Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: In $\triangle PAT,$ $\angle P=36^{\circ},$ $\angle A=56^{\circ},$ and $PA=10.$ Points $U$ and $G$ lie on sides $\overline{TP}$ and $\overline{TA},$ respectively, so that $PU=AG=1.$ Let $M$ and $N$ be the midpoints of segments $\overline{PA}$ and $\overline{UG},$ respectively. What is the degree measure of the acute angle formed by lines $MN$ and $PA?$
$\textbf{(A) } 76 \qquad  \textbf{(B) } 77 \qquad  \textbf{(C) } 78 \qquad  \textbf{(D) } 79 \qquad \textbf{(E) } 80$

Solution: Link $PN$ , extend $PN$ to $Q$ so that $QN=PN$ . Then link $QG$ and $QA$
$\because M,N$ are the midpoints of $PA$ and $PQ,$ respectively
$\therefore MN$ is the midsegment of $\bigtriangleup PAQ$
$\therefore \angle QAP=\angle NMP$
Notice that $\bigtriangleup PUN\cong \bigtriangleup QGN$
As a result, $QG=AG=UP=1$ $\angle AQG=\angle QAG$ $\angle GQN=\angle NPU$
Also, $\angle GQN+\angle QPA=\angle QPU+\angle QPA=\angle UPA=36^{\circ}$
As a result, $2\angle QAG=180^{\circ}-56^{\circ}-36^{\circ}=88^{\circ}$
Therefore, $\angle QAP=\angle QAG+\angle TAP=56^{\circ}+44^{\circ}=100^{\circ}$
Since we are asked for the acute angle between the two lines, the answer to this problem is $\boxed{80}$

Answer: E 80
Problem: In $\triangle PAT,$ $\angle P=36^{\circ},$ $\angle A=56^{\circ},$ and $PA=10.$ Points $U$ and $G$ lie on sides $\overline{TP}$ and $\overline{TA},$ respectively, so that $PU=AG=1.$ Let $M$ and $N$ be the midpoints of segments $\overline{PA}$ and $\overline{UG},$ respectively. What is the degree measure of the acute angle formed by lines $MN$ and $PA?$
$\textbf{(A) } 76 \qquad  \textbf{(B) } 77 \qquad  \textbf{(C) } 78 \qquad  \textbf{(D) } 79 \qquad \textbf{(E) } 80$

Solution: Let the mid-point of $\overline{AT}$ be $B$ and the mid-point of $\overline{GT}$ be $C$ . 
Since $BC=CG-BG$ and $CG=AB-\frac{1}{2}$ , we can conclude that $BC=\frac{1}{2}$ . 
Similarly, we can conclude that $BM-CN=\frac{1}{2}$ . Construct $\overline{ND}\parallel\overline{BC}$ and intersects $\overline{BM}$ at $D$ , which gives $MD=DN=\frac{1}{2}$ . 
Since $\angle{ABD}=\angle{BDN}$ $MD=DN$ , we can find the value of $\angle{DMN}$ , which is equal to $\frac{1}{2}\angle T=44^{\circ}$ . Since $\overline{BM}\parallel\overline{PT}$ , which means $\angle{DMN}+\angle{MNP}+\angle{P}=180^{\circ}$ , we can infer that $\angle{MNP}=100^{\circ}$ . 
As we are required to give the acute angle formed, the final answer would be $80^{\circ}$ , which is $\boxed{80}$

Answer: E 80
Problem: In $\triangle PAT,$ $\angle P=36^{\circ},$ $\angle A=56^{\circ},$ and $PA=10.$ Points $U$ and $G$ lie on sides $\overline{TP}$ and $\overline{TA},$ respectively, so that $PU=AG=1.$ Let $M$ and $N$ be the midpoints of segments $\overline{PA}$ and $\overline{UG},$ respectively. What is the degree measure of the acute angle formed by lines $MN$ and $PA?$
$\textbf{(A) } 76 \qquad  \textbf{(B) } 77 \qquad  \textbf{(C) } 78 \qquad  \textbf{(D) } 79 \qquad \textbf{(E) } 80$

