Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: A stack of $2000$ cards is labelled with the integers from $1$ to $2000,$ with different integers on different cards. The cards in the stack are not in numerical order. The top card is removed from the stack and placed on the table, and the next card is moved to the bottom of the stack. The new top card is removed from the stack and placed on the table, to the right of the card already there, and the next card in the stack is moved to the bottom of the stack. The process - placing the top card to the right of the cards already on the table and moving the next card in the stack to the bottom of the stack - is repeated until all cards are on the table. It is found that, reading from left to right, the labels on the cards are now in ascending order: $1,2,3,\ldots,1999,2000.$ In the original stack of cards, how many cards were above the card labeled $1999$

Solution: To simplify matters, we want a power of $2$ . Hence, we will add $48$ 'fake' cards which we must discard in our actual count. Using similar logic as Solution 1, we find that 1999 has position $1024$ in a $2048$ card stack, where the fake cards towards the front.
Let the fake cards have positions $1, 3, 5, \cdots, 95$ . Then, we know that the real cards filling the gaps between the fake cards must be the cards such that they go to their correct starting positions in the $2000$ card case, where all of them are below $1999$ . From this, we know that the cards from positions $1$ to $96$ alternate in fake-real-fake-real, where we have the correct order of cards once the first $96$ have moved and we can start putting real cards on the table. Hence, $1999$ is in position $1024 - 96 = 928$ , so $\boxed{927}$ cards are above it. - Spacesam

Answer:  927
Problem: A stack of $2000$ cards is labelled with the integers from $1$ to $2000,$ with different integers on different cards. The cards in the stack are not in numerical order. The top card is removed from the stack and placed on the table, and the next card is moved to the bottom of the stack. The new top card is removed from the stack and placed on the table, to the right of the card already there, and the next card in the stack is moved to the bottom of the stack. The process - placing the top card to the right of the cards already on the table and moving the next card in the stack to the bottom of the stack - is repeated until all cards are on the table. It is found that, reading from left to right, the labels on the cards are now in ascending order: $1,2,3,\ldots,1999,2000.$ In the original stack of cards, how many cards were above the card labeled $1999$

Solution: We work backwards. To reverse the process, we must move the bottom card to the top, and add a new number to the top. Let $d_n$ equal the number of cards below 1999 after $n$ process reversals. Reversing one process, our deck only has $2000$ , so we reverse again to obtain $1999, 2000$ . So, $d_2 = 1$ . When $d_{n-1} > 0$ , after a process reversal, the bottom card is moved to the top (it goes above $1999$ ), so we subtract by one to account for this (the addition of the new card doesn't matter since it goes above $1999$ ), giving us $d_n = d_{n-1} - 1$ . So, $d_3 = 0$ . Then, when $d_{n-1} = 0$ , after a process reversal, $1999$ moves to the top and one more card is added above $1999$ . Since at $d_{n}$ $n$ cards have been added, there must be $n - 2$ cards below $1999$ . So, $d_4 = 2$ . Then, consider all $d_{n-1} = 0$ . Then, $d_n = n-2$ as previously stated. After $n-2$ process reversals, we go back to $d_{2n - 2} = n - 2 - (n - 2) = 0$ . Then, $d_{2n-1} = 2n - 3$ . Next, we can simply calculate $2 \cdot 4 - 1 = 7$ $2 \cdot 7 - 1 = 13$ , and so on (which is a very simple bash, as $2000$ is a small number. If you don't want to do this, define sequence $a_n = 2a_{n-1} - 1$ , and solve for the closed form, which is very easy). Consequently, we derive $d_{1537} = 1537 - 2 = 1535$ $2000 - 1537 = 463$ steps remain. So, $d_{2000} = d_{1537} - 463 = 1072$ . Finally, there are $2000 - 1 - 1072 = \boxed{927}$ numbers above $1999$

Answer:  927
Problem: The number
can be written as $\frac mn$ where $m$ and $n$ are relatively prime positive integers.  Find $m + n$

Solution: $\frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}}$
$=\frac{\log_4{16}}{\log_4{2000^6}}+\frac{\log_5{125}}{\log_5{2000^6}}$
$=\frac{\log{16}}{\log{2000^6}}+\frac{\log{125}}{\log{2000^6}}$
$=\frac{\log{2000}}{\log{2000^6}}$
$=\frac{\log{2000}}{6\log{2000}}$
$=\frac{1}{6}$
Therefore, $m+n=1+6=\boxed{007}$

