Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Given a positive integer $n$ , it can be shown that every complex number of the form $r+si$ , where $r$ and $s$ are integers, can be uniquely expressed in the base $-n+i$ using the integers $0,1,2,\ldots,n^2$ as digits. That is, the equation
is true for a unique choice of non-negative integer $m$ and digits $a_0,a_1,\ldots,a_m$ chosen from the set $\{0,1,2,\ldots,n^2\}$ , with $a_m\ne 0$ . We write
to denote the base $-n+i$ expansion of $r+si$ . There are only finitely many integers $k+0i$ that have four-digit expansions
Find the sum of all such $k$

Solution: First, we find the first three powers of $-3+i$
$(-3+i)^1=-3+i ; (-3+i)^2=8-6i ; (-3+i)^3=-18+26i$
So we solve the diophantine equation $a_1-6a_2+26a_3=0 \Longrightarrow a_1-6a_2=-26a_3$
The minimum the left-hand side can go is -54, so $1\leq a_3 \leq 2$ since $a_3$ can't equal 0, so we try cases:
So we have four-digit integers $(292a_0)_{-3+i}$ and $(154a_0)_{-3+i}$ , and we need to find the sum of all integers $k$ that can be expressed by one of those.
$(292a_0)_{-3+i}$
We plug the first three digits into base 10 to get $30+a_0$ . The sum of the integers $k$ in that form is $345$
$(154a_0)_{-3+i}$
We plug the first three digits into base 10 to get $10+a_0$ . The sum of the integers $k$ in that form is $145$ . The answer is $345+145=\boxed{490}$ .
~minor edit by Yiyj1

Answer:  490
Problem: Point $P$ is inside $\triangle ABC$ . Line segments $APD$ $BPE$ , and $CPF$ are drawn with $D$ on $BC$ $E$ on $AC$ , and $F$ on $AB$ (see the figure below). Given that $AP=6$ $BP=9$ $PD=6$ $PE=3$ , and $CF=20$ , find the area of $\triangle ABC$

Solution: Let $[RST]$ be the area of polygon $RST$ .  We'll make use of the following fact: if $P$ is a point in the interior of triangle $XYZ$ , and line $XP$ intersects line $YZ$ at point $L$ , then $\dfrac{XP}{PL} = \frac{[XPY] + [ZPX]}{[YPZ]}.$
This is true because triangles $XPY$ and $YPL$ have their areas in ratio $XP:PL$ (as they share a common height from $Y$ ), and the same is true of triangles $ZPY$ and $LPZ$
We'll also use the related fact that $\dfrac{[XPY]}{[ZPX]} = \dfrac{YL}{LZ}$ .  This is slightly more well known, as it is used in the standard proof of Ceva's theorem
Now we'll apply these results to the problem at hand.
Since $AP = PD = 6$ , this means that $[APB] + [APC] = [BPC]$ ; thus $\triangle BPC$ has half the area of $\triangle ABC$ .  And since $PE = 3 = \dfrac{1}{3}BP$ , we can conclude that $\triangle APC$ has one third of the combined areas of triangle $BPC$ and $APB$ , and thus $\dfrac{1}{4}$ of the area of $\triangle ABC$ .  This means that $\triangle APB$ is left with $\dfrac{1}{4}$ of the area of triangle $ABC$ \[[BPC]: [APC]: [APB] = 2:1:1.\] Since $[APC] = [APB]$ , and since $\dfrac{[APC]}{[APB]} = \dfrac{CD}{DB}$ , this means that $D$ is the midpoint of $BC$
Furthermore, we know that $\dfrac{CP}{PF} = \dfrac{[APC] + [BPC]}{[APB]} = 3$ , so $CP = \dfrac{3}{4} \cdot CF = 15$
We now apply Stewart's theorem to segment $PD$ in $\triangle BPC$ —or rather, the simplified version for a median.  This tells us that \[2 BD^2 + 2 PD^2 = BP^2+ CP^2.\] Plugging in we know, we learn that \begin{align*} 2 BD^2 + 2 \cdot 36 &= 81 + 225 = 306, \\ BD^2 &= 117. \end{align*} Happily, $BP^2 + PD^2 = 81 + 36$ is also equal to 117.  Therefore $\triangle BPD$ is a right triangle with a right angle at $B$ ; its area is thus $\dfrac{1}{2} \cdot 9 \cdot 6 = 27$ .  As $PD$ is a median of $\triangle BPC$ , the area of $BPC$ is twice this, or 54.  And we already know that $\triangle BPC$ has half the area of $\triangle ABC$ , which must therefore be $\boxed{108}$