Solution: Let the bisector of $\angle ATP$ intersect $PA$ at $X.$ We have $\angle ATX = \angle PTX = 44^{\circ},$ so $\angle TXA = 80^{\circ}.$ We claim that $MN$ is parallel to this angle bisector, meaning that the acute angle formed by $MN$ and $PA$ is $80^{\circ},$ meaning that the answer is $\boxed{80}$

Answer: E 80
Problem: In $\triangle PAT,$ $\angle P=36^{\circ},$ $\angle A=56^{\circ},$ and $PA=10.$ Points $U$ and $G$ lie on sides $\overline{TP}$ and $\overline{TA},$ respectively, so that $PU=AG=1.$ Let $M$ and $N$ be the midpoints of segments $\overline{PA}$ and $\overline{UG},$ respectively. What is the degree measure of the acute angle formed by lines $MN$ and $PA?$
$\textbf{(A) } 76 \qquad  \textbf{(B) } 77 \qquad  \textbf{(C) } 78 \qquad  \textbf{(D) } 79 \qquad \textbf{(E) } 80$

Solution: Note that $X$ , the midpoint of major arc $PA$ on $(PAT)$ is the Miquel Point of $PUAG$ (Because $PU = AG$ ). Then, since $1 = \frac{UN}{NG} = \frac{PM}{MA}$ , this spiral similarity carries $M$ to $N$ . Thus, we have $\triangle XMN \sim \triangle XAG$ , so $\angle XMN = \angle XAG$
But, we have $\angle XAG = \angle PAG = \angle PAX = 56 - \frac{180 - \angle PXA}{2} =56 - \frac{180 - \angle T}{2} = 56 - \frac{\angle A + \angle P}{2} = 56 - \frac{56+36}{2} = 56 - 46 = 10$ ; thus $\angle XMN = 10$
Then, as $X$ is the midpoint of the major arc, it lies on the perpendicular bisector of $PA$ , so $\angle XMA = 90$ . Since we want the acute angle, we have $\angle NMA = \angle XMA - \angle XMN = 90 - 10 = 80$ , so the answer is $\boxed{80}$

Answer: E 80
Problem: In $\triangle PAT,$ $\angle P=36^{\circ},$ $\angle A=56^{\circ},$ and $PA=10.$ Points $U$ and $G$ lie on sides $\overline{TP}$ and $\overline{TA},$ respectively, so that $PU=AG=1.$ Let $M$ and $N$ be the midpoints of segments $\overline{PA}$ and $\overline{UG},$ respectively. What is the degree measure of the acute angle formed by lines $MN$ and $PA?$
$\textbf{(A) } 76 \qquad  \textbf{(B) } 77 \qquad  \textbf{(C) } 78 \qquad  \textbf{(D) } 79 \qquad \textbf{(E) } 80$

Solution: The argument of the average of any two unit vectors is average of the arguments of the two vectors. Thereby, the acute angle formed is \[\frac{36^\circ{} + 180^\circ{} - 56^\circ{}}{2} = \boxed{80}.\]

Answer: E 80
Problem: A line with slope $2$ intersects a line with slope $6$ at the point $(40,30)$ . What is the distance between the $x$ -intercepts of these two lines?
$\textbf{(A) } 5 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 50$

Solution: Using point slope form, we get the equations $y-30 = 6(x-40)$ and $y-30 = 2(x-40)$ . Simplifying, we get $6x-y=210$ and $2x-y=50$ . Letting $y=0$ in both equations and solving for $x$ gives the $x$ -intercepts: $x=35$ and $x=25$ , respectively. Thus the distance between them is $35-25=\boxed{10}$

Answer: B 10
Problem: A line with slope $2$ intersects a line with slope $6$ at the point $(40,30)$ . What is the distance between the $x$ -intercepts of these two lines?
$\textbf{(A) } 5 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 50$

