Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: A piece of graph paper is folded once so that $(0,2)$ is matched with $(4,0)$ , and $(7,3)$ is matched with $(m,n)$ . Find $m+n$
$\mathrm{(A) \ }6.7 \qquad \mathrm{(B) \ }6.8 \qquad \mathrm{(C) \ }6.9 \qquad \mathrm{(D) \ }7.0 \qquad \mathrm{(E) \ }8.0$

Solution: Note that the fold is the perpendicular bisector of $(0,2)$ and $(4,0)$ .  Thus, the fold goes through the midpoint $(2,1)$
The fold also has a slope of $-\frac{1}{m}$ , where the $m$ is the slope of the line connecting these two points.  We find $m = \frac{0 - 2}{4 - 0} = -\frac{1}{2}$ . Thus, the slope of the fold is $2$ and goes through $(2,1)$ , so the equation of the fold is $y = 2x - 3$
We want the line connecting $(7,3)$ and $(m,n)$ to have the same fold as a perpendicular bisector. The slope should be $-\frac{1}{2}$ , so we get $\frac{n - 3}{m - 7} = -\frac{1}{2}$ , which leads to $m = 13 -2n$
We also want $y = 2x - 3$ to bisect the segment from $(7,3)$ to $(m,n)$ . Thus, the midpoint $(x,y) = \left(\dfrac{7 + m}{2}, \frac{3 + n}{2}\right)$ must lie on the line. Plugging into the equation of the line, we find $\frac{3 + n}{2} = (7 + m) - 3$ , which simplifies to $n = 5 + 2m$
Solving the system of two equations in two variables $m = 13 - 2n$ and $n = 5 + 2m$ gives $m = \frac{3}{5}$ and $n = \frac{31}{5}$ , for a sum of $\boxed{6.8}$

Answer: B 6.8
Problem: In triangle $ABC$ , angle $C$ is a right angle and $CB > CA$ . Point $D$ is located on $\overline{BC}$ so that angle $CAD$ is twice angle $DAB$ . If $AC/AD = 2/3$ , then $CD/BD = m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
$\mathrm{(A) \ }10 \qquad \mathrm{(B) \ }14 \qquad \mathrm{(C) \ }18 \qquad \mathrm{(D) \ }22 \qquad \mathrm{(E) \ } 26$

Solution: Let $AC = 2$ $AD = 3$ $\cos \angle CAD = \frac23$
By the pythagorean theorem $CD = \sqrt{3^2-2^2} = \sqrt{5}$
$\sin \angle BDA = \sin (180^{\circ} - \angle BDA) = \sin \angle CDA = \cos \angle (90^{\circ} - CDA) = \cos \angle CAD = \frac23$
$\sin \angle BAD = \sqrt{ \frac{1-cos (2\angle BAD)}{2} } = \sqrt{ \frac{1-\cos \angle CAD}{2} } = \sqrt{ \frac{1-\frac23}{2} } = \frac{\sqrt{6}}{6}$
By the Law of Sine, $\frac{ \sin \angle BDA }{AB} = \frac{ \sin \angle BAD }{BD}$
$\frac{ \frac23 }{ \sqrt{2^2 + ( \sqrt{5} + BD)^2} } = \frac{ \frac{\sqrt{6}}{6} }{BD}$
$8BD^2 = 3(9+ 2BD \sqrt{5} + BD^2)$
$5BD^2 - 6 BD \sqrt{5} -27=0$
As $BD>0$ $BD = \frac{6 \sqrt{5} + \sqrt{ (6 \sqrt{5})^2 - 4 \cdot 5 (-27) } }{10} = \frac{9\sqrt{5}}{5}$
$\frac{CD}{BD} = \frac{\sqrt{5}}{\frac{9\sqrt{5}}{5}} = \frac59$
$5+9=\boxed{14}$

Answer: B 14
Problem: A point $(x,y)$ in the plane is called a lattice point if both $x$ and $y$ are integers. The area of the largest square that contains exactly three lattice points in its interior is closest to
$\mathrm{(A) \ } 4.0 \qquad \mathrm{(B) \ } 4.2 \qquad \mathrm{(C) \ } 4.5 \qquad \mathrm{(D) \ } 5.0 \qquad \mathrm{(E) \ }  5.6$