Answer:  108
Problem: Point $P$ is inside $\triangle ABC$ . Line segments $APD$ $BPE$ , and $CPF$ are drawn with $D$ on $BC$ $E$ on $AC$ , and $F$ on $AB$ (see the figure below). Given that $AP=6$ $BP=9$ $PD=6$ $PE=3$ , and $CF=20$ , find the area of $\triangle ABC$

Solution: Because we're given three concurrent cevians and their lengths, it seems very tempting to apply Mass points . We immediately see that $w_E = 3$ $w_B = 1$ , and $w_A = w_D = 2$ . Now, we recall that the masses on the three sides of the triangle must be balanced out, so $w_C = 1$ and $w_F = 3$ . Thus, $CP = 15$ and $PF = 5$
Recalling that $w_C = w_B = 1$ , we see that $DC = DB$ and $DP$ is a median to $BC$ in $\triangle BCP$ . Applying Stewart's Theorem $BC^2 + 12^2 = 2(15^2 + 9^2)$ , and $BC = 6\sqrt {13}$ . Now notice that $2[BCP] = [ABC]$ , because both triangles share the same base and the $h_{\triangle ABC} = 2h_{\triangle BCP}$ . Applying Heron's formula on triangle $BCP$ with sides $15$ $9$ , and $6\sqrt{13}$ $[BCP] = 54$ and $[ABC] = \boxed{108}$

Answer:  108
Problem: Point $P$ is inside $\triangle ABC$ . Line segments $APD$ $BPE$ , and $CPF$ are drawn with $D$ on $BC$ $E$ on $AC$ , and $F$ on $AB$ (see the figure below). Given that $AP=6$ $BP=9$ $PD=6$ $PE=3$ , and $CF=20$ , find the area of $\triangle ABC$

Solution: Using a different form of Ceva's Theorem , we have $\frac {y}{x + y} + \frac {6}{6 + 6} + \frac {3}{3 + 9} = 1\Longleftrightarrow\frac {y}{x + y} = \frac {1}{4}$
Solving $4y = x + y$ and $x + y = 20$ , we obtain $x = CP = 15$ and $y = FP = 5$
Let $Q$ be the point on $AB$ such that $FC \parallel QD$ .
Since $AP = PD$ and $FP\parallel QD$ $QD = 2FP = 10$ . (Stewart's Theorem)
Also, since $FC\parallel QD$ and $QD = \frac{FC}{2}$ , we see that $FQ = QB$ $BD = DC$ , etc. ( Stewart's Theorem )
Similarly, we have $PR = RB$ $= \frac12PB = 7.5$ ) and thus $RD = \frac12PC = 4.5$
$PDR$ is a $3-4-5$ right triangle , so $\angle PDR$ $\angle ADQ$ ) is $90^\circ$ .
Therefore, the area of $\triangle ADQ = \frac12\cdot 12\cdot 6 = 36$ .
Using area ratio, $\triangle ABC = \triangle ADB\times 2 = \left(\triangle ADQ\times \frac32\right)\times 2 = 36\cdot 3 = \boxed{108}$

