Answer: 20.0

Elmer's new car gives $50\%$ percent better fuel efficiency, measured in kilometers per liter, than his old car. However, his new car uses diesel fuel, which is $20\%$ more expensive per liter than the gasoline his old car used. By what percent will Elmer save money if he uses his new car instead of his old car for a long trip?
$\textbf{(A) } 20\% \qquad \textbf{(B) } 26\tfrac23\% \qquad \textbf{(C) } 27\tfrac79\% \qquad \textbf{(D) } 33\tfrac13\% \qquad \textbf{(E) } 66\tfrac23\%$

Solution: Suppose that his old car runs at $x$ km per liter. Then his new car runs at $\frac{3}{2}x$ km per liter, or $x$ km per $\frac{2}{3}$ of a liter. Let the cost of the old car's fuel be $c$ , so the trip in the old car takes $xc$ dollars, while the trip in the new car takes $\frac{2}{3}\cdot\frac{6}{5}xc = \frac{4}{5}xc$ . He saves $\frac{\frac{1}{5}xc}{xc} = \boxed{20}$

Solution: Because they do not give you a given amount of distance, we'll just make that distance $3x$ miles. Then, we find that the new car will use $2*1.2=2.4x$ . The old car will use $3x$ . Thus the answer is $(3-2.4)/3=.6/3=20/100=  \boxed{20}$

Solution: You can find that the ratio of fuel used by the old car and the new car for the same amount of distance is $3 : 2$ , and the ratio between the fuel price of these two cars is $5 : 6$ . Therefore, by multiplying these two ratios, we get that the costs of using these two cars is \[15 : 12 = 5 : 4\] So the percentage of money saved is $1 - \frac{4}{5} = \boxed{20}$

Solution: Assume WLOG that Elmer's old car's range is $100$ miles. So, Elmer's new car's range is $100 \times 1.5 = 150$ miles. Also, assume that the gas Elmer's old car uses is $$10$ , which means that diesel will cost $$12$ . Now we can deduce that Elmer's old car uses $10 \div 100 = $0.10$ per mile, and Elmer's new car uses $12 \div 150 = $0.08$ per mile. Therefore, Elmer's new car saves $\boxed{20}$ more money than his old car.


Answer: 70.0

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 $x = \boxed{70}$

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 $d = 42 + s = \boxed{70}$

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 $\frac LE = \boxed{70}.$

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 $\boxed{70}$

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: 6.0

Equiangular hexagon $ABCDEF$ has side lengths $AB=CD=EF=1$ and $BC=DE=FA=r$ . The area of $\triangle ACE$ is $70\%$ of the area of the hexagon. What is the sum of all possible values of $r$
$\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qquad \textbf{(E)}\ 6$

Solution:
It is clear that $\triangle ACE$ is an equilateral triangle. From the Law of Cosines on $\triangle ABC$ , we get that $AC^2 = r^2+1^2-2r\cos{\frac{2\pi}{3}} = r^2+r+1$ . Therefore, the area of $\triangle ACE$ is $\frac{\sqrt{3}}{4}(r^2+r+1)$ by area of an equilateral triangle.
If we extend $BC$ $DE$ and $FA$ so that $FA$ and $BC$ meet at $X$ $BC$ and $DE$ meet at $Y$ , and $DE$ and $FA$ meet at $Z$ , we find that hexagon $ABCDEF$ is formed by taking equilateral triangle $XYZ$ of side length $r+2$ and removing three equilateral triangles, $ABX$ $CDY$ and $EFZ$ , of side length $1$ . The area of $ABCDEF$ is therefore
$\frac{\sqrt{3}}{4}(r+2)^2-\frac{3\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(r^2+4r+1)$
Based on the initial conditions,
\[\frac{\sqrt{3}}{4}(r^2+r+1) = \frac{7}{10}\left(\frac{\sqrt{3}}{4}\right)(r^2+4r+1)\]
Simplifying this gives us $r^2-6r+1 = 0$ . By Vieta's Formulas we know that the sum of the possible value of $r$ is $\boxed{6}$


