Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Square $ABCD$ is inscribed in a circle . Square $EFGH$ has vertices $E$ and $F$ on $\overline{CD}$ and vertices $G$ and $H$ on the circle. If the area of square $ABCD$ is $1$ , then the area of square $EFGH$ can be expressed as $\frac {m}{n}$ where $m$ and $n$ are relatively prime positive integers and $m < n$ . Find $10n + m$

Solution: Let $O$ be the center of the circle, and $2a$ be the side length of $ABCD$ $2b$ be the side length of $EFGH$ . By the Pythagorean Theorem , the radius of $\odot O = OC = a\sqrt{2}$
Now consider right triangle $OGI$ , where $I$ is the midpoint of $\overline{GH}$ . Then, by the Pythagorean Theorem,
\begin{align*} OG^2 = 2a^2 &= OI^2 + GI^2 = (a+2b)^2 + b^2 \\ 0 &= a^2 - 4ab - 5b^2 = (a - 5b)(a + b) \end{align*}
Thus $a = 5b$ (since lengths are positive, we discard the other root). The ratio of the areas of two similar figures is the square of the ratio of their corresponding side lengths, so $\frac{[EFGH]}{[ABCD]} = \left(\frac 15\right)^2 = \frac{1}{25}$ , and the answer is $10n + m = \boxed{251}$

Answer:  251
Problem: Square $ABCD$ is inscribed in a circle . Square $EFGH$ has vertices $E$ and $F$ on $\overline{CD}$ and vertices $G$ and $H$ on the circle. If the area of square $ABCD$ is $1$ , then the area of square $EFGH$ can be expressed as $\frac {m}{n}$ where $m$ and $n$ are relatively prime positive integers and $m < n$ . Find $10n + m$

Solution: Let point $A$ be the top-left corner of square $ABCD$ and the rest of the vertices be arranged, in alphabetical order, in a clockwise arrangement from there. Let $D$ have coordinates $(0,0)$ and the side length of square $ABCD$ be $a$ . Let $DF$ $b$ and diameter $HI$ go through $J$ the midpoint of $EF$ . Since a diameter always bisects a chord perpendicular to it, $DJ$ $JC$ and since $F$ and $E$ must be symmetric around the diameter, $FJ = JE$ and it follows that $DF = EC = b.$ Hence $FE$ the side of square $EFGH$ has length $a - 2b$ $F$ has coordinates $(b,0)$ and $G$ has coordinates $(b, 2b - a).$ We know that point $G$ must be on the circle $O$ - hence it must satisfy the circle equation. Since the center of the circle is at the center of the square $(a/2, a/2)$ and has radius $a *$ $\sqrt{2} / 2$ , half the diagonal of the square, $(x - a/2)^2 + (y - a/2)^2 = 1/2a^2$ follows as the circle equation. Then substituting coordinates of $G$ into the equation, $(b - a/2)^2 + (2b - a - a/2)^2 = a^2/2$ . Simplifying and factoring, we get $2a^2-7ab+5b^2 = (2a-5b)(a-b) = 0.$ Since $a = b$ would imply $m = n$ , and $m < n$ in the problem, we must use the other factor. We get $b = 2/5a$ , meaning the ratio of areas $((a-2b)/a)^2$ $(1/5)^2$ $1/25$ $m/n.$ Then $10n + m = 25 * 10 + 1 = \boxed{251}$

Answer:  251
Problem: Let $\triangle{PQR}$ be a right triangle with $PQ = 90$ $PR = 120$ , and $QR = 150$ . Let $C_{1}$ be the inscribed circle . Construct $\overline{ST}$ with $S$ on $\overline{PR}$ and $T$ on $\overline{QR}$ , such that $\overline{ST}$ is perpendicular to $\overline{PR}$ and tangent to $C_{1}$ . Construct $\overline{UV}$ with $U$ on $\overline{PQ}$ and $V$ on $\overline{QR}$ such that $\overline{UV}$ is perpendicular to $\overline{PQ}$ and tangent to $C_{1}$ . Let $C_{2}$ be the inscribed circle of $\triangle{RST}$ and $C_{3}$ the inscribed circle of $\triangle{QUV}$ . The distance between the centers of $C_{2}$ and $C_{3}$ can be written as $\sqrt {10n}$ . What is $n$