Answer:  007
Problem: The number
can be written as $\frac mn$ where $m$ and $n$ are relatively prime positive integers.  Find $m + n$

Solution: Alternatively, we could've noted that, because $\frac 1{\log_a{b}} = \log_b{a}$
\begin{align*} \frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}} &= 2 \cdot \frac{1}{\log_4{2000^6}} + 3\cdot \frac {1}{\log_5{2000^6} }\\ &=2{\log_{2000^6}{4}} + 3{\log_{2000^6}{5}} \\ &={\log_{2000^6}{4^2}} + {\log_{2000^6}{5^3}}\\ &={\log_{2000^6}{4^2 \cdot 5^3}}\\ &={\log_{2000^6}{2000}}\\ &= {\frac{1}{6}}.\end{align*}
Therefore our answer is $1 + 6 = \boxed{007}$

Answer:  007
Problem: The number
can be written as $\frac mn$ where $m$ and $n$ are relatively prime positive integers.  Find $m + n$

Solution: We know that $2 = \log_4{16}$ and $3 = \log_5{125}$ , and by base of change formula, $\log_a{b} = \frac{\log_c{b}}{\log_c{a}}$ . Lastly, notice $\log a + \log b = \log ab$ for all bases. \begin{align*} \frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}} = \log_{2000^6}{16} + \log_{2000^6}{125} = \log_{2000^6}{2000} = \frac16 \implies \boxed{007}

Answer:  007
Problem: The number
can be written as $\frac mn$ where $m$ and $n$ are relatively prime positive integers.  Find $m + n$

Solution: \[\frac{2}{\log_4 2000^6} + \frac{3}{\log_5 2000^6}\] \[= \frac{1}{3\log_4 2000} + \frac{1}{2\log_5 2000}\] \[= \frac{1}{3} \log_{2000} 4 + \frac{1}{2} \log_{2000} 5\] \[= \log_{2000} (\sqrt[3]{4} \cdot \sqrt{5}) = x\] \[\implies 2^{4x} \cdot 5^{3x} = 2^{\frac{2}{3}} \cdot 5^{\frac{1}{2}}\] \[\implies 4x + (3\log_2 5)x = \frac{2}{3}+\frac{1}{2} \log_2 5\] \[\implies x = \frac{\frac{2}{3} + \frac{1}{2} \log_2 5}{4 + 3\log_2 5}\] \[\implies 6x = \frac{4 + 3\log_2 5}{4 + 3\log_2 5}\] \[\implies x = \frac{1}{6}\] \[\implies m + n = \boxed{007}\]

Answer:  007
Problem: A point whose coordinates are both integers is called a lattice point. How many lattice points lie on the hyperbola $x^2 - y^2 = 2000^2$

Solution: \[(x-y)(x+y)=2000^2=2^8 \cdot 5^6\]
Note that $(x-y)$ and $(x+y)$ have the same parities , so both must be even. We first give a factor of $2$ to both $(x-y)$ and $(x+y)$ . We have $2^6 \cdot 5^6$ left. Since there are $7 \cdot 7=49$ factors of $2^6 \cdot 5^6$ , and since both $x$ and $y$ can be negative, this gives us $49\cdot2=\boxed{098}$ lattice points.

Answer:  098
Problem: A deck of forty cards consists of four $1$ 's, four $2$ 's,..., and four $10$ 's.  A matching pair (two cards with the same number) is removed from the deck.  Given that these cards are not returned to the deck, let $m/n$ be the probability that two randomly selected cards also form a pair, where $m$ and $n$ are relatively prime positive integers.  Find $m + n.$

Solution: There are ${38 \choose 2} = 703$ ways we can draw two cards from the reduced deck. The two cards will form a pair if both are one of the nine numbers that were not removed, which can happen in $9{4 \choose 2} = 54$ ways, or if the two cards are the remaining two cards of the number that was removed, which can happen in $1$ way. Thus, the answer is $\frac{54+1}{703} = \frac{55}{703}$ , and $m+n = \boxed{758}$

Answer:  758
Problem: What is the smallest positive integer with six positive odd integer divisors and twelve positive even integer divisors?