Answer: 6.0

Solution:
It is clear that $\triangle ACE$ is an equilateral triangle. From the Law of Cosines on $\triangle ABC$ , we get that $AC^2 = r^2+1^2-2r\cos{\frac{2\pi}{3}} = r^2+r+1$ . Therefore, the area of $\triangle ACE$ is $\frac{\sqrt{3}}{4}(r^2+r+1)$ by area of an equilateral triangle.
If we extend $BC$ $DE$ and $FA$ so that $FA$ and $BC$ meet at $X$ $BC$ and $DE$ meet at $Y$ , and $DE$ and $FA$ meet at $Z$ , we find that hexagon $ABCDEF$ is formed by taking equilateral triangle $XYZ$ of side length $r+2$ and removing three equilateral triangles, $ABX$ $CDY$ and $EFZ$ , of side length $1$ . The area of $ABCDEF$ is therefore
$\frac{\sqrt{3}}{4}(r+2)^2-\frac{3\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(r^2+4r+1)$
Based on the initial conditions,
\[\frac{\sqrt{3}}{4}(r^2+r+1) = \frac{7}{10}\left(\frac{\sqrt{3}}{4}\right)(r^2+4r+1)\]
Simplifying this gives us $r^2-6r+1 = 0$ . By Vieta's Formulas we know that the sum of the possible value of $r$ is $\boxed{6}$


Answer: 450.0

Equilateral $\triangle ABC$ has side length $600$ . Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$ , and $QA=QB=QC$ , and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes). There is a point $O$ whose distance from each of $A,B,C,P,$ and $Q$ is $d$ . Find $d$

Solution: The inradius of $\triangle ABC$ is $100\sqrt 3$ and the circumradius is $200 \sqrt 3$ . Now, consider the line perpendicular to plane $ABC$ through the circumcenter of $\triangle ABC$ . Note that $P,Q,O$ must lie on that line to be equidistant from each of the triangle's vertices. Also, note that since $P, Q, O$ are collinear, and $OP=OQ$ , we must have $O$ is the midpoint of $PQ$ .  Now, Let $K$ be the circumcenter of $\triangle ABC$ , and $L$ be the foot of the altitude from $A$ to $BC$ . We must have $\tan(\angle KLP+ \angle QLK)= \tan(120^{\circ})$ . Setting $KP=x$ and $KQ=y$ , assuming WLOG $x>y$ , we must have $\tan(120^{\circ})=-\sqrt{3}=\dfrac{\dfrac{x+y}{100 \sqrt{3}}}{\dfrac{30000-xy}{30000}}$ . Therefore, we must have $100(x+y)=xy-30000$ . Also, we must have $\left(\dfrac{x+y}{2}\right)^{2}=\left(\dfrac{x-y}{2}\right)^{2}+120000$ by the Pythagorean theorem, so we have $xy=120000$ , so substituting into the other equation we have $90000=100(x+y)$ , or $x+y=900$ . Since we want $\dfrac{x+y}{2}$ , the desired answer is $\boxed{450}$

Solution: Draw a good diagram. Draw $CH$ as an altitude of the triangle. Scale everything down by a factor of $100\sqrt{3}$ , so that $AB=2\sqrt{3}$ . Finally, call the center of the triangle U. Draw a cross-section of the triangle via line $CH$ , which of course includes $P, Q$ . From there, we can call $OU=h$ . There are two crucial equations we can thus generate. WLOG set $PU<QU$ , then we call $PU=d-h, QU=d+h$ . First equation: using the Pythagorean Theorem on $\triangle UOB$ $h^2+2^2=d^2$ . Next, using the tangent addition formula on angles $\angle PHU, \angle UHQ$ we see that after simplifying $-d^2+h^2=-4, 2d=3\sqrt{3}$ in the numerator, so $d=\frac{3\sqrt{3}}{2}$ . Multiply back the scalar and you get $\boxed{450}$ . Not that hard, was it?

