Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: A two-digit positive integer is said to be $\emph{cuddly}$ if it is equal to the sum of its nonzero tens digit and the square of its units digit. How many two-digit positive integers are cuddly?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4$

Solution: If the tens digit is $a$ and the ones digit is $b$ then the number is $10a+b$ so we have the equation $10a + b = a + b^2$ . We can guess and check after narrowing the possible cuddly numbers down to $13,14,24,25,35,36,46,47,57,68,78,89,$ and $99$ . (We can narrow it down to these by just thinking about how $a$ 's value affects $b$ 's value and then check all the possiblities.) Checking all of these we get that there is only $\boxed{1}$ 2-digit cuddly number, and it is $89$ . Yay!!!

Answer: B 1
Problem: A school has $100$ students and $5$ teachers. In the first period, each student is taking one class, and each teacher is teaching one class. The enrollments in the classes are $50, 20, 20, 5,$ and $5$ . Let $t$ be the average value obtained if a teacher is picked at random and the number of students in their class is noted. Let $s$ be the average value obtained if a student was picked at random and the number of students in their class, including the student, is noted. What is $t-s$
$\textbf{(A)}\ {-}18.5  \qquad\textbf{(B)}\ {-}13.5 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 13.5 \qquad\textbf{(E)}\ 18.5$

Solution: The formula for expected values is \[\text{Expected Value}=\sum(\text{Outcome}\cdot\text{Probability}).\] We have \begin{align*} t &= 50\cdot\frac15 + 20\cdot\frac15 + 20\cdot\frac15 + 5\cdot\frac15 + 5\cdot\frac15 \\ &= (50+20+20+5+5)\cdot\frac15 \\ &= 100\cdot\frac15 \\ &= 20, \\ s &= 50\cdot\frac{50}{100} + 20\cdot\frac{20}{100} + 20\cdot\frac{20}{100} + 5\cdot\frac{5}{100} + 5\cdot\frac{5}{100} \\ &= 25 + 4 + 4 + 0.25 + 0.25 \\ &= 33.5. \end{align*} Therefore, the answer is $t-s=\boxed{13.5}.$

Answer: B 13.5
Problem: Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts $210$ equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts $42$ steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship?
$\textbf{(A) }70\qquad\textbf{(B) }84\qquad\textbf{(C) }98\qquad\textbf{(D) }105\qquad\textbf{(E) }126$

Solution: Let $x$ be the length of the ship.
Then, in the time that Emily walks $210$ steps, the ship moves $210-x$ steps.
Also, in the time that Emily walks $42$ steps, the ship moves $x-42$ steps.
Since the ship and Emily have the same ratio of absolute speeds in either direction, $\frac{210}{210-x} = \frac{42}{x-42}$ . Dividing both sides by $42$ and cross multiplying, we get $5(x-42) = 210-x$ , so $6x = 420$ , and $x = \boxed{70}$

Answer: A 70
Problem: Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts $210$ equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts $42$ steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship?
$\textbf{(A) }70\qquad\textbf{(B) }84\qquad\textbf{(C) }98\qquad\textbf{(D) }105\qquad\textbf{(E) }126$

Solution: Let the speed at which Emily walks be $42$ steps per hour. Let the speed at which the ship is moving be $s$ . Walking in the direction of the ship, it takes her $210$ steps, or $\frac {210}{42} = 5$ hours, to travel. We can create an equation: \[d = 5(42-s),\] where $d$ is the length of the ship. Walking in the opposite direction of the ship, it takes her $42$ steps, or $42/42 = 1$ hour. We can create a similar equation: \[d = 1(42+s).\] Now we have two variables and two equations. We can equate the expressions for $d$ and solve for $s$ \begin{align*} 210-5s &= 42 + s \\ s &= 28. \\ \end{align*} Therefore, we have $d = 42 + s = \boxed{70}$

Answer: A 70
Problem: Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts $210$ equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts $42$ steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship?
$\textbf{(A) }70\qquad\textbf{(B) }84\qquad\textbf{(C) }98\qquad\textbf{(D) }105\qquad\textbf{(E) }126$

