#1996 AHSME

##1996 AHSME Problems/Problem 1

The addition below is incorrect. What is the largest digit that can be changed to make the addition correct?

$\begin{array}{rr}&\ \texttt{6 4 1}\\ &\texttt{8 5 2}\\ &+\texttt{9 7 3}\\ \hline &\texttt{2 4 5>> 6}\end{array}​$

$\text{(A)}\ 4\qquad\text{(B)}\ 5\qquad\text{(C)}\ 6\qquad\text{(D)}\ 7\qquad\text{(E)}\ 8$

Solution

Doing the addition as is, we get $641 + 852 + 973 = 2466$. This number is $10$ larger than the desired sum of $2456$. Therefore, we must make one of the three numbers $10$ smaller.

We may either change $641 \rightarrow 631$, $852 \rightarrow 842$, or $973 \rightarrow 963$. Either change results in a valid sum. The largest digit that could be changed is thus the $7$ in the number $973$, and the answer is $\boxed{D}$.

##Problem 2

Each day Walter gets $3$ dollars for doing his chores or $5$ dollars for doing them exceptionally well. After $10$ days of doing his chores daily, Walter has received a total of $36$ dollars. On how many days did Walter do them exceptionally well?

$\text{(A)}\ 3\qquad\text{(B)}\ 4\qquad\text{(C)}\ 5\qquad\text{(D)}\ 6\qquad\text{(E)}\ 7$

Solution

If Walter had done his chores for $10$ days without doing any of them well, he would have earned $3 \cdot 10 = 30$ dollars. He got $6$ dollars more than this.

He gets a $5 - 3 = 2$ dollar bonus every day he does his chores well. Thus, he did his chores exceptionally well $\frac{6}{2} = 3$ days, and the answer is $\boxed{A}$.

##Problem 3

$\frac{(3!)!}{3!}=$

$\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 6\qquad\text{(D)}\ 40\qquad\text{(E)}\ 120$

Solution

The numerator is $(3!)! = 6!$.

The denominator is $3! = 6$.

Using the property that $6! = 6 \cdot 5!$ in the numerator, the sixes cancel, leaving $5! = 120$, which is answer $\boxed{E}$.

##Problem 4

Six numbers from a list of nine integers are $7,8,3,5,9$ and $5$. The largest possible value of the median of all nine numbers in this list is

$\text{(A)}\ 5\qquad\text{(B)}6\qquad\text{(C)}\ 7\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$

Solution

First, put the six numbers we have in order, since we are concerned with the median: $3, 5, 5, 7, 8, 9$.

We have three more numbers to insert into the list, and the median will be the $5^{th}$ highest (and $5^{th}$ lowest) number on the list. If we top-load the list by making all three of the numbers greater than $9$, the median will be the highest it can possibly be. Thus, the maximum median is the fifth piece of data in the list, which is $8$, giving an answer of $\boxed{D}$.

(In fact, as long as the three new integers are greater than $8$, the median will be $8$.)

This illustrates one important fact about medians: no matter how high the three "outlier" numbers are, the median will never be greater than $8$. The arithmetic mean is, generally speaking, more sensitive to such outliers, while the median is resistant to a small number of data that are either too high or too low.

##Problem 5

Given that $0 < a < b < c < d$, which of the following is the largest?

$\text{(A)}\ \frac{a+b}{c+d} \qquad\text{(B)}\ \frac{a+d}{b+c} \qquad\text{(C)}\ \frac{b+c}{a+d} \qquad\text{(D)}\ \frac{b+d}{a+c} \qquad\text{(E)}\ \frac{c+d}{a+b}$

Solution

Assuming that one of the above fractions is indeed always the largest, try plugging in $a=1, b=2, c=3, d=4$, since those are valid values for the variables given the constraints of the problem. The options become:

$\text{(A)}\ \frac{1+2}{3+4} \qquad\text{(B)}\ \frac{1+4}{2+3} \qquad\text{(C)}\ \frac{2+3}{1+4} \qquad\text{(D)}\ \frac{2+4}{1+3} \qquad\text{(E)}\ \frac{3+4}{1+2}$

Simplified, the options are $\frac{3}{7}, 1, 1, \frac{3}{2}, \frac{7}{3}$, respectively. Since $\frac{7}{3}$ is the only option that is greater than $2$, the answer is $\boxed{E}$.

##Problem 6

If $f(x) = x^{(x+1)}(x+2)^{(x+3)}$, then $f(0)+f(-1)+f(-2)+f(-3) =$

$\text{(A)}\ -\frac{8}{9}\qquad\text{(B)}\ 0\qquad\text{(C)}\ \frac{8}{9}\qquad\text{(D)}\ 1\qquad\text{(E)}\ \frac{10}{9}$

Solution

Plugging in $x=0$ into the function will give $0^1\cdot 2^3$. Since $0^1 = 0$, this gives $0$.

Plugging in $x=-1$ into the function will give $(-1)^0 \cdot 1^2$. Since $(-1)^0 = 1$ and $1^2 = 1$, this gives $1$.

Plugging in $x=-2$ will give a $0^1$ factor as the second term, giving an answer of $0$.

Plugging in $x=-3$ will give $(-3)^{-2}\cdot (-1)^0$. The last term is $1$, while the first term is $\frac{1}{(-3)^2} = \frac{1}{9}$

Adding up all four values, the answer is $1 + \frac{1}{9} = \frac{10}{9}$, and the right answer is $\boxed{E}$.

##Problem 7

A father takes his twins and a younger child out to dinner on the twins' birthday. The restaurant charges $4.95$ for the father and $0.45$ for each year of a child's age, where age is defined as the age at the most recent birthday. If the bill is $9.45$, which of the following could be the age of the youngest child?

$\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5$

Solution

The bill for the three children is $9.45 - 4.95 = 4.50$. Since the charge is $0.45$ per year for the children, they must have $\frac{4.50}{0.45} = 10$ years among the three of them.