Solution: We use the fact that the number of divisors of a number $n = p_1^{e_1}p_2^{e_2} \cdots p_k^{e_k}$ is $(e_1 + 1)(e_2 + 1) \cdots (e_k + 1)$ . If a number has $18 = 2 \cdot 3 \cdot 3$ factors, then it can have at most $3$ distinct primes in its factorization.
Dividing the greatest power of $2$ from $n$ , we have an odd integer with six positive divisors, which indicates that it either is ( $6 = 2 \cdot 3$ ) a prime raised to the $5$ th power, or two primes, one of which is squared. The smallest example of the former is $3^5 = 243$ , while the smallest example of the latter is $3^2 \cdot 5 = 45$
Suppose we now divide all of the odd factors from $n$ ; then we require a power of $2$ with $\frac{18}{6} = 3$ factors, namely $2^{3-1} = 4$ . Thus, our answer is $2^2 \cdot 3^2 \cdot 5 = \boxed{180}$

Answer:  180
Problem: What is the smallest positive integer with six positive odd integer divisors and twelve positive even integer divisors?

Solution: Somewhat similar to the first solution, we see that the number $n$ has two even factors for every odd factor.  Thus, if $x$ is an odd factor of $n$ , then $2x$ and $4x$ must be the two corresponding even factors.  So, the prime factorization of $n$ is $2^2 3^a 5^b 7^c...$ for some set of integers $a, b, c, ...$
Since there are $18$ factors of $n$ , we can write:
$(2+1)(a+1)(b+1)(c+1)... = 18$
$(a+1)(b+1)(c+1)... = 6$
Since $6$ only has factors from the set $1, 2, 3, 6$ , either $a=5$ and all other variables are $0$ , or $a=3$ and $b=2$ , with again all other variables equal to $0$ .  This gives the two numbers $2^2 \cdot 3^5$ and $2^2 \cdot 3^2 \cdot 5$ .  The latter number is smaller, and is equal to $\boxed{180}$

Answer:  180
Problem: Given eight distinguishable rings, let $n$ be the number of possible five-ring arrangements on the four fingers (not the thumb) of one hand. The order of rings on each finger is significant, but it is not required that each finger have a ring. Find the leftmost three nonzero digits of $n$

Solution: There are $\binom{8}{5}$ ways to choose the rings, and there are $5!$ distinct arrangements to order the rings [we order them so that the first ring is the bottom-most on the first finger that actually has a ring, and so forth]. The number of ways to distribute the rings among the fingers is equivalent the number of ways we can drop five balls into 4 urns, or similarly dropping five balls into four compartments split by three dividers. The number of ways to arrange those dividers and balls is just $\binom {8}{3}$
Multiplying gives the answer: $\binom{8}{5}\binom{8}{3}5! = 376320$ , and the three leftmost digits are $\boxed{376}$

Answer:  376
Problem: One base of a trapezoid is $100$ units longer than the other base. The segment that joins the midpoints of the legs divides the trapezoid into two regions whose areas are in the ratio $2: 3$ . Let $x$ be the length of the segment joining the legs of the trapezoid that is parallel to the bases and that divides the trapezoid into two regions of equal area. Find the greatest integer that does not exceed $x^2/100$

Solution: Let the shorter base have length $b$ (so the longer has length $b+100$ ), and let the height be $h$ . The length of the midline of the trapezoid is the average of its bases, which is $\frac{b+b+100}{2} = b+50$ . The two regions which the midline divides the trapezoid into are two smaller trapezoids, both with height $h/2$ . Then,
\[\frac{\frac 12 (h/2) (b + b+50)}{\frac 12 (h/2) (b + 50 + b + 100)} = \frac{2}{3} \Longrightarrow \frac{b + 75}{b + 25} = \frac 32 \Longrightarrow b = 75\]
Construct the perpendiculars from the vertices of the shorter base to the longer base. This splits the trapezoid into a rectangle and two triangles; it also splits the desired line segment into three partitions with lengths $x_1, 75, x_2$ . By similar triangles, we easily find that $\frac{x - 75}{100} = \frac{x_1+x_2}{100} = \frac{h_1}{h}$
The area of the region including the shorter base must be half of the area of the entire trapezoid, so
\[2 \cdot \frac 12 h_1 (75 + x) = \frac 12 h (75 + 175) \Longrightarrow x = 125 \cdot \frac{h}{h_1} - 75\]
Substituting our expression for $\frac h{h_1}$ from above, we find that
\[x = \frac{12500}{x-75} - 75 \Longrightarrow x^2 - 75x = 5625 + 12500 - 75x \Longrightarrow x^2 = 18125\]
The answer is $\left\lfloor\frac{x^2}{100}\right\rfloor = \boxed{181}$

Answer:  181
Problem: Given that
find the greatest integer that is less than $\frac N{100}$

