Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: In $\triangle ABC$ with side lengths $AB = 13,$ $BC = 14,$ and $CA = 15,$ let $M$ be the midpoint of $\overline{BC}.$ Let $P$ be the point on the circumcircle of $\triangle ABC$ such that $M$ is on $\overline{AP}.$ There exists a unique point $Q$ on segment $\overline{AM}$ such that $\angle PBQ = \angle PCQ.$ Then $AQ$ can be written as $\frac{m}{\sqrt{n}},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: It is clear that $BQCP$ is a parallelogram. By Stewart's Theorem, $AM=\sqrt{148}$ , POP on $M$ tells $PM=\frac{49}{\sqrt{148}}$
As $QM=PM, AQ=AM-PM=\frac{99}{\sqrt{148}}$ leads to $\boxed{247}$

Answer:  247
Problem: In $\triangle ABC$ with side lengths $AB = 13,$ $BC = 14,$ and $CA = 15,$ let $M$ be the midpoint of $\overline{BC}.$ Let $P$ be the point on the circumcircle of $\triangle ABC$ such that $M$ is on $\overline{AP}.$ There exists a unique point $Q$ on segment $\overline{AM}$ such that $\angle PBQ = \angle PCQ.$ Then $AQ$ can be written as $\frac{m}{\sqrt{n}},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: First, note that by Law of Sines, $\frac{\sin(\angle{QBP})}{QP}=\frac{\sin(\angle{BPQ})}{BQ}$ and that $\frac{\sin(\angle{QCP})}{QP}=\frac{\sin(\angle{QPC})}{QP}$ . Equating the 2 expressions, you get that $\frac{\sin(\angle{BPQ})}{BQ}=\frac{\sin(\angle{QPC})}{QP}$ . Now drop the altitude from $A$ to $BC$ . As it is commonly known that the dropped altitude forms a $5-12-13$ and a $9-12-15$ triangle, you get the measures of $\angle{ABC}$ and $\angle{ACB}$ respectively, which are $\arcsin(\frac{12}{13})$ and $\arcsin(\frac{4}{5})$ . However, by the inscribed angle theorem, you get that $\angle{BPQ}=\arcsin(\frac{4}{5})$ and that $\angle{QPC}=\arcsin(\frac{12}{13})$ , respectively. Therefore, by Law of Sines (as previously stated) $\frac{BQ}{CQ}=\frac{13}{15}$
Now commence coordbashing. Let $B$ be the origin, and $A$ be the point $(5,12)$ . As $AP$ passes through $A$ , which is $(5,12)$ , and $M$ , which is $(7,0)$ , it has the equation $-6x+42$ , so therefore a point on this line can be written as $(x,42-6x)$ . As we have the ratio of the lengths, which prompts us to write the lengths in terms of the distance formula, we can just plug and chug it in to get the ratio $\frac{\sqrt{37x^2-504x+1764}}{\sqrt{37x^2-532x+1960}}=\frac{13}{15}$ . This can be squared to get $\frac{37x^2-504x+1764}{37x^2-532x+1960}=\frac{169}{225}$ . This can be solved to get a solution of $x=\frac{469}{74}$ , and an extraneous solution of $5$ which obviously doesn’t work.
Plugging $x$ into the line equation gets you $y=\frac{147}{37}$ . The distance between this point and $A$ , which is $(5,12)$ is $\sqrt{\frac{9801}{148}}$ , or simplified to $\frac{99}{\sqrt{148}}\Longrightarrow99+148=\boxed{247}$