The twins must have an even number of years in total (presuming that they did not dine in the 17 minutes between the time when the first twin was born and the second twin was born). If we let the twins be $5$ years old, that leaves $10 - 2\cdot 5 = 0$ years leftover for the youngest child. But if the twins are each $4$ years old, then the youngest child could be $10 - 2\cdot 4 = 2$ years old, which is choice $\boxed{B}$.

Of note, this is the only possibility for the ages of the children. If the twins were $3$ years old, the "younger" child would be $10 - 2\cdot 3 = 4$ years old.

## Problem 8

If $3 = k\cdot 2^r$ and $15 = k\cdot 4^r$, then $r =$

$\text{(A)}\ -\log_{2}5\qquad\text{(B)}\ \log_{5}2\qquad\text{(C)}\ \log_{10}5\qquad\text{(D)}\ \log_{2}5\qquad\text{(E)}\ \frac{5}{2}$

Solution

We want to find $r$, so our strategy is to eliminate $k$.

The first equation gives $k = \frac{3}{2^r}$.

The second equation gives $k = \frac{15}{4^r}$

Setting those two equal gives $\frac{3}{2^r} = \frac{15}{4^r}$

Cross-multiplying and dividing by $3$ gives $5\cdot 2^r = 4^r$.

We know that $4^r = 2^r \cdot 2^r$, so we can divide out $2^r$ from both sides (which is legal since $2^r \neq 0$), and we get: $5 = 2^r, r = \log_2 5$, which is option $\boxed{D}$.

## Problem 9

Triangle $PAB$ and square $ABCD$ are in perpendicular planes. Given that $PA = 3, PB = 4$ and $AB = 5$, what is $PD$?

$\text{(A)}\ 5\qquad\text{(B)}\ \sqrt{34} \qquad\text{(C)}\ \sqrt{41}\qquad\text{(D)}\ 2\sqrt{13}\qquad\text{(E)}\ 8$

Solution

Since the two planes are perpendicular, it follows that $\triangle PAD$ is a right triangle. Thus, $PD = \sqrt{PA^2 + AD^2} = \sqrt{PA^2 + AB^2} = \sqrt{34}$, which is option $\boxed{\text{B}}$.

## Problem 10

How many line segments have both their endpoints located at the vertices of a given cube?

$\text{(A)}\ 12\qquad\text{(B)}\ 15\qquad\text{(C)}\ 24\qquad\text{(D)}\ 28\qquad\text{(E)}\ 56$

Solution 1

There are $8$ choices for the first endpoint of the line segment, and $7$ choices for the second endpoint, giving a total of $8\cdot 7 = 56$ segments. However, both $\overline{AB}$ and $\overline{BA}$ were counted, while they really are the same line segment. Every segment got double counted in a similar manner, so there are really $\frac{56}{2} = 28$ line segments, and the answer is $\boxed{D}$.

In shorthand notation, we're choosing $2$ endpoints from a set of $8$ endpoints, and the answer is $\binom{8}{2} = \frac{8!}{6!2!} = 28$.

Solution 2

Each segment is either an edge, a facial diagonal, or a long/main/spacial diagonal.

A cube has $12$ edges: Four on the top face, four on the bottom face, and four that connect the top face to the bottom face.

A cube has $6$ square faces, each of which has $2$ facial diagonals, for a total of $6\cdot 2 = 12$.

A cube has $4$ spacial diagonals: each diagonal goes from one of the bottom vertices to the "opposite" top vertex.

Thus, there are $12 + 12 + 4 = 28$ segments, and the answer is $\boxed{D}$.

## Problem 11

Given a circle of radius $2$, there are many line segments of length $2$ that are tangent to the circle at their midpoints. Find the area of the region consisting of all such line segments.

$\text{(A)}\ \frac{\pi} 4\qquad\text{(B)}\ 4-\pi\qquad\text{(C)}\ \frac{\pi} 2\qquad\text{(D)}\ \pi\qquad\text{(E)}\ 2\pi$

Solution

Let line segment $AB = 2$, and let it be tangent to circle $O$ at point $P$, with radius $OP = 2$. Let $AP = PB = 1$, so that $P$ is the midpoint of $AB$.

$\triangle OAP$ is a right triangle with right angle at $P$, because $AB$ is tangent to circle $O$ at point $P$, and $OP$ is a radius.

Since $AP^2 + OP^2 = OA^2$ by the Pythagorean Theorem, we can find that $OA = \sqrt{1^2 + 2^2} = \sqrt{5}$. Similarly, $OB = \sqrt{5}$ also.

Line segment $APB$ can rotate around the circle. The closest distance of this segment to the center will always be $OP = 2$, and the longest distance of this segment will always be $PA = PB = \sqrt{5}$. Thus, the region in question is the annulus of a circle with outer radius $\sqrt{5}$ and inner radius $2$. This area is $\pi \cdot 5 - \pi \cdot 4 = \pi$, and the answer is $\boxed{D}$.

## Problem 12

A function $f$ from the integers to the integers is defined as follows:

$f(n) =\begin{cases}n+3 &\text{if n is odd}\\ \ n/2 &\text{if n is even}\end{cases}$

Suppose $k$ is odd and $f(f(f(k))) = 27$. What is the sum of the digits of $k$?

$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 15$

Solution

First iteration

To get $f(k) = 27$, you could either have $f(27 - 3)$ and add $3$, or $f(27\cdot 2)$ and divide by $2$.

If you had the former, you would have $f(24)$, and the function's rule would have you divide. Thus, $k=54$ is the only number for which $f(k) = 27$.

Second iteration

Going out one step, if you have $f(f(k)) = 27$, you would have to have $f(k) = 54$. For $f(k) = 54$, you would either have $f(54-3)$ and add $3$, or $f(54\cdot 2)$ and divide by $2$.

Both are possible: $f(51)$ and $f(108)$ return values of $54$. Thus, $f(f(51)) = f(54) = 27$, and $f(f(108)) = f(54) = 27$.

Third iteration

Going out the final step, if you have $f(f(f(k))) = 27$, you would have to have $f(f(k))) = 51$ or $f(f(k)) = 108$.

