Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: A token starts at the point $(0,0)$ of an $xy$ -coordinate grid and then makes a sequence of six moves. Each move is 1 unit in a direction parallel to one of the coordinate axes. Each move is selected randomly from the four possible directions and independently of the other moves. The probability the token ends at a point on the graph of $|y|=|x|$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Denote $(x, y)_n$ the probability that starting from point $(x, y)$ , the token reaches a point on the graph of $|y| = |x|$ in exactly $n$ moves. The problem asks us to find $(0, 0)_6$ . We start by breaking this down: \[(0, 0)_6 = \frac14 \cdot ((0, 1)_5 + (0, -1)_5 + (1, 0)_5 + (-1, 0)_5)\] Notice that by symmetry, $(0, 1)_5 = (0, -1)_5 = (1, 0)_5 = (-1, 0)_5$ , so the equation simplifies to \[(0, 0)_6 = (0, 1)_5\] We now expand $(0, 1)_5$ \[(0, 1)_5 = \frac14 \cdot ((0, 0)_4 + (0, 2)_4 + 2(1, 1)_4)\] First, we find $(0, 0)_4$ \[(0, 0)_4 = (0, 1)_3\] \[(0, 1)_3 = \frac14 \cdot ((0, 0)_2 + (0, 2)_2 + 2(1, 1)_2)\] At this point, we can just count the possibilities to find $(0, 0)_2 = \frac34$ $(0, 2)_2 = \frac{7}{16}$ , and $(1, 1)_2 = \frac58$ . Therefore, \[(0, 1)_3 = \frac14 \cdot (\frac34 + \frac{7}{16} + 2 \cdot \frac58)\] \[(0, 1)_3 = \frac{39}{64}\] Next, we find $(0, 2)_4$ \[(0, 2)_4 = \frac14 \cdot ((0, 1)_3 + (0, 3)_3 + 2(1, 2)_3)\] We already calculated $(0, 1)_3$ , so we just need to find $(0, 3)_3$ and $(1, 2)_3$ \[(0, 3)_3 = \frac14 \cdot ((0, 2)_2 + (0, 4)_2 + 2(1, 3)_2)\] \[(0, 3)_3 = \frac14 \cdot (\frac{7}{16} + 0 + 2 \cdot \frac{1}{4})\] \[(0, 3)_3 = \frac{15}{64}\] \[(1, 2)_3 = \frac14 \cdot ((1, 3)_2 + (1, 1)_2 + (0, 2)_2 + (2, 2)_2)\] \[(1, 2)_3 = \frac14 \cdot (\frac14 + \frac58 + \frac{7}{16} + \frac12)\] \[(1, 2)_3 = \frac{29}{64}\] Therefore, \[(0, 2)_4 = \frac14 \cdot (\frac{39}{64} + \frac{15}{64} + 2 \cdot \frac{29}{64})\] \[(0, 2)_4 = \frac{7}{16}\] Finally, we find $(1, 1)_4$ \[(1, 1)_4 = \frac12 \cdot ((0, 1)_3 + (1, 2)_3)\] \[(1, 1)_4 = \frac12 \cdot (\frac{39}{64} + \frac{29}{64})\] \[(1, 1)_4 = \frac{17}{32}\] Putting it all together, \[(0, 0)_6 = (0, 1)_5 =\frac14 \cdot (\frac{39}{64} + \frac{7}{16} + 2 \cdot \frac{17}{32})\] \[(0, 0)_6 = \frac{135}{256}\] Thus, the answer is $135 + 256 = \boxed{391}$

Answer:  391
Problem: Let $A=\{1,2,3,4\}$ , and $f$ and $g$ be randomly chosen (not necessarily distinct) functions from $A$ to $A$ . The probability that the range of $f$ and the range of $g$ are disjoint is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m$