Solution: Let $P = (0,0)$ be at the origin. Using the formula $A = rs$ on $\triangle PQR$ , where $r_{1}$ is the inradius (similarly define $r_2, r_3$ to be the radii of $C_2, C_3$ ), $s = \frac{PQ + QR + RP}{2} = 180$ is the semiperimeter , and $A = \frac 12 bh = 5400$ is the area, we find $r_{1} = \frac As = 30$ . Or, the inradius could be directly by using the formula $\frac{a+b-c}{2}$ , where $a$ and $b$ are the legs of the right triangle and $c$ is the hypotenuse. (This formula should be used only for right triangles .) Thus $ST, UV$ lie respectively on the lines $y = 60, x = 60$ , and so $RS = 60, UQ = 30$
Note that $\triangle PQR \sim \triangle STR \sim \triangle UQV$ . Since the ratio of corresponding lengths of similar figures are the same, we have
\[\frac{r_{1}}{PR} = \frac{r_{2}}{RS} \Longrightarrow r_{2} = 15\ \text{and} \ \frac{r_{1}}{PQ} = \frac{r_{3}}{UQ} \Longrightarrow r_{3} = 10.\]
Let the centers of $\odot C_2, C_3$ be $O_2 = (0 + r_{2}, 60 + r_{2}) = (15, 75), O_3 = (60 + r_{3}, 0 + r_{3}) = (70,10)$ , respectively; then by the distance formula we have $O_2O_3 = \sqrt{55^2 + 65^2} = \sqrt{10 \cdot 725}$ . Therefore, the answer is $n = \boxed{725}$

Answer:  725
Problem: Let $\triangle{PQR}$ be a right triangle with $PQ = 90$ $PR = 120$ , and $QR = 150$ . Let $C_{1}$ be the inscribed circle . Construct $\overline{ST}$ with $S$ on $\overline{PR}$ and $T$ on $\overline{QR}$ , such that $\overline{ST}$ is perpendicular to $\overline{PR}$ and tangent to $C_{1}$ . Construct $\overline{UV}$ with $U$ on $\overline{PQ}$ and $V$ on $\overline{QR}$ such that $\overline{UV}$ is perpendicular to $\overline{PQ}$ and tangent to $C_{1}$ . Let $C_{2}$ be the inscribed circle of $\triangle{RST}$ and $C_{3}$ the inscribed circle of $\triangle{QUV}$ . The distance between the centers of $C_{2}$ and $C_{3}$ can be written as $\sqrt {10n}$ . What is $n$

Solution: We compute $r_1 = 30, r_2 = 15, r_3 = 10$ as above. Let $A_1, A_2, A_3$ respectively the points of tangency of $C_1, C_2, C_3$ with $QR$
By the Two Tangent Theorem , we find that $A_{1}Q = 60$ $A_{1}R = 90$ . Using the similar triangles, $RA_{2} = 45$ $QA_{3} = 20$ , so $A_{2}A_{3} = QR - RA_2 - QA_3 = 85$ . Thus $(O_{2}O_{3})^{2} = (15 - 10)^{2} + (85)^{2} = 7250\implies n=\boxed{725}$

Answer:  725
Problem: Let $\triangle{PQR}$ be a right triangle with $PQ = 90$ $PR = 120$ , and $QR = 150$ . Let $C_{1}$ be the inscribed circle . Construct $\overline{ST}$ with $S$ on $\overline{PR}$ and $T$ on $\overline{QR}$ , such that $\overline{ST}$ is perpendicular to $\overline{PR}$ and tangent to $C_{1}$ . Construct $\overline{UV}$ with $U$ on $\overline{PQ}$ and $V$ on $\overline{QR}$ such that $\overline{UV}$ is perpendicular to $\overline{PQ}$ and tangent to $C_{1}$ . Let $C_{2}$ be the inscribed circle of $\triangle{RST}$ and $C_{3}$ the inscribed circle of $\triangle{QUV}$ . The distance between the centers of $C_{2}$ and $C_{3}$ can be written as $\sqrt {10n}$ . What is $n$

