Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $ABCD$ be a parallelogram with area $15$ . Points $P$ and $Q$ are the projections of $A$ and $C,$ respectively, onto the line $BD;$ and points $R$ and $S$ are the projections of $B$ and $D,$ respectively, onto the line $AC.$ See the figure, which also shows the relative locations of these points.
[asy] size(350); defaultpen(linewidth(0.8)+fontsize(11)); real theta = aTan(1.25/2); pair A = 2.5*dir(180+theta), B = (3.35,0), C = -A, D = -B, P = foot(A,B,D), Q = -P, R = foot(B,A,C), S = -R; draw(A--B--C--D--A^^B--D^^R--S^^rightanglemark(A,P,D,6)^^rightanglemark(C,Q,D,6)); draw(B--R^^C--Q^^A--P^^D--S,linetype("4 4")); dot("$A$",A,dir(270)); dot("$B$",B,E); dot("$C$",C,N); dot("$D$",D,W); dot("$P$",P,SE); dot("$Q$",Q,NE); dot("$R$",R,N); dot("$S$",S,dir(270)); [/asy]
Suppose $PQ=6$ and $RS=8,$ and let $d$ denote the length of $\overline{BD},$ the longer diagonal of $ABCD.$ Then $d^2$ can be written in the form $m+n\sqrt p,$ where $m,n,$ and $p$ are positive integers and $p$ is not divisible by the square of any prime. What is $m+n+p?$
$\textbf{(A) }81 \qquad \textbf{(B) }89 \qquad \textbf{(C) }97\qquad \textbf{(D) }105 \qquad \textbf{(E) }113$

Solution: Again, Let $X$ be the intersection of diagonals $AC$ and $BD$ . Note that triangles $\triangle QXC$ and $\triangle BXR$ are similar because they are right triangles and share $\angle CXQ$ . First, call the length of $XB = \frac{d}{2}$ . By the definition of an area of a parallelogram, $CQ \cdot 2XB = 15$ , so $CQ = \frac{15}{d}$ . Using similar triangles on $\triangle QXC$ and $\triangle BXR$ $\frac{CQ}{XQ} = \frac{BR}{XR}$ . Therefore, finding $BR$ $BR = \frac{XR}{XQ} \cdot CQ = \frac{4}{3} \cdot \frac{15}{d} = \frac{20}{d}$ . Now, applying the Pythagorean theorem once, we find $(\frac{20}{d}) ^2$ $(4)^2$ $(\frac{d}{2}) ^2$ . Solving this equation for $d^2$ , we find $d^2=\frac{64+\sqrt{4096+6400}}{2}=32+8\sqrt{41} \rightarrow 32+8+41= \boxed{81}$

Answer: A 81
Problem: Let $ABCD$ be a parallelogram with area $15$ . Points $P$ and $Q$ are the projections of $A$ and $C,$ respectively, onto the line $BD;$ and points $R$ and $S$ are the projections of $B$ and $D,$ respectively, onto the line $AC.$ See the figure, which also shows the relative locations of these points.
[asy] size(350); defaultpen(linewidth(0.8)+fontsize(11)); real theta = aTan(1.25/2); pair A = 2.5*dir(180+theta), B = (3.35,0), C = -A, D = -B, P = foot(A,B,D), Q = -P, R = foot(B,A,C), S = -R; draw(A--B--C--D--A^^B--D^^R--S^^rightanglemark(A,P,D,6)^^rightanglemark(C,Q,D,6)); draw(B--R^^C--Q^^A--P^^D--S,linetype("4 4")); dot("$A$",A,dir(270)); dot("$B$",B,E); dot("$C$",C,N); dot("$D$",D,W); dot("$P$",P,SE); dot("$Q$",Q,NE); dot("$R$",R,N); dot("$S$",S,dir(270)); [/asy]
Suppose $PQ=6$ and $RS=8,$ and let $d$ denote the length of $\overline{BD},$ the longer diagonal of $ABCD.$ Then $d^2$ can be written in the form $m+n\sqrt p,$ where $m,n,$ and $p$ are positive integers and $p$ is not divisible by the square of any prime. What is $m+n+p?$
$\textbf{(A) }81 \qquad \textbf{(B) }89 \qquad \textbf{(C) }97\qquad \textbf{(D) }105 \qquad \textbf{(E) }113$