If you doubled either of these, $k$ would not be odd. So you must subtract $3$.

If you subtract $3$ from $51$, you would compute $f(48)$, which would halve it, and not add the $3$ back.

If you subtract $3$ from $108$, you would compute $f(105)$, which would add the $3$ back.

Thus, $f(f(f(105))) = f(f(108)) = f(54) = 27$, and $105$ is odd. The desired sum of the digits is $6$, and the answer is $\boxed{B}$.

## Problem 13

Sunny runs at a steady rate, and Moonbeam runs $m$ times as fast, where $m$ is a number greater than 1. If Moonbeam gives Sunny a head start of $h$ meters, how many meters must Moonbeam run to overtake Sunny?

$\text{(A)}\ hm\qquad\text{(B)}\ \frac{h}{h+m}\qquad\text{(C)}\ \frac{h}{m-1}\qquad\text{(D)}\ \frac{hm}{m-1}\qquad\text{(E)}\ \frac{h+m}{m-1}$

Solution

If Sunny runs at a rate of $s$ for $x$ meters in $t$ minutes, then $s = \frac{x}{t}$.

In that case, Moonbeam's rate is $ms$, and Moonbeam's distance is $x + h$, and the amount of time $t$ is the same. Thus, $ms = \frac{x + h}{t}$

Solving each equation for $t$, we have $t = \frac{x}{s} = \frac{x + h}{ms}$

Cross multiplying, we get $msx = xs + hs$

Solving for $x$, we get $msx - xs = hs$, which leads to $x = \frac{h}{m - 1}$.

Note that $x$ is the distance that Sunny ran. Moonbeam ran $h$ meters more, for a total of $h + \frac{h}{m-1} = \frac{h(m-1) + h}{m-1} = \frac{hm}{m-1}$. This is answer $\boxed{D}$.

## Problem 14

Let $E(n)$ denote the sum of the even digits of $n$. For example, $E(5681) = 6+8 = 14$. Find $E(1)+E(2)+E(3)+\cdots+E(100)$

$\text{(A)}\ 200\qquad\text{(B)}\ 360\qquad\text{(C)}\ 400\qquad\text{(D)}\ 900\qquad\text{(E)}\ 2250$

Solution

The problem is asking for the sum of all the even digits in the numbers $1$ to $100$. We can remove $100$ from the list, add $00$ to the list, and tack on some leading zeros to the single digit numbers without changing the sum of the even digits. This gives the list:

$00, 01, 02, 03, ..., 10, 11, ..., 98, 99$.

There are $2\cdot 100 = 200$ digits on that list, and each digit appears $\frac{200}{10} = 20$ times.

Thus, each even digit appears $20$ times, and the sum of all the even digits is $0 \cdot 20 + 2\cdot 20 + 4\cdot 20 + 6\cdot 20 + 8\cdot 20 = (0 + 2 + 4 + 6 + 8)\cdot 20 = 400$, and the correct answer is $\boxed{C}$.

## Problem 15

Two opposite sides of a rectangle are each divided into $n$ congruent segments, and the endpoints of one segment are joined to the center to form triangle $A$. The other sides are each divided into $m$ congruent segments, and the endpoints of one of these segments are joined to the center to form triangle $B$. [See figure for $n=5, m=7$.] What is the ratio of the area of triangle $A$ to the area of triangle $B$?

$\text{(A)}\ 1\qquad\text{(B)}\ m/n\qquad\text{(C)}\ n/m\qquad\text{(D)}\ 2m/n\qquad\text{(E)}\ 2n/m$

Solution 1

Place the rectangle on a coordinate grid, with diagonal vertices $(0,0)$ and $(x,y)$. Each horizontal segment of the rectangle will have length $\frac{x}{m}$, while each vertical segment of the rectangle will have length $\frac{y}{n}$.

The center of this rectangle will be $(\frac{x}{2}, \frac{y}{2})$.

Triangle $A$ has a base length of $\frac{y}{n}$, one of the vertical segments. It has an altitude of $\frac{x}{2}$, which is the perpendicular distance from the center of the square to the left side of the square. Thus, the area of triangle $A$ is $\frac{1}{2}\cdot\frac{y}{n}\cdot\frac{x}{2} = \frac{xy}{4n}$.

Triangle $B$ has a base length of $\frac{x}{m}$, one of the horizontal segments. It has an altitude of $\frac{y}{2}$, which is the perpendicular distance from the center of the square to the bottom side of the square. Thus, the area of triangle $B$ is $\frac{1}{2}\cdot\frac{x}{m}\cdot\frac{y}{2} = \frac{xy}{4m}$.

The ratio of areas is $\frac{\frac{xy}{4n}}{\frac{xy}{4m}} = \frac{m}{n}$, which is answer $\boxed{B}$.

Solution 2

Alternately, the algebra can be made simpler if you arbitrarily assume that $(x,y) = (m,n)$, and thus that each side of the rectangle is cut into unit segments. In that case, the ratio of the areas of the two triangles $A$ and $B$ that have the same base length is just the ratio of the heights. Triangle $A$ would have height $\frac{m}{2}$, while triangle $B$, while triangle $B$ would have height $\frac{n}{2}$, giving a ratio of $\frac{m}{n}$, which is answer $\boxed{B}$.

This solution makes the extra assumption that the rectangle has dimensions $m$ by $n$, instead of arbitrary $x$ by $y$ dimensions, and is not a formal "proof"; but since the answer is invariant, extra assumptions will not change the solution.

## Problem 16

A fair standard six-sided dice is tossed three times. Given that the sum of the first two tosses equal the third, what is the probability that at least one "2" is tossed?

$\text{(A)}\ \frac{1}{6}\qquad\text{(B)}\ \frac{91}{216}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{8}{15}\qquad\text{(E)}\ \frac{7}{12}$

Solution

The third die cannot be $1$, since the minimal sum on the other two dice is $2$.

If the third die is $2$, then the first two dice must be $(1,1)$.

If the third die is $3$, then the first two dice must be $(1,2)$ or $(2,1)$.