Solution: The natural way to go is casework. And the natural process is to sort $f$ and $g$ based on range size! Via Pigeonhole Principle, we see that the only real possibilities are: $f 1 g 1; f 1 g 2; f 1 g 3; f 2 g 2$ . Note that the $1, 2$ and $1, 3$ cases are symmetrical and we need just a $*2$ . Note also that the total number of cases is $4^4*4^4=4^8$
$f 1 g 1$ : clearly, we choose one number as the range for $f$ , one for $g$ , yielding $12$ possibilities.
$f 1 g 2$ with symmetry (WLOG $f$ has 1 element): start by selecting numbers for the ranges. This yields $4$ for the one number in $f$ , and $3$ options for the two numbers for $g$ . Afterwards, note that the function with 2 numbers in the range can have $4+6+4=14$ arrangements of these two numbers (1 of one, 3 of the other *2 and 2 of each). Therefore, we have $2*12*14$ possibilities, the 2 from symmetry.
$f 2 g 2$ : no symmetry, still easy! Just note that we have $6$ choices of which numbers go to $f$ and $g$ , and within each, $14*14=196$ choices for the orientation of each of the two numbers. That's $6*196$ possibilities.
$f 1 g 3$ : again, symmetrical (WLOG $f$ has one element): $4$ ways to select the single element for $f$ , and then find the number of ways to distribute the $3$ distinct numbers in the range for $g$ . The only arrangement for the frequency of each number is ${1, 1, 2}$ in some order. Therefore, we have $3$ ways to choose which number is the one represented twice, and then note that there are $12$ ways to arrange these! The number of possibilities in this situation is $2 * 4 * 3 * 12$
Total, divided by $4^8$ , gets $\frac{3 * (1 + 2 * 7^2 + 2^2 * 7 + 2^3 * 3)}{4^7}$ , with numerator $\boxed{453}$

Answer:  453
Problem: Let $A=\{1,2,3,4\}$ , and $f$ and $g$ be randomly chosen (not necessarily distinct) functions from $A$ to $A$ . The probability that the range of $f$ and the range of $g$ are disjoint is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m$

Solution: We note there are $4^4 = 256$ possibilities for each of $f$ and $g$ from $A$ to $A$ since the input of the four values of each function has four options each for an output value.
We proceed with casework to determine the number of possible $f$ with range 1, 2, etc.
There are 4 possibilities: all elements output to 1, 2, 3, or 4.
We have ${{4}\choose {2}} = 6$ ways to choose the two output elements for $f$ . At this point we have two possibilities: either $f$ has 3 of 1 element and 1 of the other, or 2 of each element. In the first case, there are 2 ways to pick the element which there are 3 copies of, and ${{4}\choose {1}} = 4$ ways to rearrange the 4 elements, for a total of $6 * 4 * 2 = 48$ ways for this case. For the second case, there are ${{4}\choose {2}} = 6$ ways to rearrange the 4 elements, for a total of $6 * 6 = 36$ ways for this case. Adding these two, we get a total of $36 + 48 = 84$ total possibilities.
We have ${{4}\choose {3}} = 4$ ways to choose the three output elements for $f$ . We know we must have 2 of 1 element and 1 of each of the others, so there are 3 ways to pick this element. Finally, there are ${{4}\choose{1}}*{{3}\choose{1}} = 12$ ways to rearrange these elements (since we can pick the locations of the 2 single elements in this many ways), and our total is $4 * 3 * 12 = 144$ ways.
Since we know the elements present, we have $4!$ ways to arrange them, or 24 ways.
(To check, $4 + 84 + 144 + 24 = 256$ , which is the total number of possibilities).
We now break $f$ down by cases, and count the number of $g$ whose ranges are disjoint from $f$ 's.
We know that there are 3 possibilities for $g$ with 1 element. Since half the possibilities for $g$ with two elements will contain the element in $f$ , there are $84/2 = 42$ possibilities for $g$ with 2 elements. Since $3/4$ the possibilities for $g$ with 3 elements will contain the element in $f$ , there are $144/4 = 36$ possibilities for $g$ with 3 elements. Clearly, no 4-element range for $g$ is possible, so the total number of ways for this case to happen is $4(3 + 42 + 36) = 324$
We know that there are 2 possibilities for $g$ with 1 element. If $g$ has 2 elements in its range, they are uniquely determined, so the total number of sets with a range of 2 elements that work for $g$ is $84/6 = 14$ . No 3-element or 4-element ranges for $g$ are possible. Thus, the total number of ways for this to happen is $84(2 + 14) = 1344$
In this case, there is only 1 possibility for $g$ - all the output values are the element that does not appear in $f$ 's range. Thus, the total number of ways for this to happen is $144$
We find that the probability of $f$ and $g$ having disjoint ranges is equal to:
$\dfrac{324 + 1344 + 144}{256^2}=\dfrac{1812}{2^{16}}= \dfrac{453}{2^{14}}$
Thus, our final answer is $\boxed{453}$