Solution:
To make numbers more feasible, we'll scale everything down by a factor of $100$ so that $\overline{AB}=\overline{BC}=\overline{AC}=6$ . We should also note that $P$ and $Q$ must lie on the line that is perpendicular to the plane of $ABC$ and also passes through the circumcenter of $ABC$ (due to $P$ and $Q$ being equidistant from $A$ $B$ $C$ ), let $D$ be the altitude from $C$ to $AB$ . We can draw a vertical cross-section of the figure then:  We let $\angle PDI=\alpha$ so $\angle QDI=120^{\circ}-\alpha$ , also note that $\overline{PO}=\overline{QO}=\overline{CO}=d$ . Because $I$ is the centroid of $ABC$ , we know that ratio of $\overline{CI}$ to $\overline{DI}$ is $2:1$ . Since we've scaled the figure down, the length of $CD$ is $3\sqrt{3}$ , from this it's easy to know that $\overline{CI}=2\sqrt{3}$ and $\overline{DI}=\sqrt{3}$ . The following two equations arise: \begin{align} \sqrt{3}\tan{\left(\alpha\right)}+\sqrt{3}\tan{\left(120^{\circ}-\alpha\right)}&=2d \\ \sqrt{3}\tan{\left(\alpha\right)} - d &= \sqrt{d^{2}-12} \end{align} Using trig identities for the tangent, we find that \begin{align*} \sqrt{3}\tan{\left(120^{\circ}-\alpha\right)}&=\sqrt{3}\left(\frac{\tan{\left(120^{\circ}\right)}+\tan{\left(\text{-}\alpha\right)}}{1-\tan{\left(120^{\circ}\right)}\tan{\left(\text{-}\alpha\right)}}\right) \\ &= \sqrt{3}\left(\frac{\text{-}\sqrt{3}+\tan{\left(\text{-}\alpha\right)}}{1+\sqrt{3}\tan{\left(\text{-}\alpha\right)}}\right) \\ &= \sqrt{3}\left(\frac{\text{-}\sqrt{3}-\tan{\left(\alpha\right)}}{1-\sqrt{3}\tan{\left(\alpha\right)}}\right) \\ &= \frac{\sqrt{3}\tan{\left(\alpha\right)}+3}{\sqrt{3}\tan{\left(\alpha\right)}-1}.\end{align*} Okay, now we can plug this into $\left(1\right)$ to get: \begin{align}\sqrt{3}\tan{\left(\alpha\right)}+\frac{\sqrt{3}\tan{\left(\alpha\right)}+3}{\sqrt{3}\tan{\left(\alpha\right)}-1}&=2d \\ \sqrt{3}\tan{\left(\alpha\right)} - d &= \sqrt{d^{2}-12} \end{align} Notice that $\alpha$ only appears in the above system of equations in the form of $\sqrt{3}\tan{\left(\alpha\right)}$ , we can set $\sqrt{3}\tan{\left(\alpha\right)}=a$ for convenience since we really only care about $d$ . Now we have \begin{align}a+\frac{a+3}{a-1}&=2d \\ a - d &= \sqrt{d^{2}-12} \end{align} Looking at $\left(2\right)$ , it's tempting to square it to get rid of the square-root so now we have: \begin{align*}a^{2}-2ad+d^{2}&=d^{2}-12 \\ a - 2ad &= \text{-}12 \end{align*} See the sneaky $2d$ in the above equation? That we means we can substitute it for $a+\frac{a+3}{a-1}$ \begin{align*}a^{2}-2ad+d^{2}&=d^{2}-12 \\ a^{2} - a\left(a+\frac{a+3}{a-1}\right) &= \text{-}12 \\ a^{2}-a^{2}-\frac{a^{2}+3a}{a-1} &=\text{-}12 \\ -\frac{a^{2}+3a}{a-1}&=\text{-}12 \\ \text{-}a^{2}-3a&=\text{-}12a+12 \\ 0 &= a^{2}-9a+12 \end{align*} Use the quadratic formula, we find that $a=\frac{9\pm\sqrt{9^{2}-4\left(1\right)\left(12\right)}}{2\left(1\right)}=\frac{9\pm\sqrt{33}}{2}$ - the two solutions were expected because $a$ can be $\angle PDI$ or $\angle QDI$ . We can plug this into $\left(1\right)$ \begin{align*}a+\frac{a+3}{a-1}&=2d \\ \frac{9\pm\sqrt{33}}{2}+\frac{\frac{9\pm\sqrt{33}}{2}+3}{\frac{9\pm\sqrt{33}}{2}-1}=2d \\ \frac{9\pm\sqrt{33}}{2}+\frac{15\pm\sqrt{33}}{7\pm\sqrt{33}} &= 2d\end{align*} I'll use $a=\frac{9+\sqrt{33}}{2}$ because both values should give the same answer for $d$ \begin{align*} \frac{9+\sqrt{33}}{2}+\frac{15+\sqrt{33}}{7+\sqrt{33}} &= 2d \\ \frac{\left(9+\sqrt{33}\right)\left(7+\sqrt{33}\right)+\left(2\right)\left(15+\sqrt{33}\right)}{\left(2\right)\left(7+\sqrt{33}\right)} &= 2d \\ \frac{63+33+16\sqrt{33}+30+2\sqrt{33}}{14+2\sqrt{33}} &= 2d \\ \frac{126+18\sqrt{33}}{14+2\sqrt{33}} &= 2d \\ 9 &= 2d \\ \frac{9}{2} &= d\end{align*} Wait! Before you get excited, remember that we scaled the entire figure by $100$ ?? That means that the answer is $d=100\times\frac{9}{2}=\boxed{450}$ .
-fatant