Solution: [asy] real e = 0.1;  dot((0,-1)); dot((1,-1)); dot((-1,0)); dot((0,0)); dot((1,0)); dot((2,0)); dot((-1,1)); dot((0,1)); dot((1,1)); dot((0,2)); dot((-1,-1)); dot((2,2)); dot((1,2)); dot((2,1)); dot((2,-1)); dot((-1,2)); draw((0.8, -1.4+e)--(1.8-e, 0.6)--(-0.2, 1.6-e)--(-1.2+e, -0.4)--cycle); [/asy] The best square's side length is slightly less than $\sqrt 5$ , yielding an answer of $\boxed{5.0}$

Answer: D 5.0
Problem: For each positive integer $n$ , let
Let $k$ denote the smallest positive integer for which the rightmost nonzero digit of $a_k$ is odd. The rightmost nonzero digit of $a_k$ is
$\mathrm{(A) \ }1 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \ }5 \qquad \mathrm{(D) \ } 7 \qquad \mathrm{(E) \ } 9$

Solution: We have $a_n = n(n+1)\dots (n+9)$
The value $a_n$ can be written as $2^{x_n} 5^{y_n} r_n$ , where $r_n$ is not divisible by 2 and 5. The number of trailing zeroes is $z_n = \min(x_n,y_n)$ . The last non-zero digit is the last digit of $2^{x_n-z_n} 5^{y_n-z_n} r_n$
Clearly, the last non-zero digit is even iff $x_n - z_n > 0$ iff $x_n > y_n$
Thus we are looking for the smallest $n$ such that the power of $5$ that divides $a_n$ is at least equal to the power of $2$ that divides $a_n$
The number $a_n$ is a product of $10$ consecutive integers. Out of these, $5$ are divisible by $2$ . Out of those $5$ , at least $2$ are divisible by $4$ , and out of those $2$ , one is divisible by $8$ . Therefore $x_n\geq 5+2+1=8$ for all $n$
On the other hand, exactly $2$ of our ten integers are divisible by $5$ , and at most one of them can be divisible by a higher power of $5$ . As we need $y_n\geq x_n\geq 8$ , one of the integers from $n$ to $n+9$ must be divisible by $5^7 = 78125$ . Therefore $n\geq 78116$
We can now take numbers starting with $78116$ , and write each of them in the form $2^x 5^y r$ . We are looking for 10 consecutive rows where the sum of $y$ s is at least equal to the sum of $x$ s.
At this point we can stop, as we just found out that $a_{78117}$ is of the form $2^8 5^8 r_{78117}$ . Therefore the $k$ we seek is $k=78117$
Now all we need to do is to compute the last non-zero digit. As the powers of $2$ and $5$ that divide $a_{78117}$ are equal, the last non-zero digit is simply the product of the last digits of the ten $z$ s. This is $7\cdot 9\cdot 9\cdot 3\cdot 1\cdot 1\cdot 3\cdot 1\cdot 1\cdot 3 \equiv\boxed{9}$

Answer:  9
Problem: If $\texttt{a}$ and $\texttt{b}$ are digits for which
$\begin{array}{ccc}& 2 & a\\ \times & b & 3\\ \hline & 6 & 9\\ 9 & 2\\ \hline 9 & 8 & 9\end{array}$
then $\texttt{a+b =}$
$\mathrm{(A)\ } 3 \qquad \mathrm{(B) \ }4 \qquad \mathrm{(C) \  } 7 \qquad \mathrm{(D) \  } 9 \qquad \mathrm{(E) \  }12$

Solution: From the units digit calculation, we see that the units digit of $a\times 3$ is $9$ .  Since $0 \le a \le 9$ and $a$ is an integer, the only value of $a$ that works is is $a=3$ .  As a double-check, that does work, since $23 \times 3 = 69$ , which is the first line of the multiplication.
The second line of the multiplication can be found by doing the multiplication $23\times b = 92$ .  Dividing both sides by $23$ gives $b=4$
Thus, $a + b = 3 + 4 = 7$ , and the answer is $\boxed{7}$