Solution: Multiplying both sides by $19!$ yields:
\[\frac {19!}{2!17!}+\frac {19!}{3!16!}+\frac {19!}{4!15!}+\frac {19!}{5!14!}+\frac {19!}{6!13!}+\frac {19!}{7!12!}+\frac {19!}{8!11!}+\frac {19!}{9!10!}=\frac {19!N}{1!18!}.\]
\[\binom{19}{2}+\binom{19}{3}+\binom{19}{4}+\binom{19}{5}+\binom{19}{6}+\binom{19}{7}+\binom{19}{8}+\binom{19}{9} = 19N.\]
Recall the Combinatorial Identity $2^{19} = \sum_{n=0}^{19} {19 \choose n}$ . Since ${19 \choose n} = {19 \choose 19-n}$ , it follows that $\sum_{n=0}^{9} {19 \choose n} = \frac{2^{19}}{2} = 2^{18}$
Thus, $19N = 2^{18}-\binom{19}{1}-\binom{19}{0}=2^{18}-19-1 = (2^9)^2-20 = (512)^2-20 = 262124$
So, $N=\frac{262124}{19}=13796$ and $\left\lfloor \frac{N}{100} \right\rfloor =\boxed{137}$

Answer:  137
Problem: Given that
find the greatest integer that is less than $\frac N{100}$

Solution: Let $f(x) = (1+x)^{19}.$ Applying the binomial theorem gives us $f(x) = \binom{19}{19} x^{19} + \binom{19}{18} x^{18} + \binom{19}{17} x^{17}+ \cdots + \binom{19}{0}.$ Since $\frac 1{2!17!}+\frac 1{3!16!}+\dots+\frac 1{8!11!}+\frac 1{9!10!} = \frac{\frac{f(1)}{2} - \binom{19}{19} - \binom{19}{18}}{19!},$ $N = \frac{2^{18}-20}{19}.$ After some fairly easy bashing, we get $\boxed{137}$ as the answer.

Answer:  137
Problem: Given that
find the greatest integer that is less than $\frac N{100}$

Solution: Convert each denominator to $19!$ and get the numerators to be $9,51,204,612,1428,2652,3978,4862$ (refer to note). Adding these up we have $13796$ therefore $\boxed{137}$ is the desired answer.

Answer:  137
Problem: In trapezoid $ABCD$ , leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD}$ , and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001}$ , find $BC^2$

Solution: Let $x = BC$ be the height of the trapezoid, and let $y = CD$ . Since $AC \perp BD$ , it follows that $\triangle BAC \sim \triangle CBD$ , so $\frac{x}{\sqrt{11}} = \frac{y}{x} \Longrightarrow x^2 = y\sqrt{11}$
Let $E$ be the foot of the altitude from $A$ to $\overline{CD}$ . Then $AE = x$ , and $ADE$ is a right triangle . By the Pythagorean Theorem
\[x^2 + \left(y-\sqrt{11}\right)^2 = 1001 \Longrightarrow x^4 - 11x^2 - 11^2 \cdot 9 \cdot 10 = 0\]
The positive solution to this quadratic equation is $x^2 = \boxed{110}$

Answer:  110
Problem: In trapezoid $ABCD$ , leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD}$ , and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001}$ , find $BC^2$

Solution: Let $BC=x$ . Dropping the altitude from $A$ and using the Pythagorean Theorem tells us that $CD=\sqrt{11}+\sqrt{1001-x^2}$ . Therefore, we know that vector $\vec{BD}=\langle \sqrt{11}+\sqrt{1001-x^2},-x\rangle$ and vector $\vec{AC}=\langle-\sqrt{11},-x\rangle$ . Now we know that these vectors are perpendicular, so their dot product is 0. \[\vec{BD}\cdot \vec{AC}=-11-\sqrt{11(1001-x^2)}+x^2=0\] \[(x^2-11)^2=11(1001-x^2)\] \[x^4-11x^2-11\cdot 990=0.\] As above, we can solve this quadratic to get the positve solution $BC^2=x^2=\boxed{110}$

Answer:  110
Problem: In trapezoid $ABCD$ , leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD}$ , and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001}$ , find $BC^2$