If the third die is $4$, then the first two dice must be $(1,3)$, $(2,2)$, or $(3,1)$.

If the third die is $5$, then the first two dice must be $(1,4)$, $(2,3)$, $(3,2)$, or $(4,1)$.

If the third die is $6$, then the first two dice must be $(1,5)$, $(2,4)$, $(3,3)$, $(4,2)$, or $(5,1)$.

There are $15$ possibilities for the three dice. Of those possibiltiies, $7$ have a $2$ in the first two dice, and $1$ has a $2$ in the third die. Therefore, the answer is $\boxed{\frac{8}{15}}$.

## Problem 17

In rectangle $ABCD$, angle $C$ is trisected by $\overline{CF}$ and $\overline{CE}$, where $E$ is on $\overline{AB}$, $F$ is on $\overline{AD}$, $BE=6$ and $AF=2$. Which of the following is closest to the area of the rectangle $ABCD$?

$\text{(A)}\ 110\qquad\text{(B)}\ 120\qquad\text{(C)}\ 130\qquad\text{(D)}\ 140\qquad\text{(E)}\ 150$

Solution

Since $\angle C = 90^\circ$, each of the three smaller angles is $30^\circ$, and $\triangle BEC$ and $\triangle CDF$ are both $30-60-90$ triangles.

Defining the variables as illustrated above, we have $x = 6\sqrt{3}$ from $\triangle BEC$

Then $x-2 = 6\sqrt{3} - 2$, and $y = \sqrt{3} (6 \sqrt{3} - 2) = 18 - 2\sqrt{3}$.

The area of the rectangle is thus $xy = 6\sqrt{3}(18 - 2\sqrt{3}) = 108\sqrt{3} - 36$.

Using the approximation $\sqrt{3} \approx 1.7$, we get an area of just under $147.6$, which is closest to answer $\boxed{\text{E}}$. (The actual area is actually greater, since $\sqrt{3} > 1.7$).

## Problem 18

A circle of radius $2$ has center at $(2,0)$. A circle of radius $1$ has center at $(5,0)$. A line is tangent to the two circles at points in the first quadrant. Which of the following is closest to the $y$-intercept of the line?

$\text{(A)}\ \sqrt{2}/4\qquad\text{(B)}\ 8/3\qquad\text{(C)}\ 1+\sqrt 3\qquad\text{(D)}\ 2\sqrt 2\qquad\text{(E)}\ 3$

Solution

The two circles are tangent to each other at the point $(4,0)$, since it is both $2$ units from $(2,0)$ and $1$ unit from $(5,0)$.

Label the x-intercept of the common tangent line $A$, and label the y-intercept of the common tangent $B$. Triangle $\triangle OAB$ is a right triangle at the origin.

Label $D$ the point of tangency to the first, big circle, and label $E$ the point of tangency to the small circle. $\angle D$ and $\angle E$ are both right angles as well.

Label the center of the big circle $F$, and the small circle $G$.

$\triangle DAF \sim \triangle EAG \sim \triangle OAB$ because they each have one right angle, and also have a common angle $A$.

The y-intercept is the length $OB$.

We know that $DF = 2$ and $EG = 1$ because they are radii of circles. From similarity, $\frac{DF}{EG} = \frac{FA}{GA}$. Thus, $FA = 2GA$.

Looking at line $FGA$, we know that $FG + GA = FA$.

Thus, $FG + GA = 2GA$, meaning $FG = GA$. Since $FG = 3$ because it's the distance of the centers of the circles, so $GA = 3$ as well. This gives point $A$ as $(8,0)$.

Since $\triangle EAG$ is a right triangle with $GA = 3$ and $GE = 1$, we know that $AE = \sqrt{GA^2 - GE^2} = \sqrt{8}$

Thus, all triangles are $1:\sqrt{8}:3$ triangles.

$\triangle OAC$ has as its middle side $OA$, which is $8$. Thus, all sides are scaled up by a factor of $\sqrt{8}$, and $OB$, the shortest side, is $\sqrt{8}$. This means the y-intercept is $\sqrt{8} = 2\sqrt{2}$, and the answer is $\boxed{D}$.

## Problem 19

The midpoints of the sides of a regular hexagon $ABCDEF$ are joined to form a smaller hexagon. What fraction of the area of $ABCDEF$ is enclosed by the smaller hexagon?

$\text{(A)}\ \frac{1}{2}\qquad\text{(B)}\ \frac{\sqrt 3}{3}\qquad\text{(C)}\ \frac{2}{3}\qquad\text{(D)}\ \frac{3}{4}\qquad\text{(E)}\ \frac{\sqrt 3}{2}$

Solution

$\triangle OCD$ is copied six times to form the hexagon, so if we find the ratio of the area of the kite inside $\triangle OCD$ to the the area of $\triangle OCD$ itself, it will be the same ratio.

Let $OC=CD=OD =2$ so that the area of the triangle is $\frac{\sqrt{3}s^2}{4} = \sqrt{3}$.

Notice that $\triangle OCD$ is made up of a kite and two $30-60-90$ triangles. The two hypotenuses of these two triangles form $CD$, so the hypotenuse of each triangle must be $\frac{2}{2} = 1$. Thus, the legs of each triangle are $\frac{1}{2}$ and $\frac{\sqrt{3}}{2}$, and the area of two of these triangles is $\frac{1}{2}\cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}$.

Subtracting the area of the two triangles from the area of the equilateral triangle, we find that the area of the kite is $\sqrt{3} - \frac{\sqrt{3}}{4} = \frac{3\sqrt{3}}{4}$.

Thus, the ratio of areas is $\frac{3}{4}$, which is option $\boxed{D}$.

## Problem 20

In the xy-plane, what is the length of the shortest path from $(0,0)$ to $(12,16)$ that does not go inside the circle $(x-6)^{2}+(y-8)^{2}= 25$?