Answer:  108
Problem: Point $P$ is inside $\triangle ABC$ . Line segments $APD$ $BPE$ , and $CPF$ are drawn with $D$ on $BC$ $E$ on $AC$ , and $F$ on $AB$ (see the figure below). Given that $AP=6$ $BP=9$ $PD=6$ $PE=3$ , and $CF=20$ , find the area of $\triangle ABC$

Solution: First, let $[AEP]=a, [AFP]=b,$ and $[ECP]=c.$ Thus, we can easily find that $\frac{[AEP]}{[BPD]}=\frac{3}{9}=\frac{1}{3} \Leftrightarrow [BPD]=3[AEP]=3a.$ Now, $\frac{[ABP]}{[BPD]}=\frac{6}{6}=1\Leftrightarrow [ABP]=3a.$ In the same manner, we find that $[CPD]=a+c.$ Now, we can find that $\frac{[BPC]}{[PEC]}=\frac{9}{3}=3 \Leftrightarrow \frac{(3a)+(a+c)}{c}=3 \Leftrightarrow c=2a.$ We can now use this to find that $\frac{[APC]}{[AFP]}=\frac{[BPC]}{[BFP]}=\frac{PC}{FP} \Leftrightarrow \frac{3a}{b}=\frac{6a}{3a-b} \Leftrightarrow a=b.$ Plugging this value in, we find that $\frac{FC}{FP}=3 \Leftrightarrow PC=15, FP=5.$ Now, since $\frac{[AEP]}{[PEC]}=\frac{a}{2a}=\frac{1}{2},$ we can find that $2AE=EC.$ Setting $AC=b,$ we can apply Stewart's Theorem on triangle $APC$ to find that $(15)(15)(\frac{b}{3})+(6)(6)(\frac{2b}{3})=(\frac{2b}{3})(\frac{b}{3})(b)+(b)(3)(3).$ Solving, we find that $b=\sqrt{405} \Leftrightarrow AE=\frac{b}{3}=\sqrt{45}.$ But, $3^2+6^2=45,$ meaning that $\angle{APE}=90 \Leftrightarrow [APE]=\frac{(6)(3)}{2}=9=a.$ Since $[ABC]=a+a+2a+2a+3a+3a=12a=(12)(9)=108,$ we conclude that the answer is $\boxed{108}$

Answer:  108
Problem: Point $P$ is inside $\triangle ABC$ . Line segments $APD$ $BPE$ , and $CPF$ are drawn with $D$ on $BC$ $E$ on $AC$ , and $F$ on $AB$ (see the figure below). Given that $AP=6$ $BP=9$ $PD=6$ $PE=3$ , and $CF=20$ , find the area of $\triangle ABC$

Solution: Firstly, since they all meet at one single point, denoting the mass of them separately. Assuming $M(A)=6;M(D)=6;M(B)=3;M(E)=9$ ; we can get that $M(P)=12;M(F)=9;M(C)=3$ ; which leads to the ratio between segments, \[\frac{CE}{AE}=2;\frac{BF}{AF}=2;\frac{BD}{CD}=1.\] Denoting that $CE=2x;AE=x; AF=y; BF=2y; CD=z; DB=z.$
Now we know three cevians' length, Applying Stewart theorem to them, getting three different equations: \begin{align} (3x)^2 \cdot 2y+(2z)^2 \cdot y&=(3y)(2y^2+400), \\ (3y)^2 \cdot z+(3x)^2 \cdot z&=(2z)(z^2+144), \\ (2z)^2 \cdot x+(3y)^2 \cdot x&=(3x)(2x^2+144). \end{align} After solving the system of equation, we get that $x=3\sqrt{5};y=\sqrt{13};z=3\sqrt{13}$
pulling $x,y,z$ back to get the length of $AC=9\sqrt{5};AB=3\sqrt{13};BC=6\sqrt{13}$ ; now we can apply Heron's formula here, which is \[\sqrt\frac{(9\sqrt{5}+9\sqrt{13})(9\sqrt{13}-9\sqrt{5})(9\sqrt{5}+3\sqrt{13})(9\sqrt{5}-3\sqrt{13})}{16}=108.\]
Our answer is $\boxed{108}$