Solution: Suppose that Emily and the ship take steps simultaneously such that Emily's steps cover a greater length than the ship's steps.
Let $L$ be the length of the ship, $E$ be Emily's step length, and $S$ be the ship's step length. We wish to find $\frac LE.$
When Emily walks from the back of the ship to the front, she walks a distance of $210E$ and the front of the ship moves a distance of $210S.$ We have $210E=L+210S$ for this scenario, which rearranges to \[210E-210S=L. \hspace{15mm}(1)\] When Emily walks in the opposite direction, she walks a distance of $42E$ and the back of the ship moves a distance of $42S.$ We have $42E=L-42S$ for this scenario, which rearranges to \[42E+42S=L. \hspace{19.125mm}(2)\] We multiply $(2)$ by $5$ and then add $(1)$ to get $420E=6L,$ from which $\frac LE = \boxed{70}.$

Answer: A 70
Problem: Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts $210$ equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts $42$ steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship?
$\textbf{(A) }70\qquad\textbf{(B) }84\qquad\textbf{(C) }98\qquad\textbf{(D) }105\qquad\textbf{(E) }126$

Solution: Every time Emily takes a step, the boat also "takes a step". Call the length of the boats step $s$ . Call the length of the boat $x$
When Emily is walking in the same direction as the boat, every time she takes a step the boat moves an additional distance of $s$ . This means that she travels a total distance of $x + 210 s$ to reach the other end of the boat.
When Emily is walking in the opposite direction of the boat, every time she takes a step the distance till the end of the boat reduces by $s$ (since the boat is coming towards her and moves a distance of $s$ ). This means that she travels a total distance of $x - 42 s$ to reach the other end of the boat.
Taking Emily's step as a unit of distance, we now have two equations \begin{align*} 210 &= x + 210 s, \\ 42 &= x - 42s. \end{align*} Solving for $x$ you get $\boxed{70}$

Answer: A 70
Problem: Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts $210$ equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts $42$ steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship?
$\textbf{(A) }70\qquad\textbf{(B) }84\qquad\textbf{(C) }98\qquad\textbf{(D) }105\qquad\textbf{(E) }126$

Solution: Call the speed of the boat $v_s$ and the speed of Emily $v_e$
Consider the scenario when Emily is walking along with the boat. Relative to an observer on the boat, her speed is $v_e-v_s$
Consider the scenario when Emily is walking in the opposite direction. Relative to an observer on the boat, her speed is $v_e+v_s$
Since Emily takes $210$ steps to walk along with the boat and $42$ steps to walk opposite the boat, that means it takes her $5$ times longer to walk the length of a stationary boat at $v_e-v_s$ compared to $v_e+v_s$
This means that $5(v_e-v_s)=v_e+v_s$ , so $v_s = \frac{2v_e}{3}$
As Emily takes $210$ steps to walk the length of the boat at a speed of $v_e- \frac{2v_e}{3}=\frac{v_e}{3}$ , she must take $\frac13$ of the time to walk the length of the boat at a speed of $v_e$ , so our answer is $\frac{210}{3} = \boxed{70}$

Answer: A 70
Problem: The base-nine representation of the number $N$ is $27006000052_{\text{nine}}.$ What is the remainder when $N$ is divided by $5?$
$\textbf{(A) } 0\qquad\textbf{(B) } 1\qquad\textbf{(C) } 2\qquad\textbf{(D) } 3\qquad\textbf{(E) }4$

Solution: Recall that $9\equiv-1\pmod{5}.$ We expand $N$ by the definition of bases: \begin{align*} N&=27006000052_9 \\ &= 2\cdot9^{10} + 7\cdot9^9 + 6\cdot9^6 + 5\cdot9 + 2 \\ &\equiv 2\cdot(-1)^{10} + 7\cdot(-1)^9 + 6\cdot(-1)^6 + 5\cdot(-1) + 2 &&\pmod{5} \\ &\equiv 2-7+6-5+2 &&\pmod{5} \\ &\equiv -2 &&\pmod{5} \\ &\equiv \boxed{3} ~Aidensharp ~Kante314 ~MRENTHUSIASM

Answer: D 3
Problem: The base-nine representation of the number $N$ is $27006000052_{\text{nine}}.$ What is the remainder when $N$ is divided by $5?$
$\textbf{(A) } 0\qquad\textbf{(B) } 1\qquad\textbf{(C) } 2\qquad\textbf{(D) } 3\qquad\textbf{(E) }4$