Answer:  247
Problem: In $\triangle ABC$ with side lengths $AB = 13,$ $BC = 14,$ and $CA = 15,$ let $M$ be the midpoint of $\overline{BC}.$ Let $P$ be the point on the circumcircle of $\triangle ABC$ such that $M$ is on $\overline{AP}.$ There exists a unique point $Q$ on segment $\overline{AM}$ such that $\angle PBQ = \angle PCQ.$ Then $AQ$ can be written as $\frac{m}{\sqrt{n}},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: We use the law of Cosine and get \[AB^2 = AM^2 + BM^2 - 2 AM \cdot BM \cos \angle AMB,\] \[AC^2 = AM^2 + CM^2 + 2 AM \cdot CM \cos \angle AMB \implies\] \[AM^2 = \frac {AB^2 + AC^2}{2}- BM^2 = \sqrt{148} \approx 12.\] We use the power of point $M$ with respect circumcircle $\triangle ABC$ and get \[AM \cdot MP = BM \cdot CM = BM^2 \implies\] \[PM = \frac {49}{\sqrt {148}} \approx \frac {48}{12} \approx 4 < AM.\] It is clear that if $Q = P,$ then $\angle PBQ = \angle PCQ = 0 \implies$
if $Q$ is symmetric to $P$ with respect $M$ then $\angle PBQ = \angle PCQ.$
There exists a unique point $Q$ on segment $\overline{AM}, PM < AM \implies$ \[PQ = AM - PM = \frac{99}{\sqrt{148}} \implies \boxed{247}.\] vladimir.shelomovskii@gmail.com, vvsss

Answer:  247
Problem: Let $A$ be an acute angle such that $\tan A = 2 \cos A.$ Find the number of positive integers $n$ less than or equal to $1000$ such that $\sec^n A + \tan^n A$ is a positive integer whose units digit is $9.$

Solution: Denote $a_n = \sec^n A + \tan^n A$ .
For any $k$ , we have \begin{align*} a_n & = \sec^n A + \tan^n A \\ & = \left( \sec^{n-k} A + \tan^{n-k} A \right) \left( \sec^k A + \tan^k A \right) - \sec^{n-k} A \tan^k A - \tan^{n-k} A \sec^k A \\ & = a_{n-k} a_k - 2^k \sec^{n-k} A \cos^k A - 2^k \tan^{n-k} A \cot^k A \\ & = a_{n-k} a_k - 2^k a_{n-2k} . \end{align*}
Next, we compute the first several terms of $a_n$
By solving equation $\tan A = 2 \cos A$ , we get $\cos A = \frac{\sqrt{2 \sqrt{17} - 2}}{4}$ .
Thus, $a_0 = 2$ $a_1 = \sqrt{\sqrt{17} + 4}$ $a_2 = \sqrt{17}$ $a_3 = \sqrt{\sqrt{17} + 4} \left( \sqrt{17} - 2 \right)$ $a_4 = 9$
In the rest of analysis, we set $k = 4$ .
Thus, \begin{align*} a_n & = a_{n-4} a_4 - 2^4 a_{n-8}  \\ & = 9 a_{n-4} - 16 a_{n-8} . \end{align*}
Thus, to get $a_n$ an integer, we have $4 | n$ .
In the rest of analysis, we only consider such $n$ . Denote $n = 4 m$ and $b_m = a_{4n}$ .
Thus, \begin{align*} b_m & = 9 b_{m-1} - 16 b_{m-2} \end{align*} with initial conditions $b_0 = 2$ $b_1 = 9$
To get the units digit of $b_m$ to be 9, we have \begin{align*} b_m \equiv -1 & \pmod{2} \\ b_m \equiv -1 & \pmod{5} \end{align*}
Modulo 2, for $m \geq 2$ , we have \begin{align*} b_m & \equiv 9 b_{m-1} - 16 b_{m-2} \\ & \equiv b_{m-1} . \end{align*}
Because $b_1 \equiv -1 \pmod{2}$ , we always have $b_m \equiv -1 \pmod{2}$ for all $m \geq 2$
Modulo 5, for $m \geq 5$ , we have \begin{align*} b_m & \equiv 9 b_{m-1} - 16 b_{m-2} \\ & \equiv - b_{m-1} - b_{m-2} . \end{align*}
We have $b_0 \equiv 2 \pmod{5}$ $b_1 \equiv -1 \pmod{5}$ $b_2 \equiv -1 \pmod{5}$ $b_3 \equiv 2 \pmod{5}$ $b_4 \equiv -1 \pmod{5}$ $b_5 \equiv -1 \pmod{5}$ $b_6 \equiv 2 \pmod{5}$ .
Therefore, the congruent values modulo 5 is cyclic with period 3.
To get $b_m \equiv -1 \pmod{5}$ , we have $3 \nmid m \pmod{3}$
From the above analysis with modulus 2 and modulus 5, we require $3 \nmid m \pmod{3}$
For $n \leq 1000$ , because $n = 4m$ , we only need to count feasible $m$ with $m \leq 250$ .
The number of feasible $m$ is \begin{align*} 250 - \left\lfloor \frac{250}{3} \right\rfloor & = 250 - 83 \\ & = \boxed{167}