Solution: Let $BQ = PD = x.$ We know that the area of the parallelogram is $15,$ so it follows that $[\triangle{BCD}] = [\triangle{BAD}] = \tfrac{15}{2}$ and the height of each triangle, which are also the lengths of $QC$ and $AP,$ is $\tfrac{15}{2(x+3)}.$ Suppose that $E = RS \cap BD.$ Because $\angle{BRE} = \angle{CQE}$ and $\angle{BER} = \angle{CQD},$ we have $\triangle{BRE} \sim \triangle{CQE}.$ The length of $CE,$ by the Pythagorean Theorem is $\sqrt{3^2+(\tfrac{15}{2(x+3)})^2}$ and the length of $BR,$ by the Pythagorean Theorem on $\triangle{BRE},$ is $\sqrt{(x+3)^2 - 4}.$ Note that \[\sin{\angle QEC} = \frac{CQ}{CE} = \frac{BR}{BE}\] Substituting in our values, \[\frac{\frac{15}{2(x+3)}}{\sqrt{9+(\frac{15}{2(x+3)})^2}} = \frac{\sqrt{(x+3)^2 - 4^2}}{x+3}\] To rid unnecessary computation, we let $(x+3)^2 = a.$ The equation simplifies, after cross multiplying, to \[\sqrt{9+\frac{15^2}{4a}} \sqrt{a-16 } = \frac{15}{2}\] \[36a^2 - 576a - 15^2\cdot 16 = 0\] \[a^2-16a-100 =0\] By the quadratic formula, $a \in \{\tfrac{16 - \sqrt{656}}{2}, \tfrac{16 + \sqrt{656}}{2}\},$ so we discard the negative solution. The value of $BD^2$ is \[BD^2 = (2x+6)^2 = 4(x+3)^2 = 4a = 4 \cdot \frac{16 + \sqrt{656}}{2} = 32+8\sqrt{41}\] and the desired answer is $32+8+41 = \boxed{81}$ ~skyscraper

Answer: A 81
Problem: Let $ABCD$ be a parallelogram with area $15$ . Points $P$ and $Q$ are the projections of $A$ and $C,$ respectively, onto the line $BD;$ and points $R$ and $S$ are the projections of $B$ and $D,$ respectively, onto the line $AC.$ See the figure, which also shows the relative locations of these points.
[asy] size(350); defaultpen(linewidth(0.8)+fontsize(11)); real theta = aTan(1.25/2); pair A = 2.5*dir(180+theta), B = (3.35,0), C = -A, D = -B, P = foot(A,B,D), Q = -P, R = foot(B,A,C), S = -R; draw(A--B--C--D--A^^B--D^^R--S^^rightanglemark(A,P,D,6)^^rightanglemark(C,Q,D,6)); draw(B--R^^C--Q^^A--P^^D--S,linetype("4 4")); dot("$A$",A,dir(270)); dot("$B$",B,E); dot("$C$",C,N); dot("$D$",D,W); dot("$P$",P,SE); dot("$Q$",Q,NE); dot("$R$",R,N); dot("$S$",S,dir(270)); [/asy]
Suppose $PQ=6$ and $RS=8,$ and let $d$ denote the length of $\overline{BD},$ the longer diagonal of $ABCD.$ Then $d^2$ can be written in the form $m+n\sqrt p,$ where $m,n,$ and $p$ are positive integers and $p$ is not divisible by the square of any prime. What is $m+n+p?$
$\textbf{(A) }81 \qquad \textbf{(B) }89 \qquad \textbf{(C) }97\qquad \textbf{(D) }105 \qquad \textbf{(E) }113$

Solution: Let the intersection of $RS$ and $BD$ be $X$
$\because$ $\angle APX = \angle DSX$ and $\angle AXP = \angle DXS$ $\triangle APX \sim \triangle DSX$ by $AA$
$\therefore$ $\frac{PA}{DS} = \frac68 = \frac34$ $DS = \frac43 \cdot PA$
By the Pythagorean theorem and the property of projection, $BD^2 = (DS+BR)^2 + RS^2 = 4DS^2 + 64 = 4(\frac43 \cdot PA)^2 + 64 = \frac{64}{9} \cdot PA^2 + 64$ $\frac{64}{9} \cdot PA^2 = BD^2 - 64$
$\because [ABCD] = PA \cdot BD = 15$ $\therefore PA = \frac{15}{BD}$
\[\frac{64}{9} (\frac{15}{BD})^2 = BD^2 - 64\]
\[\frac{1600}{BD^2} = BD^2 - 64\]
\[BD^4 - 64 BD^2 - 1600 = 0\]
\[BD^2 = \frac{64 + \sqrt{64^2 - 4 (-1600)}}{2} = 32 + 8 \sqrt{41}\]
Therefore, the answer is $32 + 8 + 41 = \boxed{81}$