Answer:  453
Problem: On square $ABCD$ , points $E,F,G$ , and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34$ . Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P$ , and the areas of the quadrilaterals $AEPH, BFPE, CGPF,$ and $DHPG$ are in the ratio $269:275:405:411.$ Find the area of square $ABCD$
[asy] pair A = (0,sqrt(850)); pair B = (0,0); pair C = (sqrt(850),0); pair D = (sqrt(850),sqrt(850)); draw(A--B--C--D--cycle); dotfactor = 3; dot("$A$",A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); dot("$D$",D,dir(45)); pair H = ((2sqrt(850)-sqrt(306))/6,sqrt(850)); pair F = ((2sqrt(850)+sqrt(306)+7)/6,0); dot("$H$",H,dir(90)); dot("$F$",F,dir(270)); draw(H--F); pair E = (0,(sqrt(850)-6)/2); pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2); dot("$E$",E,dir(180)); dot("$G$",G,dir(0)); draw(E--G); pair P = extension(H,F,E,G); dot("$P$",P,dir(60)); label("$w$", intersectionpoint( A--P, E--H )); label("$x$", intersectionpoint( B--P, E--F )); label("$y$", intersectionpoint( C--P, G--F )); label("$z$", intersectionpoint( D--P, G--H ));[/asy]

Solution: $269+275+405+411=1360$ , a multiple of $17$ . In addition, $EG=FH=34$ , which is $17\cdot 2$ .
Therefore, we suspect the square of the "hypotenuse" of a right triangle, corresponding to $EG$ and $FH$ must be a multiple of $17$ . All of these triples are primitive:
\[17=1^2+4^2\] \[34=3^2+5^2\] \[51=\emptyset\] \[68=\emptyset\text{ others}\] \[85=2^2+9^2=6^2+7^2\] \[102=\emptyset\] \[119=\emptyset \dots\]
The sides of the square can only equal the longer leg, or else the lines would have to extend outside of the square. Substituting $EG=FH=34$ \[\sqrt{17}\rightarrow 34\implies 8\sqrt{17}\implies A=\textcolor{red}{1088}\] \[\sqrt{34}\rightarrow 34\implies 5\sqrt{34}\implies A=850\] \[\sqrt{85}\rightarrow 34\implies \{18\sqrt{85}/5,14\sqrt{85}/5\}\implies A=\textcolor{red}{1101.6,666.4}\]
Thus, $\boxed{850}$ is the only valid answer.

Answer:  850
Problem: On square $ABCD$ , points $E,F,G$ , and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34$ . Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P$ , and the areas of the quadrilaterals $AEPH, BFPE, CGPF,$ and $DHPG$ are in the ratio $269:275:405:411.$ Find the area of square $ABCD$
[asy] pair A = (0,sqrt(850)); pair B = (0,0); pair C = (sqrt(850),0); pair D = (sqrt(850),sqrt(850)); draw(A--B--C--D--cycle); dotfactor = 3; dot("$A$",A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); dot("$D$",D,dir(45)); pair H = ((2sqrt(850)-sqrt(306))/6,sqrt(850)); pair F = ((2sqrt(850)+sqrt(306)+7)/6,0); dot("$H$",H,dir(90)); dot("$F$",F,dir(270)); draw(H--F); pair E = (0,(sqrt(850)-6)/2); pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2); dot("$E$",E,dir(180)); dot("$G$",G,dir(0)); draw(E--G); pair P = extension(H,F,E,G); dot("$P$",P,dir(60)); label("$w$", intersectionpoint( A--P, E--H )); label("$x$", intersectionpoint( B--P, E--F )); label("$y$", intersectionpoint( C--P, G--F )); label("$z$", intersectionpoint( D--P, G--H ));[/asy]