Answer:  167
Problem: Let $A$ be an acute angle such that $\tan A = 2 \cos A.$ Find the number of positive integers $n$ less than or equal to $1000$ such that $\sec^n A + \tan^n A$ is a positive integer whose units digit is $9.$

Solution: \[\tan A = 2 \cos A \implies \sin A = 2 \cos^2 A \implies  \sin^2 A + \cos^2 A = 4  \cos^4 A + \cos^2 A = 1\] \[\implies   \cos^2 A = \frac {\sqrt {17} - 1}{8}.\] \[c_n = \sec^n A + \tan^n A = \frac {1}{\cos^n A} + 2^n \cos^n A = (4\cos^2 A +1)^{\frac {n}{2}}+(4 \cos^2 A)^{\frac {n}{2}} =\] \[= \left(\frac {\sqrt {17} + 1}{2}\right)^{\frac {n}{2}}+ \left(\frac {\sqrt {17} - 1}{2}\right)^{\frac {n}{2}}.\]
It is clear, that $c_n$ is not integer if $n \ne 4k, k > 0.$
Denote $x = \frac {\sqrt {17} + 1}{2}, y = \frac {\sqrt {17} - 1}{2} \implies$ \[x \cdot y = 4, x + y = \sqrt{17}, x - y = 1 \implies x^2 + y^2 = (x - y)^2 + 2xy = 9 = c_4.\]
\[c_8 = x^4 + y^4 = (x^2 + y^2)^2 - 2x^2 y^2 = 9^2 - 2 \cdot 16 = 49.\] \[c_{4k+4} = x^{4k+4} + y^{4k+4} = (x^{4k} + y^{4k})(x^2+y^2)- (x^2 y^2)(x^{4k-2}+y^{4k-2}) = 9 c_{4k}- 16 c_{4k – 4} \implies\] \[c_{12} = 9 c_8 - 16 c_4 = 9 \cdot 49 - 16 \cdot 9 = 9 \cdot 33 = 297.\] \[c_{16} = 9 c_{12} - 16 c_8 = 9 \cdot 297 - 16 \cdot 49 = 1889.\] \[c_{12m + 4} \pmod{10} = 9 \cdot c_{12m} \pmod{10} - 16 \pmod{10} \cdot c_{12m - 4}  \pmod{10} =\] \[= (9 \cdot 7 - 6 \cdot 9)  \pmod{10} = (3 - 4)  \pmod{10} = 9.\] \[c_{12m + 8}\pmod{10} = 9 \cdot c_{12m+4} \pmod{10} - 16 \pmod{10} \cdot c_{12m } \pmod{10} =\] \[= (9 \cdot 9 - 6 \cdot 7) \pmod{10} = (1 - 2)\pmod{10} = 9.\] \[c_{12m + 12} \pmod{10} = 9 \cdot c_{12m + 8} \pmod{10} - 16 \pmod{10} \cdot c_{12m + 4} \pmod{10} =\] \[= (9 \cdot 9 - 6 \cdot 9) \pmod{10} = (1 - 4) \pmod{10} = 7 \implies\]
The condition is satisfied iff $n = 12 k + 4$ or $n = 12k + 8.$
If $n \le N$ then the number of possible n is $\left\lfloor \frac{N}{4} \right\rfloor - \left\lfloor \frac{N}{12} \right\rfloor.$
For $N = 1000$ we get $\left\lfloor \frac{1000}{4} \right\rfloor - \left\lfloor \frac{1000}{12} \right\rfloor = 250 - 83 = \boxed{167}.$

