Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Two long cylindrical tubes of the same length but different diameters lie parallel to each other on a flat surface . The larger tube has radius $72$ and rolls along the surface toward the smaller tube, which has radius $24$ . It rolls over the smaller tube and continues rolling along the flat surface until it comes to rest on the same point of its circumference as it started, having made one complete revolution. If the smaller tube never moves, and the rolling occurs with no slipping, the larger tube ends up a distance $x$ from where it starts. The distance $x$ can be expressed in the form $a\pi+b\sqrt{c},$ where $a,$ $b,$ and $c$ are integers and $c$ is not divisible by the square of any prime . Find $a+b+c.$

Solution: 2007 AIME II-11.png
If it weren’t for the small tube, the larger tube would travel $144\pi$ . Consider the distance from which the larger tube first contacts the smaller tube, until when it completely loses contact with the smaller tube.
Drawing the radii as shown in the diagram, notice that the hypotenuse of the right triangle in the diagram has a length of $72 + 24 = 96$ . The horizontal line divides the radius of the larger circle into $72 - 24 = 48$ on the top half, which indicates that the right triangle has leg of 48 and hypotenuse of 96, a $30-60-90 \triangle$
Find the length of the purple arc in the diagram (the distance the tube rolled, but not the horizontal distance). The sixty degree central angle indicates to take $\frac{60}{360} = \frac 16$ of the circumference of the larger circle (twice), while the $180 - 2(30) = 120^{\circ}$ central angle in the smaller circle indicates to take $\frac{120}{360} = \frac 13$ of the circumference. This adds up to $2 \cdot \frac 16 144\pi + \frac 13 48\pi = 64\pi$
The actual horizontal distance it takes can be found by using the $30-60-90 \triangle$ s. The missing leg is equal in length to $48\sqrt{3}$ . Thus, the total horizontal distance covered is $96\sqrt{3}$
Thus, we get $144\pi - 64\pi + 96\sqrt{3} = 80\pi + 96\sqrt{3}$ , and our answer is $\boxed{179}$

Answer:  179
Problem: The increasing geometric sequence $x_{0},x_{1},x_{2},\ldots$ consists entirely of integral powers of $3.$ Given that
$\sum_{n=0}^{7}\log_{3}(x_{n}) = 308$ and $56 \leq \log_{3}\left ( \sum_{n=0}^{7}x_{n}\right ) \leq 57,$
find $\log_{3}(x_{14}).$

Solution: Suppose that $x_0 = a$ , and that the common ratio between the terms is $r$
The first conditions tells us that $\log_3 a + \log_3 ar + \ldots + \log_3 ar^7 = 308$ . Using the rules of logarithms , we can simplify that to $\log_3 a^8r^{1 + 2 + \ldots + 7} = 308$ . Thus, $a^8r^{28} = 3^{308}$ . Since all of the terms of the geometric sequence are integral powers of $3$ , we know that both $a$ and $r$ must be powers of 3. Denote $3^x = a$ and $3^y = r$ . We find that $8x + 28y = 308$ . The possible positive integral pairs of $(x,y)$ are $(35,1),\ (28,3),\ (21,5),\ (14,7),\ (7,9),\ (0,11)$
The second condition tells us that $56 \le \log_3 (a + ar + \ldots ar^7) \le 57$ . Using the sum formula for a geometric series and substituting $x$ and $y$ , this simplifies to $3^{56} \le 3^x \frac{3^{8y} - 1}{3^y-1} \le 3^{57}$ . The fractional part $\approx \frac{3^{8y}}{3^y} = 3^{7y}$ . Thus, we need $\approx 56 \le x + 7y \le 57$ . Checking the pairs above, only $(21,5)$ is close.
Our solution is therefore $\log_3 (ar^{14}) = \log_3 3^x + \log_3 3^{14y} = x + 14y = \boxed{091}$

Answer:  091
Problem: The increasing geometric sequence $x_{0},x_{1},x_{2},\ldots$ consists entirely of integral powers of $3.$ Given that
$\sum_{n=0}^{7}\log_{3}(x_{n}) = 308$ and $56 \leq \log_{3}\left ( \sum_{n=0}^{7}x_{n}\right ) \leq 57,$
find $\log_{3}(x_{14}).$