$\text{(A)}\ 10\sqrt 3\qquad\text{(B)}\ 10\sqrt 5\qquad\text{(C)}\ 10\sqrt 3+\frac{ 5\pi}{3}\qquad\text{(D)}\ 40\frac{\sqrt{3}}{3}\qquad\text{(E)}\ 10+5\pi$

Solution

The pathway from $A(0,0)$ to $D(12,16)$ will consist of three segments:

1) $\overline{AB}$, where $AB$ is tangent to the circle at point $B$.

2) $\overline{CD}$, where $CD$ is tangent to the circle at point $C$.

3) $\widehat {BC}$, where $BC$ is an arc around the circle.

The actual path will go $A \rightarrow B \rightarrow C \rightarrow D$, so the acutal segments will be in order $1, 3, 2$.

Let $O$ be the center of the circle at $(6,8)$.

$OA = 10$ and $OB = 5$ since $B$ is on the circle. Since $\triangle OAB$ is a right triangle with right angle $B$, we find that $AB = \sqrt{10^2 - 5^2} = 5\sqrt{3}$. This means that $\triangle OAB$ is a $30-60-90$ triangle with sides $5:5\sqrt{3}:10$.

Notice that $OAD$ is a line, since all points are on $y = 2x$. In fact, it is a line that makes a $60^\circ$ angle with the positive x-axis. Thus, $\angle DOC = 60^\circ$, and $\angle AOB = 60^\circ$. These are two parts of the stright line $OAD$. The third angle is $\angle BOC$, which must be $60^\circ$ as well. Thus, the arc that we travel is a $60^\circ$ arc, and we travel $\frac{C}{6} = \frac{2\pi r}{6} = \frac{2\pi \cdot 5}{6} = \frac{5\pi}{3}$ around the circle.

Thus, $AB = 5\sqrt{3}$, $\widehat {BC} = \frac{5\pi}{3}$, and ${CD} = 5\sqrt{3}$. The total distance is $10\sqrt{3} + \frac{5\pi}{3}$, which is option $\boxed{C}$.

## Problem 21

Triangles $ABC$ and $ABD$ are isosceles with $AB=AC=BD$, and $BD$ intersects $AC$ at $E$. If $BD$ is perpendicular to $AC$, then $\angle C+\angle D$ is

$\text{(A)}\ 115^\circ\qquad\text{(B)}\ 120^\circ\qquad\text{(C)}\ 130^\circ\qquad\text{(D)}\ 135^\circ\qquad\text{(E)}\ \text{not uniquely determined}$

Solution 1

Redraw the figure as a concave pentagon $ADECB$:

The angles of the pentagon will still sum to $180^\circ \cdot 3 = 540^\circ$, regardless of whether the pentagon is concave or not. As a quick proof, note that the nine angles of three original triangles $\triangle AEB$, $\triangle CBE$, and $\triangle DEA$ all make up the angles of the pentagon without overlap.

Since reflex $\angle E = 270^\circ$, we have:

$\angle C + \angle B + \angle A + \angle D = 540^\circ - 270^\circ = 270^\circ$.

From isosceles $\triangle ABC$, we get $\angle B = \angle C$, so:

$2\angle C + \angle D + \angle A = 270^\circ$

From isosceles $\triangle ABD$, we get $\angle A = \angle D$, so:

$2\angle C + 2\angle D = 270^\circ$

$\angle C + \angle D = 135^\circ$, which is answer $\boxed{D}$

Solution 2

Let $m\angle ABC = x$. By the isosceles triangle theorem, we have $m\angle ACB = x$ and $m\angle BAD = m\angle BDA$. Because the angles of a triangle sum to $\pi$, we have $m\angle BAC = \pi - 2x$, then $m\angle ABD=\frac{\pi}{2}-(\pi-2x)=2x-\frac{\pi}{2}$. Then we have $m\angle BAD + m\angle BDA = \pi - m\angle ABD$. Substituting, this becomes $2m\angle BDA = \pi-\left(\frac{\pi}{2}-(\pi-2x)\right) \to m \angle BDA = \frac{3 \pi}{4} - x$. Adding $x$, which is $m\angle ACB$, we have $m\angle BDA + m\angle ACB = \frac{3 \pi}{4} = 135^\circ \to \boxed{\textbf{D}}$

## Problem 22

Four distinct points, $A$, $B$, $C$, and $D$, are to be selected from $1996$ points evenly spaced around a circle. All quadruples are equally likely to be chosen. What is the probability that the chord $AB$ intersects the chord $CD$?

$\text{(A)}\ \frac{1}{4}\qquad\text{(B)}\ \frac{1}{3}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{2}{3}\qquad\text{(E)}\ \frac{3}{4}$

Solution

Let $WXYZ$ be a convex cyclic quadrilateral inscribed in a circle. There are $\frac{\binom{4}{2}}{2} = 3$ ways to divide the points into two groups of two.

If you pick $WX$ and $YZ$, you have two sides of the quadrilateral, which do not intersect.

If you pick $XY$ and $ZW$, you have the other two sides of the quadrilateral, which do not intersect.

If you pick $WY$ and $XZ$, you have the diagonals of the quadrilateral, which do intersect.

Any four points on the original circle of $1996$ can be connected to form such a convex quarilateral $WXYZ$, and only placing $A$ and $C$ as one of the diagonals of the figure will form intersecting chords. Thus, the answer is $\frac{1}{3}$, which is option $\boxed{\text{B}}$.

Notice that $1996$ is irrelevant to the solution of the problem; in fact, you may pick points from the entire circumference of the circle.

## Problem 23

The sum of the lengths of the twelve edges of a rectangular box is $140$, and the distance from one corner of the box to the farthest corner is $21$. The total surface area of the box is

$\text{(A)}\ 776\qquad\text{(B)}\ 784\qquad\text{(C)}\ 798\qquad\text{(D)}\ 800\qquad\text{(E)}\ 812$

Solution

Let $x, y$, and $z$ be the unique lengths of the edges of the box. Each box has $4$ edges of each length, so: $4x + 4y + 4z = 140 \ \Longrightarrow \ x + y + z = 35.$ The spacial diagonal (longest distance) is given by $\sqrt{x^2 + y^2 + z^2}$. Thus, we have $\sqrt{x^2 + y^2 + z^2} = 21$, so $x^2 + y^2 + z^2 = 21^2$.

