Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: A finite set $\mathcal{S}$ of distinct real numbers has the following properties: the mean of $\mathcal{S}\cup\{1\}$ is $13$ less than the mean of $\mathcal{S}$ , and the mean of $\mathcal{S}\cup\{2001\}$ is $27$ more than the mean of $\mathcal{S}$ . Find the mean of $\mathcal{S}$

Solution: Since this is a weighted average problem, the mean of $S$ is $\frac{13}{27}$ as far from $1$ as it is from $2001$ . Thus, the mean of $S$ is $1 + \frac{13}{13 + 27}(2001 - 1) = \boxed{651}$

Answer:  651
Problem: Find the sum of the roots , real and non-real, of the equation $x^{2001}+\left(\frac 12-x\right)^{2001}=0$ , given that there are no multiple roots.

Solution: From Vieta's formulas , in a polynomial of the form $a_nx^n + a_{n-1}x^{n-1} + \cdots + a_0 = 0$ , then the sum of the roots is $\frac{-a_{n-1}}{a_n}$
From the Binomial Theorem , the first term of $\left(\frac 12-x\right)^{2001}$ is $-x^{2001}$ , but $x^{2001}+-x^{2001}=0$ , so the term with the largest degree is $x^{2000}$ . So we need the coefficient of that term, as well as the coefficient of $x^{1999}$
\begin{align*}\binom{2001}{1} \cdot (-x)^{2000} \cdot \left(\frac{1}{2}\right)^1&=\frac{2001x^{2000}}{2}\\ \binom{2001}{2} \cdot (-x)^{1999} \cdot \left(\frac{1}{2}\right)^2 &=\frac{-x^{1999}*2001*2000}{8}=-2001 \cdot 250x^{1999} \end{align*}
Applying Vieta's formulas, we find that the sum of the roots is $-\frac{-2001 \cdot 250}{\frac{2001}{2}}=250 \cdot 2=\boxed{500}$

Answer:  500
Problem: Find the sum of the roots , real and non-real, of the equation $x^{2001}+\left(\frac 12-x\right)^{2001}=0$ , given that there are no multiple roots.

Solution: We find that the given equation has a $2000^{\text{th}}$ degree polynomial. Note that there are no multiple roots. Thus, if $\frac{1}{2} - x$ is a root, $x$ is also a root. Thus, we pair up $1000$ pairs of roots that sum to $\frac{1}{2}$ to get a sum of $\boxed{500}$

Answer:  500
Problem: Find the sum of the roots , real and non-real, of the equation $x^{2001}+\left(\frac 12-x\right)^{2001}=0$ , given that there are no multiple roots.

Solution: Note that if $r$ is a root, then $\frac{1}{2}-r$ is a root and they sum up to $\frac{1}{2}.$ We make the substitution $y=x-\frac{1}{4}$ so \[(\frac{1}{4}+y)^{2001}+(\frac{1}{4}-y)^{2001}=0.\] Expanding gives \[2\cdot\frac{1}{4}\cdot\binom{2001}{1}y^{2000}-0y^{1999}+\cdots\] so by Vieta, the sum of the roots of $y$ is 0.  Since $x$ has a degree of 2000, then $x$ has 2000 roots so the sum of the roots is \[2000(\sum_{n=1}^{2000} y+\frac{1}{4})=2000(0+\frac{1}{4})=\boxed{500}.\]

Answer:  500
Problem: In triangle $ABC$ , angles $A$ and $B$ measure $60$ degrees and $45$ degrees, respectively. The bisector of angle $A$ intersects $\overline{BC}$ at $T$ , and $AT=24$ . The area of triangle $ABC$ can be written in the form $a+b\sqrt{c}$ , where $a$ $b$ , and $c$ are positive integers, and $c$ is not divisible by the square of any prime. Find $a+b+c$