Solution: We need to first convert $N$ into a regular base- $10$ number: \[N = 27006000052_9 = 2\cdot9^{10} + 7\cdot9^9 + 6\cdot9^6 + 5\cdot9 + 2.\]
Now, consider how the last digit of $9$ changes with changes of the power of $9:$ \begin{align*} 9^0&=1, \\ 9^1&=9, \\ 9^2&=\ldots 1, \\ 9^3&=\ldots 9, \\ 9^4&=\ldots 1, \\ & \ \vdots \end{align*} Note that if $x$ is odd, then $9^x \equiv 4\pmod{5}.$ On the other hand, if $x$ is even, then $9^x \equiv 1\pmod{5}.$
Therefore, we have \begin{align*} N&\equiv 2\cdot(1) + 7\cdot(4) + 6\cdot(1) + 5\cdot(4) + 2\cdot(1) &&\pmod{5} \\ &\equiv 2+28+6+20+2 &&\pmod{5} \\ &\equiv 58 &&\pmod{5} \\ &\equiv \boxed{3}$ may simplify the process further, as given by Solution 1.

Answer: D 3
Problem: How many ordered pairs $(x,y)$ of real numbers satisfy the following system of equations? \begin{align*} x^2+3y&=9 \\ (|x|+|y|-4)^2 &= 1 \end{align*} $\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 7$

Solution: The second equation is $(|x|+|y| - 4)^2 = 1$ . We know that the graph of $|x| + |y|$ is a very simple diamond shape, so let's see if we can reduce this equation to that form: \[(|x|+|y| - 4)^2 = 1 \implies |x|+|y| - 4 = \pm 1 \implies |x|+|y| = \{3,5\}.\] We now have two separate graphs for this equation and one graph for the first equation, so let's put it on the coordinate plane: [asy]  Label f;  f.p=fontsize(6);  xaxis(-8,8,Ticks(f, 1.0,0.5));  yaxis(-8,8,Ticks(f, 1.0,0.5));  real f(real x)  {  return 3-x;  }  draw(graph(f,0,3));  real f(real x)  {  return 3+x;  }  draw(graph(f,0,-3)); real f(real x)  {  return x-3;  }  draw(graph(f,0,3)); real f(real x)  {  return -x-3;  }  draw(graph(f,0,-3)); real f(real x)  {  return 5-x;  }  draw(graph(f,0,5));  real f(real x)  {  return 5+x;  }  draw(graph(f,0,-5)); real f(real x)  {  return x-5;  }  draw(graph(f,0,5)); real f(real x)  {  return -x-5;  }  draw(graph(f,0,-5));  real f(real x)  {  return 3-x;  }  draw(graph(f,0,3));  real f(real x)  {  return 3+x;  }  draw(graph(f,0,-3)); real f(real x)  {  return x-3;  }  draw(graph(f,0,3)); real f(real x)  {  return (-x^2)/3+3;  }  draw(graph(f,-5,5)); [/asy] We see from the graph that there are $5$ intersections, so the answer is $\boxed{5}$

Answer: D 5
Problem: Isosceles triangle $ABC$ has $AB = AC = 3\sqrt6$ , and a circle with radius $5\sqrt2$ is tangent to line $AB$ at $B$ and to line $AC$ at $C$ . What is the area of the circle that passes through vertices $A$ $B$ , and $C?$
$\textbf{(A) }24\pi\qquad\textbf{(B) }25\pi\qquad\textbf{(C) }26\pi\qquad\textbf{(D) }27\pi\qquad\textbf{(E) }28\pi$