Solution: In order for the line with slope $2$ to travel "up" $30$ units (from $y=0$ ), it must have traveled $30/2=15$ units to the right. Thus, the $x$ -intercept is at $x=40-15=25$ . As for the line with slope $6$ , in order for it to travel "up" $30$ units it must have traveled $30/6=5$ units to the right. Thus its $x$ -intercept is at $x=40-5=35$ . Then the distance between them is $35-25=\boxed{10}$

Answer: B 10
Problem: A circle has a chord of length $10$ , and the distance from the center of the circle to the chord is $5$ . What is the area of the circle?
$\textbf{(A) }25\pi \qquad \textbf{(B) }50\pi \qquad \textbf{(C) }75\pi \qquad \textbf{(D) }100\pi \qquad \textbf{(E) }125\pi \qquad$

Solution: Let $O$ be the center of the circle, $\overline{AB}$ be the chord, and $M$ be the midpoint of $\overline{AB},$ as shown below. [asy] /* Made by MRENTHUSIASM */ size(200); pair O, A, B, M; O = (0,0); A = (-5,5); B = (5,5); M = midpoint(A--B);  draw(Circle(O,5sqrt(2)));  dot("$O$", O, 1.5*S, linewidth(4.5)); dot("$A$", A, 1.5*NW, linewidth(4.5)); dot("$B$", B, 1.5*NE, linewidth(4.5)); dot("$M$", M, 1.5*N, linewidth(4.5)); draw(A--B^^M--O^^A--O^^M--O^^B--O); label("$5$", midpoint(A--M), 1.5*N); label("$5$", midpoint(B--M), 1.5*N); label("$5$", midpoint(O--M), 1.5*E); label("$r$", midpoint(O--A), 1.5*SW); label("$r$", midpoint(O--B), 1.5*SE); [/asy] Note that $\overline{OM}\perp\overline{AB}.$ Since $OM=AM=BM=5,$ we conclude that $\triangle OMA$ and $\triangle OMB$ are congruent isosceles right triangles. It follows that $r=5\sqrt2,$ so the area of $\odot O$ is $\pi r^2=\boxed{50}$

Answer: B 50
Problem: How many subsets of $\{2,3,4,5,6,7,8,9\}$ contain at least one prime number?
$\textbf{(A) } 128 \qquad \textbf{(B) } 192 \qquad \textbf{(C) } 224 \qquad \textbf{(D) } 240 \qquad \textbf{(E) } 256$

Solution: Since an element of a subset is either in or out, the total number of subsets of the $8$ -element set is $2^8 = 256$ . However, since we are only concerned about the subsets with at least $1$ prime in it, we can use complementary counting to count the subsets without a prime and subtract that from the total. Because there are $4$ non-primes, there are $2^8 -2^4 = \boxed{240}$ subsets with at least $1$ prime.

Answer: D 240
Problem: How many subsets of $\{2,3,4,5,6,7,8,9\}$ contain at least one prime number?
$\textbf{(A) } 128 \qquad \textbf{(B) } 192 \qquad \textbf{(C) } 224 \qquad \textbf{(D) } 240 \qquad \textbf{(E) } 256$

Solution: We can construct our subset by choosing which primes are included and which composites are included. There are $2^4-1$ ways to select the primes (total subsets minus the empty set) and $2^4$ ways to select the composites. Thus, there are $15\cdot16=\boxed{240}$ ways to choose a subset of the eight numbers.

Answer: D 240
Problem: How many subsets of $\{2,3,4,5,6,7,8,9\}$ contain at least one prime number?
$\textbf{(A) } 128 \qquad \textbf{(B) } 192 \qquad \textbf{(C) } 224 \qquad \textbf{(D) } 240 \qquad \textbf{(E) } 256$

Solution: We complement count. We know that we have $2^8$ subsets in all, including the empty subset. We subtract $\dbinom{4}{0}+\dbinom{4}{1}+\dbinom{4}{2}+\dbinom{4}{3}+\dbinom{4}{4} = 2^4 = 16$ We have $2^8-16 = 256-16 = \boxed{240}$ ways to choose a subset of the eight numbers that include a prime number. ~hh99754539