Answer: C 7
Problem: The adjacent sides of the decagon shown meet at right angles.  What is its perimeter?
[asy] defaultpen(linewidth(.8pt)); dotfactor=4; dot(origin);dot((12,0));dot((12,1));dot((9,1));dot((9,7));dot((7,7));dot((7,10));dot((3,10));dot((3,8));dot((0,8)); draw(origin--(12,0)--(12,1)--(9,1)--(9,7)--(7,7)--(7,10)--(3,10)--(3,8)--(0,8)--cycle); label("$8$",midpoint(origin--(0,8)),W); label("$2$",midpoint((3,8)--(3,10)),W); label("$12$",midpoint(origin--(12,0)),S);[/asy]
$\mathrm{(A)\ } 22 \qquad \mathrm{(B) \ }32 \qquad \mathrm{(C) \  } 34 \qquad \mathrm{(D) \  } 44 \qquad \mathrm{(E) \  }50$

Solution: The three unlabelled vertical sides have the same sum as the two labelled vertical sides, which is $10$
The four unlabelled horizontal sides have the same sum as the one large horizontal side, which is $12$
Thus, the perimeter is $2(12+10) = 44$ , which is option $\boxed{44}$

Answer: D 44
Problem: If $x$ $y$ , and $z$ are real numbers such that
then $x + y + z =$
$\mathrm{(A)\ } -12 \qquad \mathrm{(B) \ }0 \qquad \mathrm{(C) \  } 8 \qquad \mathrm{(D) \  } 12 \qquad \mathrm{(E) \  }50$

Solution: If the sum of three squared expressions is zero, then each expression itself must be zero, since $a^2 \ge 0$ with the equality iff $a=0$
In this case, $x-3=0$ $y-4=0$ , and $z-5=0$ .  Adding the three equations and moving the constant to the right gives $x + y + z = 12$ , and the answer is $\boxed{12}$

Answer: D 12
Problem: If $a$ is $50\%$ larger than $c$ , and $b$ is $25\%$ larger than $c$ , then $a$ is what percent larger than $b$
$\mathrm{(A)\ } 20\% \qquad \mathrm{(B) \ }25\% \qquad \mathrm{(C) \  } 50\% \qquad \mathrm{(D) \  } 100\% \qquad \mathrm{(E) \  }200\%$

Solution: Translating each sentence into an equation, $a = 1.5c$ and $b = 1.25c$
We want a relationship between $a$ and $b$ .  Dividing the second equation into the first will cancel the $c$ , so we try that and get:
$\frac{a}{b} = \frac{1.5}{1.25}$
$\frac{a}{b} = \frac{150}{125}$
$\frac{a}{b} = \frac{6}{5}$
$a = 1.2b$
In this case, $a$ is $1.2 - 1 = 0.2 = 20\%$ bigger than $b$ , and the answer is $\boxed{20}$

Answer: A 20
Problem: If $a$ is $50\%$ larger than $c$ , and $b$ is $25\%$ larger than $c$ , then $a$ is what percent larger than $b$
$\mathrm{(A)\ } 20\% \qquad \mathrm{(B) \ }25\% \qquad \mathrm{(C) \  } 50\% \qquad \mathrm{(D) \  } 100\% \qquad \mathrm{(E) \  }200\%$

Solution: Arbitrarily assign a value to one of the variables.  Since $c$ is the smallest variable, let $c = 100$
If $a$ is $50\%$ larger than $c$ , then $a = 150$
If $b$ is $25\%$ larger than $c$ , then $b = 125$
We see that $\frac{a}{b} = \frac{150}{125} = 1.2$ So, $a$ is $20\%$ bigger than $b$ , and the answer is $\boxed{20}$

Answer: A 20
Problem: A rectangle with perimeter $176$ is divided into five congruent rectangles as shown in the diagram.  What is the perimeter of one of the five congruent rectangles? [asy] defaultpen(linewidth(.8pt)); draw(origin--(0,3)--(4,3)--(4,0)--cycle); draw((0,1)--(4,1)); draw((2,0)--midpoint((0,1)--(4,1))); real r = 4/3; draw((r,3)--foot((r,3),(0,1),(4,1))); draw((2r,3)--foot((2r,3),(0,1),(4,1)));[/asy]
$\mathrm{(A)\ } 35.2 \qquad \mathrm{(B) \ }76 \qquad \mathrm{(C) \  } 80 \qquad \mathrm{(D) \  } 84 \qquad \mathrm{(E) \  }86$