Our target expression is the surface area of the box:

$S = 2xy + 2xz + 2yz.$

Since $S$ is a symmetric polynomial of degree $2$, we try squaring the first equation to get:

$35^2 = (x + y + z)^2 = x^2 + y^2 + z^2 + 2xy +2yz + 2xz = 35^2.$

Substituting in our long diagonal and surface area expressions, we get: $21^2 + S = 35^2$, so $S = (35 + 21)(35 - 21) = 56\cdot 14 = 784$, which is option $\boxed{(\text{B})}$.

## Problem 24

The sequence $1, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2,\ldots$ consists of $1$’s separated by blocks of $2$’s with $n$ $2$’s in the $n^{th}$ block. The sum of the first $1234$ terms of this sequence is

$\text{(A)}\ 1996\qquad\text{(B)}\ 2419\qquad\text{(C)}\ 2429\qquad\text{(D)}\ 2439\qquad\text{(E)}\ 2449$

Solution

The sum of the first $1$ numbers is $1$

The sum of the next $2$ numbers is $2 + 1$

The sum of the next $3$ numbers is $2 + 2 + 1$

In genereal, we can write "the sum of the next $n$ numbers is $1 + 2(n-1)$", where the word "next" follows the pattern established above.

Thus, we first want to find what triangular numbers $1234$ is between. By plugging in various values of $n$ into $f(n) = \frac{n(n+1)}{2}$, we find:

$f(50) = 1275$

$f(49) = 1225$

Thus, we want to add up all those sums from "next $1$ number" to the "next $49$ numbers", which will give us all the numbers up to and including the $1225^{th}$ number. Then, we can manually tack on the remaining $2$s to hit $1234$.

We want to find:

$\sum_{n=1}^{49} 1 + 2(n-1)$

$\sum_{n=1}^{49} 2n - 1$

$\sum_{n=1}^{49} 2n - \sum_{n=1}^{49} 1$

$2 \sum_{n=1}^{49} n - 49$

$2\cdot \frac{49\cdot 50}{2} - 49$

$49^2$

$2401$

Thus, the sum of the first $1225$ terms is $2401$. We have to add $9$ more $2$s to get to the $1234^{th}$ term, which gives us $2419$, or option $\boxed{B}$.

Note: If you notice that the above sums form $1 + 3 + 5 + 7... + (2n-1) = n^2$, the fact that $49^2$ appears at the end should come as no surprise.

Solution 2

We note that the number of elements in each block are

$2, 3, 4,$ and so on.

Let $n$ be the number of blocks up to $1234$ terms.

We have $\frac {2+n+1}{2} \cdot (n)=1234$

Solving for $n$, we get about $48.2$.

We have $48$ full blocks and $1$ partial block. We find that there are a total of $49$ $1$'s

Now, we change every number in the sequence to $2$, and then add. We get $2468$. Since we added $49$ by changing all $49$ $1$'s to $2$, we must subract that $49$. Giving us $2468-49=2419$ $\boxed{B}$

## Problem 25

Given that $x^2 + y^2 = 14x + 6y + 6$, what is the largest possible value that $3x + 4y$ can have?

$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$

Solution

Complete the square to get $(x-7)^2 + (y-3)^2 = 64$. Applying Cauchy-Schwarz directly, $64*25=(3^2+4^2)((x-7)^2 + (y-3)^2) \ge (3(x-7)+4(y-3))^2$. $40 \ge 3x+4y-33$ $3x+4y \le 73$. Thus our answer is $\boxed{(B)}$.

## Problem 26

An urn contains marbles of four colors: red, white, blue, and green. When four marbles are drawn without replacement, the following events are equally likely:

(a) the selection of four red marbles;

(b) the selection of one white and three red marbles;

(c) the selection of one white, one blue, and two red marbles; and

(d) the selection of one marble of each color.

What is the smallest number of marbles satisfying the given condition?

$\text{(A)}\ 19\qquad\text{(B)}\ 21\qquad\text{(C)}\ 46\qquad\text{(D)}\ 69\qquad\text{(E)}\ \text{more than 69}$

Solution

Let the bag contain $n$ marbles total, with $r, w, b, g$ representing the number of red, white, blue, and green marbles, respectively. Note that $r + w + b + g = n$.

The number of ways to select four red marbles out of the set of marbles without replacement is:

$\binom{r}{4} = \frac{r!}{24\cdot (r -4)!}$

The number of ways to select one white and three red marbles is:

$\binom{w}{1}\binom{r}{3} = \frac{w\cdot r!}{6\cdot (r - 3)!}$

The number of ways to select one white, one blue, and two red marbles is:

$\binom{w}{1}\binom{b}{1} \binom{r}{2} = \frac{wb\cdot r!}{2(r-2)!}$

The number of ways to select one marble of each colors is:

$\binom{w}{1}\binom{b}{1} \binom{g}{1}\binom{r}{1} = wbgr$

Setting the first and second statements equal, we find:

$\frac{r!}{24\cdot (r -4)!} = \frac{w\cdot r!}{6\cdot (r - 3)!}$

$r - 3 = 4w$

Setting the first and third statements equal, we find:

$\frac{r!}{24\cdot (r -4)!} = \frac{wb\cdot r!}{2(r-2)!}$

$(r-3)(r-2) = 12wb$

Setting the last two statements equal, we find:

$\frac{wb\cdot r!}{2(r-2)!} = wbgr$

$r - 1 = 2g$

These are all the "linking equations" that are needed; the transitive property of equality makes the other three equalities unnecessary.

From the first equation, we know that $r$ must be $3$ more than a multiple of $4$, or that $r \equiv 3 \mod 4$

Putting the first equation into the second equation, we find $r-2 = 3b$. Therefore, $r \equiv 2 \mod 3$. Using the Chinese Remainder Theorem, we find that $r \equiv 11 \mod 12$.