Solution: Let $\odot O_1$ be the circle with radius $5\sqrt2$ that is tangent to $\overleftrightarrow{AB}$ at $B$ and to $\overleftrightarrow{AC}$ at $C.$ Note that $\angle ABO_1 = \angle ACO_1 = 90^\circ.$ Since the opposite angles of quadrilateral $ABO_1C$ are supplementary, quadrilateral $ABO_1C$ is cyclic.
Let $\odot O_2$ be the circumcircle of quadrilateral $ABO_1C.$ It follows that $\odot O_2$ is also the circumcircle of $\triangle ABC,$ as shown below: [asy] /* Made by MRENTHUSIASM */ size(200); pair A, B, C, D, O1, O2; A = (0,2sqrt(26)); O1 = (0,0); B = intersectionpoints(Circle(A,3sqrt(6)),Circle(O1,5sqrt(2)))[0]; C = intersectionpoints(Circle(A,3sqrt(6)),Circle(O1,5sqrt(2)))[1]; O2 = midpoint(A--O1); fill(A--B--C--cycle, yellow); dot("$A$",A,1.5*N,linewidth(4)); dot("$B$",B,1.5*W,linewidth(4)); dot("$C$",C,1.5*E,linewidth(4)); dot("$O_1$",O1,1.5*S,linewidth(4)); dot("$O_2$",O2,1.5*N,linewidth(4)); label("$3\sqrt6$",midpoint(A--B),scale(0.5)*rotate(90)*dir(midpoint(A--B)--A),red+fontsize(10)); label("$3\sqrt6$",midpoint(A--C),scale(0.5)*rotate(90)*dir(midpoint(A--C)--C),red+fontsize(10)); label("$5\sqrt2$",midpoint(O1--B),0.5*SW,red+fontsize(10)); label("$5\sqrt2$",midpoint(O1--C),0.5*SE,red+fontsize(10)); markscalefactor=0.05; draw(rightanglemark(A,B,O1)^^rightanglemark(A,C,O1),red); draw(A--B--O1--C--cycle^^B--C^^circumcircle(A,B,C)); [/asy] By the Inscribed Angle Theorem, we conclude that $\overline{AO_1}$ is the diameter of $\odot O_2.$ By the Pythagorean Theorem on right $\triangle ABO_1,$ we have \[AO_1 = \sqrt{AB^2 + BO_1^2} = 2\sqrt{26}.\] Therefore, the area of $\odot O_2$ is $\pi\cdot\left(\frac{AO_1}{2}\right)^2=\boxed{26}.$

Answer: C 26
Problem: Isosceles triangle $ABC$ has $AB = AC = 3\sqrt6$ , and a circle with radius $5\sqrt2$ is tangent to line $AB$ at $B$ and to line $AC$ at $C$ . What is the area of the circle that passes through vertices $A$ $B$ , and $C?$
$\textbf{(A) }24\pi\qquad\textbf{(B) }25\pi\qquad\textbf{(C) }26\pi\qquad\textbf{(D) }27\pi\qquad\textbf{(E) }28\pi$

Solution: [asy] import olympiad; unitsize(50); pair A,B,C,D,E,I,F,G,O; A=origin; B=(2,3); C=(-2,3); D=(4.3,6.3); E=(-4.3,6.3); F=(1,1.5); G=(-1,1.5); O=circumcenter(A,B,C); // olympiad - circumcenter I=incenter(A,D,E); draw(A--B--C--cycle); dot(O); dot(I); dot(F); dot(G); draw(circumcircle(A,B,C)); // olympiad - circumcircle draw(incircle(A,D,E)); draw(I--B); draw(I--C); draw(I--A); draw(rightanglemark(A,C,I));  draw(rightanglemark(A,B,I)); draw(O--F); draw(O--G); draw(rightanglemark(A,F,O));  draw(rightanglemark(A,G,O));  label("$O$",O,W); label("$A$",A,S); label("$B$",B,N); label("$C$",C,W); label("$D$",F,S); label("$E$",G,W);  label("$3\sqrt{6}$",(1.5,1.5),S); label("$3\sqrt{6}$",(-1.5,1.5),S); label("$5\sqrt{2}$",(1,3.625),N); label("$5\sqrt{2}$",(-1,3.625),N); label("$I$",I,N); label("$r$",(-0.25,1.5),E); label("$r$",(0.5,2.125),S); add(pathticks(A--F,1,0.5,0,2)); add(pathticks(F--B,1,0.5,0,2));  add(pathticks(A--G,1,0.5,0,2)); add(pathticks(G--C,1,0.5,0,2)); [/asy] Because circle $I$ is tangent to $\overline{AB}$ at $B, \angle{ABI} \cong 90^{\circ}$ . Because $O$ is the circumcenter of $\bigtriangleup ABC, \overline{OD}$ is the perpendicular bisector of $\overline{AB}$ , and $\angle{BAI} \cong \angle{DAO}$ , so therefore $\bigtriangleup ADO \sim \bigtriangleup ABI$ by AA similarity. Then we have $\frac{AD}{AB} = \frac{DO}{BI} \implies \frac{1}{2} = \frac{r}{5\sqrt{2}} \implies r = \frac{5\sqrt{2}}{2}$ . We also know that $\overline{AD} = \frac{3\sqrt{6}}{2}$ because of the perpendicular bisector, so the hypotenuse of $\bigtriangleup ADO$ is \[\sqrt{\left(\frac{5\sqrt{2}}{2}\right)^2+\left(\frac{3\sqrt{6}}{2}\right)^2} = \sqrt{\frac{25}{2}+\frac{27}{2}} = \sqrt{26}.\] This is the radius of the circumcircle of $\bigtriangleup ABC$ , so the area of this circle is $\boxed{26}$