Solution: Let $l$ represent the length of one of the smaller rectangles, and let $w$ represent the width of one of the smaller rectangles, with $w < l$
From the large rectangle, we see that the top has length $3w$ , the right has length $l + w$ , the bottom has length $2l$ , and the left has length $l + 2$
Since the perimeter of the large rectangle is $176$ , we know that $172 = 3w + l + w + 2l + l + w$ , or $172 = 5w + 4l$
From the top and bottom of the large rectangle, we know that $3w = 2l$ , or $l = 1.5w$
Plugging that into the first equation, we get $176 = 5w + 4(1.5)w$
$176 = 11w$
$w = 16$
$l = 1.5w = 24$
$P = 2l + 2w = 2(16 + 24) = 80$ , and the answer is $\boxed{80}$

Answer: C 80
Problem: Consider the sequence
$1,-2,3,-4,5,-6,\ldots,$
whose $n$ th term is $(-1)^{n+1}\cdot n$ . What is the average of the first $200$ terms of the sequence?
$\textbf{(A)}-\!1\qquad\textbf{(B)}-\!0.5\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 0.5\qquad\textbf{(E)}\ 1$

Solution: The average of a list is the sum of all numbers divided by the size of the list.
The sum of the list can be found by adding the numbers in pairs: $(1 + -2) + (3 + -4) + ... + (199 + -200)$
The sum of each pair is $-1$ , and there are $100$ pairs, so the total sum is $-100$
There are $200$ numbers on the list, so the average is $\frac{-100}{200} = -0.5$ , and the answer is $\boxed{0.5}$

Answer: B 0.5
Problem: The sum of seven integers is $-1$ . What is the maximum number of the seven integers that can be larger than $13$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$

Solution: If the first six integers are $14$ , the last number can be $(-14\cdot 6) - 1 = -85$ .  The sum of all seven integers will be $-1$
However, if all seven integers are over $13$ , the smallest possible sum is $14\cdot 7 = 98$
Thus, the answer is $6$ , which is option $\boxed{6}$

Answer: D 6
Problem: Mientka Publishing Company prices its bestseller Where's Walter? as follows:
$C(n) =\left\{\begin{matrix}12n, &\text{if }1\le n\le 24\\ 11n, &\text{if }25\le n\le 48\\ 10n, &\text{if }49\le n\end{matrix}\right.$
where $n$ is the number of books ordered, and $C(n)$ is the cost in dollars of $n$ books. Notice that $25$ books cost less than $24$ books. For how many values of $n$ is it cheaper to buy more than $n$ books than to buy exactly $n$ books?
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 8$

Solution: Clearly, the areas of concern are where the piecewise function shifts value.
Since $C(25) = 11\cdot 25 = 275$ , we want to find the least value of $n$ for which $C(n) > 275$
If $n \le 24$ , then $C(n) = 12n$ , so for $C(n) > 275$ $12n > 275$ , which is equivalent to $n > 22.91$ .  Thus, both $n=23$ and $n=24$ will be more expensive than $n=25$
Since $C(49) = 10\cdot 49 = 490$ , we want to find the least value of $n$ for which $C(n) > 490$
If $25 \le n \le 48$ , then $C(n) = 11n$ , so for $C(n) > 490$ , we have $11n > 490$ , leading to $n > 44.5$ .  Thus, $n=45, 46, 47, 48$ will be more expensive than $n=49$
Thus, there are $2 + 4 = \boxed{6}$

Answer: D 6
Problem: In the sixth, seventh, eighth, and ninth basketball games of the season, a player scored $23$ $14$ $11$ , and $20$ points, respectively. Her points-per-game average was higher after nine games than it was after the first five games. If her average after ten games was greater than $18$ , what is the least number of points she could have scored in the tenth game?
$\textbf{(A)}\ 26\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 29\qquad\textbf{(E)}\ 30$