Solution: The radius of an incircle is $r=A_t/\text{semiperimeter}$ . The area of the triangle is equal to $\frac{90\times120}{2} = 5400$ and the semiperimeter is equal to $\frac{90+120+150}{2} = 180$ . The radius, therefore, is equal to $\frac{5400}{180} = 30$ . Thus using similar triangles the dimensions of the triangle circumscribing the circle with center $C_2$ are equal to $120-2(30) = 60$ $\frac{1}{2}(90) = 45$ , and $\frac{1}{2}\times150 = 75$ . The radius of the circle inscribed in this triangle with dimensions $45\times60\times75$ is found using the formula mentioned at the very beginning. The radius of the incircle is equal to $15$
Defining $P$ as $(0,0)$ $C_2$ is equal to $(60+15,15)$ or $(75,15)$ . Also using similar triangles, the dimensions of the triangle circumscribing the circle with center $C_3$ are equal to $90-2(30)$ $\frac{1}{3}\times120$ $\frac{1}{3}\times150$ or $30,40,50$ . The radius of $C_3$ by using the formula mentioned at the beginning  is $10$ . Using $P$ as $(0,0)$ $C_3$ is equal to $(10, 60+10)$ or $(10,70)$ . Using the distance formula, the distance between $C_2$ and $C_3$ $\sqrt{(75-10)^2 +(15-70)^2}$ this equals $\sqrt{7250}$ or $\sqrt{725\times10}$ , thus $n$ is $\boxed{725}$

Answer:  725
Problem: Let $\triangle{PQR}$ be a right triangle with $PQ = 90$ $PR = 120$ , and $QR = 150$ . Let $C_{1}$ be the inscribed circle . Construct $\overline{ST}$ with $S$ on $\overline{PR}$ and $T$ on $\overline{QR}$ , such that $\overline{ST}$ is perpendicular to $\overline{PR}$ and tangent to $C_{1}$ . Construct $\overline{UV}$ with $U$ on $\overline{PQ}$ and $V$ on $\overline{QR}$ such that $\overline{UV}$ is perpendicular to $\overline{PQ}$ and tangent to $C_{1}$ . Let $C_{2}$ be the inscribed circle of $\triangle{RST}$ and $C_{3}$ the inscribed circle of $\triangle{QUV}$ . The distance between the centers of $C_{2}$ and $C_{3}$ can be written as $\sqrt {10n}$ . What is $n$

Solution: We can calculate the inradius of each triangle as with the previous solutions. Now, notice that the hexagon $PSTVU$ is a square with its corner cut off. We literally complete the square and mark the last corner as point X. Now, construct right triangle with $C_3C_2$ as its hypotenuse. The right angle will be at point $Y$ . We will now find the length of each leg and use the Pythagorean Theorem to compute the desired length. We see that the length of $C_3Y$ is $50 + 15 = 65$ , as seen by the inradius of $C_2$ and $10$ less than the square's side length. $C_2Y$ is $45 + 10 = 55$ , which is $15$ less than the square plus the inradius of $C_3$ . Our final answer is $\sqrt{65^2 + 55^2} = \sqrt{7250} = \sqrt{(10)(725)} \rightarrow \boxed{725}$

Answer:  725
Problem: A certain function $f$ has the properties that $f(3x) = 3f(x)$ for all positive real values of $x$ , and that $f(x) = 1-|x-2|$ for $1\le x \le 3$ . Find the smallest $x$ for which $f(x) = f(2001)$