Solution: After chasing angles, $\angle ATC=75^{\circ}$ and $\angle TCA=75^{\circ}$ , meaning $\triangle TAC$ is an isosceles triangle and $AC=24$
Using law of sines on $\triangle ABC$ , we can create the following equation:
$\frac{24}{\sin(\angle ABC)}$ $=$ $\frac{BC}{\sin(\angle BAC)}$
$\angle ABC=45^{\circ}$ and $\angle BAC=60^{\circ}$ , so $BC = 12\sqrt{6}$
We can then use the Law of Sines area formula $\frac{1}{2} \cdot BC \cdot AC \cdot \sin(\angle BCA)$ to find the area of the triangle.
$\sin(75)$ can be found through the sin addition formula.
$\sin(75)$ $=$ $\frac{\sqrt{6} + \sqrt{2}}{4}$
Therefore, the area of the triangle is $\frac{\sqrt{6} + \sqrt{2}}{4}$ $\cdot$ $24$ $\cdot$ $12\sqrt{6}$ $\cdot$ $\frac{1}{2}$
$72\sqrt{3} + 216$
$72 + 3 + 216 =$ $\boxed{291}$

Answer:  291
Problem: In triangle $ABC$ , angles $A$ and $B$ measure $60$ degrees and $45$ degrees, respectively. The bisector of angle $A$ intersects $\overline{BC}$ at $T$ , and $AT=24$ . The area of triangle $ABC$ can be written in the form $a+b\sqrt{c}$ , where $a$ $b$ , and $c$ are positive integers, and $c$ is not divisible by the square of any prime. Find $a+b+c$

Solution: First, draw a good diagram.
We realize that $\angle C = 75^\circ$ , and $\angle CAT = 30^\circ$ .  Therefore, $\angle CTA = 75^\circ$ as well, making $\triangle CAT$ an isosceles triangle. $AT$ and $AC$ are congruent, so $AC=24$ . We now drop an altitude from $C$ , and call the foot this altitude point $D$
By 30-60-90 triangles, $AD=12$ and $CD=12\sqrt{3}$
We also notice that $\triangle CDB$ is an isosceles right triangle. $CD$ is congruent to $BD$ , which makes $BD=12\sqrt{3}$ .  The base $AB$ is $12+12\sqrt{3}$ , and the altitude $CD=12\sqrt{3}$ .  We can easily find that the area of triangle $ABC$ is $216+72\sqrt{3}$ , so $a+b+c=\boxed{291}$

Answer:  291
Problem: An equilateral triangle is inscribed in the ellipse whose equation is $x^2+4y^2=4$ . One vertex of the triangle is $(0,1)$ , one altitude is contained in the y-axis, and the square of the length of each side is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Denote the vertices of the triangle $A,B,$ and $C,$ where $B$ is in quadrant 4 and $C$ is in quadrant $3.$
Note that the slope of $\overline{AC}$ is $\tan 60^\circ = \sqrt {3}.$ Hence, the equation of the line containing $\overline{AC}$ is \[y = x\sqrt {3} + 1.\] This will intersect the ellipse when \begin{eqnarray*}4 = x^{2} + 4y^{2} & = & x^{2} + 4(x\sqrt {3} + 1)^{2} \\ & = & x^{2} + 4(3x^{2} + 2x\sqrt {3} + 1) \implies x(13x+8\sqrt 3)=0\implies x = \frac { - 8\sqrt {3}}{13}. \end{eqnarray*} We ignore the $x=0$ solution because it is not in quadrant 3.
Since the triangle is symmetric with respect to the y-axis, the coordinates of $B$ and $C$ are now $\left(\frac {8\sqrt {3}}{13},y_{0}\right)$ and $\left(\frac { - 8\sqrt {3}}{13},y_{0}\right),$ respectively, for some value of $y_{0}.$
It is clear that the value of $y_{0}$ is irrelevant to the length of $BC$ . Our answer is \[BC = 2*\frac {8\sqrt {3}}{13}=\sqrt {4\left(\frac {8\sqrt {3}}{13}\right)^{2}} = \sqrt {\frac {768}{169}}\implies m + n = \boxed{937}.\]

Answer:  937
Problem: An equilateral triangle is inscribed in the ellipse whose equation is $x^2+4y^2=4$ . One vertex of the triangle is $(0,1)$ , one altitude is contained in the y-axis, and the square of the length of each side is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Solving for $y$ in terms of $x$ gives $y=\sqrt{4-x^2}/2$ , so the two other points of the triangle are $(x,\sqrt{4-x^2}/2)$ and $(-x,\sqrt{4-x^2}/2)$ , which are a distance of $2x$ apart. Thus $2x$ equals the distance between $(x,\sqrt{4-x^2}/2)$ and $(0,1)$ , so by the distance formula we have
\[2x=\sqrt{x^2+(1-\sqrt{4-x^2}/2)^2}.\]
Squaring both sides and simplifying through algebra yields $x^2=192/169$ , so $2x=\sqrt{768/169}$ and the answer is $\boxed{937}$