Solution: The sum of the scores for games $6$ through $9$ is $68$ .  The average in these four games is $\frac{68}{4} = 17$
The total points in all ten games is greater than $10\cdot 18 = 180$ .  Thus, it must be at least $181$
There are at least $181 - 68 = 113$ points in the other six games:  games $1-5$ and game $10$
Games $1-5$ must have an average of less than $17$ .  Thus we cannot put more than $16 + 17 + 17 + 17 + 17 = 84$ points in those five games.
Thus, the tenth game must have at least $113 - 84 = 29$ points, and the answer is $\boxed{29}$

Answer: D 29
Problem: How many two-digit positive integers $N$ have the property that the sum of $N$ and the number obtained by reversing the order of the digits of is a perfect square?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$

Solution: Let $N = 10t + u$ , where $t$ is the tens digit and $u$ is the units digit.
The condition of the problem is that $10t + u + 10u + t$ is a perfect square.
Simplifying and factoring, we want $11(t+u)$ to be a perfect square.
Thus, $t+u$ must at least be a multiple of $11$ , and since $t$ and $u$ are digits, the only multiple of $11$ that works is $11$ itself.
Thus, $(t,u) = (2,9)$ is the first solution, and $(t,u) = (9,2)$ is the last solution.  There are $8$ solutions in total, leading to answer $\boxed{8}$

Answer: E 8
Problem: The number of geese in a flock increases so that the difference between the populations in year $n+2$ and year $n$ is directly proportional to the population in year $n+1$ . If the populations in the years $1994$ $1995$ , and $1997$ were $39$ $60$ , and $123$ , respectively, then the population in $1996$ was
$\textbf{(A)}\ 81\qquad\textbf{(B)}\ 84\qquad\textbf{(C)}\ 87\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ 102$

Solution: Let $x$ be the population in $1996$ , and let $k$ be the constant of proportionality.
If $n=1994$ , then the difference in population between $1996$ and $1994$ is directly proportional to the population in $1995$
Translating this sentence, $(x - 39) = k(60)$
Similarly, letting $n=1995$ gives the sentence $(123 - 60) = kx$
Since $kx = 63$ , we have $k = \frac{63}{x}$
Plugging this into the first equation, we have:
$(x - 39) = \frac{60\cdot 63}{x}$
$x - 39 = \frac{3780}{x}$
$x^2 - 39x - 3780 = 0$
$(x - 84)(x + 45) = 0$
Since $x>0$ , we must have $x=84$ , and the answer is $\boxed{84}$

Answer: B 84
Problem: Medians $BD$ and $CE$ of triangle $ABC$ are perpendicular, $BD=8$ , and $CE=12$ .  The area of triangle $ABC$ is
[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (1.25,1); pair C = (2,0); pair D = midpoint(A--C); pair E = midpoint(A--B); pair G = intersectionpoint(E--C,B--D); dot(A);dot(B);dot(C);dot(D);dot(E);dot(G); label("$A$",A,S);label("$B$",B,N);label("$C$",C,S);label("$D$",D,S);label("$E$",E,NW);label("$G$",G,NE); draw(A--B--C--cycle); draw(B--D); draw(E--C); draw(rightanglemark(C,G,D,3));[/asy]
$\textbf{(A)}\ 24\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 48\qquad\textbf{(D)}\ 64\qquad\textbf{(E)}\ 96$

Solution: [asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (1.25,1); pair C = (2,0); pair D = midpoint(A--C); pair E = midpoint(A--B); pair F = midpoint(B--C); pair G = intersectionpoint(E--C,B--D); dot(A);dot(B);dot(C);dot(D);dot(E);dot(G);dot(F); label("$A$",A,S);label("$B$",B,N);label("$C$",C,S);label("$D$",D,S);label("$E$",E,NW);label("$G$",G,NE);label("$F$",F,NE); draw(A--B--C--cycle); draw(B--D); draw(E--C); draw(A--F); draw(rightanglemark(B,G,E,3));[/asy]
One median divides a triangle into $2$ equal areas, so all three medians will divide a triangle into $6$ equal areas.
The median $CE$ is divided into a $2:1$ ratio at centroid $G$ , so $GE = \frac{1}{3}\cdot CE = \frac{1}{3}\cdot 12 = 4$
Similarly, $BG = \frac{2}{3}\cdot 8 = \frac{16}{3}$
The area of the right triangle $\triangle BEG$ is $\frac{1}{2}\cdot\frac{16}{3}\cdot 4$
The area of the whole figure is $6\cdot \frac{1}{2}\cdot\frac{16}{3}\cdot 4 = 64$ , and the correct answer is $\boxed{64}$