Solution: Iterating the condition $f(3x) = 3f(x)$ , we find that $f(x) = 3^kf\left(\frac{x}{3^k}\right)$ for positive integers $k$ . We know the definition of $f(x)$ from $1 \le x \le 3$ , so we would like to express $f(2001) = 3^kf\left(\frac{2001}{3^k}\right),\ 1 \le \frac{2001}{3^k} \le 3 \Longrightarrow k = 6$ . Indeed,
\[f(2001) = 729\left[1 - \left| \frac{2001}{729} - 2\right|\right] = 186.\]
We now need the smallest $x$ such that $f(x) = 3^kf\left(\frac{x}{3^k}\right) = 186$ . The range of $f(x),\ 1 \le x \le 3$ , is $0 \le f(x) \le 1$ . So when $1 \le \frac{x}{3^k} \le 3$ , we have $0 \le f\left(\frac{x}{3^k}\right) = \frac{186}{3^k} \le 1$ . Multiplying by $3^k$ $0 \le 186 \le 3^k$ , so the smallest value of $k$ is $k = 5$ . Then,
\[186 = {3^5}f\left(\frac{x}{3^5}\right).\]
Because we forced $1 \le \frac{x}{3^5} \le 3$ , so
\[186 = {3^5}f\left(\frac{x}{3^5}\right) = 243\left[1 - \left| \frac{x}{243} - 2\right|\right] \Longrightarrow x = \pm 57 + 2  \cdot 243.\]
We want the smaller value of $x = \boxed{429}$

Answer:  429
Problem: A certain function $f$ has the properties that $f(3x) = 3f(x)$ for all positive real values of $x$ , and that $f(x) = 1-|x-2|$ for $1\le x \le 3$ . Find the smallest $x$ for which $f(x) = f(2001)$

Solution: First, we start by graphing the function when $1\leq{x}\leq3$ , which consists of the lines $y=x-1$ and $y=3-x$ that intersect at $(2,1)$ . Similarly, using $f(3x)=3f(x)$ , we get a dilation of our initial figure by a factor of 3 for the next interval and so on.
Observe that the intersection of two lines always has coordinates $(2y,y)$ where $y=3^a$ for some $a$ . First, we compute $f(2001)$ . The nearest intersection point is $(1458,729)$ when $a=7$ . Therefore, we can safely assume that $f(2001)$ is somewhere on the line with a slope of $-1$ that intersects at that nearest point. Using the fact that the slope of the line is $-1$ , we compute $f(2001)=729-543=186$ . However, we want the minimum value such that $f(x)=186$ and we see that there is another intersection point on the left which has a $y>186$ , namely $(486,243)$ . Therefore, we want the point that lies on the line with slope $1$ that intersects this point. Once again, since the slope of the line is $1$ , we get $x=486-57=\boxed{429}$

Answer:  429
Problem: Each unit square of a 3-by-3 unit-square grid is to be colored either blue or red. For each square, either color is equally likely to be used. The probability of obtaining a grid that does not have a 2-by-2 red square is $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: We can use complementary counting , counting all of the colorings that have at least one red $2\times 2$ square.
By the Principle of Inclusion-Exclusion , there are (alternatively subtracting and adding) $128-40+8-1=95$ ways to have at least one red $2 \times 2$ square.
There are $2^9=512$ ways to paint the $3 \times 3$ square with no restrictions, so there are $512-95=417$ ways to paint the square with the restriction. Therefore, the probability of obtaining a grid that does not have a $2 \times 2$ red square is $\frac{417}{512}$ , and $417+512=\boxed{929}$

Answer:  929
Problem: Each unit square of a 3-by-3 unit-square grid is to be colored either blue or red. For each square, either color is equally likely to be used. The probability of obtaining a grid that does not have a 2-by-2 red square is $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: We consider how many ways we can have 2*2 grid
$(1)$ : All the grids are red-- $1$ case
$(2)$ : One unit square is blue--The blue lies on the center of the bigger square, makes no 2*2 grid $9-1=8$ cases
$(3)$ : Two unit squares are blue--one of the squares lies in the center of the bigger square, makes no 2*2 grid, $8$ cases. 
Or, two squares lie on second column, first row, second column third row; second row first column, second row third column, 2 extra cases. $\binom 9 2-8-2=26$ cases
$(4)$ Three unit squares are blue. We find that if a 2*2 square is formed, there are 5 extra unit squares can be painted. But cases that three squares in the same column or same row is overcomunted. So in this case, there are $4\cdot (\binom 5 3)-4=36$
$(5)$ Four unit squares are blue, no overcomunted case will be considered. there are $4\cdot \binom 5 4=20$
$(6)$ Five unit squares are blue, $4$ cases in all
Sum up those cases, there are $1+8+26+36+20+4=95$ cases that a 2*2 grid can be formed.
In all, there are $2^9=512$ possible ways to paint the big square, so the answer is $1-\frac{95}{512}=\frac{417}{512}$ leads to $\boxed{929}$