Answer: D 240
Problem: Suppose $S$ cans of soda can be purchased from a vending machine for $Q$ quarters. Which of the following expressions describes the number of cans of soda that can be purchased for $D$ dollars, where $1$ dollar is worth $4$ quarters?
$\textbf{(A) } \frac{4DQ}{S} \qquad \textbf{(B) } \frac{4DS}{Q} \qquad \textbf{(C) } \frac{4Q}{DS} \qquad \textbf{(D) } \frac{DQ}{4S} \qquad \textbf{(E) } \frac{DS}{4Q}$

Solution: Each can of soda costs $\frac QS$ quarters, or $\frac{Q}{4S}$ dollars. Therefore, $D$ dollars can purchase $\frac{D}{\left(\tfrac{Q}{4S}\right)}=\boxed{4}$ cans of soda.

Answer: B 4
Problem: Suppose $S$ cans of soda can be purchased from a vending machine for $Q$ quarters. Which of the following expressions describes the number of cans of soda that can be purchased for $D$ dollars, where $1$ dollar is worth $4$ quarters?
$\textbf{(A) } \frac{4DQ}{S} \qquad \textbf{(B) } \frac{4DS}{Q} \qquad \textbf{(C) } \frac{4Q}{DS} \qquad \textbf{(D) } \frac{DQ}{4S} \qquad \textbf{(E) } \frac{DS}{4Q}$

Solution: Note that $S$ is in the unit of $\text{can}.$ On the other hand, $Q$ and $D$ are both in the unit of $\text{cost}.$
The units of $\textbf{(A)},\textbf{(B)},\textbf{(C)},\textbf{(D)},$ and $\textbf{(E)}$ are $\frac{\text{cost}^2}{\text{can}},\text{can},\frac{1}{\text{can}},\frac{\text{cost}^2}{\text{can}},$ and $\text{can},$ respectively. Since the answer is in the unit of $\text{can},$ we eliminate $\textbf{(A)},\textbf{(C)},$ and $\textbf{(D)}.$ Moreover, it is clear that $D$ dollars can purchase more than $S=\frac{DS}{4Q}$ cans of soda, so we eliminate $\textbf{(E)}.$
Finally, the answer is $\boxed{4}.$

Answer: B 4
Problem: What is the value of \[\log_37\cdot\log_59\cdot\log_711\cdot\log_913\cdots\log_{21}25\cdot\log_{23}27?\] $\textbf{(A) } 3 \qquad \textbf{(B) } 3\log_{7}23 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 10$

Solution: From the Change of Base Formula, we have \[\frac{\prod_{i=3}^{13} \log (2i+1)}{\prod_{i=1}^{11}\log (2i+1)} = \frac{\log 25 \cdot \log 27}{\log 3 \cdot \log 5} = \frac{(2\log 5)\cdot(3\log 3)}{\log 3 \cdot \log 5} = \boxed{6}.\]

Answer: C 6
Problem: What is the value of \[\log_37\cdot\log_59\cdot\log_711\cdot\log_913\cdots\log_{21}25\cdot\log_{23}27?\] $\textbf{(A) } 3 \qquad \textbf{(B) } 3\log_{7}23 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 10$

Solution: Using the chain rule of logarithms $\log _{a} b \cdot \log _{b} c = \log _{a} c,$ we get \begin{align*} \log_37\cdot\log_59\cdot\log_711\cdot\log_913\cdots\log_{21}25\cdot\log_{23}27 &= (\log _{3} 7 \cdot \log _{7} 11 \cdots \log _{23} 27) \cdot (\log _{5} 9 \cdot \log _{9} 13 \cdots \log _{21} 25) \\ &= \log _{3} 27 \cdot \log _{5} 25 \\ &= 3 \cdot 2 \\ &= \boxed{6}

Answer: C 6
Problem: What is \[\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) ?\]
$\textbf{(A) }100100 \qquad \textbf{(B) }500500\qquad \textbf{(C) }505000 \qquad \textbf{(D) }1001000 \qquad \textbf{(E) }1010000 \qquad$