Answer: A 81
Problem: Carlos took $70\%$ of a whole pie. Maria took one third of the remainder. What portion of the whole pie was left?
$\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 15\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 30\%\qquad\textbf{(E)}\ 35\%$

Solution: If Carlos took $70\%$ of the pie, there must be $(100 - 70)\% = 30\%$ left. After Maria takes $\frac{1}{3}$ of the remaining $30\%, \ 1 - \frac{1}{3} = \frac{2}{3}$ of the remaining $30\%$ is left.
Therefore, the answer is $30\% \cdot \frac{2}{3} = \boxed{20}.$

Answer: C 20
Problem: Carlos took $70\%$ of a whole pie. Maria took one third of the remainder. What portion of the whole pie was left?
$\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 15\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 30\%\qquad\textbf{(E)}\ 35\%$

Solution: Like solution 1, it is clear that there is $30\%$ of the pie remaining. Since Maria takes $\frac{1}{3}$ of the remainder, she takes $\frac{1}{3} \cdot 30\% = 10\%,$ meaning that there is $30\% - 10\% = \boxed{20}$ left.

Answer: C 20
Problem: Carlos took $70\%$ of a whole pie. Maria took one third of the remainder. What portion of the whole pie was left?
$\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 15\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 30\%\qquad\textbf{(E)}\ 35\%$

Solution: We have \[\left(100\%-70\%\right)\cdot\left(1-\frac13\right)=30\%\cdot\frac23=\boxed{20}\] of the whole pie left.

Answer: C 20
Problem: In the plane figure shown below, $3$ of the unit squares have been shaded. What is the least number of additional unit squares that must be shaded so that the resulting figure has two lines of symmetry?
[asy] import olympiad; unitsize(25); filldraw((1,3)--(1,4)--(2,4)--(2,3)--cycle, gray(0.7)); filldraw((2,1)--(2,2)--(3,2)--(3,1)--cycle, gray(0.7)); filldraw((4,0)--(5,0)--(5,1)--(4,1)--cycle, gray(0.7)); for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { pair A = (j,i); } } for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { if (j != 5) { draw((j,i)--(j+1,i)); } if (i != 4) { draw((j,i)--(j,i+1)); } } } [/asy]
$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$

Solution: The two lines of symmetry must be horizontally and vertically through the middle. We can then fill the boxes in like so:
[asy] import olympiad; unitsize(25); filldraw((1,3)--(1,4)--(2,4)--(2,3)--cycle, gray(0.7)); filldraw((0,3)--(0,4)--(1,4)--(1,3)--cycle, gray(0.9)); filldraw((3,3)--(3,4)--(4,4)--(4,3)--cycle, gray(0.9)); filldraw((4,3)--(4,4)--(5,4)--(5,3)--cycle, gray(0.9)); filldraw((2,2)--(2,3)--(3,3)--(3,2)--cycle, gray(0.9)); filldraw((2,1)--(2,2)--(3,2)--(3,1)--cycle, gray(0.7)); filldraw((3,0)--(4,0)--(4,1)--(3,1)--cycle, gray(0.9)); filldraw((1,0)--(2,0)--(2,1)--(1,1)--cycle, gray(0.9)); filldraw((4,0)--(5,0)--(5,1)--(4,1)--cycle, gray(0.7)); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray(0.9)); for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { pair A = (j,i);  } } for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { if (j != 5) { draw((j,i)--(j+1,i)); } if (i != 4) { draw((j,i)--(j,i+1)); } } } [/asy]
where the light gray boxes are the ones we have filled. Counting these, we get $\boxed{7}$ total boxes.