Answer:  167
Problem: A cube-shaped container has vertices $A,$ $B,$ $C,$ and $D,$ where $\overline{AB}$ and $\overline{CD}$ are parallel edges of the cube, and $\overline{AC}$ and $\overline{BD}$ are diagonals of faces of the cube, as shown. Vertex $A$ of the cube is set on a horizontal plane $\mathcal{P}$ so that the plane of the rectangle $ABDC$ is perpendicular to $\mathcal{P},$ vertex $B$ is $2$ meters above $\mathcal{P},$ vertex $C$ is $8$ meters above $\mathcal{P},$ and vertex $D$ is $10$ meters above $\mathcal{P}.$ The cube contains water whose surface is parallel to $\mathcal{P}$ at a height of $7$ meters above $\mathcal{P}.$ The volume of water is $\frac{m}{n}$ cubic meters, where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: Denote $h(X)$ the distance from point $X$ to $\mathcal{P},  h(A) = 0, h(B) = 2,$ $h(C) = 8, h(D) = 10, h(G) = h(I) = h(H) = 7,  AB = a, AC = a \sqrt{2}.$
Let slope $AB$ to $\mathcal{P}$ be $\alpha.$ Notation is shown in the diagram. \[\tan \alpha = \frac {\sin \alpha}{\cos \alpha} = \frac {h(B)}{AB}\cdot \frac {AC}{h(C)} = \frac {\sqrt{2}}{4} \implies a = 6.\] Let $S = GI \cap CD \implies h(S) = h(G) = 7.$ \[h(C) – h(G) = 8 - 7 = 1, h(D)- h(I) = 10 - 7 = 3.\] \[h(E) = h(F) = \frac {h(D) +h(B)}{2} = 6 \implies\] \[\frac {DI}{DE} = \frac {h(D) - h(I)}{h(D)-h(E)} = \frac {3}{4} \implies DI = DH = \frac {9}{2}.\]
Similarly $CG = \frac {3}{2} \implies SD = 9.$
Let the volume without water be $V,$ volume of the pyramid $SCGJ$ be $U.$
It is clear that $U + V = 27U = \frac {SD}{6} \cdot DI^2 = \frac {243}{8} \implies$ $V = \frac {243 \cdot 26}{8 \cdot 27 } = \frac {117}{4} = 6^3 - \frac {747}{4}$ from which $\boxed{751}.$

Answer:  751
Problem: For each positive integer $n$ let $a_n$ be the least positive integer multiple of $23$ such that $a_n \equiv 1 \pmod{2^n}.$ Find the number of positive integers $n$ less than or equal to $1000$ that satisfy $a_n = a_{n+1}.$

Solution: Denote $a_n = 23 b_n$ .
Thus, for each $n$ , we need to find smallest positive integer $k_n$ , such that \[ 23 b_n = 2^n k_n + 1 . \]
Thus, we need to find smallest $k_n$ , such that \[ 2^n k_n \equiv - 1 \pmod{23} . \]
Now, we find the smallest $m$ , such that $2^m \equiv 1 \pmod{23}$ .
By Fermat's Theorem, we must have $m | \phi \left( 23 \right)$ . That is, $m | 22$ .
We find $m = 11$
Therefore, for each $n$ , we need to find smallest $k_n$ , such that \[ 2^{{\rm Rem} \left( n , 11 \right)} k_n \equiv - 1 \pmod{23} . \]
We have the following results:
If \({\rm Rem} \left( n , 11 \right) = 0\), then \(k_n = 22\) and \(b_n = 1\).
If \({\rm Rem} \left( n , 11 \right) = 1\), then \(k_n = 11\) and \(b_n = 1\).
If \({\rm Rem} \left( n , 11 \right) = 2\), then \(k_n = 17\) and \(b_n = 3\).
If \({\rm Rem} \left( n , 11 \right) = 3\), then \(k_n = 20\) and \(b_n = 7\).
If \({\rm Rem} \left( n , 11 \right) = 4\), then \(k_n = 10\) and \(b_n = 7\).
If \({\rm Rem} \left( n , 11 \right) = 5\), then \(k_n = 5\) and \(b_n = 7\).
If \({\rm Rem} \left( n , 11 \right) = 6\), then \(k_n = 14\) and \(b_n = 39\).
If \({\rm Rem} \left( n , 11 \right) = 7\), then \(k_n = 7\) and \(b_n = 39\).
If \({\rm Rem} \left( n , 11 \right) = 8\), then \(k_n = 15\) and \(b_n = 167\).
If \({\rm Rem} \left( n , 11 \right) = 9\), then \(k_n = 19\) and \(b_n = 423\).
If \({\rm Rem} \left( n , 11 \right) = 10\), then \(k_n = 21\) and \(b_n = 935\).
Therefore, in each cycle, $n \in \left\{ 11 l , 11l + 1 , \cdots , 11l + 10 \right\}$ , we have $n = 11l$ $11l + 3$ $11l + 4$ $11l + 6$ , such that $b_n = b_{n+1}$ . That is, $a_n = a_{n+1}$ .
At the boundary of two consecutive cycles, $b_{11L + 10} \neq b_{11 \left(l + 1 \right)}$
We have $1000 = 90 \cdot 11 + 10$ .
Therefore, the number of feasible $n$ is $91 \cdot 4 - 1 = \boxed{363}$