Answer:  929
Problem: How many positive integer multiples of $1001$ can be expressed in the form $10^{j} - 10^{i}$ , where $i$ and $j$ are integers and $0\leq i < j \leq 99$

Solution: Observation: We see that there is a pattern with $10^k \pmod{1001}$ \[10^0 \equiv 1 \pmod{1001}\] \[10^1 \equiv 10 \pmod{1001}\] \[10^2 \equiv 100 \pmod{1001}\] \[10^3 \equiv -1 \pmod{1001}\] \[10^4 \equiv -10 \pmod{1001}\] \[10^5 \equiv -100 \pmod{1001}\] \[10^6 \equiv 1 \pmod{1001}\] \[10^7 \equiv 10 \pmod{1001}\] \[10^8 \equiv 100 \pmod{1001}\]
So, this pattern repeats every 6.
Also, $10^j-10^i \equiv 0 \pmod{1001}$ , so $10^j \equiv 10^i \pmod{1001}$ , and thus, \[j \equiv i \pmod{6}\] . Continue with the 2nd paragraph of solution 1, and we get the answer of $\boxed{784}$

Answer:  784
Problem: How many positive integer multiples of $1001$ can be expressed in the form $10^{j} - 10^{i}$ , where $i$ and $j$ are integers and $0\leq i < j \leq 99$

Solution: Note that $1001=7\cdot 11\cdot 13,$ and note that $10^3 \equiv \pmod{p}$ for prime $p | 1001$ ; therefore, the order of 10 modulo $7,11$ , and $13$ must divide 6. A quick check on 7 reveals that it is indeed 6. Therefore we note that $i-j=6k$ for some natural number k. From here, we note that for $j=0,1,2,3,$ we have 16 options and we have 15,14,...,1 option(s) for the next 90 numbers (6 each), so our total is $4\cdot 16 + 6 \cdot \frac{15 \cdot 16}{2} = \boxed{784}$

Answer:  784
Problem: How many positive integer multiples of $1001$ can be expressed in the form $10^{j} - 10^{i}$ , where $i$ and $j$ are integers and $0\leq i < j \leq 99$

Solution: $10^j - 10^i \equiv 0 \pmod{1001} \iff 10^{j - i} - 1 \equiv 0 \pmod{1001} \iff 10^{j - i} \equiv 1 \pmod{1001} \iff j \equiv i \pmod 6$ . If $j \equiv i \equiv n \pmod 6$ for $n = 0, 1, 2, 3$ , there are $17$ choices for each value of $n$ , yielding $4 \cdot \dbinom{17}{2} = 544$ . However, if $n = 4, 5$ , there are only $16$ choices, giving us $2 \cdot \dbinom{16}{2} = 240$ . So, our final answer is $544 + 240 = \boxed{784}$ .
~Puck_0

Answer:  784
Problem: Club Truncator is in a soccer league with six other teams, each of which it plays once. In any of its 6 matches, the probabilities that Club Truncator will win, lose, or tie are each $\frac {1}{3}$ . The probability that Club Truncator will finish the season with more wins than losses is $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: Note that the probability that Club Truncator will have more wins than losses is equal to the probability that it will have more losses than wins; the only other possibility is that they have the same number of wins and losses. Thus, by the complement principle , the desired probability is half the probability that Club Truncator does not have the same number of wins and losses.
The possible ways to achieve the same number of wins and losses are $0$ ties, $3$ wins and $3$ losses; $2$ ties, $2$ wins, and $2$ losses; $4$ ties, $1$ win, and $1$ loss; or $6$ ties. Since there are $6$ games, there are $\frac{6!}{3!3!}$ ways for the first, and $\frac{6!}{2!2!2!}$ $\frac{6!}{4!}$ , and $1$ ways for the rest, respectively, out of a total of $3^6$ . This gives a probability of $141/729$ . Then the desired answer is $\frac{1 - \frac{141}{729}}{2} = \frac{98}{243}$ , so the answer is $m+n = \boxed{341}$