Answer: D 7
Problem: In the plane figure shown below, $3$ of the unit squares have been shaded. What is the least number of additional unit squares that must be shaded so that the resulting figure has two lines of symmetry?
[asy] import olympiad; unitsize(25); filldraw((1,3)--(1,4)--(2,4)--(2,3)--cycle, gray(0.7)); filldraw((2,1)--(2,2)--(3,2)--(3,1)--cycle, gray(0.7)); filldraw((4,0)--(5,0)--(5,1)--(4,1)--cycle, gray(0.7)); for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { pair A = (j,i); } } for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { if (j != 5) { draw((j,i)--(j+1,i)); } if (i != 4) { draw((j,i)--(j,i+1)); } } } [/asy]
$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$

Solution: We label the three shaded unit squares $A,B,$ and $C,$ then construct the two lines of symmetry of the resulting figure, as shown below: [asy] import olympiad; unitsize(25); filldraw((1,3)--(1,4)--(2,4)--(2,3)--cycle, gray(0.7)); filldraw((2,1)--(2,2)--(3,2)--(3,1)--cycle, gray(0.7)); filldraw((4,0)--(5,0)--(5,1)--(4,1)--cycle, gray(0.7)); for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { pair A = (j,i); } } for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { if (j != 5) { draw((j,i)--(j+1,i)); } if (i != 4) { draw((j,i)--(j,i+1)); } } } draw((-1,2)--(6,2),linewidth(2)+red); draw((2.5,-1)--(2.5,5),linewidth(2)+red); label("$A$",(1.5,3.5)); label("$B$",(2.5,1.5)); label("$C$",(4.5,0.5)); [/asy] Note that:
The shaded unit squares contributed by $A,B,$ and $C$ are all distinct, so we need to shade at least $4+4+2-3=\boxed{7} draw((-1,2)--(6,2),linewidth(2)+red); draw((2.5,-1)--(2.5,5),linewidth(2)+red); label("$A$",(1.5,3.5)); label("$B$",(2.5,1.5)); label("$C$",(4.5,0.5)); label("$A'$",(3.5,3.5)); label("$A'$",(1.5,0.5)); label("$A'$",(3.5,0.5)); label("$B'$",(2.5,2.5)); label("$C'$",(0.5,0.5)); label("$C'$",(0.5,3.5)); label("$C'$",(4.5,3.5)); [/asy] ~MRENTHUSIASM

Answer: D 7
Problem: How many solutions does the equation $\tan(2x)=\cos(\tfrac{x}{2})$ have on the interval $[0,2\pi]?$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution: We count the intersections of the graphs of $y=\tan(2x)$ and $y=\cos\left(\frac x2\right):$
The graphs of $y=\tan(2x)$ and $y=\cos\left(\frac x2\right)$ intersect once on each of the five branches of $y=\tan(2x),$ as shown below: [asy] /* Made by MRENTHUSIASM */ size(800,200);   real f(real x) { return tan(2*x); }  real g(real x) { return cos(x/2); }  draw(graph(f,0,atan(3)/2),red,"$y=\tan(2x)$"); draw(graph(f,-atan(3)/2+pi/2,atan(3)/2+pi/2),red); draw(graph(f,-atan(3)/2+2*pi/2,atan(3)/2+2*pi/2),red); draw(graph(f,-atan(3)/2+3*pi/2,atan(3)/2+3*pi/2),red); draw(graph(f,-atan(3)/2+4*pi/2,2*pi),red); draw(graph(g,0,2pi),blue,"$y=\cos\left(\frac x2\right)$");  real xMin = 0; real xMax = 9/4*pi; real yMin = -3; real yMax = 3;  //Draws the horizontal gridlines void horizontalLines() {   for (real i = yMin+1; i < yMax; ++i)   {     draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4));   } }  //Draws the vertical gridlines void verticalLines() {   for (real i = xMin+pi/2; i < xMax; i+=pi/2)   {     draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4));   } }  //Draws the horizontal ticks void horizontalTicks() {   for (real i = yMin+1; i < yMax; ++i)   {     draw((-1/8,i)--(1/8,i), black+linewidth(1));   } }  //Draws the vertical ticks void verticalTicks() {   for (real i = xMin+pi/2; i < xMax; i+=pi/2)   {     draw((i,-1/8)--(i,1/8), black+linewidth(1));   } }  horizontalLines(); verticalLines(); horizontalTicks(); verticalTicks(); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2));  pair A[], B[]; A[0] = (2pi,0); A[1] = (0,2); A[2] = (0,0); A[3] = (0,-2); B[0] = intersectionpoints(graph(f,0,atan(3)/2),graph(g,0,2pi))[0]; B[1] = intersectionpoints(graph(f,-atan(3)/2+pi/2,atan(3)/2+pi/2),graph(g,0,2pi))[0]; B[2] = intersectionpoints(graph(f,-atan(3)/2+2*pi/2,atan(3)/2+2*pi/2),graph(g,0,2pi))[0]; B[3] = intersectionpoints(graph(f,-atan(3)/2+3*pi/2,atan(3)/2+3*pi/2),graph(g,0,2pi))[0]; B[4] = intersectionpoints(graph(f,-atan(3)/2+4*pi/2,atan(3)/2+4*pi/2),graph(g,0,2pi))[0];  label("$2\pi$",A[0],(0,-2.5)); label("$2$",A[1],(-2.5,0)); label("$0$",A[2],(-2.5,0)); label("$-2$",A[3],(-2.5,0));  for (int i = 0; i < 5; ++i) { 	dot(B[i],black+linewidth(5)); }  add(legend(),point(E),60E,UnFill); [/asy] Therefore, the answer is $\boxed{5}.$