Solution: Let $BC=x$ and $CD=y+\sqrt{11}$ . From Pythagoras with $AD$ , we obtain $x^2+y^2=1001$ . Since $AC$ and $BD$ are perpendicular diagonals of a quadrilateral, then $AB^2+CD^2=BC^2+AD^2$ , so we have \[\left(y+\sqrt{11}\right)^2+11=x^2+1001.\] Substituting $x^2=1001-y^2$ and simplifying yields \[y^2+\sqrt{11}y-990=0,\] and the quadratic formula gives $y=9\sqrt{11}$ . Then from $x^2+y^2=1001$ , we plug in $y$ to find $x^2=\boxed{110}$

Answer:  110
Problem: In trapezoid $ABCD$ , leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD}$ , and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001}$ , find $BC^2$

Solution: Let $E$ be the intersection of the diagonals. Since the diagonals are perpendicular, applying the Pythagorean Theorem multiple times we have \begin{align*} BC^2&=BE^2+CE^2 \\ &=(AB^2-AE^2)+(CD^2-DE^2) \\ &=CD^2+\sqrt{11}^2-(AE^2+DE^2) \\ &=CD^2+11-AD^2 \\ &=CD^2-990 \end{align*} Followed by dropping the perpendicular like in solution 1, we obtain system of equation \[BC^2=CD^2-990\] \[BC^2+CD^2-2\sqrt{11}CD=990\] Rearrange the first equation yields \[BC^2-CD^2=990\] Equating it with the second equation we have \[BC^2-CD^2=BC^2+CD^2-2\sqrt{11}CD\] Which gives $CD^2=\frac{BC^2}{11}$ .
Substituting into equation 1 obtains the quadratic in terms of $BC^2$ \[(BC^2)^2-11BC^2-11\cdot990=0\] Solving the quadratic to obtain $BC^2=\boxed{110}$

Answer:  110
Problem: Given that $z$ is a complex number such that $z+\frac 1z=2\cos 3^\circ$ , find the least integer that is greater than $z^{2000}+\frac 1{z^{2000}}$

Solution: Using the quadratic equation on $z^2 - (2 \cos 3 )z + 1 = 0$ , we have $z = \frac{2\cos 3 \pm \sqrt{4\cos^2 3 - 4}}{2} = \cos 3 \pm i\sin 3 = \text{cis}\,3^{\circ}$
There are other ways we can come to this conclusion. Note that if $z$ is on the unit circle in the complex plane, then $z = e^{i\theta} = \cos \theta + i\sin \theta$ and $\frac 1z= e^{-i\theta} = \cos \theta - i\sin \theta$ . We have $z+\frac 1z = 2\cos \theta = 2\cos 3^\circ$ and $\theta = 3^\circ$ . Alternatively, we could let $z = a + bi$ and solve to get $z=\cos 3^\circ + i\sin 3^\circ$
Using De Moivre's Theorem we have $z^{2000} = \cos 6000^\circ + i\sin 6000^\circ$ $6000 = 16(360) + 240$ , so $z^{2000} = \cos 240^\circ + i\sin 240^\circ$
We want $z^{2000}+\frac 1{z^{2000}} = 2\cos 240^\circ = -1$
Finally, the least integer greater than $-1$ is $\boxed{000}$

Answer:  000
Problem: Given that $z$ is a complex number such that $z+\frac 1z=2\cos 3^\circ$ , find the least integer that is greater than $z^{2000}+\frac 1{z^{2000}}$

Solution: Let $z=re^{i\theta}$ . Notice that we have $2\cos(3^{\circ})=e^{i\frac{\pi}{60}}+e^{-i\frac{\pi}{60}}=re^{i\theta}+\frac{1}{r}e^{-i\theta}.$
$r$ must be $1$ (or else if you take the magnitude would not be the same). Therefore, $z=e^{i\frac{\pi}{\theta}}$ and plugging into the desired expression, we get $e^{i\frac{100\pi}{3}}+e^{-i\frac{100\pi}{3}}=2\cos{\frac{100\pi}{3}}=-1$ . Therefore, the least integer greater is $\boxed{000}.$

Answer:  000
Problem: Given that $z$ is a complex number such that $z+\frac 1z=2\cos 3^\circ$ , find the least integer that is greater than $z^{2000}+\frac 1{z^{2000}}$