Answer: C 26
Problem: Isosceles triangle $ABC$ has $AB = AC = 3\sqrt6$ , and a circle with radius $5\sqrt2$ is tangent to line $AB$ at $B$ and to line $AC$ at $C$ . What is the area of the circle that passes through vertices $A$ $B$ , and $C?$
$\textbf{(A) }24\pi\qquad\textbf{(B) }25\pi\qquad\textbf{(C) }26\pi\qquad\textbf{(D) }27\pi\qquad\textbf{(E) }28\pi$

Solution: Denote by $O$ the center of the circle that is tangent to line $AB$ at $B$ and to line $AC$ at $C$
Because this circle is tangent to line $AB$ at $B$ , we have $OB \perp AB$ and $OB = 5 \sqrt{2}$
Because this circle is tangent to line $AC$ at $C$ , we have $OC \perp AC$ and $OC = 5 \sqrt{2}$
Because $AB = AC$ $OB = OC$ $AO = AO$ , we get $\triangle ABO \cong \triangle ACO$ . Hence, $\angle BAO = \angle CAO$
Let $AO$ and $BC$ meet at point $D$ .
Because $AB = AC$ $\angle BAO = \angle CAO$ $AD = AD$ , we get $\triangle ABD \cong \triangle ACD$ . Hence, $BD = CD$ and $\angle ADB = \angle ADC = 90^\circ$
Denote $\theta = \angle BAO$ . Hence, $\angle BAC = 2 \theta$
Denote by $R$ the circumradius of $\triangle ABC$ .
In $\triangle ABC$ , following from the law of sines, $2 R = \frac{BC}{\sin \angle BAC}$
Therefore, the area of the circumcircle of $\triangle ABC$ is \begin{align*} \pi R^2 & = \pi \left( \frac{BC}{2 \sin \angle BAC} \right)^2 \\ & = \pi \left( \frac{2 BD}{2 \sin \angle BAC} \right)^2 \\ & = \pi \left( \frac{BD}{\sin 2 \theta} \right)^2 \\ & = \pi \left( \frac{AB \sin \theta }{\sin 2 \theta} \right)^2 \\ & = \pi \left( \frac{AB \sin \theta }{2 \sin \theta \cos \theta} \right)^2 \\ & = \pi \left( \frac{AB }{2 \cos \theta} \right)^2 \\ & = \pi \left( \frac{AO}{2} \right)^2 \\ & = \frac{\pi}{4} \left( AB^2 + OB^2 \right) \\ & = \boxed{26} ~Steven Chen (www.professorchenedu.com)

Answer: C 26
Problem: An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$ , which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of pillars at $A$ $B$ , and $C$ are $12$ $9$ , and $10$ meters, respectively. What is the height, in meters, of the pillar at $E$
$\textbf{(A) }9 \qquad\textbf{(B) } 6\sqrt{3} \qquad\textbf{(C) } 8\sqrt{3} \qquad\textbf{(D) } 17 \qquad\textbf{(E) }12\sqrt{3}$

Solution: The pillar at $B$ has height $9$ and the pillar at $A$ has height $12.$ Since the solar panel is flat, the inclination from pillar $B$ to pillar $A$ is $3.$ Call the center of the hexagon $G.$ Since $\overrightarrow{CG}\parallel\overrightarrow{BA},$ it follows that the solar panel has height $13$ at $G.$ Since the solar panel is flat, the heights of the solar panel at $B,G,$ and $E$ are collinear. Therefore, the pillar at $E$ has height $9+4+4=\boxed{17}.$

Answer: D 17
Problem: An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$ , which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of pillars at $A$ $B$ , and $C$ are $12$ $9$ , and $10$ meters, respectively. What is the height, in meters, of the pillar at $E$
$\textbf{(A) }9 \qquad\textbf{(B) } 6\sqrt{3} \qquad\textbf{(C) } 8\sqrt{3} \qquad\textbf{(D) } 17 \qquad\textbf{(E) }12\sqrt{3}$