Answer: E 5
Problem: There is a unique positive integer $n$ such that \[\log_2{(\log_{16}{n})} = \log_4{(\log_4{n})}.\] What is the sum of the digits of $n?$
$\textbf{(A) } 4 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 13$

Solution: We can use the fact that $\log_{a^b} c = \frac{1}{b} \log_a c.$ This can be proved by using change of base formula to base $a.$
So, the original equation $\log_2{(\log_{2^4}{n})} = \log_{2^2}{(\log_{2^2}{n})}$ becomes \[\log_2\left({\frac{1}{4}\log_{2}{n}}\right) = \frac{1}{2}\log_2\left({\frac{1}{2}\log_2{n}}\right).\] Using log property of addition, we expand both sides and then simplify: \begin{align*} \log_2{\frac{1}{4}}+\log_2{(\log_{2}{n}}) &= \frac{1}{2}\left[\log_2{\frac{1}{2}} +\log_{2}{(\log_2{n})}\right] \\ \log_2{\frac{1}{4}}+\log_2{(\log_{2}{n}}) &= \frac{1}{2}\left[-1 +\log_{2}{(\log_2{n})}\right] \\ -2+\log_2{(\log_{2}{n}}) &= -\frac{1}{2}+ \frac{1}{2}(\log_{2}{(\log_2{n})}). \end{align*} Subtracting $\frac{1}{2}(\log_{2}{(\log_2{n})})$ from both sides and adding $2$ to both sides gives us \[\frac{1}{2}(\log_{2}{(\log_2{n})}) = \frac{3}{2}.\] Multiplying by $2,$ exponentiating, and simplifying gives us \begin{align*} \log_{2}{(\log_2{n})} &= 3 \\ \log_2{n}&=8 \\ n&=256. \end{align*} Adding the digits together, we have $2+5+6=\boxed{13}.$

Answer: E 13
Problem: There is a unique positive integer $n$ such that \[\log_2{(\log_{16}{n})} = \log_4{(\log_4{n})}.\] What is the sum of the digits of $n?$
$\textbf{(A) } 4 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 13$

Solution: We will apply the following logarithmic identity: \[\log_{p^k}{q^k}=\log_{p}{q},\] which can be proven by the Change of Base Formula \[\log_{p^k}{q^k}=\frac{\log_{p}{q^k}}{\log_{p}{p^k}}=\frac{k\log_{p}{q}}{k}=\log_{p}{q}.\] Note that $\log_{16}{n}\neq0,$ so we rewrite the original equation as follows: \begin{align*} \log_4{(\log_{16}{n})^2}&=\log_4{(\log_4{n})} \\ (\log_{16}{n})^2&=\log_4{n} \\ (\log_{16}{n})^2&=\log_{16}{n^2} \\ (\log_{16}{n})^2&=2\log_{16}{n} \\ \log_{16}{n}&=2, \end{align*} from which $n=16^2=256.$ The sum of its digits is $2+5+6=\boxed{13}.$

Answer: E 13
Problem: There is a unique positive integer $n$ such that \[\log_2{(\log_{16}{n})} = \log_4{(\log_4{n})}.\] What is the sum of the digits of $n?$
$\textbf{(A) } 4 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 13$

