Difficulty rating: 1470

Small Hint:

Factor $1-\frac{1}{n^2}$ as a difference of squares

Big Hint:

Separate the product into factors $\frac{n-1}{n}$ and $\frac{n+1}{n}$

Solution:

For each $n,$ $1-\frac1{n^2}=\frac{n-1}{n}\cdot\frac{n+1}{n}.$ Therefore the product telescopes: $\begin{aligned} \prod_{n=2}^{10}\left(1-\frac1{n^2}\right) &=\left(\frac12\right) \left(\frac{11}{10}\right)\\ &=\frac{11}{20}. \end{aligned}$

Thus the correct answer is C.

10.

The $120$ permutations of AHSME are arranged in dictionary order, as if each were an ordinary five-letter word. The last letter of the $86$ th word in this list is

$A$

$H$

$S$

$M$

$E$

Concepts:permutation lexicographic order counting

Difficulty rating: 1590

Small Hint:

Group the words into blocks of $4!$ according to their first letter

Big Hint:

Within the $M$ -block, group by the second letter and then list only the needed small block

Solution:

Each first letter occupies a block of $4!=24$ words. Positions $73$ through $96$ begin with $M,$ so the $86$ th word is the $14$ th word in that block. The first $6$ begin with $MA,$ the next $6$ with $ME,$ and the $13$ th through $18$ th begin with $MH.$ The first two of those are $MHAES$ and $MHASE.$ Thus the $86$ th word ends in E.

Therefore the correct answer is E.

11.

In $\triangle ABC,$ $AB=13,$ $BC=14$ and $CA=15.$ Also, $M$ is the midpoint of side $AB$ and $H$ is the foot of the altitude from $A$ to $BC.$ The length of $HM$ is

$6$

$6.5$

$7$

$7.5$

$8$

Concepts:right triangle midpoint of hypotenuse median

Difficulty rating: 1440

Small Hint:

Focus on right triangle $AHB$

Big Hint:

Recall the distance from the midpoint of a right triangle’s hypotenuse to each vertex

Solution:

Triangle $AHB$ is right at $H,$ and $M$ is the midpoint of its hypotenuse $AB.$ The midpoint of a right triangle’s hypotenuse is equidistant from all three vertices, so $HM=\frac{AB}{2}=\frac{13}{2}=6.5.$

Thus the correct answer is B.

12.

John scores $93$ on this year’s AHSME. Had the old scoring system still been in effect, he would score only $84$ for the same answers. How many questions does he leave unanswered? (In the new scoring system that year, one received $5$ points for each correct answer, $0$ points for each wrong answer, and $2$ points for each problem left unanswered. In the previous scoring system, one started with $30$ points, received $4$ more for each correct answer, lost $1$ point for each wrong answer, and neither gained nor lost points for unanswered questions. There are $30$ questions in the $1986$ AHSME.)

$6$

$9$

$11$

$14$

not uniquely determined

Concepts:system of equations scoring elimination

Difficulty rating: 1560

Small Hint:

Let $c,w,u$ be the numbers correct, wrong and unanswered

Big Hint:

Use $c+w+u=30$ to rewrite the old score before comparing it with the new score

Solution:

Let $c,w,u$ be the numbers correct, wrong and unanswered. The old score and total number of questions give $\begin{aligned} 30+4c-w&=84,\\ c+w+u&=30. \end{aligned}$ Eliminating $w$ yields $5c+u=84.$ The new score is $5c+2u=93.$ Subtracting these equations gives $u=9.$

Thus the correct answer is B.

13.

A parabola $y=ax^2+bx+c$ has vertex $(4,2).$ If $(2,0)$ is on the parabola, then $abc$ equals

$-12$

$-6$

$0$

$6$

$12$

Concepts:parabola vertex form polynomial coefficient

Difficulty rating: 1300

Small Hint:

Write the parabola in vertex form $y=a(x-4)^2+2$

Big Hint:

Use the given point to find $a,$ then expand to read $b$ and $c$

Solution:

Write the equation as $y=a(x-4)^2+2.$ Substituting $(2,0)$ gives $0=4a+2,$ so $a=-\frac{1}{2}.$ Expanding, $y=-\frac12x^2+4x-6.$ Thus $b=4,$ $c=-6,$ and $abc=(-\frac{1}{2})(4)(-6)=12.$

Therefore the correct answer is E.

14.

Suppose hops, skips and jumps are specific units of length. If $b$ hops equals $c$ skips, $d$ jumps equals $e$ hops, and $f$ jumps equals $g$ meters, then one meter equals how many skips?