Answer:  341
Problem: Club Truncator is in a soccer league with six other teams, each of which it plays once. In any of its 6 matches, the probabilities that Club Truncator will win, lose, or tie are each $\frac {1}{3}$ . The probability that Club Truncator will finish the season with more wins than losses is $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: At first, it wins $6$ games, only one way
Secondly, it wins $5$ games, the other game can be either win or loss, there are $\binom{6}{5}\cdot 2=12$ ways
Thirdly, it wins $4$ games, still the other two games can be either win or loss, there are $\binom{6}{4}\cdot 2^2=60$ ways
Fourthly, it wins $3$ games, this time, it can't lose $3$ games but other arrangements of the three non-winning games are fine, there are $\binom{6}{3}\cdot (2^3-1)=140$ ways
Fifth case, it wins $2$ games, only $0/1$ lose and $4/3$ draw is ok, so there are $\binom{6}{2}(1+\binom{4}{1})=75$ cases
Last case, it only wins $1$ game so the rest games must be all draw, $1$ game
The answer is $\frac{1+12+60+140+75+6}{3^6}=\frac{98}{243}$ leads to $\boxed{341}$

Answer:  341
Problem: Given a triangle , its midpoint triangle is obtained by joining the midpoints of its sides. A sequence of polyhedra $P_{i}$ is defined recursively as follows: $P_{0}$ is a regular tetrahedron whose volume is 1. To obtain $P_{i + 1}$ , replace the midpoint triangle of every face of $P_{i}$ by an outward-pointing regular tetrahedron that has the midpoint triangle as a face. The volume of $P_{3}$ is $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: On the first construction, $P_1$ , four new tetrahedra will be constructed with side lengths $\frac 12$ of the original one. Since the ratio of the volume of similar polygons is the cube of the ratio of their corresponding lengths, it follows that each of these new tetrahedra will have volume $\left(\frac 12\right)^3 = \frac 18$ . The total volume added here is then $\Delta P_1 = 4 \cdot \frac 18 = \frac 12$
We now note that for each midpoint triangle we construct in step $P_{i}$ , there are now $6$ places to construct new midpoint triangles for step $P_{i+1}$ . The outward tetrahedron for the midpoint triangle provides $3$ of the faces, while the three equilateral triangles surrounding the midpoint triangle provide the other $3$ . This is because if you read this question carefully, it asks to add new tetrahedra to each face of $P_{i}$ which also includes the ones that were left over when we did the previous addition of tetrahedra. However, the volume of the tetrahedra being constructed decrease by a factor of $\frac 18$ . Thus we have the recursion $\Delta P_{i+1} = \frac{6}{8} \Delta P_i$ , and so $\Delta P_i = \frac 12 \cdot \left(\frac{3}{4}\right)^{i-1} P_1$
The volume of $P_3 = P_0 + \Delta P_1 + \Delta P_2 + \Delta P_3 = 1 + \frac 12 + \frac 38 + \frac 9{32} = \frac{69}{32}$ , and $m+n=\boxed{101}$ . Note that the summation was in fact a geometric series