Solution: All these integral powers of $3$ are all different, thus in base $3$ the sum of these powers would consist of $1$ s and $0$ s. Thus the largest value $x_7$ must be $3^{56}$ in order to preserve the givens. Then we find by the given that $x_7x_6x_5\dots x_0 = 3^{308}$ , and we know that the exponents of $x_i$ are in an arithmetic sequence. Thus $56+(56-r)+(56-2r)+\dots +(56-7r) = 308$ , and $r = 5$ . Thus $\log_3 (x_{14}) = \boxed{091}$

Answer:  091
Problem: The increasing geometric sequence $x_{0},x_{1},x_{2},\ldots$ consists entirely of integral powers of $3.$ Given that
$\sum_{n=0}^{7}\log_{3}(x_{n}) = 308$ and $56 \leq \log_{3}\left ( \sum_{n=0}^{7}x_{n}\right ) \leq 57,$
find $\log_{3}(x_{14}).$

Solution: Like above, we use logarithmic identities to convert the problem into workable equations. We begin by labelling the powers of 3. Call $x_0$ $x_1$ $x_2$ ..., as $3^n$ $3^{n+m}$ , and $3^{n+2m}$ ... respectively. With this format we can rewrite the first given equation as $n + n + m + n+2m + n+3m+...+n+7m = 308$ . Simplify to get $2n+7m=77$ . (1)
Now, rewrite the second given equation as $3^{56} \leq \left( \sum_{n=0}^{7}x_{n} \right) \leq 3^{57}$ . Obviously, $x_7$ , aka $3^{n+7m}$ $<3^{57}$ because there are some small fractional change that is left over. This means $n+7m$ is $\leq56$ . Thinking about the geometric sequence, it's clear each consecutive value of $x_i$ will be either a power of three times smaller or larger. In other words, the earliest values of $x_i$ will be negligible compared to the last values of $x_i$ . Even in the best case scenario, where the common ratio is 3, the values left of $x_7$ are not enough to sum to a value greater than 2 times $x_7$ (amount needed to raise the power of 3 by 1). This confirms that $3^{n+7m} = 3^{56}$ . (2)
Use equations 1 and 2 to get $m=5$ and $n=21$ $\log_{3}{x_{14}} = \log_{3}{3^{21+14*5}} = 21+14*5 = \boxed{091}$

Answer:  091
Problem: The increasing geometric sequence $x_{0},x_{1},x_{2},\ldots$ consists entirely of integral powers of $3.$ Given that
$\sum_{n=0}^{7}\log_{3}(x_{n}) = 308$ and $56 \leq \log_{3}\left ( \sum_{n=0}^{7}x_{n}\right ) \leq 57,$
find $\log_{3}(x_{14}).$

Solution: Proceed as in Solution 3 for the first few steps. We have the sequence $3^{a},3^{a+n},3^{a+2n}...$ . As stated above, we then get that $a+a+n+...+a+7n=308$ , from which we simplify to $2a+7n=77$ . From here, we just go brute force using the second statement (that $3^{56}\leq 3^{a}+...+3^{a+7n}\leq 3^{57}$ ). Rearranging the equation from earlier, we get \[n=11-\frac{2a}{7}\] from which it is clear that $a$ is a multiple of $7$ . Testing the first few values of $a$ , we get:
Case 1 ( $a=7,n=9$ )
The sequence is then $3^{7}+...+3^{70}$ , which breaks the upper bound.
Case 2 ( $a=14,n=7$ )
The sequence is then $3^{14}+...+3^{63}$ , which also breaks the upper bound.
Case 3 ( $a=21,n=5$ )
This is the first reasonable one, giving $3^{21}+...+3^{56}$ . It seems like this would break the upper bound, but from some testing we get: \[3^{21}+...+3^{56}<3^{57}\] \[1+...+3^{35}<3^{36}\] \[1+...+3^{30}<2*3^{35}\] \[3^{5}+...+3^{30}<2*3^{35}-1\] \[1+...+3^{25}<2*3^{30}-3^{-5}\] Repeating this process over and over, we eventually get (while ignoring the extremely small fractions left over) \[1<2*3^{5}\] which confirms that this satisfies our upper bound. Thus $a=21,n=5$ , so $x_14=3^{a+14n}\rightarrow3^{91}$ . We then get the requested answer, $\log_3(3^{91})=\boxed{091}$ ~ amcrunner