Answer:  363
Problem: Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients $2$ and $-2,$ respectively. The graphs of both polynomials pass through the two points $(16,54)$ and $(20,53).$ Find $P(0) + Q(0).$

Solution: Let $R(x)=P(x)+Q(x).$ Since the $x^2$ -terms of $P(x)$ and $Q(x)$ cancel, we conclude that $R(x)$ is a linear polynomial.
Note that \begin{alignat*}{8} R(16) &= P(16)+Q(16) &&= 54+54 &&= 108, \\ R(20) &= P(20)+Q(20) &&= 53+53 &&= 106, \end{alignat*} so the slope of $R(x)$ is $\frac{106-108}{20-16}=-\frac12.$
It follows that the equation of $R(x)$ is \[R(x)=-\frac12x+c\] for some constant $c,$ and we wish to find $R(0)=c.$
We substitute $x=20$ into this equation to get $106=-\frac12\cdot20+c,$ from which $c=\boxed{116}.$

Answer:  116
Problem: Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients $2$ and $-2,$ respectively. The graphs of both polynomials pass through the two points $(16,54)$ and $(20,53).$ Find $P(0) + Q(0).$

Solution: Let \begin{alignat*}{8} P(x) &= &2x^2 + ax + b, \\ Q(x) &= &\hspace{1mm}-2x^2 + cx + d, \end{alignat*} for some constants $a,b,c$ and $d.$
We are given that \begin{alignat*}{8} P(16) &= &512 + 16a + b &= 54, \hspace{20mm}&&(1) \\ Q(16) &= &\hspace{1mm}-512 + 16c + d &= 54, &&(2) \\ P(20) &= &800 + 20a + b &= 53,  &&(3) \\ Q(20) &= &\hspace{1mm}-800 + 20c + d &= 53, &&(4) \end{alignat*} and we wish to find \[P(0)+Q(0)=b+d.\] We need to cancel $a$ and $c.$ Since $\operatorname{lcm}(16,20)=80,$ we subtract $4\cdot[(3)+(4)]$ from $5\cdot[(1)+(2)]$ to get \[b+d=5\cdot(54+54)-4\cdot(53+53)=\boxed{116}.\] ~MRENTHUSIASM

Answer:  116
Problem: Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients $2$ and $-2,$ respectively. The graphs of both polynomials pass through the two points $(16,54)$ and $(20,53).$ Find $P(0) + Q(0).$

Solution: Like Solution 2, we can begin by setting $P$ and $Q$ to the quadratic above, giving us \begin{alignat*}{8} P(16) &= &512 + 16a + b &= 54, \hspace{20mm}&&(1) \\ Q(16) &= &\hspace{1mm}-512 + 16c + d &= 54, &&(2) \\ P(20) &= &800 + 20a + b &= 53,  &&(3) \\ Q(20) &= &\hspace{1mm}-800 + 20c + d &= 53, &&(4) \end{alignat*} We can first add $(1)$ and $(2)$ to obtain $16(a-c) + (b+d) = 108.$
Next, we can add $(3)$ and $(4)$ to obtain $20(a-c) + (b+d) = 106.$ By subtracting these two equations, we find that $4(a-c) = -2,$ so substituting this into equation $[(1) + (2)],$ we know that $4 \cdot (-2) + (b+d) = 108,$ so therefore $b+d = \boxed{116}.$

Answer:  116
Problem: Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients $2$ and $-2,$ respectively. The graphs of both polynomials pass through the two points $(16,54)$ and $(20,53).$ Find $P(0) + Q(0).$