Answer: D 64
Problem: Medians $BD$ and $CE$ of triangle $ABC$ are perpendicular, $BD=8$ , and $CE=12$ .  The area of triangle $ABC$ is
[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (1.25,1); pair C = (2,0); pair D = midpoint(A--C); pair E = midpoint(A--B); pair G = intersectionpoint(E--C,B--D); dot(A);dot(B);dot(C);dot(D);dot(E);dot(G); label("$A$",A,S);label("$B$",B,N);label("$C$",C,S);label("$D$",D,S);label("$E$",E,NW);label("$G$",G,NE); draw(A--B--C--cycle); draw(B--D); draw(E--C); draw(rightanglemark(C,G,D,3));[/asy]
$\textbf{(A)}\ 24\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 48\qquad\textbf{(D)}\ 64\qquad\textbf{(E)}\ 96$

Solution: Notice that if you were to draw in line ED, you would get an orthodiagonal quadrilateral with diagonals 8 and 12. The area is going to be equal to 48. Now we need to examine the triangle AED. If the area we are trying to find is denoted as A, we can tell that the area of AEC is A/2. The area of AED is going to be half of that since AD = DC so it would be A/4. This means that 48 is 3/4 of A, so naturally A is going to be 64. Giving $\boxed{64}$

Answer: D 64
Problem: The three row sums and the three column sums of the array
\[\left[\begin{matrix}4 & 9 & 2\\ 8 & 1 & 6\\ 3 & 5 & 7\end{matrix}\right]\]
are the same. What is the least number of entries that must be altered to make all six sums different from one another?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution: If you change $3$ numbers, then you either change one number in each column and row (ie sudoku-style):
\[\left[\begin{matrix}* & 9 & 2\\ 8 & * & 6\\ 3 & 5 & *\end{matrix}\right]\]
Or you leave at least one row and one column unchanged:
\[\left[\begin{matrix}* & 9 & 2\\ * & * & 6\\ 3 & 5 & 7\end{matrix}\right]\]
In the first case, you are changing just one common number in two sums, so you wind up with three pairs of sums.  (In the example given, the sum in row $x$ is the same as in column $x$ .)
In the second case, since two of the sums are unchanged, and the sums started out equal, they must remain equal.  (In the second example given, row $3$ and column $3$ are untouched.)
Either way, $3$ changes is not enough. However, building on the second example, if you change either the untouched column or the untouched row, you will get a possible answer:
\[\left[\begin{matrix}* & 9 & 2\\ * & * & 6\\ 3 & * & 7\end{matrix}\right]\]
Letting the $*$ be a zero does indeed give $6$ different sums, so the answer is $4$ , which is option $\boxed{4}$

Answer: D 4
Problem: A line $x=k$ intersects the graph of $y=\log_5 x$ and the graph of $y=\log_5 (x + 4)$ . The distance between the points of intersection is $0.5$ . Given that $k = a + \sqrt{b}$ , where $a$ and $b$ are integers, what is $a+b$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$

Solution: Since the line $x=k$ is vertical, we are only concerned with vertical distance.
In other words, we want to find the value of $k$ for which the distance $|\log_5 x - \log_5 (x+4)| = \frac{1}{2}$
Since $\log_5 x$ is a strictly increasing function, we have:
$\log_5 (x + 4) - \log_5 x = \frac{1}{2}$
$\log_5 (\frac{x+4}{x}) = \frac{1}{2}$
$\frac{x+4}{x} = 5^\frac{1}{2}$
$x + 4 = x\sqrt{5}$
$x\sqrt{5} - x = 4$
$x = \frac{4}{\sqrt{5} - 1}$
$x = \frac{4(\sqrt{5} + 1)}{5 - 1^2}$
$x = 1 + \sqrt{5}$
The desired quantity is $1 + 5 = 6$ , and the answer is $\boxed{6}$