The third equation gives no new restrictions on $r$; it is already odd by the first equation.

Thus, the minimal positive value of $r$ is $11$. This requires $g=\frac{r - 1}{2} = 5$ by the third equation, and $w = \frac{r-3}{4} = 2$ by the first equation. Finally, the second equation gives $b = \frac{(r-3)(r-2)}{12w} = 3$.

The minimal total number of marbles is $11 + 5 + 2 + 3 = \boxed{21}$, which is option $\boxed{B}$.

## Problem 27

Consider two solid spherical balls, one centered at $(0, 0,\frac{21}{2})$ with radius $6$, and the other centered at $(0, 0, 1)$ with radius $\frac{9}{2}$. How many points with only integer coordinates (lattice points) are there in the intersection of the balls?

$\text{(A)}\ 7\qquad\text{(B)}\ 9\qquad\text{(C)}\ 11\qquad\text{(D)}\ 13\qquad\text{(E)}\ 15$

Solution

The two equations of the balls are

$x^2 + y^2 + \left(z - \frac{21}{2}\right)^2 \le 36$

$x^2 + y^2 + (z - 1)^2 \le \frac{81}{4}$

Note that along the $z$ axis, the first ball goes from $10.5 \pm 6$, and the seocnd ball goes from $1 \pm 4.5$. The only integer value that $z$ can be is $z=5$.

Plugging that in to both equations, we get:

$x^2 + y^2 \le \frac{23}{4}$

$x^2 + y^2 \le \frac{17}{4}$

The second inequality implies the first inequality, so the only condition that matters is the second inequality.

From here, we do casework, noting that $|x|, |y| \le 3$:

For $x=\pm 2$, we must have $y=0$. This gives $2$ points.

For $x = \pm 1$, we can have $y\in \{-1, 0, 1\}$. This gives $2\cdot 3 = 6$ points.

For $x = 0$, we can have $y \in \{-2, -1, 0, 1, 2\}$. This gives $5$ points.

Thus, there are $\boxed{13}$ possible points, giving answer $\boxed{D}$.

## Problem 28

On a $4\times 4\times 3$ rectangular parallelepiped, vertices $A$, $B$, and $C$ are adjacent to vertex $D$. The perpendicular distance from $D$ to the plane containing $A$, $B$, and $C$ is closest to

$\text{(A)}\ 1.6\qquad\text{(B)}\ 1.9\qquad\text{(C)}\ 2.1\qquad\text{(D)}\ 2.7\qquad\text{(E)}\ 2.9$

Solution 1

By placing the cube in a coordinate system such that $D$ is at the origin, $A(0,0,3)$, $B(4,0,0)$, and $C(0,4,0)$, we find that the equation of plane $ABC$ is:

$\frac{x}{4} + \frac{y}{4} + \frac{z}{3} = 1$, so $3x + 3y + 4z - 12 = 0.$ The equation for the distance of a point $(a,b,c)$ to a plane $Ax + By + Cz + D = 0$ is given by: $\frac{Aa + Bb + Cc + D}{\sqrt{A^2 + B^2 + C^2}}.$

Note that the capital letters are coefficients, while the lower case is the point itself. Thus, the distance from the origin (where $a=b=c=0$) to the plane is given by: $\frac{D}{\sqrt{A^2 + B^2 + C^2}} = \frac{12}{\sqrt{9 + 9 + 16}} = \frac{12}{\sqrt{34}}.$

Since $\sqrt{34} < 6$, this number should be just a little over $2$, and the correct answer is $\boxed{\text{(C)}}$.

Note that the equation above for the distance from a point to a plane is a 3D analogue of the 2D case of the distance formula, where you take the distance from a point to a line. In the 2D case, both $c$ and $C$ are set equal to $0$.

## Problem 29

If $n$ is a positive integer such that $2n$ has $28$ positive divisors and $3n$ has $30$ positive divisors, then how many positive divisors does $6n$ have?

$\text{(A)}\ 32\qquad\text{(B)}\ 34\qquad\text{(C)}\ 35\qquad\text{(D)}\ 36\qquad\text{(E)}\ 38$

Solution 1

Working with the second part of the problem first, we know that $3n$ has $30$ divisors. We try to find the various possible prime factorizations of $3n$ by splitting $30$ into various products of $1, 2$ or $3$ integers.

$30 \rightarrow p^{29}$

$2 \cdot 15 \rightarrow pq^{14}$

$3\cdot 10 \rightarrow p^2q^9$

$5\cdot 6 \rightarrow p^4q^5$

$2\cdot 3\cdot 5 \rightarrow pq^2r^4$

The variables $p, q, r$ are different prime factors, and one of them must be $3$. We now try to count the factors of $2n$, to see which prime factorization is correct and has $28$ factors.

In the first case, $p=3$ is the only possibility. This gives $2n = 2\cdot p^{28}$, which has $2\cdot {29}$ factors, which is way too many.

In the second case, $p=3$ gives $2n = 2q^{14}$. If $q=2$, then there are $16$ factors, while if $q\neq 2$, there are $2\cdot 15 = 30$ factors.

In the second case, $q=3$ gives $2n = 2p3^{13}$. If $p=2$, then there are $3\cdot 13$ factors, while if $p\neq 2$, there are $2\cdot 2 \cdot 13$ factors.

In the third case, $p=3$ gives $2n = 2\cdot 3\cdot q^9$. If $q=2$, then there are $11\cdot 2 = 22$ factors, while if $q \neq 2$, there are $2\cdot 2\cdot 10$ factors.

In the third case, $q=3$ gives $2n = 2\cdot p^2\cdot 3^8$. If $p=2$, then there are $4\cdot 9$ factors, while if $p \neq 2$, there are $2\cdot 3\cdot 9$ factors.

In the fourth case, $p=3$ gives $2n = 2\cdot 3^3\cdot q^5$. If $q=2$, then there are $7\cdot 4= 28$ factors. This is the factorization we want.