Solution: Let \begin{alignat*}{8} P(x) &= &2x^2 + ax + b, \\ Q(x) &= &\hspace{1mm}-2x^2 + cx + d, \end{alignat*} By substituting $(16, 54)$ and $(20, 53)$ into these equations, we can get: \begin{align*} 2(16)^2 + 16a + b &= 54, \\ 2(20)^2 + 20a + b &= 53. \end{align*} Hence, $a = -72.25$ and $b = 698.$
Similarly, \begin{align*} -2(16)^2 + 16c + d &= 54, \\ -2(20)^2 + 20c + d &= 53. \end{align*} Hence, $c = 71.75$ and $d = -582.$
Notice that $b = P(0)$ and $d = Q(0).$ Therefore \[P(0) + Q(0) = 698 + (-582) = \boxed{116}.\] ~Littlemouse

Answer:  116
Problem: Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients $2$ and $-2,$ respectively. The graphs of both polynomials pass through the two points $(16,54)$ and $(20,53).$ Find $P(0) + Q(0).$

Solution: Add the equations of the polynomials $y=2x^2+ax+b$ and $y=-2x^2+cx+d$ to get $2y=(a+c)x+(b+d)$ . This equation must also pass through the two points $(16,54)$ and $(20,53)$
Let $m=a+c$ and $n=b+d$ . We then have two equations: \begin{align*} 108&=16m+n, \\ 106&=20m+n. \end{align*} We are trying to solve for $n=P(0)$ . Using elimination: \begin{align*} 540&=80m+5n, \\ 424&=80m+4n. \end{align*} Subtracting both equations, we find that $n=\boxed{116}$

Answer:  116
Problem: Find the three-digit positive integer $\underline{a}\,\underline{b}\,\underline{c}$ whose representation in base nine is $\underline{b}\,\underline{c}\,\underline{a}_{\,\text{nine}},$ where $a,$ $b,$ and $c$ are (not necessarily distinct) digits.

Solution: We are given that \[100a + 10b + c = 81b + 9c + a,\] which rearranges to \[99a = 71b + 8c.\] Taking both sides modulo $71,$ we have \begin{align*} 28a &\equiv 8c \pmod{71} \\ 7a &\equiv 2c \pmod{71}. \end{align*} The only solution occurs at $(a,c)=(2,7),$ from which $b=2.$
Therefore, the requested three-digit positive integer is $\underline{a}\,\underline{b}\,\underline{c}=\boxed{227}.$

Answer:  227
Problem: Find the three-digit positive integer $\underline{a}\,\underline{b}\,\underline{c}$ whose representation in base nine is $\underline{b}\,\underline{c}\,\underline{a}_{\,\text{nine}},$ where $a,$ $b,$ and $c$ are (not necessarily distinct) digits.

Solution: As shown in Solution 1, we get $99a = 71b+8c$
Note that $99$ and $71$ are large numbers comparatively to $8$ , so we hypothesize that $a$ and $b$ are equal and $8c$ fills the gap between them. The difference between $99$ and $71$ is $28$ , which is a multiple of $4$ . So, if we multiply this by $2$ , it will be a multiple of $8$ and thus the gap can be filled. Therefore, the only solution is $(a,b,c)=(2,2,7)$ , and the answer is $\underline{a}\,\underline{b}\,\underline{c}=\boxed{227}$

Answer:  227
Problem: Find the three-digit positive integer $\underline{a}\,\underline{b}\,\underline{c}$ whose representation in base nine is $\underline{b}\,\underline{c}\,\underline{a}_{\,\text{nine}},$ where $a,$ $b,$ and $c$ are (not necessarily distinct) digits.

Solution: As shown in Solution 1, we get $99a = 71b+8c.$
We list a few multiples of $99$ out: \[99,198,297,396.\] Of course, $99$ can't be made of just $8$ 's. If we use one $71$ , we get a remainder of $28$ , which can't be made of $8$ 's either. So $99$ doesn't work. $198$ can't be made up of just $8$ 's. If we use one $71$ , we get a remainder of $127$ , which can't be made of $8$ 's. If we use two $71$ 's, we get a remainder of $56$ , which can be made of $8$ 's. 
Therefore we get $99\cdot2=71\cdot2+8\cdot7$ so $a=2,b=2,$ and $c=7$ . Plugging this back into the original problem shows that this answer is indeed correct. Therefore, $\underline{a}\,\underline{b}\,\underline{c}=\boxed{227}.$

Answer:  227
Problem: Find the three-digit positive integer $\underline{a}\,\underline{b}\,\underline{c}$ whose representation in base nine is $\underline{b}\,\underline{c}\,\underline{a}_{\,\text{nine}},$ where $a,$ $b,$ and $c$ are (not necessarily distinct) digits.