Solution: Continue in the same way as solution 1 to get that $POK$ has area $3a$ , and $OK = \frac{d}{10}$ . You can then find $PK$ has length $\frac 32$
Then, if we drop a perpendicular from $H$ to $BC$ at $L$ , We get $\triangle HLF \sim \triangle OPK$
Thus, $LF = \frac{15\cdot 34}{d}$ , and we know $HL = d$ , and $HF = 34$ . Thus, we can set up an equation in terms of $d$ using the Pythagorean theorem.
\[\frac{15^2 \cdot 34^2}{d^2} + d^2 = 34^2\]
\[d^4 - 34^2 d^2 + 15^2 \cdot 34^2 = 0\]
\[(d^2 - 34 \cdot 25)(d^2 - 34 \cdot 9) = 0\]
$d^2 = 34 \cdot 9$ is extraneous, so $d^2 = 34 \cdot 25$ . Since the area is $d^2$ , we have it is equal to $34 \cdot 25 = \boxed{850}$

Answer:  850
Problem: Let $m$ be the largest real solution to the equation
$\frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}=x^2-11x-4$
There are positive integers $a$ $b$ , and $c$ such that $m=a+\sqrt{b+\sqrt{c}}$ . Find $a+b+c$

Solution: The first step is to notice that the 4 on the right hand side can simplify the terms on the left hand side. If we distribute 1 to $\frac{3}{x-3}$ , then the fraction becomes of the form $\frac{x}{x - 3}$ . A similar cancellation happens with the other four terms. If we assume $x = 0$ is not the highest solution (which is true given the answer format) we can cancel the common factor of $x$ from both sides of the equation.
$\frac{1}{x - 3} + \frac{1}{x - 5} + \frac{1}{x - 17} + \frac{1}{x - 19} = x - 11$
Then, if we make the substitution $y = x - 11$ , we can further simplify.
$\frac{1}{y + 8} + \frac{1}{y + 6} + \frac{1}{y - 6} + \frac{1}{y - 8} =y$
If we group and combine the terms of the form $y - n$ and $y + n$ , we get this equation:
$\frac{2y}{y^2 - 64} + \frac{2y}{y^2 - 36} = y$
Then, we can cancel out a $y$ from both sides, knowing that $x = 11$ is not a possible solution given the answer format.
After we do that, we can make the final substitution $z = y^2$
$\frac{2}{z - 64} + \frac{2}{z - 36} = 1$
$2z - 128 + 2z - 72 = (z - 64)(z - 36)$
$4z -  200 = z^2 - 100z + 64(36)$
$z^2 - 104z + 2504 = 0$
Using the quadratic formula, we get that the largest solution for $z$ is $z = 52 + 10\sqrt{2}$ . Then, repeatedly substituting backwards, we find that the largest value of $x$ is $11 + \sqrt{52 + \sqrt{200}}$ . The answer is thus $11 + 52 + 200 = \boxed{263}$

Answer:  263
Problem: In $\triangle ABC$ $AB = 3$ $BC = 4$ , and $CA = 5$ . Circle $\omega$ intersects $\overline{AB}$ at $E$ and $B$ $\overline{BC}$ at $B$ and $D$ , and $\overline{AC}$ at $F$ and $G$ . Given that $EF=DF$ and $\frac{DG}{EG} = \frac{3}{4}$ , length $DE=\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. Find $a+b+c$