Answer:  108
Problem: Point $P$ is inside $\triangle ABC$ . Line segments $APD$ $BPE$ , and $CPF$ are drawn with $D$ on $BC$ $E$ on $AC$ , and $F$ on $AB$ (see the figure below). Given that $AP=6$ $BP=9$ $PD=6$ $PE=3$ , and $CF=20$ , find the area of $\triangle ABC$

Solution: In Solution 5, instead of finding all of $x, y, z$ , we only need $y, z$ . This is because after we solve for $y, z$ , we can notice that $\triangle BAD$ is isosceles with $AB = BD$ . Because $P$ is the midpoint of the base, $BP$ is an altitude of $\triangle BAD$ . Therefore, $[BAD] = \frac{(AD)(BP)}{2} = \frac{12 \cdot 9}{2} = 54$ . Using the same altitude property, we can find that $[ABC] = 2[BAD] = 2 \cdot 54 = \boxed{108}$

Answer:  108
Problem: Point $P$ is inside $\triangle ABC$ . Line segments $APD$ $BPE$ , and $CPF$ are drawn with $D$ on $BC$ $E$ on $AC$ , and $F$ on $AB$ (see the figure below). Given that $AP=6$ $BP=9$ $PD=6$ $PE=3$ , and $CF=20$ , find the area of $\triangle ABC$

Solution: Set $AF=x,$ and use mass points to find that $PF=5$ and $BF=2x.$ Using Stewart's Theorem on $APB,$ we find that $AB=3\sqrt{13}.$ Then we notice that $APB$ is right, which means the area of $APB$ is $27.$ Because $CF=4\cdot PF,$ the area of $ABC$ is $4$ times the area of $APB,$ which means the area of $ABC=4\cdot 27=\boxed{108}.$

Answer:  108
Problem: Point $P$ is inside $\triangle ABC$ . Line segments $APD$ $BPE$ , and $CPF$ are drawn with $D$ on $BC$ $E$ on $AC$ , and $F$ on $AB$ (see the figure below). Given that $AP=6$ $BP=9$ $PD=6$ $PE=3$ , and $CF=20$ , find the area of $\triangle ABC$

Solution: We try to solve this using only elementary concepts. Let the areas of triangles $BCP$ $ACP$ and $ABP$ be $X$ $Y$ and $Z$ respectively. Then $\frac{X}{Y+Z}=\frac{6}{6}=1$ and $\frac{Y}{X+Z}=\frac{3}{9}=\frac{1}{3}$ . Hence $\frac{X}{2}=Y=Z$ . Similarly $\frac{FP}{PC}=\frac{Z}{X+Y}=\frac{1}{3}$ and since $CF=20$ we then have $FP=5$ . Additionally we now see that triangles $FPE$ and $CPB$ are similar, so $FE \parallel BC$ and $\frac{FE}{BC} = \frac{1}{3}$ . Hence $\frac{AF}{FB}=\frac{1}{2}$ . Now construct a point $K$ on segment $BP$ such that $BK=6$ and $KP=3$ , we will have $FK \parallel AP$ , and hence $\frac{FK}{AP} = \frac{FK}{6} = \frac{2}{3}$ , giving $FK=4$ . Triangle $FKP$ is therefore a 3-4-5 triangle! So $FK \perp BE$ and so $AP \perp BE$ . Then it is easy to calculate that $Z = \frac{1}{2} \times 6 \times 9 = 27$ and the area of triangle $ABC = X+Y+Z = 4Z = 4 \times 27 = \boxed{108}$ .
~Leole

Answer:  108
Problem: Point $P$ is inside $\triangle ABC$ . Line segments $APD$ $BPE$ , and $CPF$ are drawn with $D$ on $BC$ $E$ on $AC$ , and $F$ on $AB$ (see the figure below). Given that $AP=6$ $BP=9$ $PD=6$ $PE=3$ , and $CF=20$ , find the area of $\triangle ABC$