Answer:  937
Problem: An equilateral triangle is inscribed in the ellipse whose equation is $x^2+4y^2=4$ . One vertex of the triangle is $(0,1)$ , one altitude is contained in the y-axis, and the square of the length of each side is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Since the altitude goes along the $y$ axis, this means that the base is a horizontal line, which means that the endpoints of the base are $(x,y)$ and $(-x,y)$ , and WLOG, we can say that $x$ is positive.
Now, since all sides of an equilateral triangle are the same, we can do this (distance from one of the endpoints of the base to the vertex and the length of the base):
$\sqrt{x^2 + (y-1)^2} = 2x$
Square both sides,
$x^2 + (y-1)^2 = 4x^2\implies (y-1)^2 = 3x^2$
Now, with the equation of the ellipse: $x^2 + 4y^2 = 4$
$x^2 = 4-4y^2$
$3x^2 = 12-12y^2$
Substituting,
$12-12y^2 = y^2 - 2y +1$
Moving stuff around and solving:
$y = \frac{-11}{13}, 1$
The second is found to be extraneous, so, when we go back and figure out $x$ and then $2x$ (which is the side length), we find it to be:
$\sqrt{\frac{768}{169}}$
and so we get the desired answer of $\boxed{937}$

Answer:  937
Problem: An equilateral triangle is inscribed in the ellipse whose equation is $x^2+4y^2=4$ . One vertex of the triangle is $(0,1)$ , one altitude is contained in the y-axis, and the square of the length of each side is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Denote $(0,1)$ as vertex $A,$ $B$ as the vertex to the left of the $y$ -axis and $C$ as the vertex to the right of the $y$ -axis. Let $D$ be the intersection of $BC$ and the $y$ -axis.
Let $x_0$ be the $x$ -coordinate of $C.$ This implies \[C=\left(x_0 , \sqrt{\frac{4-x_0^2}{4}}\right)\] and \[B=\left(-x_0 , \sqrt{\frac{4-x_0^2}{4}}\right).\] Note that $BC=2x_0$ and \[\frac{BC}{\sqrt3}=AD=1-\sqrt{\frac{4-x_0^2}{4}}.\] This yields \[\frac{2x_0}{\sqrt3}=1-\sqrt{\frac{4-x_0^2}{4}}.\] Re-arranging and squaring, we have \[\frac{4-x_0^2}{4}=\frac{4x_0^2}{3}-\frac{4x_0}{\sqrt3} +1.\] Simplifying and solving for $x_0$ , we have \[x_0=\frac{48}{13\sqrt 3}.\] As the length of each side is $2x_0,$ our desired length is \[4x_0^2=\frac{768}{169}\] which means our desired answer is \[768+169=\boxed{937}\]

Answer:  937
Problem: An equilateral triangle is inscribed in the ellipse whose equation is $x^2+4y^2=4$ . One vertex of the triangle is $(0,1)$ , one altitude is contained in the y-axis, and the square of the length of each side is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Notice that $x^2+4y^2=4$ can be rewritten as $(x)^2+(2y)^2=2^2$ . The points of the triangle are $(0, 1)$ $(-x, 1-x\sqrt{3})$ , and $(x, 1-x\sqrt{3})$ . When plugging the second coordinate into the equation, we get $x^2+4-8x\sqrt{3}+12x^2=4$ , which equals $13x^2-8x\sqrt{3}=0$ . 
This yields $x(13x-8\sqrt{3})=0$ . Obviously x can't be 0, so $x=\frac{8\sqrt{3}}{13}$ . The side length of the equilateral triangle is twice of this, so $\frac{16\sqrt{3}}{13}$ .
This can be rewritten as $\sqrt{\frac{256\cdot3}{169}}=\sqrt{\frac{768}{169}}$ $768+169=\boxed{937}$ .
~ MC413551