Solution: Recall that the sum of the first $100$ positive integers is $\sum^{100}_{k=1} k = \frac{101\cdot100}{2}=5050.$ It follows that \begin{align*} \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) &= \sum^{100}_{i=1} \sum^{100}_{j=1}i + \sum^{100}_{i=1} \sum^{100}_{j=1}j \\ &= \sum^{100}_{i=1} (100i) + 100 \sum^{100}_{j=1}j \\ &= 100 \sum^{100}_{i=1}i + 100 \sum^{100}_{j=1}j \\ &= 100\cdot5050 + 100\cdot5050 \\ &= \boxed{1010000} ~MRENTHUSIASM

Answer: E 1010000
Problem: What is \[\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) ?\]
$\textbf{(A) }100100 \qquad \textbf{(B) }500500\qquad \textbf{(C) }505000 \qquad \textbf{(D) }1001000 \qquad \textbf{(E) }1010000 \qquad$

Solution: Recall that the sum of the first $100$ positive integers is $\sum^{100}_{k=1} k = \frac{101\cdot100}{2}=5050.$ It follows that \begin{align*} \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) &= \sum^{100}_{i=1} \left(\sum^{100}_{j=1} i + \sum^{100}_{j=1} j\right) \\ &= \sum^{100}_{i=1} (100i+5050) \\ &= 100\sum^{100}_{i=1} i + \sum^{100}_{i=1} 5050 \\ &= 100\cdot5050+5050\cdot100 \\ &= \boxed{1010000} ~Vfire ~MRENTHUSIASM

Answer: E 1010000
Problem: What is \[\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) ?\]
$\textbf{(A) }100100 \qquad \textbf{(B) }500500\qquad \textbf{(C) }505000 \qquad \textbf{(D) }1001000 \qquad \textbf{(E) }1010000 \qquad$

Solution: Recall that the sum of the first $100$ positive integers is $\sum^{100}_{k=1} k = \frac{101\cdot100}{2}=5050.$ Since the nested summation is symmetric with respect to $i$ and $j,$ it follows that \begin{align*} \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) &= \sum^{100}_{i=1} \sum^{100}_{i=1} (2i) \\ &= 2\sum^{100}_{i=1} \sum^{100}_{i=1} i \\ &= 2\sum^{100}_{i=1} 5050 \\ &= 2\cdot(5050\cdot100) \\ &= \boxed{1010000} ~Vfire ~MRENTHUSIASM

Answer: E 1010000
Problem: What is \[\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) ?\]
$\textbf{(A) }100100 \qquad \textbf{(B) }500500\qquad \textbf{(C) }505000 \qquad \textbf{(D) }1001000 \qquad \textbf{(E) }1010000 \qquad$

Solution: The sum contains $100\cdot100=10000$ terms, and the average value of both $i$ and $j$ is $\frac{101}{2}.$ Therefore, the sum becomes \[10000\cdot\left(\frac{101}{2}+\frac{101}{2}\right)=\boxed{1010000}.\] ~Rejas ~MRENTHUSIASM

Answer: E 1010000
Problem: What is \[\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) ?\]
$\textbf{(A) }100100 \qquad \textbf{(B) }500500\qquad \textbf{(C) }505000 \qquad \textbf{(D) }1001000 \qquad \textbf{(E) }1010000 \qquad$

Solution: We start by writing out the first few terms: \[\begin{array}{ccccccccc} (1+1) &+ &(1+2) &+ &(1+3) &+ &\cdots &+ &(1+100) \\ (2+1) &+ &(2+2) &+ &(2+3) &+ &\cdots &+ &(2+100) \\ (3+1) &+ &(3+2) &+ &(3+3) &+ &\cdots &+ &(3+100) \\ [-1ex] &&&&\vdots&&&& \\ (100+1) &+ &(100+2) &+ &(100+3) &+ &\cdots &+ &(100+100). \end{array}\] From the first terms in the parentheses, the sum $1+2+3+\cdots+100$ occurs $100$ times vertically.
From the second terms in the parentheses, the sum $1+2+3+\cdots+100$ occurs $100$ times horizontally.
Recall that the sum of the first $100$ positive integers is $1+2+3+\cdots+100=\frac{101\cdot100}{2}=5050.$ Therefore, the answer is \[2\cdot\left(5050\cdot100\right)=\boxed{1010000}.\] ~RandomPieKevin ‎~MRENTHUSIASM