Solution: We start with mass points as in Solution 2, and receive $BF:AF = 2$ $BD:CD = 1$ $CE:AE = 2$ Law of Cosines on triangles $ADB$ and $ADC$ with $\theta = \angle ADB$ and $BD=DC=x$ gives \[36+x^2-12x\cos \theta = 81\] \[36+x^2-12x\cos (180-\theta) = 36+x^2+12x\cos \theta = 225\] Adding them: $72+2x^2=306 \implies x=3\sqrt{13}$ , so $BC = 6\sqrt{13}$ . Similarly, $AB = 3\sqrt{13}$ and $AC = 9\sqrt{5}$ . Using Heron's, \[[ \triangle ABC ]= \sqrt{\left(\dfrac{9\sqrt{13}+9\sqrt{5}}{2}\right)\left(\dfrac{9\sqrt{13}09\sqrt{5}}{2}\right)\left(\dfrac{3\sqrt{13}+9\sqrt{5}}{2}\right)\left(\dfrac{-3\sqrt{13}+9\sqrt{5}}{2}\right)} = \boxed{108}.\]

Answer:  108
Problem: One commercially available ten-button lock may be opened by pressing -- in any order -- the correct five buttons. The sample shown below has $\{1,2,3,6,9\}$ as its combination . Suppose that these locks are redesigned so that sets of as many as nine buttons or as few as one button could serve as combinations. How many additional combinations would this allow?
1988-1.png

Solution: Currently there are ${10 \choose 5}$ possible combinations.
With any integer $x$ from $1$ to $9$ , the number of ways to choose a set of $x$ buttons is $\sum^{9}_{k=1}{10 \choose k}$ .
Now we can use the identity $\sum^{n}_{k=0}{n \choose k}=2^{n}$ .
So the number of additional combinations is just $2^{10}-{10\choose 0}-{10\choose 10}-{10 \choose 5}=1024-1-1-252=\boxed{770}$

Answer:  770
Problem: For any positive integer $k$ , let $f_1(k)$ denote the square of the sum of the digits of $k$ .  For $n \ge 2$ , let $f_n(k) = f_1(f_{n - 1}(k))$ .  Find $f_{1988}(11)$

Solution: We see that $f_{1}(11)=4$
$f_2(11) = f_1(4)=16$
$f_3(11) = f_1(16)=49$
$f_4(11) = f_1(49)=169$
$f_5(11) = f_1(169)=256$
$f_6(11) = f_1(256)=169$
Note that this revolves between the two numbers. Since $1988$ is even, we thus have $f_{1988}(11) = f_{4}(11) = \boxed{169}$

Answer:  169
Problem: Find $(\log_2 x)^2$ if $\log_2 (\log_8 x) = \log_8 (\log_2 x)$

Solution: Raise both as exponents with base 8:
\begin{align*} 8^{\log_2 (\log_8 x)} &= 8^{\log_8 (\log_2 x)}\\ 2^{3 \log_2(\log_8x)} &= \log_2x\\ (\log_8x)^3 &= \log_2x\\ \left(\frac{\log_2x}{\log_28}\right)^3 &= \log_2x\\ (\log_2x)^2 &= (\log_28)^3 = \boxed{027}

Answer:  027
Problem: Find $(\log_2 x)^2$ if $\log_2 (\log_8 x) = \log_8 (\log_2 x)$

Solution: We wish to convert this expression into one which has a uniform base. Let's scale down all the powers of 8 to 2.
\begin{align*} {\log_2 (\frac{1}{3}\log_2 x)} &= \frac{1}{3}{\log_2 (\log_2 x)}\\ {\log_2 x = y}\\ {\log_2 (\frac{1}{3}y)} &= \frac{1}{3}{\log_2 (y)}\\ {3\log_2 (\frac{1}{3}y)} &= {\log_2 (y)}\\  {\log_2 (\frac{1}{3}y)^3} &= {\log_2 (y)}\\  \end{align*} Solving, we get $y^2 = 27$ , which is what we want. $\boxed{27}$