Solution: Let the height of the pillar at $D$ be $x.$ Notice that the difference between the heights of pillar $C$ and pillar $D$ is equal to the difference between the heights of pillar $A$ and pillar $F.$ So, the height at $F$ is $x+2.$ Now, doing the same thing for pillar $E$ we get the height is $x+3.$ Therefore, we can see the difference between the heights at pillar $C$ and pillar $D$ is half the difference between the heights at $B$ and $E,$ so \begin{align*} x+3-9&=2 \cdot (x-10) \\  x-6&=2 \cdot (x-10) \\ x&=14. \end{align*} The answer is $x+3=\boxed{17}.$

Answer: D 17
Problem: An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$ , which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of pillars at $A$ $B$ , and $C$ are $12$ $9$ , and $10$ meters, respectively. What is the height, in meters, of the pillar at $E$
$\textbf{(A) }9 \qquad\textbf{(B) } 6\sqrt{3} \qquad\textbf{(C) } 8\sqrt{3} \qquad\textbf{(D) } 17 \qquad\textbf{(E) }12\sqrt{3}$

Solution: We can extend $BA$ and $BC$ to $G$ and $H$ , respectively, such that $AG = CH$ and $E$ lies on $\overline{GH}$ [asy] unitsize(1cm); pair A = (-sqrt(3),1); pair B = (0,2); pair C = (sqrt(3),1); pair D = (sqrt(3),-1); pair E = (0,-2); pair F = (-sqrt(3),-1); draw(A--B--C--D--E--F--cycle); label("$A, 12$", A, NW); label("$B, 9$", B, N); label("$C, 10$", C, NE); label("$D$", D, SE); label("$E$", E, S); label("$F$", F, SW);  pair G = (-4*sqrt(3),-2); pair H = (4*sqrt(3),-2); label("$G, 21$", G, S); label("$H, 13$", H, S); draw(A--G, dashed); draw(C--H, dashed); dot(A^^B^^C^^D^^E^^F^^G^^H,linewidth(4.5)); [/asy] Because of hexagon proportions, $\frac{BA}{AG} = \frac{1}{3}$ and $\frac{BC}{CH} = \frac{1}{3}$ . Let $g$ be the height of $G$ . Because $A$ $B$ and $G$ lie on the same line, $\frac{12-9}{g-12} = \frac{1}{3}$ , so $g-12 = 9$ and $g = 21$ . Similarly, the height of $H$ is $13$ $E$ is the midpoint of $GH$ , so we can take the average of these heights to get our answer, $\boxed{17}$

Answer: D 17
Problem: An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$ , which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of pillars at $A$ $B$ , and $C$ are $12$ $9$ , and $10$ meters, respectively. What is the height, in meters, of the pillar at $E$
$\textbf{(A) }9 \qquad\textbf{(B) } 6\sqrt{3} \qquad\textbf{(C) } 8\sqrt{3} \qquad\textbf{(D) } 17 \qquad\textbf{(E) }12\sqrt{3}$

Solution: Denote by $h_X$ the height of any point $X$
Denote by $M$ the midpoint of $A$ and $C$ .
Hence, \[h_M = \frac{h_A + h_C}{2} = 11.\] Denote by $O$ the center of $ABCDEF$ . Because $ABCDEF$ is a regular hexagon, $O$ is the midpoint of $B$ and $E$ .
Hence, \[h_O = \frac{h_E + h_B}{2} = \frac{h_E + 9}{2}.\] Because $ABCDEF$ is a regular hexagon, $M$ is the midpoint of $B$ and $O$ .
Hence, \[h_M = \frac{h_B + h_O}{2} = \frac{9 + h_O}{2}.\] Solving these equations, we get $h_E = \boxed{17}$

Answer: D 17
Problem: An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$ , which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of pillars at $A$ $B$ , and $C$ are $12$ $9$ , and $10$ meters, respectively. What is the height, in meters, of the pillar at $E$
$\textbf{(A) }9 \qquad\textbf{(B) } 6\sqrt{3} \qquad\textbf{(C) } 8\sqrt{3} \qquad\textbf{(D) } 17 \qquad\textbf{(E) }12\sqrt{3}$

Solution: In this solution, we define rise as the change of height (in meters) from the solar panel to the ground. It follows that the rise from $B$ to $A$ is $12-9=3,$ and the rise from $B$ to $C$ is $10-9=1.$ Note that $\vec{BE}=2\vec{BA}+2\vec{BC},$ so the rise from $B$ to $E$ is $2\cdot3+2\cdot1=8.$
Together, the height of the pillar at $E$ is $9+8=\boxed{17}$ meters.

Answer: D 17