Answer: E 1010000
Problem: What is \[\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) ?\]
$\textbf{(A) }100100 \qquad \textbf{(B) }500500\qquad \textbf{(C) }505000 \qquad \textbf{(D) }1001000 \qquad \textbf{(E) }1010000 \qquad$

Solution: When we expand the nested summation as shown in Solution 5, note that:
Together, the nested summation becomes \begin{align*} \sum^{100}_{k=1}\left[(k+1)k\right] + \sum^{99}_{k=1}\left[(k+101)(100-k)\right] &= \sum^{100}_{k=1}\left[k^2+k\right] + \sum^{99}_{k=1}\left[-k^2-k+10100\right] \\ &= \sum^{100}_{k=1}k^2 + \sum^{100}_{k=1}k - \sum^{99}_{k=1}k^2 - \sum^{99}_{k=1}k + \sum^{99}_{k=1}10100 \\ &= \left(\sum^{100}_{k=1}k^2 - \sum^{99}_{k=1}k^2\right) + \left(\sum^{100}_{k=1}k - \sum^{99}_{k=1}k\right) + \sum^{99}_{k=1}10100 \\ &= 100^2+100+10100\cdot99 \\ &= \boxed{1010000} ~MRENTHUSIASM

Answer: E 1010000
Problem: Side $\overline{AB}$ of $\triangle ABC$ has length $10$ . The bisector of angle $A$ meets $\overline{BC}$ at $D$ , and $CD = 3$ . The set of all possible values of $AC$ is an open interval $(m,n)$ . What is $m+n$
$\textbf{(A) }16 \qquad \textbf{(B) }17 \qquad \textbf{(C) }18 \qquad \textbf{(D) }19 \qquad \textbf{(E) }20 \qquad$

Solution: Let $AC=x.$ By Angle Bisector Theorem, we have $\frac{AB}{AC}=\frac{BD}{CD},$ from which $BD=CD\cdot\frac{AB}{AC}=\frac{30}{x}.$
Recall that $x>0.$ We apply the Triangle Inequality to $\triangle ABC:$
Taking the intersection of the solutions gives \[(m,n)=(0,\infty)\cap(0,15)\cap(3,\infty)=(3,15),\] so the answer is $m+n=\boxed{18}.$

Answer: C 18
Problem: Square $ABCD$ has side length $30$ . Point $P$ lies inside the square so that $AP = 12$ and $BP = 26$ . The centroids of $\triangle{ABP}$ $\triangle{BCP}$ $\triangle{CDP}$ , and $\triangle{DAP}$ are the vertices of a convex quadrilateral. What is the area of that quadrilateral?
[asy] unitsize(120); pair B = (0, 0), A = (0, 1), D = (1, 1), C = (1, 0), P = (1/4, 2/3); draw(A--B--C--D--cycle); dot(P); defaultpen(fontsize(10pt)); draw(A--P--B); draw(C--P--D); label("$A$", A, W); label("$B$", B, W); label("$C$", C, E); label("$D$", D, E); label("$P$", P, N*1.5+E*0.5); dot(A); dot(B); dot(C); dot(D); [/asy]
$\textbf{(A) }100\sqrt{2}\qquad\textbf{(B) }100\sqrt{3}\qquad\textbf{(C) }200\qquad\textbf{(D) }200\sqrt{2}\qquad\textbf{(E) }200\sqrt{3}$