Answer:  27
Problem: Find $(\log_2 x)^2$ if $\log_2 (\log_8 x) = \log_8 (\log_2 x)$

Solution: First we have \begin{align*} \log_2(\log_8x)&=\log_8(\log_2x)\\ \frac{\log_2(\log_8x)}{\log_8(\log_2x)}&=1 \end{align*} Changing the base in the numerator yields \begin{align*} \frac{3\log_8(\log_8x)}{\log_8(\log_2x)}&=1\\ \frac{\log_8(\log_8x)}{\log_8(\log_2x)}&=\frac{1}{3}\\ \end{align*} Using the property $\frac{\log_ab}{\log_ac}=\log_cb$ yields \begin{align*} \log_{\log_2x}(\log_8x)&=\frac{1}{3}\\ (\log_2x)^\frac{1}{3}&=\log_8x\\ \sqrt[3]{\log_2x}&=\frac{\log_2x}{3} \end{align*} Now setting $y=\log_2x$ , we have \[\sqrt[3]{y}=\frac{y}{3}\] Solving gets $y=\log_2x=3\sqrt{3}\Longrightarrow(\log_2x)^2=(3\sqrt{3})^2=\boxed{27}$

Answer:  27
Problem: Find $(\log_2 x)^2$ if $\log_2 (\log_8 x) = \log_8 (\log_2 x)$

Solution: Say that $\log_{2^3}x=a$ and $\log_2x=b$ so we have $\log_2a=\log_{2^3}b$ . And we want $b^2$
$\\ \log_2a=\frac13 \log_{2}b \ \ \text{\tiny{(step 1)}}\\ \frac{\log_2a}{\log_{2}b}=\log_ba=\frac13\\ b^{1/3}=a.$
Because $3a=b$ (as $2^{3a}=x$ and $2^b=x$ from our setup), we have that
$b^{1/3}=\frac{b}{3}\\ b^{-2/3}=\frac13\\ b=3^{3/2}\\ \\b^2=3^3=\boxed{27}$

Answer:  27
Problem: Suppose that $|x_i| < 1$ for $i = 1, 2, \dots, n$ .  Suppose further that $|x_1| + |x_2| + \dots + |x_n| = 19 + |x_1 + x_2 + \dots + x_n|.$ What is the smallest possible value of $n$

Solution: Since $|x_i| < 1$ then
\[|x_1| + |x_2| + \dots + |x_n| = 19 + |x_1 + x_2 + \dots + x_n| < n.\]
So $n \ge 20$ . We now just need to find an example where $n = 20$ : suppose $x_{2k-1} = \frac{19}{20}$ and $x_{2k} = -\frac{19}{20}$ ; then on the left hand side we have $\left|\frac{19}{20}\right| + \left|-\frac{19}{20}\right| + \dots + \left|-\frac{19}{20}\right| = 20\left(\frac{19}{20}\right) = 19$ . On the right hand side, we have $19 + \left|\frac{19}{20} - \frac{19}{20} + \dots - \frac{19}{20}\right| = 19 + 0 = 19$ , and so the equation can hold for $n = \boxed{020}$

Answer:  020
Problem: Suppose that $|x_i| < 1$ for $i = 1, 2, \dots, n$ .  Suppose further that $|x_1| + |x_2| + \dots + |x_n| = 19 + |x_1 + x_2 + \dots + x_n|.$ What is the smallest possible value of $n$

Solution: Straight off, we notice that the RHS must be greater than or equal to 19, because an absolute value only gives nonnegative values. It then becomes clear that $|x_1|+|x_2|+...+|x_n|\ge19$ . If each $x_n$ were equal to 1, then $n=19$ . However, $x_n<1$ , so there must be at least one extra term to satisfy the inequality. Therefore, $n=\boxed{020}$