Solution: Using the change of base formula on the RHS of the initial equation yields \[\log_2{(\log_{16}{n})} = \frac{\log_2{(\log_4{n})}}{\log_2{4}}.\] This means we can multiply each side by $2$ for \[\log_2{(\log_{16}{n})^2} = \log_2{(\log_4{n})}.\] Canceling out the logs gives \[(\log_{16}{n})^2=\log_4{n}.\] We use change of base on the RHS to see that \begin{align*} (\log_{16}{n})^2&=\frac{ \log_{16}{n}}{\log_{16}{4}} \\ (\log_{16}{n})^2&=2 \log_{16}{n}. \end{align*} Substituting in $m = \log_{16}{n}$ gives $m^2=2m,$ so $m$ is either $0$ or $2.$ Since $m=0$ yields no solution for $n$ (since this would lead to use taking the log of $0$ ), we get $2 = \log_{16}{n},$ or $n=16^2=256,$ for the digit-sum of $2 + 5 + 6 = \boxed{13}.$

Answer: E 13
Problem: There is a unique positive integer $n$ such that \[\log_2{(\log_{16}{n})} = \log_4{(\log_4{n})}.\] What is the sum of the digits of $n?$
$\textbf{(A) } 4 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 13$

Solution: Suppose $\log_2(\log_{16}n)=k\implies\log_{16}n=2^k\implies n=16^{2^k}.$ Similarly, we have $\log_4(\log_4 n)=k\implies \log_4 n=4^k\implies n=4^{4^k}.$ Thus, we have \[16^{2^k}=(4^2)^{2^k}=4^{2^{k+1}}\] and \[4^{4^k}=4^{2^{2k}},\] so $k+1=2k\implies k=1.$ Plugging this in to either one of the expressions for $n$ gives $256$ , and the requested answer is $2+5+6=\boxed{13}.$

Answer: E 13
Problem: There is a unique positive integer $n$ such that \[\log_2{(\log_{16}{n})} = \log_4{(\log_4{n})}.\] What is the sum of the digits of $n?$
$\textbf{(A) } 4 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 13$

Solution: We know that, as the answer is an integer, $n$ must be some power of $16.$ Testing $16$ yields \begin{align*} \log_2{(\log_{16}{16})} &= \log_4{(\log_4{16})} \\ \log_2{1} &= \log_4{2} \\ 0 &= \frac{1}{2}, \end{align*} which does not work. We then try $256,$ giving us \begin{align*} \log_2{(\log_{16}{256})} &= \log_4{(\log_4{256})} \\ \log_2{2} &= \log_4{4} \\ 1 &= 1, \end{align*} which holds true. Thus, $n = 256,$ so the answer is $2 + 5 + 6 = \boxed{13}.$

Answer: E 13
Problem: Line $l$ in the coordinate plane has equation $3x-5y+40=0$ . This line is rotated $45^{\circ}$ counterclockwise about the point $(20,20)$ to obtain line $k$ . What is the $x$ -coordinate of the $x$ -intercept of line $k?$
$\textbf{(A) } 10 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 30$

Solution: The slope of the line is $\frac{3}{5}$ . We must transform it by $45^{\circ}$
$45^{\circ}$ creates an isosceles right triangle, since the sum of the angles of the triangle must be $180^{\circ}$ and one angle is $90^{\circ}$ . This means the last leg angle must also be $45^{\circ}$
In the isosceles right triangle, the two legs are congruent. We can therefore construct an isosceles right triangle with a line of $\frac{3}{5}$ slope on graph paper. That line with $\frac{3}{5}$ slope starts at $(0,0)$ and will go to $(5,3)$ , the vector $<5, 3 >$
Construct another line from $(0,0)$ to $(3,-5)$ , the vector $<3,-5>$ . This is $\perp$ and equal to the original line segment. The difference between the two vectors is $<2,8>$ , which is the slope $4$ , and that is the slope of line $k$
Furthermore, the equation $3x-5y+40=0$ passes straight through $(20,20)$ since $3(20)-5(20)+40=60-100+40=0$ , which means that any rotations about $(20,20)$ would contain $(20,20)$ . We can create a line of slope $4$ through $(20,20)$ . The $x$ -intercept is therefore $20-\frac{20}{4} = \boxed{15.}$ ~lopkiloinm ~ShawnX (diagram)