Solution: We use the diagram from solution 3. From basic angle chasing, \[180=\angle{QOC}+\angle{CO}P=2\angle{OCP}+2\angle{OCQ}=2\angle{QCP}\] so triangle QCP is a right triangle. This means that triangles $CQI$ and $CPI$ are similar. If we let $\angle{IDQ}=x$ and $\angle{PDI}=y$ , then we know $x+y=120$ and \[\frac{PG}{GC}=\frac{GC}{GQ}\Rightarrow\frac{100\sqrt{3}\tan{y}}{200\sqrt{3}}=\frac{200\sqrt{3}}{100\sqrt{3}\tan{x}}\Rightarrow\tan{x}\tan{y}=4\] We also know that \[PQ=2d=100\sqrt{3}(\tan{x}+\tan{y})\] \[d=50\sqrt{3}(\tan{x}+\tan{y})\] \[\frac{d}{1-\tan{x}\tan{y}}=50\sqrt{3}\cdot\frac{\tan{x}+\tan{y}}{1-\tan{x}\tan{y}}\] \[\frac{d}{-3}=50\sqrt{3}\tan{(x+y)}\] \[d=-150\sqrt{3}\tan{120}=-150\sqrt{3}(-\sqrt{3})=\boxed{450}\]

Solution:
We use the diagram from solution 3.
Let $BP = a$ and $BQ = b$ . Then, by Stewart's on $BPQ$ , we find \[2x^3 + 2x^3 = a^2x + b^2x \implies a^2 + b^2 = 4x^2.\]
The altitude from $P$ to $ABC$ is $\sqrt{a^2 - (200\sqrt{3})^2}$ so \[PQ = 2x = \sqrt{a^2 - (200\sqrt{3})^2} + \sqrt{b^2 - (200\sqrt{3})^2}.\]
Furthermore, the altitude from $P$ to $AB$ is $\sqrt{a^2 - 300^2}$ , so, by LoC and the dihedral condition, \[a^2 - 300^2 + b^2 - 300^2 + \sqrt{a^2 - 300^2}\sqrt{b^2-300^2} = 4x^2.\]
Squaring the equation for $PQ$ and substituting $a^2 + b^2  = 4x^2$ yields \[2\sqrt{a^2 - (200\sqrt{3})^2}\sqrt{b^2 - (200\sqrt{3})^2} = 6\cdot 200^2.\]
Substituting $a^2 + b^2 = 4x^2$ into the other equation, \[\sqrt{a^2 - 300^2}\sqrt{b^2-300^2} = 2\cdot 300^2.\]
Squaring both of these gives \[a^2b^2-3\cdot 200^2(a^2 + b^2) + 9\cdot 200^4 = 9\cdot 200^4\] \[a^2b^2 - 300^2(a^2+b^2) + 300^4 = 4\cdot 300^4.\]
Substituting $a^2 + b^2 = 4x^2$ and solving for $x$ gives $\boxed{450}$ , as desired.

Solution:
Let $AB = a, M$ be midpoint $BC, I$ be the center of equilateral $\triangle ABC,$ $IM = b = \frac {a}{2\sqrt{3}}, O$ be the center of sphere $ABCPQ.$ Then \[AI = 2b, AO = BO = PO =QO = d.\] \[QA=QB=QC,PA=PB=PC \implies\] \[POIQ\perp ABC, \angle PMQ = 120^\circ.\] (See upper diagram).
We construct the circle PQMD, use the formulas for  intersecting chords and get \[DI = 5b, FI = EO = \frac{3b}{2}\] \[\implies FM = \frac{5b}{2}.\] (See lower diagram).
We apply the Law of Sine to $\triangle PMQ$ and get \[2EM \sin 120^\circ =PQ\] \[\implies r \sqrt{3} = 2d\] \[\implies 3r^2 = 4d^2.\] We apply the Pythagorean Law on $\triangle AOI$ and $\triangle EFM$ and get \[d^2 = 4b^2 + OI^2, r^2 = \frac {25b^2}{4} + EF^2 \implies\] \[r = 3b\implies d = \frac {3a}{2} = \boxed{450}.\] vladimir.shelomovskii@gmail.com, vvsss

Solution:
Let $M$ be the midpoint of $\overline{AB}$ and $X$ the center of $\triangle ABC$ . Then \[P, O, Q, M, X, C\] all lie in the same vertical plane. We can make the following observations:
To make calculations easier, we will denote $100\sqrt{3}=m$ , so that $MX=m$ and $XC=2m$