Answer:  091
Problem: triangular array of squares has one square in the first row, two in the second, and in general, $k$ squares in the $k$ th row for $1 \leq k \leq 11.$ With the exception of the bottom row, each square rests on two squares in the row immediately below (illustrated in the given diagram). In each square of the eleventh row, a $0$ or a $1$ is placed. Numbers are then placed into the other squares, with the entry for each square being the sum of the entries in the two squares below it. For how many initial distributions of $0$ 's and $1$ 's in the bottom row is the number in the top square a multiple of $3$
[asy] for (int i=0; i<12; ++i){  for (int j=0; j<i; ++j){    //dot((-j+i/2,-i));    draw((-j+i/2,-i)--(-j+i/2+1,-i)--(-j+i/2+1,-i+1)--(-j+i/2,-i+1)--cycle);  } } [/asy]

Solution: Label each of the bottom squares as $x_0, x_1 \ldots x_9, x_{10}$
Through induction , we can find that the top square is equal to ${10\choose0}x_0 + {10\choose1}x_1 + {10\choose2}x_2 + \ldots {10\choose10}x_{10}$ . (This also makes sense based on a combinatorial argument: the number of ways a number can "travel" to the top position going only up is equal to the number of times it will be counted in the final sum.)
Examine the equation $\mod 3$ . All of the coefficients from $x_2 \ldots x_8$ will be multiples of $3$ (since the numerator will have a $9$ ). Thus, the expression boils down to $x_0 + 10x_1 + 10x_9 + x_{10} \equiv 0 \mod 3$ . Reduce to find that $x_0 + x_1 + x_9 + x_{10} \equiv 0 \mod 3$ . Out of $x_0,\ x_1,\ x_9,\ x_{10}$ , either all are equal to $0$ , or three of them are equal to $1$ . This gives ${4\choose0} + {4\choose3} = 1 + 4 = 5$ possible combinations of numbers that work.
The seven terms from $x_2 \ldots x_8$ can assume either $0$ or $1$ , giving us $2^7$ possibilities. The answer is therefore $5 \cdot 2^7 = \boxed{640}$

Answer:  640
Problem: Let $f(x)$ be a polynomial with real coefficients such that $f(0) = 1,$ $f(2)+f(3)=125,$ and for all $x$ $f(x)f(2x^{2})=f(2x^{3}+x).$ Find $f(5).$

Solution: If the leading term of $f(x)$ is $ax^m$ , then the leading term of $f(x)f(2x^2) = ax^m \cdot a(2x^2)^m = 2^ma^2x^{3m}$ , and the leading term of $f(2x^3 + x) = 2^max^{3m}$ . Hence $2^ma^2 = 2^ma$ , and $a = 1$ . Because $f(0) = 1$ , the product of all the roots
of $f(x)$ is $\pm 1$ . If $f(\lambda) = 0$ , then $f(2\lambda^3 + \lambda) = 0$ . Assume that there exists a root $\lambda$ with $|\lambda| \neq 1$ . Then there must be such a root $\lambda_1$ with $|\lambda_1| > 1$ . Then \[|2\lambda^3 + \lambda| \ge 2|\lambda|^3 - |\lambda| > 2|\lambda| - |\lambda| = |\lambda|.\] But then $f(x)$ would have infinitely many roots, given by $\lambda_{k+1} = 2\lambda_k^3 + \lambda_k$ , for $k \ge 1$ . 
Therefore $|\lambda| = 1$ for all of the roots of the polynomial. Thus $\lambda\overline{\lambda}=1$ , and $(2\lambda^3 + \lambda)\overline{(2\lambda^3 + \lambda)} = 1$ .
Solving these equations simultaneously for $\lambda = a + bi$ yields $a = 0$ $b^2 = 1$ , and so $\lambda^2 = -1$ . Because the polynomial has real coefficients, the polynomial must have the form $f(x) = (1 + x^2)^n$ for some integer $n \ge 1$ . The condition $f(2) + f(3) = 125$ implies $n = 2$ , giving $f(5) = \boxed{676}$