Solution: As shown below, let $M_1,M_2,M_3,M_4$ be the midpoints of $\overline{AB},\overline{BC},\overline{CD},\overline{DA},$ respectively, and $G_1,G_2,G_3,G_4$ be the centroids of $\triangle{ABP},\triangle{BCP},\triangle{CDP},\triangle{DAP},$ respectively. [asy] /* Made by MRENTHUSIASM */ unitsize(210); pair B = (0, 0), A = (0, 1), D = (1, 1), C = (1, 0), P = (1/4, 2/3); pair M1 = midpoint(A--B); pair M2 = midpoint(B--C); pair M3 = midpoint(C--D); pair M4 = midpoint(D--A); pair G1 = centroid(A,B,P); pair G2 = centroid(B,C,P); pair G3 = centroid(C,D,P); pair G4 = centroid(D,A,P); filldraw(M1--M2--P--cycle,red); filldraw(M2--M3--P--cycle,yellow); filldraw(M3--M4--P--cycle,green); filldraw(M4--M1--P--cycle,lightblue); draw(A--B--C--D--cycle); draw(M1--M2--M3--M4--cycle); draw(G1--G2--G3--G4--cycle); dot(P); defaultpen(fontsize(10pt)); draw(A--P--B); draw(C--P--D); label("$A$", A, W); label("$B$", B, W); label("$C$", C, E); label("$D$", D, E); label("$P$", P, N); label("$M_1$", M1, W); label("$M_2$", M2, S); label("$M_3$", M3, E); label("$M_4$", M4, N); label("$G_1$", G1, 1.5S); label("$G_2$", G2, 1.5E); label("$G_3$", G3, 1.5NE); label("$G_4$", G4, 1.5E); dot(A); dot(B); dot(C); dot(D); dot(M1); dot(M2); dot(M3); dot(M4); dot(G1); dot(G2); dot(G3); dot(G4); [/asy] By SAS, we conclude that $\triangle G_1G_2P\sim\triangle M_1M_2P, \triangle G_2G_3P\sim\triangle M_2M_3P, \triangle G_3G_4P\sim\triangle M_3M_4P,$ and $\triangle G_4G_1P\sim\triangle M_4M_1P.$ By the properties of centroids, the ratio of similitude for each pair of triangles is $\frac23.$
Note that quadrilateral $M_1M_2M_3M_4$ is a square of side-length $15\sqrt2.$ It follows that:
Together, quadrilateral $G_1G_2G_3G_4$ is a square of side-length $10\sqrt2,$ so its area is $\left(10\sqrt2\right)^2=\boxed{200}.$

Answer: C 200
Problem: Square $ABCD$ has side length $30$ . Point $P$ lies inside the square so that $AP = 12$ and $BP = 26$ . The centroids of $\triangle{ABP}$ $\triangle{BCP}$ $\triangle{CDP}$ , and $\triangle{DAP}$ are the vertices of a convex quadrilateral. What is the area of that quadrilateral?
[asy] unitsize(120); pair B = (0, 0), A = (0, 1), D = (1, 1), C = (1, 0), P = (1/4, 2/3); draw(A--B--C--D--cycle); dot(P); defaultpen(fontsize(10pt)); draw(A--P--B); draw(C--P--D); label("$A$", A, W); label("$B$", B, W); label("$C$", C, E); label("$D$", D, E); label("$P$", P, N*1.5+E*0.5); dot(A); dot(B); dot(C); dot(D); [/asy]
$\textbf{(A) }100\sqrt{2}\qquad\textbf{(B) }100\sqrt{3}\qquad\textbf{(C) }200\qquad\textbf{(D) }200\sqrt{2}\qquad\textbf{(E) }200\sqrt{3}$

Solution: This solution refers to the diagram in Solution 1.
By SAS, we conclude that $\triangle G_1G_3P\sim\triangle M_1M_3P$ and $\triangle G_2G_4P\sim\triangle M_2M_4P.$ By the properties of centroids, the ratio of similitude for each pair of triangles is $\frac23.$
Note that quadrilateral $M_1M_2M_3M_4$ is a square of diagonal-length $30,$ so $\overline{M_1M_3}\perp\overline{M_2M_4}.$ Since $\overline{G_1G_3}\parallel\overline{M_1M_3}$ and $\overline{G_2G_4}\parallel\overline{M_2M_4}$ by the Converse of the Corresponding Angles Postulate, we have $\overline{G_1G_3}\perp\overline{G_2G_4}.$
Therefore, the area of quadrilateral $G_1G_2G_3G_4$ is \[\frac12\cdot G_1G_3\cdot G_2G_4 = \frac12\cdot\left(\frac23\cdot M_1M_3\right)\cdot\left(\frac23\cdot M_2M_4\right)=\boxed{200}.\] ~Funnybunny5246 ~MRENTHUSIASM