Denote $PX=p$ and $QX=q$ , where the tangent addition formula on $\triangle PMQ$ yields \[\frac{\tan\measuredangle PMX+\tan\measuredangle QMX}{1-\tan\measuredangle PMX\tan\measuredangle QMX}=\tan(120^{\circ})=-\sqrt{3}.\] Using $\tan\measuredangle PMX=\frac{p}{m}$ and $\tan\measuredangle QMX=\frac{q}{m}$ , we have \[\frac{\frac{p}{m}+\frac{q}{m}}{1-\frac{p}{m}\cdot\frac{q}{m}}=-\sqrt{3}.\] After multiplying both numerator and denominator by $m^{2}$ we have \[\frac{(p+q)m}{m^{2}-pq}=-\sqrt{3}.\] But note that $pq=(2m)(2m)=4m^{2}$ by power of a point at $X$ , where we deduce by symmetry that $MM^{\prime}=MX=m$ on the diagram below:
Thus \begin{align*} \frac{(p+q)m}{m^{2}-4m^{2}}=-\sqrt{3} \\ \frac{p+q}{-3m}=-\sqrt{3} \\ p+q=\left(-\sqrt{3}\right)\left(-3m\right) \\ p+q=3\sqrt{3}\cdot m.\end{align*} Earlier we assigned the variable $m$ to the length $100\sqrt{3}$ which implies $PQ=\left(3\sqrt{3}\right)\left(100\sqrt{3}\right)=900$ . Thus the distance $d$ is equal to $\frac{PQ}{2}=\boxed{450}$

Solution:
Let $Z$ be the center of $\triangle ABC$ . Let $A’$ be the midpoint of $BC$ . Let $ZA’ = c = 100\sqrt{3}$ and $ZA = 2c = 200\sqrt{3}$ . Let $PZ = a$ and $QZ = b$ . We will be working in the plane that contains the points: $A$ $P$ $A’$ $Q$ $O$ , and $Z$
Since $P$ $O$ , and $Q$ are collinear and $PO = QO = AO$ $\triangle PAQ$ is a right triangle with $\angle PAQ = 90^{\circ}$ . Since $AZ \perp PQ$ $(PZ)(QZ) = (AZ)^2 = ab = (2c)^2 = 120000$
$PA’ = \sqrt{a^2 + c^2}$ $QA’ = \sqrt{b^2 + c^2}$ $PQ = a + b$ , and $\angle PAQ = 120^{\circ}$ . By Law of Cosines \[(a + b)^2 = a^2 + b^2 + 2c^2 + \sqrt{a^2b^2 + a^2c^2 + b^2c^2 + c^4}\] . Substituting $4c^2$ for $ab$ and simplifying, we get \[6c = \sqrt{17c^2 + a^2 + b^2}\] . Squaring and simplifying, we get \[a^2 + b^2 = 19c^2 = 570000\] . Adding $2ab = 8c^2$ to both sides we get $PQ = a + b = 900$ . Since $O$ is the midpoint of $PQ$ $d = PO = \boxed{450}$


Answer: 677.0

Equilateral $\triangle ABC$ has side length $\sqrt{111}$ . There are four distinct triangles $AD_1E_1$ $AD_1E_2$ $AD_2E_3$ , and $AD_2E_4$ , each congruent to $\triangle ABC$ ,
with $BD_1 = BD_2 = \sqrt{11}$ . Find $\sum_{k=1}^4(CE_k)^2$

Solution:
Screen Shot 2020-02-17 at 2.24.50 PM.png
We create a diagram. Each of the red quadrilaterals are actually 60 degree rotations about point A. We easily get that $\overline{C E_1} = \sqrt{11}$ , and $\overline{C E_3} = \sqrt{11}$ . If we set $\angle B A D_2 = \theta$ , we can start angle chasing. In particular, we will like to find $\angle D E_4 C$ , and $\angle D E_2 C$ , since then we will be able to set up some Law Of Cosines. $\angle D E_4 C = \angle D E_4 A + \angle A E_4 C = 90 - \frac{\theta}{2} + 30 + \frac{\theta}{2} = 120^{\circ}$ That was convenient! We can do it with the other angle as well. $\angle D E_2 C = \angle D E_2 A - \angle C E_2 A = 90 - \frac{\theta}{2} - (30 - \frac{\theta}{2}) = 60^{\circ}$ . That means we are able to set up Law of Cosines, on triangles $\triangle D E_4 C$ and $\triangle D E_2 C$ , with some really convenient angles.
Let $CE_2 = x$ , and $CE_4 = y$ \[333 = 11 + x^2 - \sqrt{11} x\] \[333 = 11 + y^2 + \sqrt{11} y\] We subtract and get: \[0 = (x+y)(x-y-\sqrt{11})\] $x+y$ obviously can't be 0, so $x-y = \sqrt{11}$ We add and get: \[666 = 22 + x^2 + y^2 + \sqrt{11} (y-x)\] $y-x = -\sqrt{11}$ . Thus, we can fill in and solve. \[666 = 22 + x^2 + y^2 - 11\] \[655 = x^2 + y^2\] Thus our answer is $C E_1^2 + C E_2^2 + C E_2^2 + C E_4^2 = 11 + 11 + C E_2^2 + C E_4^2 = 11 + 11 + x^2 + y^2 = 11 + 11 + 655 = \boxed{677}$