$\frac{bdg}{cef}$

$\frac{cdf}{beg}$

$\frac{cdg}{bef}$

$\frac{cef}{bdg}$

$\frac{ceg}{bdf}$

Concepts:unit conversion ratio and proportion

Difficulty rating: 1360

Small Hint:

Turn each equality into a conversion factor for one unit

Big Hint:

Convert meters to jumps, jumps to hops, and hops to skips in that order

Solution:

From the three relations, $\begin{aligned} 1\text{ meter}&=\frac fg\text{ jumps},\\ 1\text{ jump}&=\frac ed\text{ hops},\\ 1\text{ hop}&=\frac cb\text{ skips}. \end{aligned}$ Multiplying the conversion factors gives $\frac fg\cdot\frac ed\cdot\frac cb=\frac{cef}{bdg}$ skips.

Thus the correct answer is D.

15.

A student attempted to compute the average, $A,$ of $x,$ $y$ and $z$ by computing the average of $x$ and $y,$ and then computing the average of the result and $z.$ Whenever $x\lt y\lt z,$ the student’s final result is

correct

always less than $A$

always greater than $A$

sometimes less than $A$ and sometimes equal to $A$

sometimes greater than $A$ and sometimes equal to $A$

Concepts:average inequality algebraic comparison

Difficulty rating: 1420

Small Hint:

Write both the true average and the student’s result as algebraic expressions

Big Hint:

Subtract the true average and use the order $x\lt y\lt z$ to determine the sign

Solution:

The student’s result is $\frac{x+y+2z}{4},$ while the true average is $\frac{x+y+z}{3}.$ Their difference is $\begin{aligned} &\frac{x+y+2z}{4}-\frac{x+y+z}{3}\\ &\qquad=\frac{2z-x-y}{12}. \end{aligned}$ Since $z\gt x$ and $z\gt y,$ the numerator is positive. The student’s result is therefore always greater than $A.$

Thus the correct answer is C.

16.

In $\triangle ABC,$ $AB=8,$ $BC=7,$ $CA=6$ and side $BC$ is extended, as shown in the figure, to a point $P$ so that $\triangle PAB$ is similar to $\triangle PCA.$ The length of $PC$ is

$7$

$8$

$9$

$10$

$11$

Difficulty rating: 1610

Small Hint:

Use the vertex order in $\triangle PAB\sim\triangle PCA$ to match corresponding sides

Big Hint:

Set $PC=t$ and use both $PB=PC+7$ and the repeated similarity ratio

Solution:

The stated order of similarity gives $\frac{PA}{PC}=\frac{PB}{PA}=\frac{AB}{CA}=\frac43.$ Let $PC=t.$ Then $PA=\frac{4t}{3}$ and $PB=\frac{16t}{9}.$ Since $PB=PC+CB=t+7,$ $\frac{16t}{9}=t+7,$ which gives $t=9.$

Thus the correct answer is C.

17.

A drawer in a darkened room contains $100$ red socks, $80$ green socks, $60$ blue socks and $40$ black socks. A youngster selects socks one at a time from the drawer but is unable to see the color of the socks drawn. What is the smallest number of socks that must be selected to guarantee that the selection contains at least $10$ pairs? (A pair of socks is two socks of the same color. No sock may be counted in more than one pair.)

$21$

$23$

$24$

$30$

$50$

Concepts:pigeonhole principle parity worst case

Difficulty rating: 1850

Small Hint:

For each color, at most one selected sock can remain unpaired

Big Hint:

Use the parity of the total to sharpen the four-unpaired bound, then construct a near-miss

Solution:

With $23$ selected socks, the number of colors having an odd count must itself be odd, so it is at most $3.$ Thus at most $3$ socks are unpaired, leaving at least $20$ socks in $10$ pairs. But $22$ socks do not suffice: color counts $7,5,5,5$ produce only $3+2+2+2=9$ pairs. Therefore the minimum is $23.$

Thus the correct answer is B.

18.

A plane intersects a right circular cylinder of radius $1$ forming an ellipse. If the major axis of the ellipse is $50\%$ longer than the minor axis, the length of the major axis is

$1$

$\frac32$

$2$

$\frac94$

$3$

Concepts:cylinder ellipse spatial geometry

Difficulty rating: 1830

Small Hint:

The minor axis of such an elliptical section is a diameter of the cylinder

Big Hint:

Increase that diameter by $50\%$ to obtain the major axis

Solution:

The minor axis of an elliptical plane section of a right circular cylinder is a diameter of the cylinder. Its length is therefore $2.$ The major axis is $50\%$ longer, so its length is $2+0.50(2)=3.$

Thus the correct answer is E.