Answer:  937
Problem: An equilateral triangle is inscribed in the ellipse whose equation is $x^2+4y^2=4$ . One vertex of the triangle is $(0,1)$ , one altitude is contained in the y-axis, and the square of the length of each side is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Consider the transformation $(x,y)$ to $(x/2, y).$ This sends the ellipse to the unit circle. If we let $n$ be one-fourth of the side length of the triangle, the equilateral triangle is sent to an isosceles triangle with side lengths $2n, n\sqrt{13}, n\sqrt{13}.$ Let the triangle be $ABC$ such that $AB=AC.$ Let the foot of the altitude from A be $X.$ Then $BX=n,$ and $AX=2n\sqrt{3}.$ Let $C$ be a point such that $AC$ is a diameter of the unit circle. Then $XC=2-2n\sqrt{3}.$ Using power of a point on X, \[n^2=2n\sqrt{3}(2-2n\sqrt{3})\] Simplifying gets us to $13n^2=4n\sqrt{3}.$ Then, $n=\dfrac{4\sqrt{3}}{13}$ which means the side length is $\dfrac{16\sqrt{3}}{13}=\sqrt{\dfrac{768}{169}}.$ Thus, the answer is $768+169=\boxed{937}.$

Answer:  937
Problem: A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers . Find $m + n$

Solution: Recast the problem entirely as a block-walking problem.  Call the respective dice $a, b, c, d$ .  In the diagram below, the lowest $y$ -coordinate at each of $a$ $b$ $c$ , and $d$ corresponds to the value of the roll.
AIME01IN6.png
The red path corresponds to the sequence of rolls $2, 3, 5, 5$ .  This establishes a bijection between valid dice roll sequences and block walking paths.
The solution to this problem is therefore $\dfrac{\binom{9}{4}}{6^4} = {\dfrac{7}{72}}$ . So the answer is $\boxed{079}$

Answer:  079
Problem: A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers . Find $m + n$

Solution: Let $a, b, c,$ and $d$ be the results of rolling the four dice respectively. We have the range $1\leq a\leq b\leq c\leq d\leq 6$ , and there are $6^4=1296$ total outcomes from rolling the dice. To transfer the inequality into a strictly increasing inequality, we can transform it into $1\leq a<b+1<c+2<d+3\leq 9$ . Now, lets suppose $a'=a, b'=b+1, c'=c+2,$ and $d'=d+3$ . Then, $1\leq a'<b'<c'<d' \leq 9$ . Clearly, there are $\binom{9}{4}=126$ values that satisfy this equation, so our answer is $\dfrac{\binom{9}{4}}{6^4} = {\dfrac{7}{72}}\implies \boxed{079}$

Answer:  079
Problem: A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers . Find $m + n$

Solution: If we take any combination of four numbers, there is only one way to order them in a non-decreasing order. It suffices now to find the number of combinations for four numbers from $\{1,2,3,4,5,6\}$ . We can visualize this as taking the four dice and splitting them into 6 slots (each slot representing one of {1,2,3,4,5,6}), or dividing them amongst 5 separators. Thus, there are ${9\choose4} = 126$ outcomes of four dice. The solution is therefore $\frac{126}{6^4} = \frac{7}{72}$ , and $7 + 72 = \boxed{079}$

Answer:  079
Problem: A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers . Find $m + n$

Solution: Call the dice rolls $a, b, c, d$ . The difference between the $a$ and $d$ distinguishes the number of possible rolls there are.
Continuing, we see that the sum is equal to $\sum_{i = 0}^{5}{i+2\choose2}(6 - i) = 6 + 15 + 24 + 30 + 30 + 21 = 126$ . The requested probability is $\frac{126}{6^4} = \frac{7}{72}$ and our answer is $\boxed{79}$

Answer:  79
Problem: A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers . Find $m + n$

Solution: The dice rolls can be in the form
where A, B, C, D are some possible value of the dice rolls. (These forms are not keeping track of whether or not the dice are in ascending order, just the possible outcomes.)
There are a total of $6^4$ possible dice rolls.
Thus,
Therefore, our answer is $\boxed{079}$

Answer:  079
Problem: A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers . Find $m + n$