Answer:  101
Problem: In quadrilateral $ABCD$ $\angle{BAD}\cong\angle{ADC}$ and $\angle{ABD}\cong\angle{BCD}$ $AB = 8$ $BD = 10$ , and $BC = 6$ . The length $CD$ may be written in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: Extend $\overline{AD}$ and $\overline{BC}$ to meet at $E$ . Then, since $\angle BAD = \angle ADC$ and $\angle ABD = \angle DCE$ , we know that $\triangle ABD \sim \triangle DCE$ . Hence $\angle ADB = \angle DEC$ , and $\triangle BDE$ is isosceles . Then $BD = BE = 10$
Using the similarity, we have:
\[\frac{AB}{BD} = \frac 8{10} = \frac{CD}{CE} = \frac{CD}{16} \Longrightarrow CD = \frac{64}5\]
The answer is $m+n = \boxed{069}$

Answer:  069
Problem: In quadrilateral $ABCD$ $\angle{BAD}\cong\angle{ADC}$ and $\angle{ABD}\cong\angle{BCD}$ $AB = 8$ $BD = 10$ , and $BC = 6$ . The length $CD$ may be written in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: Draw a line from $B$ , parallel to $\overline{AD}$ , and let it meet $\overline{CD}$ at $M$ . Note that $\triangle{DAB}$ is similar to $\triangle{BMC}$ by AA similarity, since $\angle{ABD}=\angle{MCB}$ and since $BM$ is parallel to $CD$ then $\angle{BMC}=\angle{ADM}=\angle{DAB}$ . Now since $ADMB$ is an isosceles trapezoid, $MD=8$ . By the similarity, we have $MC=AB\cdot \frac{BC}{BD}=8\cdot \frac{6}{10}=\frac{24}{5}$ , hence $CD=MC+MD=\frac{24}{5}+8=\frac{64}{5}\implies 64+5=\boxed{069}$

Answer:  069
Problem: In quadrilateral $ABCD$ $\angle{BAD}\cong\angle{ADC}$ and $\angle{ABD}\cong\angle{BCD}$ $AB = 8$ $BD = 10$ , and $BC = 6$ . The length $CD$ may be written in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: Since $\angle{BAD}=\angle{ADM}$ , if we extend AB and DC, they must meet at one point to form a isosceles triangle $\triangle{ADM}$ .Now, since the problem told that $\angle{ABD}=\angle{BCD}$ , we can imply that $\angle{DBM}=\angle{BCM}$ Since $\angle{M}=\angle{M}$ , so $\triangle{CBM}\sim\triangle{BDM}$ . Assume the length of $BM=x$ ;Since $\frac{BC}{MB}=\frac{DB}{MD}$ we can get $\frac{6}{x}=\frac{10}{8+x}$ , we get that $x=12$ .So $AM=DM=20$ similarly, we use the same pair of similar triangle we get $\frac{CM}{BM}=\frac{BM}{DM}$ , we get that $CM=\frac{36}{5}$ . Finally, $CD=MD-MC=\frac{64}{5}\implies 64+5=69=\boxed{069}$ ~bluesoul

Answer:  069
Problem: In quadrilateral $ABCD$ $\angle{BAD}\cong\angle{ADC}$ and $\angle{ABD}\cong\angle{BCD}$ $AB = 8$ $BD = 10$ , and $BC = 6$ . The length $CD$ may be written in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution: Denote $\angle{BAD}=\angle{CDA}=x$ , and $\angle{ABD}=\angle{BCD}=y$ . Note that $\angle{ADB}=180^\circ-x-y$ , and $\angle{DBC}=360^\circ-2x-2y$ . This motivates us to draw the angle bisector of $\angle{DBC}$ because $\angle{DBC} = 2 \angle{ADB}$ , so we do so and consider the intersection with $CD$ as $E$ . By the angle bisector theorem, we have $\frac{CE}{DE} = \frac{BC}{BD} = \frac{3}{5}$ , so we write $CE=3z$ and $DE=5z$ . We also know that $\angle{EBC}=\angle{ADB}$ and $\angle{BCE}=\angle{DBA}$ , so $\triangle{ADB} \sim \triangle{EBC}$ . Hence, $\frac{CE}{BC}=\frac{AB}{BD}$ , so we have $3z=\frac{24}{5}$ . As $CD=8z$ , it must be that $CD=\frac{64}{5}$ , so the final answer is $\boxed{069}$

Answer:  069