Thus, $3n = 3^4 \cdot 2^5$, which has $5\cdot 6 = 30$ factors, and $2n = 3^3 \cdot 2^6$, which has $4\cdot 7 = 28$ factors.

In this case, $6n = 3^4\cdot 2^6$, which has $5\cdot 7 = 35$ factors, and the answer is $\boxed{C}$

## Problem 30

A hexagon inscribed in a circle has three consecutive sides each of length 3 and three consecutive sides each of length 5. The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length 3 and the other with three sides each of length 5, has length equal to $m/n​$, where $m​$ and $n​$ are relatively prime positive integers. Find $m + n​$.

$\textbf{(A)}\ 309 \qquad \textbf{(B)}\ 349 \qquad \textbf{(C)}\ 369 \qquad \textbf{(D)}\ 389 \qquad \textbf{(E)}\ 409$

Solution 1

In hexagon $ABCDEF$, let $AB=BC=CD=3$ and let $DE=EF=FA=5$. Since arc $BAF$ is one third of the circumference of the circle, it follows that $\angle BCF = \angle BEF=60^{\circ}$. Similarly, $\angle CBE =\angle CFE=60^{\circ}$. Let $P$ be the intersection of $\overline{BE}$ and $\overline{CF}$, $Q$ that of $\overline{BE}$ and $\overline{AD}$, and $R$ that of $\overline{CF}$ and $\overline{AD}$. Triangles $EFP$ and $BCP$ are equilateral, and by symmetry, triangle $PQR$ is isosceles and thus also equilateral.

Furthermore, $\angle BAD$ and $\angle BED$ subtend the same arc, as do $\angle ABE$ and $\angle ADE$. Hence triangles $ABQ$ and $EDQ$ are similar. Therefore, $\frac{AQ}{EQ}=\frac{BQ}{DQ}=\frac{AB}{ED}=\frac{3}{5}.$ It follows that $\frac{\frac{AD-PQ}{2}}{PQ+5} =\frac{3}{5}\quad \mbox {and}\quad \frac{3-PQ}{\frac{AD+PQ}{2}}=\frac{3}{5}.$ Solving the two equations simultaneously yields $AD=360/49,$ so $m+n=\boxed{409}. \blacksquare$

Solution 2

All angle measures are in degrees. Let the first trapezoid be $ABCD$, where $AB=BC=CD=3$. Then the second trapezoid is $AFED$, where $AF=FE=ED=5$. We look for $AD$.

Since $ABCD$ is an isosceles trapezoid, we know that $\angle BAD=\angle CDA$ and, since $AB=BC$, if we drew $AC$, we would see $\angle BCA=\angle BAC$. Anyway, $\widehat{AB}=\widehat{BC}=\widehat{CD}$ ($\widehat{AB}$ means arc AB). Using similar reasoning, $\widehat{AF}=\widehat{FE}=\widehat{ED}$.

Let $\widehat{AB}=2\phi$ and $\widehat{AF}=2\theta$. Since $6\theta+6\phi=360$ (add up the angles), $2\theta+2\phi=120$ and thus $\widehat{AB}+\widehat{AF}=\widehat{BF}=120$. Therefore, $\angle FAB=\frac{1}{2}\widehat{BDF}=\frac{1}{2}(240)=120$. $\angle CDE=120$ as well.

Now I focus on triangle $FAB$. By the Law of Cosines, $BF^2=3^2+5^2-30\cos{120}=9+25+15=49$, so $BF=7$. Seeing $\angle ABF=\theta$ and $\angle AFB=\phi$, we can now use the Law of Sines to get: $\sin{\phi}=\frac{3\sqrt{3}}{14}\;\text{and}\;\sin{\theta}=\frac{5\sqrt{3}}{14}.$

Now I focus on triangle $AFD$. $\angle AFD=3\phi$ and $\angle ADF=\theta$, and we are given that $AF=5$, so $\frac{\sin{\theta}}{5}=\frac{\sin{3\phi}}{AD}.$ We know $\sin{\theta}=\frac{5\sqrt{3}}{14}$, but we need to find $\sin{3\phi}$. Using various identities, we see

$\begin{align*}\sin{3\phi}&=\sin{(\phi+2\phi)}=\sin{\phi}\cos{2\phi}+\cos{\phi}\sin{2\phi}\\ &=\sin{\phi}(1-2\sin^2{\phi})+2\sin{\phi}\cos^2{\phi}\\ &=\sin{\phi}\left(1-2\sin^2{\phi}+2(1-\sin^2{\phi})\right)\\ &=\sin{\phi}(3-4\sin^2{\phi})\\ &=\frac{3\sqrt{3}}{14}\left(3-\frac{27}{49}\right)=\frac{3\sqrt{3}}{14}\left(\frac{120}{49}\right)=\frac{180\sqrt{3}}{343} \end{align*}$

Returning to finding $AD$, we remember $\frac{\sin{\theta}}{5}=\frac{\sin{3\phi}}{AD}\;\text{so}\;AD=\frac{5\sin{3\phi}}{\sin{\theta}}.$ Plugging in and solving, we see $AD=\frac{360}{49}$. Thus, the answer is $360 + 49 = 409$, which is answer choice $\boxed{\textbf{(E)}}$.

Solution 2

Let $x$ be the desired length. One can use Parameshvara's circumradius formula, which states that for a cyclic quadrilateral with sides $a, b, c, d$ the circumradius $R$ satisfies $R^2=\frac{1}{16}\cdot\frac{(ab+cd)(ac+bd)(ad+bc)}{(s-a)(s-b)(s-c)(s-d)},$ where $s$ is the semiperimeter. Applying this to the trapezoid with sides $3, 3, 3, x$, we see that many terms cancel and we are left with $R^2=\frac{27}{9-x}$ Similar canceling occurs for the trapezoid with sides $5, 5, 5, x$, and since the two quadrilaterals share the same circumradius, we can equate: $\frac{27}{9-x}=\frac{125}{15-x}$ Solving for $x$ gives $x=\frac{360}{49}$, so the answer is $\fbox{(E) 409}$.