Solution: In a manner similar to the above solution, instead consider breaking it down into two sets of two dice rolls. The first subset must have a maximum value which is $\le$ the minimum value of the second subset.
Thus, the number of combinations is $\sum_{i=1}^6 i \cdot \left(\frac{(7 - i)(8 - i)}{2}\right) = 21 + 30 + 30 + 24 + 15 + 6 = 126$ , and the probability again is $\frac7{72}$ , giving $m+n=\boxed{079}$

Answer:  079
Problem: A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers . Find $m + n$

Solution: Lets try casework and observe the cases. Notice that if the last roll is a $1$ , then the only dice rolls may be $1-1-1-1$ , which is only $1$ possibility. Observe that if the last roll is $2$ , then there are $4 = 1 + 3$ possibilities. When the last roll is a $3$ , there are $10 = 1 + 3 + 6$ possibilities. Notice when the last roll is $n$ , the number of cases is the sum of the first $n$ positive triangular numbers. This is easily provable when observing the numbers of possibilities after assigning a value to the last and second to last rolls.
So there are a total of $1 + 1 + 3 + 1 + 3 + 6 + 1 + 3 + 6 + 10 + 1 + 3 + 6 + 10 + 15 + 1 + 3 + 6 + 10 + 15 + 21 = 126$ possibilities. So the probability is $\frac{126}{6^4} = \frac{7}{72}$ and $7 + 72 = \boxed{079}$ . ~skyscraper

Answer:  079
Problem: A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers . Find $m + n$

Solution: This is equivalent to picking a four-element sequence of $\{1, 2, 3, 4, 5, 6\}$ with repetition. Notice that once the sequence is picked, there is one and only one way to order these so that they form a sequence of rolls satisfying our conditions.
Now count the number of such four-element sequences, let $a$ be the number of $1$ s in the sequence, $b$ be the number of $2$ s, $c$ $3$ s, $d$ $4$ s, $e$ $5$ s, and $f$ $6$ s. Now we see that we must have \[a + b + c + d + e + f = 4\] with $a, b, c, d, e, f$ being nonnegative integers since there are a total of $4$ numbers picked. The number of solutions to this is $\dbinom{9}{4},$ so our total number is equal to $\dfrac{\binom{9}{4}}{6^4} = \dfrac{7}{72},$ making our answer $\boxed{079}.$

Answer:  079
Problem: A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers . Find $m + n$

Solution: Let the rolls be $a,b,c$ and $d\newline$ let $z=a-1, e=b-a, f=c-b, g=d-c, h=6-d\newline$ $z+e+f+g+h=5\newline$ This equation has $C(5+5-1, 5-1)=126$ integer solutions $\newline$ $126/1296=7/72\newline$ $7+72=\boxed{79}$ ~ryanbear

Answer:  79
Problem: A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers . Find $m + n$

Solution: Let's say the four dice values are all different. These can only be arranged in one way to satisfy our conditions, so there are $\binom{6}{4}=15$ ways. If there are three different values, there are $\binom{6}{3}$ to choose the numbers and $\binom{3}{1}$ to choose which number will have the repeat, so $20\cdot3=60$ ways. If there are two different values, there are $\binom{6}{2}$ ways to choose the numbers. If there are two of each, then there is one way. If there are three of one and one of another, then there are $\binom{2}{1}$ ways. Therefore $15\cdot(1+2)=45$ . Now if all the numbers are the same, there are $\binom{6}{1}=6$ ways. Altogether we have $15+60+45+6=126$ ways. $\frac{126}{6^4}=\frac{21}{216}=\frac{7}{72}$ $7+72=\boxed{79}$ . 
~MC413551

Answer:  79
Problem: Triangle $ABC$ has $AB=21$ $AC=22$ and $BC=20$ . Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$ , respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$ . Then $DE=m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Let $I$ be the incenter of $\triangle ABC$ , so that $BI$ and $CI$ are angle bisectors of $\angle ABC$ and $\angle ACB$ respectively. Then, $\angle BID = \angle CBI = \angle DBI,$ so $\triangle BDI$ is isosceles , and similarly $\triangle CEI$ is isosceles. It follows that $DE = DB + EC$ , so the perimeter of $\triangle ADE$ is $AD + AE + DE = AB + AC = 43$ . Hence, the ratio of the perimeters of $\triangle ADE$ and $\triangle ABC$ is $\frac{43}{63}$ , which is the scale factor between the two similar triangles, and thus $DE = \frac{43}{63} \times 20 = \frac{860}{63}$ . Thus, $m + n = \boxed{923}$