Answer: B 15.
Problem: Line $l$ in the coordinate plane has equation $3x-5y+40=0$ . This line is rotated $45^{\circ}$ counterclockwise about the point $(20,20)$ to obtain line $k$ . What is the $x$ -coordinate of the $x$ -intercept of line $k?$
$\textbf{(A) } 10 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 30$

Solution: Since the slope of the line is $\frac{3}{5}$ , and the angle we are rotating around is x, then $\tan x = \frac{3}{5}$ $\tan(x+45^{\circ}) = \frac{\tan x + \tan(45^{\circ})}{1-\tan x*\tan(45^{\circ})} = \frac{0.6+1}{1-0.6} = \frac{1.6}{0.4} = 4$
Hence, the slope of the rotated line is $4$ . Since we know the line intersects the point $(20,20)$ , then we know the line is $y=4x-60$ . Set $y=0$ to find the x-intercept, and so $x=\boxed{15}$

Answer:  15
Problem: Line $l$ in the coordinate plane has equation $3x-5y+40=0$ . This line is rotated $45^{\circ}$ counterclockwise about the point $(20,20)$ to obtain line $k$ . What is the $x$ -coordinate of the $x$ -intercept of line $k?$
$\textbf{(A) } 10 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 30$

Solution: [asy] draw((0,0)--(20, 0)--(20, 20)--(0, 20)--cycle); draw((20, 20)--(0, 8)); draw((15, 0)--(20, 20));  dot("$P$", (20, 20)); dot("$A$", (0, 8), dir(75)); dot("$B$", (15, 0), dir(45)); dot("$X$", (20, 0)); dot("$Y$", (0, 20), dir(50)); [/asy]
Let $P$ be $(20, 20)$ and $X, Y$ be $(20, 0)$ and $(0, 20)$ respectively. Since the slope of the line is $3/5$ we know that $\tan{\angle{YPA}} = 3/5.$ Segments $\overline{PA}$ and $\overline{PB}$ represent the before and after of rotating $l$ by 45 counterclockwise. Thus, $\angle{XPB} = 45 - \angle{YPA}$ and \[BX = 20 \tan{\angle{XPB}} = 20 \cdot \frac{1 - 3/5}{1 + 3/5} = 5\] by tangent addition formula. Since $BX$ is 5 and the sidelength of the square is 20 the answer is $20 - 5 \implies \boxed{15}.$

Answer: B 15
Problem: Line $l$ in the coordinate plane has equation $3x-5y+40=0$ . This line is rotated $45^{\circ}$ counterclockwise about the point $(20,20)$ to obtain line $k$ . What is the $x$ -coordinate of the $x$ -intercept of line $k?$
$\textbf{(A) } 10 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 30$

Solution: Using the protractor you brought, carefully graph the equation and rotate the given line $45^{\circ}$ counter-clockwise about the point $(20,20)$ . Scaling everything down by a factor of 5 makes this process easier.
It should then become fairly obvious that the x intercept is $x=\boxed{15}$ (only use this as a last resort).

Answer:  15
Problem: Line $l$ in the coordinate plane has equation $3x-5y+40=0$ . This line is rotated $45^{\circ}$ counterclockwise about the point $(20,20)$ to obtain line $k$ . What is the $x$ -coordinate of the $x$ -intercept of line $k?$
$\textbf{(A) } 10 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 30$

Solution: First note that the given line goes through $(20,20)$ with a slope of $\frac{3}{5}$ . This means that $(25,23)$ is on the line. Now consider translating the graph so that $(20,20)$ goes to the origin, then $(25,23)$ becomes $(5,3)$ . We now rotate the line $45^\circ$ about the origin using a rotation matrix. This maps $(5,3)$ to \[\begin{bmatrix} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\end{bmatrix}\begin{bmatrix} 5 \\ 3\end{bmatrix}=\begin{bmatrix}\sqrt{2} \\ 4\sqrt{2}\end{bmatrix}\] The line through the origin and $(\sqrt{2}, 4\sqrt{2})$ has slope $4$ . Translating this line so that the origin is mapped to $(20,20)$ , we find that the equation for the new line is $4x-60$ , meaning that the $x$ -intercept is $x=\frac{60}{4}=\boxed{15}$

Answer: B 15