19.

A park is in the shape of a regular hexagon $2$ km on a side. Starting at a corner, Alice walks along the perimeter of the park for a distance of $5$ km. How many kilometers is she from her starting point?

$\sqrt{13}$

$\sqrt{14}$

$\sqrt{15}$

$\sqrt{16}$

$\sqrt{17}$

Concepts:regular hexagon displacement Pythagorean theorem

Difficulty rating: 1620

Small Hint:

The walk consists of two complete sides and half of the next side

Big Hint:

Resolve the three directed segments into horizontal and vertical components

Solution:

Choose the first side in the horizontal direction. The three directed portions of the walk have vectors $\begin{gathered} (2,0),\qquad (1,\sqrt3),\\ \left(-\frac12,\frac{\sqrt3}{2}\right). \end{gathered}$ Their sum is $(\frac{5}{2},\frac{3\sqrt3}{2}).$ Its squared length is $\left(\frac52\right)^2+\left(\frac{3\sqrt3}{2}\right)^2=13.$ The distance is therefore $\sqrt{13}$ km.

Thus the correct answer is A.

20.

Suppose $x$ and $y$ are inversely proportional and positive. If $x$ increases by $p\%,$ then $y$ decreases by

$p\%$

$\frac{p}{1+p}\%$

$\frac{100}{p}\%$

$\frac{p}{100+p}\%$

$\frac{100p}{100+p}\%$

Concepts:inverse proportion percent change algebra

Difficulty rating: 1530

Small Hint:

An increase by $p\%$ multiplies $x$ by $\frac{100+p}{100}$

Big Hint:

Inverse proportionality divides $y$ by that factor; compare the new value with the old one

Solution:

The new value of $x$ is $\frac{x(100+p)}{100}.$ Hence the new value of $y$ is $y\cdot\frac{100}{100+p}.$ The fractional decrease is $1-\frac{100}{100+p}=\frac{p}{100+p}.$ Expressed as a percentage, this is $\frac{100p}{100+p}\%.$

Thus the correct answer is E.

21.

In the configuration below, $\theta$ is measured in radians, $C$ is the center of the circle, $BCD$ and $ACE$ are line segments, and $AB$ is tangent to the circle at $A.$

A necessary and sufficient condition for the equality of the two shaded areas, given $0\lt\theta\lt\frac{\pi}{2},$ is

$\tan\theta=\theta$

$\tan\theta=2\theta$

$\tan\theta=4\theta$

$\tan2\theta=\theta$

$\tan\frac{\theta}{2}=\theta$

Concepts:sector area tangent right triangle equation

Difficulty rating: 2110

Small Hint:

Let the circle’s radius be $r=AC$ and compare a sector with triangle $ABC$

Big Hint:

Equality of the two shaded pieces means the whole triangle has twice the sector’s area

Solution:

Let $r=AC.$ The upper shaded sector has area $\frac{\theta r^2}{2}.$ The lower shaded region is triangle $ABC$ with an equal sector removed. Thus the two shaded regions are equal exactly when $2\left(\frac{\theta r^2}{2}\right) =\frac12r\cdot AB.$ This is equivalent to $\frac{AB}{r}=2\theta.$ Since $AB$ is tangent at $A,$ triangle $ABC$ is right at $A,$ and $\frac{AB}{r}=\tan\theta.$ Therefore the condition is $\tan\theta=2\theta.$

Thus the correct answer is B.

22.

Six distinct integers are picked at random from $\{1,2,3,\ldots,10\}.$ What is the probability that, among those selected, the second smallest is $3?$

$\frac1{60}$

$\frac16$

$\frac13$

$\frac12$

none of these

Concepts:combination probability order statistic

Difficulty rating: 1740

Small Hint:

Count all six-element subsets of the ten integers

Big Hint:

For second-smallest $3,$ choose one element below $3$ and four above $3$

Solution:

There are $\binom{10}{6}=210$ possible sets. If the second-smallest element is $3,$ then $3$ is selected, one element is chosen from $\{1,2\},$ and four are chosen from $\{4,5,\ldots,10\}.$ This gives $2\binom74=70$ favorable sets. The probability is $\frac{70}{210}=\frac{1}{3}.$

Thus the correct answer is C.

23.