Answer:  923
Problem: Triangle $ABC$ has $AB=21$ $AC=22$ and $BC=20$ . Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$ , respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$ . Then $DE=m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: The semiperimeter of $ABC$ is $s = \frac{20 + 21 + 22}{2} = \frac{63}{2}$ . By Heron's formula , the area of the whole triangle is $A = \sqrt{s(s-a)(s-b)(s-c)} = \frac{21\sqrt{1311}}{4}$ . Using the formula $A = rs$ , we find that the inradius is $r = \frac{A}{s} = \frac{\sqrt{1311}}6$ . Since $\triangle ADE \sim \triangle ABC$ , the ratio of the heights of triangles $ADE$ and $ABC$ is equal to the ratio between sides $DE$ and $BC$ . From $A=\frac{1}{2}bh$ , we find $h_{ABC} = \frac{21\sqrt{1311}}{40}$ . Thus, we have
Solving for $DE$ gives $DE=\frac{860}{63},$ so the answer is $m+n=\boxed{923}$

Answer:  923
Problem: Triangle $ABC$ has $AB=21$ $AC=22$ and $BC=20$ . Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$ , respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$ . Then $DE=m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Let $P$ be the incenter ; then it is be the intersection of all three angle bisectors . Draw the bisector $AP$ to where it intersects $BC$ , and name the intersection $F$
Using the angle bisector theorem , we know the ratio $BF:CF$ is $21:22$ , thus we shall assign a weight of $22$ to point $B$ and a weight of $21$ to point $C$ , giving $F$ a weight of $43$ . In the same manner, using another bisector, we find that $A$ has a weight of $20$ . So, now we know $P$ has a weight of $63$ , and the ratio of $FP:PA$ is $20:43$ . Therefore, the smaller similar triangle $ADE$ is $43/63$ the height of the original triangle $ABC$ . So, $DE$ is $43/63$ the size of $BC$ . Multiplying this ratio by the length of $BC$ , we find $DE$ is $860/63 = m/n$ . Therefore, $m+n=\boxed{923}$

Answer:  923
Problem: Triangle $ABC$ has $AB=21$ $AC=22$ and $BC=20$ . Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$ , respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$ . Then $DE=m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: More directly than Solution 2, we have \[DE=BC\left(\frac{h_a-r}{h_a}\right)=20\left(1-\frac{r}{\frac{[ABC]}{\frac{BC}{2}}}\right)=20\left(1-\frac{10r}{sr}\right)=20\left(1-\frac{10}{\frac{63}{2}}\right)=\frac{860}{63}\implies \boxed{923}.\]

Answer:  923
Problem: Triangle $ABC$ has $AB=21$ $AC=22$ and $BC=20$ . Points $D$ and $E$ are located on $\overline{AB}$ and $\overline{AC}$ , respectively, such that $\overline{DE}$ is parallel to $\overline{BC}$ and contains the center of the inscribed circle of triangle $ABC$ . Then $DE=m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Diagram borrowed from Solution 3.
Let the angle bisector of $\angle{A}$ intersects $BC$ at $F$
Applying the Angle Bisector Theorem on $\angle{A}$ we have \[\frac{AB}{BF}=\frac{AC}{CF}\] \[BF=BC\cdot(\frac{AB}{AB+AC})\] \[BF=\frac{420}{43}\] Since $BP$ is the angle bisector of $\angle{B}$ , we can once again apply the Angle Bisector Theorem on $\angle{B}$ which gives \[\frac{BA}{AP}=\frac{BF}{FP}\] \[\frac{AP}{PF}=\frac{AB}{BF}=\frac{41}{20}\] Since $\bigtriangleup ADE\sim\bigtriangleup ABC$ we have \[\frac{DE}{BC}=\frac{AP}{AF}\] \[DE=BC\cdot(\frac{AP}{(\frac{61}{41})\cdot AP})\] Solving gets $DE=\frac{860}{63}$ . Thus $m+n=860+63=\boxed{923}$

Answer:  923