Solution: Since $\angle DBE = 90^\circ$ $DE$ is the diameter of $\omega$ . Then $\angle DFE=\angle DGE=90^\circ$ . But $DF=FE$ , so $\triangle DEF$ is a 45-45-90 triangle. Letting $DG=3x$ , we have that $EG=4x$ $DE=5x$ , and $DF=EF=\frac{5x}{\sqrt{2}}$
Note that $\triangle DGE \sim \triangle ABC$ by SAS similarity, so $\angle BAC = \angle GDE$ and $\angle ACB = \angle DEG$ . Since $DEFG$ is a cyclic quadrilateral, $\angle BAC = \angle GDE=180^\circ-\angle EFG = \angle AFE$ and $\angle ACB = \angle DEG = \angle GFD$ , implying that $\triangle AFE$ and $\triangle CDF$ are isosceles. As a result, $AE=CD=\frac{5x}{\sqrt{2}}$ , so $BE=3-\frac{5x}{\sqrt{2}}$ and $BD =4-\frac{5x}{\sqrt{2}}$
Finally, using the Pythagorean Theorem on $\triangle BDE$ \[\left(3-\frac{5x}{\sqrt{2}}\right)^2 + \left(4-\frac{5x}{\sqrt{2}}\right)^2 = (5x)^2\] Solving for $x$ , we get that $x=\frac{5\sqrt{2}}{14}$ , so $DE= 5x =\frac{25 \sqrt{2}}{14}$ . Thus, the answer is $25+2+14=\boxed{041}$

Answer:  041
Problem: In $\triangle ABC$ $AB = 3$ $BC = 4$ , and $CA = 5$ . Circle $\omega$ intersects $\overline{AB}$ at $E$ and $B$ $\overline{BC}$ at $B$ and $D$ , and $\overline{AC}$ at $F$ and $G$ . Given that $EF=DF$ and $\frac{DG}{EG} = \frac{3}{4}$ , length $DE=\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. Find $a+b+c$

Solution: [asy] pair A = (0,3); pair B = (0,0); pair C = (4,0); draw(A--B--C--cycle); dotfactor = 3; dot("$A$",A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); pair D = (2.21, 0); pair E = (0, 1.21); pair F = (1.71, 1.71); pair G = (2, 1.5); dot("$D$",D,dir(270)); dot("$E$",E,dir(180)); dot("$F$",F,dir(90)); dot("$G$",G,dir(0)); draw(Circle((1.109, 0.609), 1.28)); draw(D--E); draw(E--F); draw(D--F); draw(E--G); draw(D--G); draw(B--F); draw(B--G); [/asy]
First we note that $\triangle DEF$ is an isosceles right triangle with hypotenuse $\overline{DE}$ the same as the diameter of $\omega$ . We also note that $\triangle DGE \sim \triangle ABC$ since $\angle EGD$ is a right angle and the ratios of the sides are $3:4:5$
From congruent arc intersections, we know that $\angle GED \cong \angle GBC$ , and that from similar triangles $\angle GED$ is also congruent to $\angle GCB$ . Thus, $\triangle BGC$ is an isosceles triangle with $BG = GC$ , so $G$ is the midpoint of $\overline{AC}$ and $AG = GC = 5/2$ . Similarly, we can find from angle chasing that $\angle ABF = \angle EDF = \frac{\pi}4$ . Therefore, $\overline{BF}$ is the angle bisector of $\angle B$ . From the angle bisector theorem, we have $\frac{AF}{AB} = \frac{CF}{CB}$ , so $AF = 15/7$ and $CF = 20/7$
Lastly, we apply power of a point from points $A$ and $C$ with respect to $\omega$ and have $AE \times AB=AF \times AG$ and $CD \times CB=CG \times CF$ , so we can compute that $EB = \frac{17}{14}$ and $DB = \frac{31}{14}$ . From the Pythagorean Theorem, we result in $DE = \frac{25 \sqrt{2}}{14}$ , so $a+b+c=25+2+14= \boxed{041}$

Answer:  041
Problem: In $\triangle ABC$ $AB = 3$ $BC = 4$ , and $CA = 5$ . Circle $\omega$ intersects $\overline{AB}$ at $E$ and $B$ $\overline{BC}$ at $B$ and $D$ , and $\overline{AC}$ at $F$ and $G$ . Given that $EF=DF$ and $\frac{DG}{EG} = \frac{3}{4}$ , length $DE=\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. Find $a+b+c$