Solution: For this solution, we assume that $z^{2000} + 1/z^{2000}$ and $z^{2048} + 1/z^{2048}$ have the same least integer greater than their solution. we have $z + 1/z = 2\cos 3$ . Since $\cos 3 < 1$ $2\cos 3 < 2$ . If we square the equation $z + 1/z = 2\cos 3$ , we get $z^2 + 2 + 1/(z^2) = 4\cos^2 3$ , or $z^2 + 1/(z^2) = 4\cos^2 3 - 2$ $4\cos^2 3 - 2$ is is less than $2$ , since $4\cos^2 3$ is less than $4$ . if we square the equation again, we get $z^4 + 1/(z^4) = (4\cos^2 3 - 2)^2 -2$ . since $4\cos^2 3 - 2$ is less than 2, $(4\cos^2 3 - 2)^2$ is less than 4, and $(4\cos^2 3 - 2)^2 -2$ is less than 2. However $(4\cos^2 3 - 2)^2 -2$ is also less than $4\cos^2 3 - 2$ . we can see that every time we square the equation, the right hand side gets smaller, and into the negatives. Since the smallest integer that is allowed as an answer is 0, thus smallest integer greater is $\boxed{000}.$

Answer:  000
Problem: Given that $z$ is a complex number such that $z+\frac 1z=2\cos 3^\circ$ , find the least integer that is greater than $z^{2000}+\frac 1{z^{2000}}$

Solution: First, let $z = a+bi$ where $a$ and $b$ are real numbers. We now have that \[a+bi + \frac{a-bi}{a^2+b^2} = 2 \cos{3^{\circ}}\] given the coniditons of the problem. Equating imaginary coefficients, we have that \[b \left( 1 - \frac{1}{a^2+b^2}\right) = 0\] giving us that either $b=0$ or $|z| = 1$ . Let's consider the latter case for now.
We now know that $a^2+b^2=1$ , so when we equate real coefficients we have that $2a = 2 \cos{3^{\circ}}$ , therefore $a = \cos{3^{\circ}}$ . So, $b = \cos{3^{\circ}}$ and then we can write $z = \text{cis}(3)^{\circ}$
By De Moivre's Theorem, \[z^{2000} + \frac{1}{z^{2000}} = \text{cis} (6000)^{\circ} + \text{cis} (-6000)^{\circ}\] . The imaginary parts cancel, leaving us with $2 \cos{6000^{\circ}}$ which is $240 \pmod{360}$ . Therefore, we have it being $-1$ and our answer is $\boxed{000}$

Answer:  000
Problem: circle is inscribed in quadrilateral $ABCD$ tangent to $\overline{AB}$ at $P$ and to $\overline{CD}$ at $Q$ . Given that $AP=19$ $PB=26$ $CQ=37$ , and $QD=23$ , find the square of the radius of the circle.

Solution: Call the center of the circle $O$ . By drawing the lines from $O$ tangent to the sides and from $O$ to the vertices of the quadrilateral, four pairs of congruent right triangles are formed.
Thus, $\angle{AOP}+\angle{POB}+\angle{COQ}+\angle{QOD}=180$ , or $(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=180$
Take the $\tan$ of both sides and use the identity for $\tan(A+B)$ to get \[\tan(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+\tan(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=n\cdot0=0.\]
Use the identity for $\tan(A+B)$ again to get \[\frac{\tfrac{45}{r}}{1-19\cdot\tfrac{26}{r^2}}+\frac{\tfrac{60}{r}}{1-37\cdot\tfrac{23}{r^2}}=0.\]
Solving gives $r^2=\boxed{647}$

Answer:  647
Problem: circle is inscribed in quadrilateral $ABCD$ tangent to $\overline{AB}$ at $P$ and to $\overline{CD}$ at $Q$ . Given that $AP=19$ $PB=26$ $CQ=37$ , and $QD=23$ , find the square of the radius of the circle.

Solution: Just use the area formula for tangential quadrilaterals. The numbers are really big. A terrible problem to work on ( $a, b, c,$ and $d$ are the tangent lengths, not the side lengths). \[A = \sqrt{(a+b+c+d)(abc+bcd+cda+dab)} = 105\sqrt{647}\] $r^2=\frac{A^2}{(a+b+c+d)^2} = \boxed{647}$

Answer:  647
Problem: circle is inscribed in quadrilateral $ABCD$ tangent to $\overline{AB}$ at $P$ and to $\overline{CD}$ at $Q$ . Given that $AP=19$ $PB=26$ $CQ=37$ , and $QD=23$ , find the square of the radius of the circle.

Solution: Using the formulas established in solution 2, one notices: \[r^2=\frac{A^2}{(a+b+c+d)^2}\] \[r^2=\frac{(a+b+c+d)(abc+bcd+cda+abd) }{(a+b+c+d)^2}\] \[r^2=\frac{abc+bcd+acd+abd}{a+b+c+d}\] \[r^2=\boxed{647}\]

Answer:  647