Answer: A 6
Problem: A list of integers has mode $32$ and mean $22$ . The smallest number in the list is $10$ . The median $m$ of the list is a member of the list. If the list member $m$ were replaced by $m+10$ , the mean and median of the new list would be $24$ and $m+10$ , respectively. If were $m$ instead replaced by $m-8$ , the median of the new list would be $m-4$ . What is $m$
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 20$

Solution: Let there be $n$ integers on the list.  The list of $n$ integers has mean $22$ , so the sum of the integers is $22n$
Replacing $m$ with $m+10$ will increase the sum of the list from $22n$ to $22n + 10$
The new mean of the list is $24$ , so the new sum of the list is also $24n$
Thus, we get $22n + 10 = 24n$ , leading to $n=5$ numbers on the list.
If there are $5$ numbers on the list with mode $32$ and smallest number $10$ , then the list is $\{10, x, m, 32, 32\}$
Since replacing $m$ with $m-8$ gives a new median of $m-4$ , and $m-4$ must be on the list of $5$ integers since $5$ is odd, $x = m-4$ , and the list is now $\{10, m-4, m, 32, 32\}$
The sum of the numbers on this list is $22n = 22\cdot 5 = 110$ , so we get:
$10 + m - 4 + m + 32 + 32 = 110$
$70 + 2m = 110$
$m = 20$ , giving answer $\boxed{20}$

Answer: E 20
Problem: A circle with center $O$ is tangent to the coordinate axes and to the hypotenuse of the $30^\circ$ $60^\circ$ $90^\circ$ triangle $ABC$ as shown, where $AB=1$ . To the nearest hundredth, what is the radius of the circle?
[asy] defaultpen(linewidth(.8pt)); dotfactor=3; pair A = origin; pair B = (1,0); pair C = (0,sqrt(3)); pair O = (2.33,2.33); dot(A);dot(B);dot(C);dot(O); label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,W);label("$O$",O,NW); label("$1$",midpoint(A--B),S);label("$60^\circ$",B,2W + N); draw((3,0)--A--(0,3)); draw(B--C); draw(Arc(O,2.33,163,288.5));[/asy]
$\textbf{(A)}\ 2.18\qquad\textbf{(B)}\ 2.24\qquad\textbf{(C)}\ 2.31\qquad\textbf{(D)}\ 2.37\qquad\textbf{(E)}\ 2.41$

Solution: [asy] defaultpen(linewidth(.8pt)); dotfactor=3; pair A = origin; pair B = (1,0); pair C = (0,sqrt(3)); pair O = (2.33,2.33); pair D = (2.33,0); pair E = (0, 2.33); pair F = (0.35, 1.1); dot(A);dot(B);dot(C);dot(O);dot(D);dot(E);dot(F); label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,W);label("$O$",O,NW);label("$D$",D,SW);label("$E$",E,SW);label("$F$", F, W); label("$1$",midpoint(A--B),S);label("$60^\circ$",B,2W + N);  draw((3,0)--A--(0,3)); draw(B--C);draw(O--E);draw(O--D); draw(Arc(O,2.33,163,288.5));[/asy]
Draw radii $OE$ and $OD$ to the axes, and label the point of tangency to triangle $ABC$ point $F$ .  Let the radius of the circle $O$ be $r$ .  Square $OEAD$ has side length $r$
Because $BD$ and $BF$ are tangents from a common point $B$ $BD = BF$
$AD = AB + BD$
$r = 1 + BD$
$r = 1 + BF$
Similarly, $CF = CE$ , and we can write:
$AE = AC + CE$
$r = \sqrt{3} + CF$
Equating the radii lengths, we have $1 + BF = \sqrt{3} + CF$
This means $BF - CF = \sqrt{3} - 1$
$BF + CF = 2$ by the 30-60-90 triangle.
Therefore, $2BF = 2 + \sqrt{3} - 1$ , and we get $BF = \frac{1}{2} + \frac{\sqrt{3}}{2}$
The radius of the circle is $AD$ , which is $BF + 1 = \frac{3}{2} + \frac{\sqrt{3}}{2}$
Using decimal approximations, $r \approx 1.5 + \frac{1.73^+}{2} \approx 2.37$ , and the answer is $\boxed{2.37}$

Answer: D 2.37