Let $\begin{aligned} N={}&69^5+5\cdot69^4+10\cdot69^3\\ &{}+10\cdot69^2+5\cdot69+1. \end{aligned}$ How many positive integers are factors of $N?$

$3$

$5$

$69$

$125$

$216$

Concepts:binomial theorem prime factorization divisor count

Difficulty rating: 1890

Small Hint:

Recognize the coefficients $1,5,10,10,5,1$

Big Hint:

After using the binomial theorem, factor the resulting base into primes

Solution:

By the binomial theorem, $\begin{aligned} N&=(69+1)^5=70^5\\ &=2^5\cdot5^5\cdot7^5. \end{aligned}$ A divisor independently chooses an exponent from $0$ through $5$ for each of the three primes. Thus $N$ has $6^3=216$ positive divisors.

Therefore the correct answer is E.

24.

Let $p(x)=x^2+bx+c,$ where $b$ and $c$ are integers. If $p(x)$ is a factor of both

$x^4+6x^2+25$

and

$3x^4+4x^2+28x+5,$ what is $p(1)?$

$0$

$1$

$2$

$4$

$8$

Concepts:polynomial factor linear combination Gauss lemma

Difficulty rating: 2200

Small Hint:

A common factor divides every integer linear combination of the two polynomials

Big Hint:

Subtract the second polynomial from three times the first

Solution:

The common factor $p(x)$ divides $\begin{aligned} &3(x^4+6x^2+25)\\ &\quad{}-(3x^4+4x^2+28x+5)\\ &\qquad=14(x^2-2x+5). \end{aligned}$ Because $p(x)$ is monic with integer coefficients, Gauss’s lemma implies that it divides $x^2-2x+5.$ The two polynomials are monic and have the same degree, so $p(x)=x^2-2x+5.$ Hence $p(1)=1-2+5=4.$

Thus the correct answer is D.

25.

If $\lfloor x\rfloor$ is the greatest integer less than or equal to $x,$ then $\sum_{N=1}^{1024}\lfloor\log_2N\rfloor=$

$8192$

$8204$

$9218$

$\lfloor\log_2(1024!)\rfloor$

none of these

Concepts:floor function logarithm geometric series

Difficulty rating: 2300

Small Hint:

Group the integers $N$ according to the interval $2^k\le N\lt2^{k+1}$

Big Hint:

There are $2^k$ integers in the $k$ th group; handle $N=1024$ separately

Solution:

For $0\le k\le9,$ exactly $2^k$ integers $N$ satisfy $2^k\le N\lt2^{k+1},$ and each contributes $k.$ The final integer $1024=2^{10}$ contributes $10.$ Hence $\sum_{N=1}^{1024}\lfloor\log_2N\rfloor =\sum_{k=0}^{9}k2^k+10.$ The finite geometric-sum identity gives $\sum_{k=0}^{9}k2^k=8194,$ so the requested sum is $8204.$

Thus the correct answer is B.

26.

It is desired to construct a right triangle in the coordinate plane so that its legs are parallel to the $x$ and $y$ axes and so that the medians to the midpoints of the legs lie on the lines $y=3x+1$ and $y=mx+2.$ The number of different constants $m$ for which such a triangle exists is

$0$

$1$

$2$

$3$

more than $3$

Concepts:coordinate geometry median slope construction

Difficulty rating: 2230

Small Hint:

Place an axis-aligned right triangle at convenient coordinates and compute the slopes of the two medians to its legs

Big Hint:

The two slopes differ by a factor of $4;$ remember that either given line could be the steeper one

Solution:

Place the right-angle vertex at $(0,0)$ and the other vertices at $(a,0)$ and $(0,b).$ The medians to the legs have slopes $-\frac{2b}{a}$ and $-\frac{b}{2a},$ whose ratio is $4.$ Therefore, if one median has slope $3,$ the other can have slope $12$ or $\frac{3}{4}.$ Both occur: choose a triangle with the required pair of slopes and translate its centroid to the intersection of the two specified lines. Thus there are two possible values of $m.$

Therefore the correct answer is C.

27.