Answer: C 200
Problem: Square $ABCD$ has side length $30$ . Point $P$ lies inside the square so that $AP = 12$ and $BP = 26$ . The centroids of $\triangle{ABP}$ $\triangle{BCP}$ $\triangle{CDP}$ , and $\triangle{DAP}$ are the vertices of a convex quadrilateral. What is the area of that quadrilateral?
[asy] unitsize(120); pair B = (0, 0), A = (0, 1), D = (1, 1), C = (1, 0), P = (1/4, 2/3); draw(A--B--C--D--cycle); dot(P); defaultpen(fontsize(10pt)); draw(A--P--B); draw(C--P--D); label("$A$", A, W); label("$B$", B, W); label("$C$", C, E); label("$D$", D, E); label("$P$", P, N*1.5+E*0.5); dot(A); dot(B); dot(C); dot(D); [/asy]
$\textbf{(A) }100\sqrt{2}\qquad\textbf{(B) }100\sqrt{3}\qquad\textbf{(C) }200\qquad\textbf{(D) }200\sqrt{2}\qquad\textbf{(E) }200\sqrt{3}$

Solution: This solution refers to the diagram in Solution 1.
We place the diagram in the coordinate plane: Let $A=(0,30),B=(0,0),C=(30,0),D=(30,30),$ and $P=(3x,3y).$
Recall that for any triangle in the coordinate plane, the coordinates of its centroid are the averages of the coordinates of its vertices. It follows that $G_1=(x,y+10),G_2=(x+10,y),G_3=(x+20,y+10),$ and $G_4=(x+10,y+20).$
Note that $G_1G_3=G_2G_4=20$ and $\overline{G_1G_3}\perp\overline{G_2G_4}.$ Therefore, the area of quadrilateral $G_1G_2G_3G_4$ is \[\frac12\cdot G_1G_3\cdot G_2G_4=\boxed{200}.\]

Answer: C 200
Problem: Square $ABCD$ has side length $30$ . Point $P$ lies inside the square so that $AP = 12$ and $BP = 26$ . The centroids of $\triangle{ABP}$ $\triangle{BCP}$ $\triangle{CDP}$ , and $\triangle{DAP}$ are the vertices of a convex quadrilateral. What is the area of that quadrilateral?
[asy] unitsize(120); pair B = (0, 0), A = (0, 1), D = (1, 1), C = (1, 0), P = (1/4, 2/3); draw(A--B--C--D--cycle); dot(P); defaultpen(fontsize(10pt)); draw(A--P--B); draw(C--P--D); label("$A$", A, W); label("$B$", B, W); label("$C$", C, E); label("$D$", D, E); label("$P$", P, N*1.5+E*0.5); dot(A); dot(B); dot(C); dot(D); [/asy]
$\textbf{(A) }100\sqrt{2}\qquad\textbf{(B) }100\sqrt{3}\qquad\textbf{(C) }200\qquad\textbf{(D) }200\sqrt{2}\qquad\textbf{(E) }200\sqrt{3}$

Solution: Let $X,Y,Z,W$ be the midpoints of sides $AB,BC,CD,DE$ , respectively.
Notice that a homothety centered at P with ratio $\frac{2}{3}$ will send $XYZW$ to $G_{1}G_{2}G_{3}G_{4}$ , so $G_{1}G_{2}G_{3}G_{4}$ is a square with area $\left(\frac{2}{3}\right)^2 [XYZW]$ , but $[XYZW]=\frac{1}{2}[ABCD]$ so our desired area is \[\frac{4}{9}\cdot\frac{1}{2}\cdot900=\boxed{200}\]

Answer: C 200