Answer:  020
Problem: Let $m/n$ , in lowest terms, be the probability that a randomly chosen positive divisor of $10^{99}$ is an integer multiple of $10^{88}$ .  Find $m + n$

Solution: $10^{99} = 2^{99}5^{99}$ , so it has $(99 + 1)(99 + 1) = 10000$ factors. Out of these, we only want those factors of $10^{99}$ which are divisible by $10^{88}$ ; it is easy to draw a bijection to the number of factors that $10^{11} = 2^{11}5^{11}$ has, which is $(11 + 1)(11 + 1) = 144$ . Our probability is $\frac{m}{n} = \frac{144}{10000} = \frac{9}{625}$ , and $m + n = \boxed{634}$

Answer:  634
Problem: It is possible to place positive integers into the vacant twenty-one squares of the $5 \times 5$ square shown below so that the numbers in each row and column form arithmetic sequences.  Find the number that must occupy the vacant square marked by the asterisk (*).
1988 AIME-6.png

Solution: Let the coordinates of the square at the bottom left be $(0,0)$ , the square to the right $(1,0)$ , etc.
Label the leftmost column (from bottom to top) $0, a, 2a, 3a, 4a$ and the bottom-most row (from left to right) $0, b, 2b, 3b, 4b$ . Our method will be to use the given numbers to set up equations to solve for $a$ and $b$ , and then calculate $(*)$
$\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a & & & * & \\ \hline 3a & 74 & & & \\ \hline 2a & & & & 186 \\ \hline a & & 103 & & \\ \hline 0 & b & 2b & 3b & 4b \\ \hline \end{tabular}$
We can compute the squares at the intersections of two existing numbers in terms of $a$ and $b$ ; two such equations will give us the values of $a$ and $b$ . On the fourth row from the bottom, the common difference is $74 - 3a$ , so the square at $(2,3)$ has a value of $148 - 3a$ . On the third column from the left, the common difference is $103 - 2b$ , so that square also has a value of $2b + 3(103 - 2b) = 309 - 4b$ . Equating, we get $148 - 3a = 309 - 4b \Longrightarrow 4b - 3a = 161$
Now we compute the square $(2,2)$ . By rows, this value is simply the average of $2a$ and $186$ , so it is equal to $\frac{2a + 186}{2} = a + 93$ . By columns, the common difference is $103 - 2b$ , so our value is $206 - 2b$ . Equating, $a + 93 = 206 - 2b \Longrightarrow a + 2b = 113$
Solving \begin{align*}4b - 3a &= 161\\ a + 2b &= 113 \end{align*}
gives $a = 13$ $b = 50$ . Now it is simple to calculate $(4,3)$ . One way to do it is to see that $(2,2)$ has $206 - 2b = 106$ and $(4,2)$ has $186$ , so $(3,2)$ has $\frac{106 + 186}{2} = 146$ . Now, $(3,0)$ has $3b = 150$ , so $(3,2) = \frac{(3,0) + (3,4)}{2} \Longrightarrow (3,4) = * = \boxed{142}$

Answer:  142
Problem: It is possible to place positive integers into the vacant twenty-one squares of the $5 \times 5$ square shown below so that the numbers in each row and column form arithmetic sequences.  Find the number that must occupy the vacant square marked by the asterisk (*).
1988 AIME-6.png