Answer: 865.0

Equilateral $\triangle ABC$ is inscribed in a circle of radius $2$ . Extend $\overline{AB}$ through $B$ to point $D$ so that $AD=13,$ and extend $\overline{AC}$ through $C$ to point $E$ so that $AE = 11.$ Through $D,$ draw a line $l_1$ parallel to $\overline{AE},$ and through $E,$ draw a line $l_2$ parallel to $\overline{AD}.$ Let $F$ be the intersection of $l_1$ and $l_2.$ Let $G$ be the point on the circle that is collinear with $A$ and $F$ and distinct from $A.$ Given that the area of $\triangle CBG$ can be expressed in the form $\frac{p\sqrt{q}}{r},$ where $p, q,$ and $r$ are positive integers, $p$ and $r$ are relatively prime , and $q$ is not divisible by the square of any prime, find $p+q+r.$

Solution:
Notice that $\angle{E} = \angle{BGC} = 120^\circ$ because $\angle{A} = 60^\circ$ . Also, $\angle{GBC} = \angle{GAC} = \angle{FAE}$ because they both correspond to arc ${GC}$ . So $\Delta{GBC} \sim \Delta{EAF}$
\[[EAF] = \frac12 (AE)(EF)\sin \angle AEF  = \frac12\cdot11\cdot13\cdot\sin{120^\circ} = \frac {143\sqrt3}4.\]
Because the ratio of the area of two similar figures is the square of the ratio of the corresponding sides, $[GBC] = \frac {BC^2}{AF^2}\cdot[EAF] = \frac {12}{11^2 + 13^2 - 2\cdot11\cdot13\cdot\cos120^\circ}\cdot\frac {143\sqrt3}4 = \frac {429\sqrt3}{433}$ . Therefore, the answer is $429+433+3=\boxed{865}$

Solution:
Solution by e_power_pi_times_i/edited by srisainandan6
Let the center of the circle be $O$ and the origin. Then, $A (0,2)$ $B (-\sqrt{3}, -1)$ $C (\sqrt{3}, -1)$ $D$ and $E$ can be calculated easily knowing $AD$ and $AE$ $D (-\dfrac{13}{2}, \dfrac{-13\sqrt{3}+4}{2})$ $E (\dfrac{11}{2}, \dfrac{-11\sqrt{3}+4}{2})$ . As $DF$ and $EF$ are parallel to $AE$ and $AD$ $F (-1, -12\sqrt{3}+2)$ $G$ and $A$ is the intersection between $AF$ and circle $O$ . Therefore $G (-\dfrac{48\sqrt{3}}{433}, -\dfrac{862}{433})$ . Using the Shoelace Theorem, $[CBG] = \dfrac{429\sqrt{3}}{433}$ , so the answer is $\boxed{865}$ . Note that although the solution may appear short, actually getting all the coordinates take a while as there is plenty of computation.

Solution: Lines $l_1$ and $l_2$ are constructed such that $AEFD$ is a parallelogram, hence $DF = 13$ . Since $BAC$ is equilateral with angle of $60^{\circ}$ , angle $D$ is $120^{\circ}$ . Use law of cosines to find $AF = \sqrt{433}$ . Then use law of sines to find angle $BAG$ and $GAC$ . Next we use Ptolemy's Theorem on $ABGC$ to find that $CG + BG = AG$ . Next we use law of cosine on triangles $BAG$ and $GAC$ , solving for BG and CG respectively. Subtract the two equations and divide out a $BG + CG$ to find the value of $CG - BG$ . Next, $AG = 2\cdot R \cos{\theta}$ , where R is radius of circle $= 2$ and $\theta =$ angle $BAG$ . We already know sine of the angle so find cosine, hence we have found $AG$ . At this point it is system of equation yielding $CG = \frac{26\sqrt{3}}{\sqrt{433}}$ and $BG = \frac{22\sqrt{3}}{\sqrt{433}}$ . Given $[CBG] = \frac{BC \cdot CG \cdot BG}{4R}$ , and $BC = 2\sqrt{3}$ by $30-60-90$ triangle, we can evaluate to find $[CBG] = \frac{429\sqrt{3}}{433}$ , to give answer = $\boxed{865}$