Answer:  676
Problem: Let $f(x)$ be a polynomial with real coefficients such that $f(0) = 1,$ $f(2)+f(3)=125,$ and for all $x$ $f(x)f(2x^{2})=f(2x^{3}+x).$ Find $f(5).$

Solution: Let $r$ be a root of $f(x)$ . Then we have $f(r)f(2r^2)=f(2r^3+r)$ ; since $r$ is a root, we have $f(r)=0$ ; therefore $2r^3+r$ is also a root. Thus, if $r$ is real and non-zero, $|2r^3+r|>r$ , so $f(x)$ has infinitely many roots. Since $f(x)$ is a polynomial (thus of finite degree) and $f(0)$ is nonzero, $f(x)$ has no real roots.
Note that $f(x)$ is not constant. We then find two complex roots: $r = \pm i$ . We find that $f(i)f(-2) = f(-i)$ , and that $f(-i)f(-2) = f(i)$ . This means that $f(i)f(-i)f(-2)^2 = f(i)f(-i) \Longrightarrow f(i)f(-i)(f(-2)^2 - 1) = 0$ . Thus, $\pm i$ are roots of the polynomial, and so $(x - i)(x + i) = x^2 + 1$ will be a factor of the polynomial. (Note: This requires the assumption that $f(-2)\neq1$ . Clearly, $f(-2)\neq-1$ , because that would imply the existence of a real root.)
The polynomial is thus in the form of $f(x) = (x^2 + 1)g(x)$ . Substituting into the given expression, we have
\[(x^2+1)g(x)(4x^4+1)g(2x^2)=((2x^3+x)^2+1)g(2x^3+x)\] \[(4x^6+4x^4+x^2+1)g(x)g(2x^2)=(4x^6+4x^4+x^2+1)g(2x^3+x)\]
Thus either $4x^6+4x^4+x^2+1=(4x^4+1)(x^2+1)$ is 0 for any $x$ , or $g(x)$ satisfies the same constraints as $f(x)$ . Continuing, by infinite descent, $f(x) = (x^2 + 1)^n$ for some $n$
Since $f(2)+f(3)=125=5^n+10^n$ for some $n$ , we have $n=2$ ; so $f(5) = \boxed{676}$

Answer:  676
Problem: Let $f(x)$ be a polynomial with real coefficients such that $f(0) = 1,$ $f(2)+f(3)=125,$ and for all $x$ $f(x)f(2x^{2})=f(2x^{3}+x).$ Find $f(5).$

Solution: Let $r$ be a root of $f(x).$ This means that $f(r)f(2r^2)=f(2r^3+r).$ In other words, $2r^3+r$ is a root of $f(x)$ too. Since $f(x)$ can't have infinitely many roots, \[Q(x)=P(P(\dotsb P(P(r)) \dotsb))\] is cyclic, where $P(x)=2x^3+x.$ Now, we will do casework.
Case 1: $\deg f\geq1$
Subcase 1: $|r|>1$
This means that \[|2r^3+r|\geq|2r^3|-|r|=|r|(2|r|^2-1)>|r|(2\cdot1^2-1)=|r|.\] It follows that $|2r^3+r|>|r|$ for all $r.$ This implies that $Q(r)$ can't be cyclic. Thus, it is impossible for $|r|>1$ to be true.
Subcase 2: $|r|<1$
This means that $|2r^3+r|\geq2|r^3|-|r|=|r|(|2r^2|-1)<|r|.$ It follows that $|2r^3+r|<|r|$ for all $r.$ This implies that $Q(r)$ can't be cyclic. Thus, it is impossible for $|r|>1$ to be true.
Subcase 3: $|r|=1.$
Since $|r|$ is not greater than or less than 1, $|r|=1.$ This means that all the roots of the polynomial have a magnitude of $1.$ More specifically, $|2r^3+r|$ has a magnitude of one. Since this would mean an equality condition from the triangle inequality, $2r^3$ and $r$ are collinear with the origin in the complex plane. In other words, $\frac{2r^3}{r}=\pm c\Leftrightarrow cr=2r^3\Leftrightarrow 2r^2=c\Leftrightarrow r=\pm\sqrt{\pm\frac{c}{2}},$ for some real constant $c.$ Now, from $|r|=1,$ we find that $\left|\pm\sqrt{\pm\frac{c}{2}}\right|=1\Leftrightarrow \sqrt{\pm\frac{c}{2}}=1\Leftrightarrow \pm\frac{c}{2}=1\Leftrightarrow c=\pm2.$ Putting this back into the equation, we find that $r=1,-1,i,-i.$ Now, this means that $2r^3+r=3,-3,i,-i.$ $3$ and $-3$ obviously doesn't have a magnitude of $1.$ Thus, $i,-i$ are the only possible roots of the polynomial. Since roots come in conjugate pairs, $f(x)=[(x-i)(x+i)]^n=(x^2+1)^n,$ works for all constants $n\neq0.$
Case 2: $\deg f=0.$
This means that $f(x)=c,$ for some constant $c.$ In other words, $c^2=c.$ We can easily find that this means that $c=0,1.$ Combining all the cases, we conclude that $f(x)=(x^2+1)^n,0,1$ are the only polynomials that satisfy this equation.
Now, we can test! $f(x)=0,1$ obviously don't satisfy $f(2)+f(3)=125.$ Thus, $f(x)=(x^2+1)^n.$ Substituting, we find that $5^n+10^n=125\Leftrightarrow n=2.$ We conclude that $f(5)=(5^2+1)^2=26^2=\boxed{676}.$