Solution: First, let $a =$ the number to be placed in the first column, fourth row.  Let $b =$ the number to be placed in the second column, fifth row.  We can determine the entire first column and fifth row in terms of $a$ and $b$
$\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a & & & & \\ \hline 3a & & & & \\ \hline 2a & & & & \\ \hline a & & & & \\ \hline 0 & b & 2b & 3b & 4b \\ \hline \end{tabular}$
Next, let $a + b + c =$ the number to be placed in the second column, fourth row.  We can determine the entire second column and fourth row in terms of $a$ $b$ , and $c$
$\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a & 4a + b + 4c & & & \\ \hline 3a & 3a + b + 3c & & & \\ \hline 2a & 2a + b + 2c & & & \\ \hline a & a + b + c & a + 2b + 2c & a + 3b + 3c & a + 4b + 4c \\ \hline 0 & b & 2b & 3b & 4b \\ \hline \end{tabular}$
We have now determined at least two values in each row and column.  We can finish the table without introducing any more variables:
$\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a & 4a + b + 4c & 4a + 2b + 8c & 4a + 3b + 12c & 4a + 4b + 16c \\ \hline 3a & 3a + b + 3c & 3a + 2b + 6c & 3a + 3b + 9c & 3a + 4b + 12c \\ \hline 2a & 2a + b + 2c & 2a + 2b + 4c & 2a + 3b + 6c & 2a + 4b + 8c \\ \hline a & a + b + c & a + 2b + 2c & a + 3b + 3c & a + 4b + 4c \\ \hline 0 & b & 2b & 3b & 4b \\ \hline \end{tabular}$
We now have a system of equations.
Solving, we find that $(a,b,c) = (13,50, - 5)$ .  The number in the square marked by the asterisk is $4a + 3b + 12c = \boxed{142}$

Answer:  142
Problem: It is possible to place positive integers into the vacant twenty-one squares of the $5 \times 5$ square shown below so that the numbers in each row and column form arithmetic sequences.  Find the number that must occupy the vacant square marked by the asterisk (*).
1988 AIME-6.png

Solution: We begin with the table that was given to us and add in the following arithmetic progression on the bottom:
$\begin{tabular}[b]{|c|c|c|c|c|}\hline & & & * & \\ \hline & 74 & & & \\ \hline & & & & 186 \\ \hline & & 103 & & \\ \hline 0 & x & 2x & 3x & 4x \\ \hline \end{tabular}$
Since all the rows and columns satisfy an arithmetic progression, we have the following:
$\begin{tabular}[b]{|c|c|c|c|c|}\hline & & 412 - 6x & 392 - 5x & 372 - 4x \\ \hline & 74 & 309 - 4x  & 294 - 3x  & 279 - 2x \\ \hline & & 206 - 2x & 196 - x & 186 \\ \hline & & 103 & 98 + x & 2x + 93 \\ \hline 0 & x & 2x & 3x & 4x \\ \hline \end{tabular}$
We can solve for $x$ in the 2nd row, namely $324 - 5x = 74$ because the arithmetic progression from left to right has difference $x - 15$ . Therefore, we have $x = 50$ , and because the desired asterisk is $392 - 5x$ , the answer is $392 - 250$ $\boxed{142}$

Answer:  142
Problem: In triangle $ABC$ $\tan \angle CAB = 22/7$ , and the altitude from $A$ divides $BC$ into segments of length 3 and 17.  What is the area of triangle $ABC$

Solution: Call $\angle BAD$ $\alpha$ and $\angle CAD$ $\beta$ . So, $\tan \alpha = \frac {17}{h}$ and $\tan \beta = \frac {3}{h}$ . Using the tangent addition formula $\tan (\alpha + \beta) = \dfrac {\tan \alpha + \tan \beta}{1 - \tan \alpha \cdot \tan \beta}$ , we get $\tan (\alpha + \beta) = \dfrac {\frac {20}{h}}{\frac {h^2 - 51}{h^2}} = \frac {22}{7}$
Simplifying, we get $\frac {20h}{h^2 - 51} = \frac {22}{7}$ . Cross-multiplying and simplifying, we get $11h^2-70h-561 = 0$ . Factoring, we get $(11h+51)(h-11) = 0$ , so we take the positive positive solution, which is $h = 11$ . Therefore, the answer is $\frac {20 \cdot 11}{2} = 110$ , so the answer is $\boxed{110}$

Answer:  110