Solution:
Note that $AB=2\sqrt3$ $DF=11$ , and $EF=13$ . If we take a homothety of the parallelogram with respect to $A$ , such that $F$ maps to $G$ , we see that $\frac{[ABG]}{[ACG]}=\frac{11}{13}$ . Since $\angle AGB=\angle AGC=60^{\circ}$ , from the sine area formula we have $\frac{BG}{CG}=\frac{11}{13}$ . Let $BG=11k$ and $CG=13k$
By Law of Cosines on $\triangle BGC$ , we have \[12=k^2(11^2+11\cdot13+13^2)=433k^2\implies k^2=\frac{12}{433}\] Thus, $[CBG]=\frac12 (11k)(13k)\sin 120^{\circ} = \frac{\sqrt3}{4}\cdot 143\cdot \frac{12}{433}=\frac{429\sqrt3}{433}\implies\boxed{865}$


Answer: 336.0

Equilateral triangle $ABC$ has side length $840$ . Point $D$ lies on the same side of line $BC$ as $A$ such that $\overline{BD} \perp \overline{BC}$ . The line $\ell$ through $D$ parallel to line $BC$ intersects sides $\overline{AB}$ and $\overline{AC}$ at points $E$ and $F$ , respectively. Point $G$ lies on $\ell$ such that $F$ is between $E$ and $G$ $\triangle AFG$ is isosceles, and the ratio of the area of $\triangle AFG$ to the area of $\triangle BED$ is $8:9$ . Find $AF$ [asy] pair A,B,C,D,E,F,G; B=origin; A=5*dir(60); C=(5,0); E=0.6*A+0.4*B; F=0.6*A+0.4*C; G=rotate(240,F)*A; D=extension(E,F,B,dir(90)); draw(D--G--A,grey); draw(B--0.5*A+rotate(60,B)*A*0.5,grey); draw(A--B--C--cycle,linewidth(1.5)); dot(A^^B^^C^^D^^E^^F^^G); label("$A$",A,dir(90)); label("$B$",B,dir(225)); label("$C$",C,dir(-45)); label("$D$",D,dir(180)); label("$E$",E,dir(-45)); label("$F$",F,dir(225)); label("$G$",G,dir(0)); label("$\ell$",midpoint(E--F),dir(90)); [/asy]

Solution:
By angle chasing, we conclude that $\triangle AGF$ is a $30^\circ\text{-}30^\circ\text{-}120^\circ$ triangle, and $\triangle BED$ is a $30^\circ\text{-}60^\circ\text{-}90^\circ$ triangle.
Let $AF=x.$ It follows that $FG=x$ and $EB=FC=840-x.$ By the side-length ratios in $\triangle BED,$ we have $DE=\frac{840-x}{2}$ and $DB=\frac{840-x}{2}\cdot\sqrt3.$
Let the brackets denote areas. We have \[[AFG]=\frac12\cdot AF\cdot FG\cdot\sin{\angle AFG}=\frac12\cdot x\cdot x\cdot\sin{120^\circ}=\frac12\cdot x^2\cdot\frac{\sqrt3}{2}\] and \[[BED]=\frac12\cdot DE\cdot DB=\frac12\cdot\frac{840-x}{2}\cdot\left(\frac{840-x}{2}\cdot\sqrt3\right).\]
We set up and solve an equation for $x:$ \begin{align*} \frac{[AFG]}{[BED]}&=\frac89 \\ \frac{\frac12\cdot x^2\cdot\frac{\sqrt3}{2}}{\frac12\cdot\frac{840-x}{2}\cdot\left(\frac{840-x}{2}\cdot\sqrt3\right)}&=\frac89 \\ \frac{2x^2}{(840-x)^2}&=\frac89 \\ \frac{x^2}{(840-x)^2}&=\frac49. \end{align*} Since $0<x<840,$ it is clear that $\frac{x}{840-x}>0.$ Therefore, we take the positive square root for both sides: \begin{align*} \frac{x}{840-x}&=\frac23 \\ 3x&=1680-2x \\ 5x&=1680 \\ x&=\boxed{336}