Solution: As shown in Solution 1, we get $99a = 71b+8c$
We can see that $99$ is $28$ larger than $71$ , and we have an $8c$ . We can clearly see that $56$ is a multiple of $8$ , and any larger than $56$ would result in $c$ being larger than $9$ . Therefore, our only solution is $a = 2, b = 2, c = 7$ . Our answer is $\underline{a}\,\underline{b}\,\underline{c}=\boxed{227}$

Answer:  227
Problem: Find the three-digit positive integer $\underline{a}\,\underline{b}\,\underline{c}$ whose representation in base nine is $\underline{b}\,\underline{c}\,\underline{a}_{\,\text{nine}},$ where $a,$ $b,$ and $c$ are (not necessarily distinct) digits.

Solution: As shown in Solution 1, we have that $99a = 71b + 8c$
Note that by the divisibility rule for $9$ , we have $a+b+c \equiv a \pmod{9}$ .  Since $b$ and $c$ are base- $9$ digits, we can say that $b+c = 0$ or $b+c=9$ .  The former possibility can be easily eliminated, and thus $b+c=9$ .  Next, we write the equation from Solution 1 as $99a = 63b + 8(b+c)$ , and dividing this by $9$ gives $11a = 7b+8$ .  Taking both sides modulo $7$ , we have $4a \equiv 1 \pmod{7}$ .  Multiplying both sides by $2$ gives $a\equiv 2 \pmod{7}$ , which implies $a=2$ .  From here, we can find that $b=2$ and $c=7$ , giving an answer of $\boxed{227}$

Answer:  227
Problem: In isosceles trapezoid $ABCD$ , parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$ , respectively, and $AD=BC=333$ . The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$ , and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$ . Find $PQ$

Solution: We have the following diagram: [asy] /* Made by MRENTHUSIASM , modified by Cytronical */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); X = intersectionpoint(B--5*(Q-B)+B,C--D); Y = (0,6*sqrt(731)); Z = intersectionpoint(A--4*(P-A)+A,B--4*(Q-B)+B); W = intersectionpoint(A--5*(P-A)+A,C--D); draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red); draw(anglemark(A,B,Q,1000),red); draw(anglemark(Q,B,C,1000),red); draw(anglemark(P,D,A,1000),red); draw(anglemark(C,D,P,1000),red); draw(anglemark(Q,C,D,1000),red); draw(anglemark(B,C,Q,1000),red); add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$P$",P,1.5*(-1,0),linewidth(4)); dot("$Q$",Q,1.5*E,linewidth(4)); dot("$X$",X,1.5*dir(-105),linewidth(4)); dot("$Y$",Y,1.5*N,linewidth(4)); dot("$Z$",Z,4.5*dir(75),linewidth(4)); dot("$W$",W,1.5*dir(-75),linewidth(4)); draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); draw(P--W^^Q--X^^Y--Z,dashed); [/asy]
Let $X$ and $W$ be the points where $AP$ and $BQ$ extend to meet $CD$ , and $YZ$ be the height of $\triangle AZB$ . As proven in Solution 2, triangles $APD$ and $DPW$ are congruent right triangles. Therefore, $AD = DW = 333$ . We can apply this logic to triangles $BCQ$ and $XCQ$ as well, giving us $BC = CX = 333$ . Since $CD = 650$ $XW = DW + CX - CD = 16$
Additionally, we can see that $\triangle XZW$ is similar to $\triangle PQZ$ and $\triangle AZB$ . We know that $\frac{XW}{AB} = \frac{16}{500}$ . So, we can say that the height of the triangle $AZB$ is $500u$ while the height of the triangle $XZW$ is $16u$ . After that, we can figure out the distance from $Y$ to $PQ: \frac{500+16}{2} = 258u$ and the height of triangle $PZQ: 500-258 = 242u$
Finally, since the ratio between the height of $PZQ$ to the height of $AZB$ is $242:500$ and $AB$ is $500$ $PQ = \boxed{242}.$

Answer:  242