In the adjoining figure, $AB$ is a diameter of the circle, $CD$ is a chord parallel to $AB,$ and $AC$ intersects $BD$ at $E,$ with $\angle AED=\alpha.$ The ratio of the area of $\triangle CDE$ to that of $\triangle ABE$ is

$\cos\alpha$

$\sin\alpha$

$\cos^2\alpha$

$\sin^2\alpha$

$1-\sin\alpha$

Concepts:circle intersecting chords triangle area trigonometry

Difficulty rating: 2320

Small Hint:

Express each triangle’s area using the two sides meeting at $E$

Big Hint:

Use the intersecting-chords theorem, then draw $AD$ and use the right triangle created by diameter $AB$

Solution:

The triangles use the same angle at $E,$ so $\frac{[CDE]}{[ABE]} =\frac{CE\cdot DE}{AE\cdot BE}.$ Intersecting chords give $AE\cdot CE=BE\cdot DE,$ so this ratio becomes $(\frac{DE}{AE})^2.$ Draw $AD.$ Since $AB$ is a diameter, $\angle ADB=90^\circ,$ and $D,E,B$ are collinear. Thus triangle $ADE$ is right at $D,$ and $\frac{DE}{AE}=\cos\alpha.$ The required ratio is $\cos^2\alpha.$

Thus the correct answer is C.

28.

$ABCDE$ is a regular pentagon. $AP,$ $AQ$ and $AR$ are the perpendiculars dropped from $A$ onto $CD,$ $CB$ extended and $DE$ extended, respectively. Let $O$ be the center of the pentagon. If $OP=1,$ then $AO+AQ+AR$ equals

$3$

$1+\sqrt5$

$4$

$2+\sqrt5$

$5$

Concepts:regular pentagon area decomposition altitude

Difficulty rating: 2320

Small Hint:

Let $s$ be the side length and compute the pentagon’s area from its five central triangles

Big Hint:

Also split the pentagon into triangles $ABC,$ $ACD$ and $ADE$ with altitudes $AQ,AP,AR$

Solution:

Let the side length be $s.$ Since the apothem $OP=1,$ the five central triangles give pentagon area $\frac{5s}{2}.$ The same pentagon is the union of triangles $ABC,$ $ACD$ and $ADE,$ whose respective altitudes to side-length bases are $AQ,AP,AR.$ Hence $\frac{s}{2}(AQ+AP+AR)=\frac{5s}{2},$ so $AQ+AP+AR=5.$ Also $AP=AO+OP=AO+1.$ Therefore $AO+AQ+AR=4.$

Thus the correct answer is C.

29.

Two of the altitudes of the scalene triangle $ABC$ have length $4$ and $12.$ If the length of the third altitude is also an integer, what is the biggest it can be?

$4$

$5$

$6$

$7$

none of these

Concepts:triangle inequality altitude area reciprocal relation

Difficulty rating: 2230

Small Hint:

For a fixed triangle area, each side is inversely proportional to its corresponding altitude

Big Hint:

Apply the triangle inequalities to side lengths proportional to $\frac{1}{4},\frac{1}{12},\frac{1}{h}$

Solution:

Let the third altitude be $h.$ Since each side equals twice the common area divided by its altitude, the side lengths are proportional to $\frac14,\qquad\frac1{12},\qquad\frac1h.$ The two nontrivial triangle inequalities give $\begin{aligned} \frac1h&\lt\frac14+\frac1{12}=\frac13,\\ \frac1h&\gt\frac14-\frac1{12}=\frac16. \end{aligned}$ Thus $3\lt h\lt6.$ The largest integral possibility is $5,$ and its three altitudes are distinct as required for a scalene triangle.

Therefore the correct answer is B.

30.

The number of real solutions $(x,y,z,w)$ of the simultaneous equations $\begin{aligned} 2y&=x+\frac{17}{x},\\ 2z&=y+\frac{17}{y},\\ 2w&=z+\frac{17}{z},\\ 2x&=w+\frac{17}{w}. \end{aligned}$ is

$1$

$2$

$4$

$8$

$16$

Concepts:cyclic system AM-GM inequality monotonicity sign symmetry

Difficulty rating: 2420

Small Hint:

The equations force all four variables to have the same sign

Big Hint:

For positive $t,$ study $f(t)=\frac{t+\frac{17}{t}}{2}$ relative to $\sqrt{17}$ and to $t$

Solution:

Each expression $t+\frac{17}{t}$ has the same sign as $t,$ so all four variables have the same sign. Suppose first that they are positive. By AM-GM, every variable is at least $\sqrt{17}.$ For $t\gt\sqrt{17},$ $\frac{t+\frac{17}{t}}{2}\lt t.$ If any variable exceeded $\sqrt{17},$ the equations would give the impossible strict cycle $x\gt y\gt z\gt w\gt x.$ Hence the only positive solution is $x=y=z=w=\sqrt{17}.$ Negating all four variables preserves the system, giving exactly one negative solution, with all variables equal to $-\sqrt{17}.$ Thus there are two real solutions.

Therefore the correct answer is B.