Solution: Call $DE=x$ and as a result $DF=EF=\frac{x\sqrt{2}}{2}, EG=\frac{4x}{5}, GD=\frac{3x}{5}$ . Since $EFGD$ is cyclic we just need to get $DG$ and using LoS(for more detail see the $2$ nd paragraph of Solution $2$ ) we get $AG=\frac{5}{2}$ and using a similar argument(use LoS again) and subtracting you get $FG=\frac{5}{14}$ so you can use Ptolemy to get $x=\frac{25\sqrt{2}}{14} \implies \boxed{041}$ .
~First

Answer:  041
Problem: In $\triangle ABC$ $AB = 3$ $BC = 4$ , and $CA = 5$ . Circle $\omega$ intersects $\overline{AB}$ at $E$ and $B$ $\overline{BC}$ at $B$ and $D$ , and $\overline{AC}$ at $F$ and $G$ . Given that $EF=DF$ and $\frac{DG}{EG} = \frac{3}{4}$ , length $DE=\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. Find $a+b+c$

Solution: See inside the $\triangle DEF$ , we can find that $AG>AF$ since if $AG<AF$ , we can see that Ptolemy Theorem inside cyclic quadrilateral $EFGD$ doesn't work. Now let's see when $AG>AF$ , since $\frac{DG}{EG} = \frac{3}{4}$ , we can assume that $EG=4x;GD=3x;ED=5x$ , since we know $EF=FD$ so $\triangle EFD$ is isosceles right triangle. We can denote $DF=EF=\frac{5x\sqrt{2}}{2}$ .Applying Ptolemy Theorem inside the cyclic quadrilateral $EFGD$ we can get the length of $FG$ can be represented as $\frac{x\sqrt{2}}{2}$ . After observing, we can see $\angle AFE=\angle EDG$ , whereas $\angle A=\angle EDG$ so we can see $\triangle AEF$ is isosceles triangle. Since $\triangle ABC$ is a $3-4-5$ triangle so we can directly know that the length of AF can be written in the form of $3x\sqrt{2}$ . Denoting a point $J$ on side $AC$ with that $DJ$ is perpendicular to side $AC$ . Now with the same reason, we can see that $\triangle DJG$ is a isosceles right triangle, so we can get $GJ=\frac{3x\sqrt{2}}{2}$ while the segment $CJ$ is $2x\sqrt{2}$ since its 3-4-5 again. Now adding all those segments together we can find that $AC=5=7x\sqrt{2}$ and $x=\frac{5\sqrt{2}}{14}$ and the desired $ED=5x=\frac{25\sqrt{2}}{14}$ which our answer is $\boxed{041}$ ~bluesoul

Answer:  041
Problem: In $\triangle ABC$ $AB = 3$ $BC = 4$ , and $CA = 5$ . Circle $\omega$ intersects $\overline{AB}$ at $E$ and $B$ $\overline{BC}$ at $B$ and $D$ , and $\overline{AC}$ at $F$ and $G$ . Given that $EF=DF$ and $\frac{DG}{EG} = \frac{3}{4}$ , length $DE=\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. Find $a+b+c$

Solution: The main element of the solution is the proof that $BF$ is bisector of $\angle B.$
Let $O$ be the midpoint of $DE.$ $\angle EBF = 90^\circ \implies$
$O$ is the center of the circle $BDGFE.$ $\angle EOF = 90^\circ \implies \overset{\Large\frown} {EF} = 90^\circ \implies \angle EBF = 45^\circ \implies$ BF is bisector of $\angle ABC\implies BF = \frac {2AB \cdot BC}{AB+BC} \cos 45^\circ =\frac {12 \cdot \sqrt{2}}{7}.$ \[\angle EGD = 90^\circ, \frac {EG}{GD}=\frac{4}{3} \implies\] \[\angle GED = \angle GCD =\gamma \implies  \overset{\Large\frown} {DG} = 2\gamma.\] \[2\angle ACB =  \overset{\Large\frown} {BEF} -  \overset{\Large\frown} {DG} \implies  \overset{\Large\frown} {BEF} = 4 \gamma \implies\] \[\angle BOF = 4 \gamma \implies \angle OBF = \angle OFB = 90^\circ – 2 \gamma.\] Let $BO = EO = DO = r \implies BF = 2 r \cos(90^\circ – 2\gamma) =$ \[=2 r \sin 2\gamma = 4r \sin \gamma \cdot \cos \gamma = 4 r\cdot \frac {3}{5} \cdot \frac {4}{5} = \frac {48}{25} = \frac {12 \cdot \sqrt{2}}{7}\implies\] \[r = \frac {25 \cdot \sqrt{2}}{28}\implies ED = 2r =  \frac {25 \cdot \sqrt{2}}{14}\implies  \boxed{041}.\] vladimir.shelomovskii@gmail.com, vvsss