Answer:  676
Problem: Four circles $\omega,$ $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ with the same radius are drawn in the interior of triangle $ABC$ such that $\omega_{A}$ is tangent to sides $AB$ and $AC$ $\omega_{B}$ to $BC$ and $BA$ $\omega_{C}$ to $CA$ and $CB$ , and $\omega$ is externally tangent to $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ . If the sides of triangle $ABC$ are $13,$ $14,$ and $15,$ the radius of $\omega$ can be represented in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: [asy] defaultpen(fontsize(12)+0.8); size(350);  pair A,B,C,X,Y,Z,P,Q,R,Zp; B=origin; C=15*right; A=IP(CR(B,13),CR(C,14)); P=incenter(A,B,C); real r=260/129; Q=r*(rotate(-90)*A/length(A)); X=extension(A,P,Q,Q+A); Y=extension(B,P,Q,Q+A); R=rotate(90,C)*(C+r*(A-C)/length(A-C)); Z=extension(C,P,R,R+A-C); Zp=circumcenter(X,Y,Z); draw(A--B--C--A); draw(CR(X,r)^^CR(Y,r)^^CR(Z,r), gray); draw(A--P--B^^P--C^^X--Y--Z--X, royalblue); draw(X--foot(X,A,C)^^Z--foot(Z,A,C),royalblue); draw(CR(Zp,r), gray); draw(incircle(A,B,C)^^incircle(X,Y,Z)^^P--foot(P,A,C), heavygreen+0.6); draw(rightanglemark(X,foot(X,A,C),C,10),linewidth(0.6)); draw(rightanglemark(Z,foot(Z,A,C),A,10),linewidth(0.6)); label("$A$",A,N); label("$B$",B,0.15*(B-P)); label("$C$",C,0.1*(C-P)); pen p=fontsize(10)+linewidth(3); dot("$O_A$",X,right,p); dot("$O_B$",Y,left+up,p); dot("$O_C$",Z,down,p); dot("$O$",Zp,dir(-45),p+red); dot("$I$",P,left+up,p); [/asy]
First, apply Heron's formula to find that $[ABC] = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84$ . The semiperimeter is $21$ , so the inradius is $\frac{A}{s} = \frac{84}{21} = 4$
Now consider the incenter $I$ of $\triangle ABC$ . Let the radius of one of the small circles be $r$ . Let the centers of the three little circles tangent to the sides of $\triangle ABC$ be $O_A$ $O_B$ , and $O_C$ . Let the center of the circle tangent to those three circles be $O$ . The homothety $\mathcal{H}\left(I, \frac{4-r}{4}\right)$ maps $\triangle ABC$ to $\triangle XYZ$ ; since $OO_A = OO_B = OO_C = 2r$ $O$ is the circumcenter of $\triangle XYZ$ and $\mathcal{H}$ therefore maps the circumcenter of $\triangle ABC$ to $O$ . Thus, $2r = R \cdot \frac{4 - r}{4}$ , where $R$ is the circumradius of $\triangle ABC$ . Substituting $R = \frac{abc}{4[ABC]} = \frac{65}{8}$ $r = \frac{260}{129}$ and the answer is $\boxed{389}$