Answer:  041
Problem: In $\triangle ABC$ $AB = 3$ $BC = 4$ , and $CA = 5$ . Circle $\omega$ intersects $\overline{AB}$ at $E$ and $B$ $\overline{BC}$ at $B$ and $D$ , and $\overline{AC}$ at $F$ and $G$ . Given that $EF=DF$ and $\frac{DG}{EG} = \frac{3}{4}$ , length $DE=\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. Find $a+b+c$

Solution: The main element of the solution is the proof that $G$ is midpoint of $AC.$
As in Solution 5 we get $\angle GED = \angle DBG =\gamma \implies$
$\triangle BCG$ is isosceles triangle with $BG=CG.$
Similarly $BG = AG \implies AG = CG = BG = \frac {AC}{2} =\frac {5}{2}.$
\[\overset{\Large\frown} {FG} = 90^\circ – \overset{\Large\frown} {GD} =  90^\circ – 2\gamma \implies\] \[\overset{\Large\frown} {BFG} = 4\gamma + 90^\circ – 2\gamma = 90^\circ + 2\gamma \implies\] \[\angle BOG = 90^\circ + 2\gamma \implies \angle BGO = \angle GBO = 45^\circ - \gamma.\] Let $\hspace{10mm} BO = EO = DO = r \implies$ \[BG = 2 r \cos(45^\circ – \gamma) = 2 r (\sin \gamma + \cos \gamma)\frac {\sqrt {2}}{2} =\] \[r \biggl(\frac {3}{5} + \frac {4}{5}\biggr) \sqrt {2} = r \frac {7 \sqrt{2}}{5} = \frac {5}{2}\implies\] \[r = \frac {25 \cdot \sqrt{2}}{28}\implies ED = \frac {25 \cdot \sqrt{2}}{14}\implies  \boxed{041}.\] vladimir.shelomovskii@gmail.com, vvsss

Answer:  041
Problem: In $\triangle ABC$ $AB = 3$ $BC = 4$ , and $CA = 5$ . Circle $\omega$ intersects $\overline{AB}$ at $E$ and $B$ $\overline{BC}$ at $B$ and $D$ , and $\overline{AC}$ at $F$ and $G$ . Given that $EF=DF$ and $\frac{DG}{EG} = \frac{3}{4}$ , length $DE=\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. Find $a+b+c$

Solution: Let $(BEFGD) = \omega$ .
By Incenter-Excenter(Fact $5$ ), $F$ is the angle bisector of $\angle B$ .
Then by Ratio Lemma we have \[\frac{AG}{CG} = \frac{\sin(ABG)}{\sin(CBG)} \cdot \frac{AB}{BC} = \frac{\sin(GDE)}{\sin(DEG)} \cdot \frac{3}{4} = 1\] Thus, $G$ is the midpoint of $AC$
We can calculate $AF$ and $CF$ to be $\frac{15}{7}$ and $\frac{20}{7}$ respectively.
And then by Power of a Point, we have $\newline$ \[\operatorname{Pow}_{\omega}(A) = AE \cdot AB = AF \cdot AG \implies AE = \frac{25}{14}\] And then similarly, we have $CD = AE = \frac{25}{14}$ $\newline$
Then $EB = \frac{17}{14}$ and $DB = \frac{31}{14}$ and by Pythagorean we have $DE = \frac{25\sqrt{2}}{14}$ , so our answer is $\boxed{041}.$

Answer:  041