Answer:  389
Problem: Four circles $\omega,$ $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ with the same radius are drawn in the interior of triangle $ABC$ such that $\omega_{A}$ is tangent to sides $AB$ and $AC$ $\omega_{B}$ to $BC$ and $BA$ $\omega_{C}$ to $CA$ and $CB$ , and $\omega$ is externally tangent to $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ . If the sides of triangle $ABC$ are $13,$ $14,$ and $15,$ the radius of $\omega$ can be represented in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: 2007 AIME II-15b.gif
Consider a 13-14-15 triangle. $A=84.$ [By Heron's Formula or by 5-12-13 and 9-12-15 right triangles.]
The inradius is $r=\frac{A}{s}=\frac{84}{21}=4$ , where $s$ is the semiperimeter.  Scale the triangle with the inradius by a linear scale factor, $u.$
The circumradius is $R=\frac{abc}{4rs}=\frac{13\cdot 14\cdot 15}{4\cdot 4\cdot 21}=\frac{65}{8},$ where $a,$ $b,$ and $c$ are the side-lengths.  Scale the triangle with the circumradius by a linear scale factor, $v$
Cut and combine the triangles, as shown.  Then solve for $4u$
The solution is $260+129=\boxed{389}$

Answer:  389
Problem: Four circles $\omega,$ $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ with the same radius are drawn in the interior of triangle $ABC$ such that $\omega_{A}$ is tangent to sides $AB$ and $AC$ $\omega_{B}$ to $BC$ and $BA$ $\omega_{C}$ to $CA$ and $CB$ , and $\omega$ is externally tangent to $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ . If the sides of triangle $ABC$ are $13,$ $14,$ and $15,$ the radius of $\omega$ can be represented in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: Let $A'$ $B'$ $C'$ , and $O$ be the centers of circles $\omega_{A}$ $\omega_{B}$ $\omega_{C}$ $\omega$ , respectively, and let $x$ be their radius.
Now, triangles $ABC$ and $A'B'C'$ are similar by parallel sides, so we can find ratios of two quantities in each triangle and set them equal to solve for $x$
Since $OA'=OB'=OC'=2x$ $O$ is the circumcenter of triangle $A'B'C'$ and its circumradius is $2x$ . Let $I$ denote the incenter of triangle $ABC$ and $r$ the inradius of $ABC$ . Then the inradius of $A'B'C'=r-x$ , so now we compute r. Computing the inradius by $A=rs$ , we find that the inradius of $ABC$ is $4$ . Additionally, using the circumradius formula $R=\frac{abc}{4K}$ where $K$ is the area of $ABC$ and $R$ is the circumradius, we find $R=\frac{65}{8}$ . Now we can equate the ratio of circumradius to inradius in triangles $ABC$ and $A'B'C'$
\[\frac{\frac{65}{8}}{4}=\frac{2x}{4-x}\]
Solving, we get $x=\frac{260}{129}$ , so our answer is $260+129=\boxed{389}$

Answer:  389
Problem: Four circles $\omega,$ $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ with the same radius are drawn in the interior of triangle $ABC$ such that $\omega_{A}$ is tangent to sides $AB$ and $AC$ $\omega_{B}$ to $BC$ and $BA$ $\omega_{C}$ to $CA$ and $CB$ , and $\omega$ is externally tangent to $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ . If the sides of triangle $ABC$ are $13,$ $14,$ and $15,$ the radius of $\omega$ can be represented in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: According to the diagram, it is easily to see that there is a small triangle made by the center of three circles which aren't in the middle. The circumradius of them is $2r$ . Now denoting $AB=13;BC=14;AC=15$ , and centers of circles tangent to $AB,AC;AC,BC;AB,BC$ are relatively $M,N,O$ with $OJ,NK$ both perpendicular to $BC$ . It is easy to know that $tanB=\frac{12}{5}$ , so $tan\angle OBJ=\frac{2}{3}$ according to half angle formula. Similarly, we can find $tan\angle NCK=\frac{1}{2}$ . So we can see that $JK=ON=14-\frac{7x}{2}$ . Obviously, $\frac{2x}{14-\frac{7x}{2}}=\frac{65}{112}$ . After solving, we get $x=\frac{260}{129}$ , so our answer is $260+129=\boxed{389}$ . ~bluesoul

Answer:  389
Problem: In quadrilateral $ABCD$ $\angle B$ is a right angle diagonal $\overline{AC}$ is perpendicular to $\overline{CD}$ $AB=18$ $BC=21$ , and $CD=14$ . Find the perimeter of $ABCD$

Solution: From the problem statement, we construct the following diagram:
Using the Pythagorean Theorem
Substituting $(AB)^2 + (BC)^2$ for $(AC)^2$
Plugging in the given information:
So the perimeter is $18+21+14+31=84$ , and the answer is $\boxed{084}$

Answer:  084
Problem: Let set $\mathcal{A}$ be a 90- element subset of $\{1,2,3,\ldots,100\},$ and let $S$ be the sum of the elements of $\mathcal{A}.$ Find the number of possible values of $S.$

Solution: The smallest $S$ is $1+2+ \ldots +90 = 91 \cdot 45 = 4095$ . The largest $S$ is $11+12+ \ldots +100=111\cdot 45=4995$ . All numbers between $4095$ and $4995$ are possible values of S, so the number of possible values of S is $4995-4095+1=901$
Alternatively, for ease of calculation, let set $\mathcal{B}$ be a 10-element subset of $\{1,2,3,\ldots,100\}$ , and let $T$ be the sum of the elements of $\mathcal{B}$ . Note that the number of possible $S$ is the number of possible $T=5050-S$ . The smallest possible $T$ is $1+2+ \ldots +10 = 55$ and the largest is $91+92+ \ldots + 100 = 955$ , so the number of possible values of T, and therefore S, is $955-55+1=\boxed{901}$

Answer:  901
Problem: Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is $\frac{1}{29}$ of the original integer.

Solution: Suppose the original number is $N = \overline{a_na_{n-1}\ldots a_1a_0},$ where the $a_i$ are digits and the first digit, $a_n,$ is nonzero. Then the number we create is $N_0 = \overline{a_{n-1}\ldots a_1a_0},$ so \[N = 29N_0.\] But $N$ is $N_0$ with the digit $a_n$ added to the left, so $N = N_0 + a_n \cdot 10^n.$ Thus, \[N_0 + a_n\cdot 10^n = 29N_0\] \[a_n \cdot 10^n = 28N_0.\] The right-hand side of this equation is divisible by seven, so the left-hand side must also be divisible by seven. The number $10^n$ is never divisible by $7,$ so $a_n$ must be divisible by $7.$ But $a_n$ is a nonzero digit, so the only possibility is $a_n = 7.$ This gives \[7 \cdot 10^n = 28N_0\] or \[10^n = 4N_0.\] Now, we want to minimize both $n$ and $N_0,$ so we take $N_0 = 25$ and $n = 2.$ Then \[N = 7 \cdot 10^2 + 25 = \boxed{725},\] and indeed, $725 = 29 \cdot 25.$ $\square$

Answer:  725
Problem: Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is $\frac{1}{29}$ of the original integer.

Solution: Let $N$ be the required number, and $N'$ be $N$ with the first digit deleted. Now, we know that $N<1000$ (because this is an AIME problem). Thus, $N$ has $1,$ $2$ or $3$ digits. Checking the other cases, we see that it must have $3$ digits.
Let $N=\overline{abc}$ , so $N=100a+10b+c$ . Thus, $N'=\overline{bc}=10b+c$ . By the constraints of the problem, we see that $N=29N'$ , so \[100a+10b+c=29(10b+c).\] Now, we subtract and divide to get \[100a=28(10b+c)\] \[25a=70b+7c.\] Clearly, $c$ must be a multiple of $5$ because both $25a$ and $70b$ are multiples of $5$ . Thus, $c=5$ . Now, we plug that into the equation: \[25a=70b+7(5)\] \[25a=70b+35\] \[5a=14b+7.\] By the same line of reasoning as earlier, $a=7$ . We again plug that into the equation to get \[35=14b+7\] \[b=2.\] Now, since $a=7$ $b=2$ , and $c=5$ , our number $N=100a+10b+c=\boxed{725}$

Answer:  725