Solution:
We express the areas of $\triangle BED$ and $\triangle AFG$ in terms of $AF$ in order to solve for $AF.$
We let $x = AF.$ Because $\triangle AFG$ is isosceles and $\triangle AEF$ is equilateral, $AF = FG = EF = AE = x.$
Let the height of $\triangle ABC$ be $h$ and the height of $\triangle AEF$ be $h'.$ Then we have that $h = \frac{\sqrt{3}}{2}(840) = 420\sqrt{3}$ and $h' = \frac{\sqrt{3}}{2}(EF) = \frac{\sqrt{3}}{2}x.$
Now we can find $DB$ and $BE$ in terms of $x.$ $DB = h - h' = 420\sqrt{3} - \frac{\sqrt{3}}{2}x,$ $BE = AB - AE = 840 - x.$ Because we are given that $\angle DBC = 90,$ $\angle DBE = 30.$ This allows us to use the sin formula for triangle area: the area of $\triangle BED$ is $\frac{1}{2}(\sin 30)\left(420\sqrt{3} - \frac{\sqrt{3}}{2}x\right)(840-x).$ Similarly, because $\angle AFG = 120,$ the area of $\triangle AFG$ is $\frac{1}{2}(\sin 120)(x^2).$
Now we can make an equation: \begin{align*} \frac{\triangle AFG}{\triangle BED} &= \frac{8}{9} \\ \frac{\frac{1}{2}(\sin 120)(x^2)}{\frac{1}{2}(\sin 30)\left(420\sqrt{3} - \frac{\sqrt{3}}{2}x\right)(840-x)} &= \frac{8}{9} \\ \frac{x^2}{\left(420 - \frac{x}{2}\right)(840-x)} &= \frac{8}{9}. \end{align*} To make further calculations easier, we scale everything down by $420$ (while keeping the same variable names, so keep that in mind). \begin{align*} \frac{x^2}{\left(1-\frac{x}{2}\right)(2-x)} &= \frac{8}{9} \\ 8\left(1-\frac{x}{2}\right)(2-x) &= 9x^2 \\ 16-16x + 4x^2 &= 9x^2 \\ 5x^2 + 16x -16 &= 0 \\ (5x-4)(x+4) &= 0. \end{align*} Thus $x = \frac{4}{5}.$ Because we scaled down everything by $420,$ the actual value of $AF$ is $\frac{4}{5}(420) = \boxed{336}.$

Solution:
$\angle AFE = \angle AEF = \angle EAF = 60^{0} \Rightarrow \angle AFG = 120^{0}$ So, If $\Delta AFG$ is isosceles, it means that $AF = FG$
Let $AF = FG = AE = EF = x$
So, $[\Delta AFG] = \frac{1}{2} \cdot x^{2} \textup{sin} 120^{0} = \frac{\sqrt{3}}{4}x^{2}$
In $\Delta BED$ $BE = 840 - x$ , Hence $DE = \frac{840 - x}{2}$ (because $\textup{sin} 30^{0} = \frac{1}{2}$
Therefore, $[\Delta BED] = \frac{1}{2} (840 - x) \left (\frac{840-x}{2} \right) \textup{sin} 60^{0}$
So, $[\Delta BED] = \frac{\sqrt{3}}{4} (840 - x) \left (\frac{840-x}{2} \right) = \frac{\sqrt{3}}{8} (840 - x)^{2}$
Now, as we know that the ratio of the areas of $\Delta AFG$ and $\Delta BED$ is $8:9$
Substituting the values, we get
$\frac{\frac{\sqrt{3}}{4}x^{2}}{\frac{\sqrt{3}}{8} (840 - x)^{2}} = \frac{8}{9} \Rightarrow \left (\frac{x}{840 - x} \right)^{2} = \frac{4}{9}$ Hence, $\frac{x}{840 - x} = \frac{2}{3}$ . Solving this, we easily get $x = 336$
We have taken $AF = x$ , Hence, $AF = \boxed{336}$

Solution:
Since $\triangle AFG$ is isosceles, $AF = FG$ , and since $\triangle AEF$ is equilateral, $AF = EF$ . Thus, $EF = FG$ , and since these triangles share an altitude, they must have the same area.
Drop perpendiculars from $E$ and $F$ to line $BC$ ; call the meeting points $P$ and $Q$ , respectively. $\triangle BEP$ is clearly congruent to both $\triangle BED$ and $\triangle FQC$ , and thus each of these new triangles has the same area as $\triangle BED$ . But we can "slide" $\triangle BEP$ over to make it adjacent to $\triangle FQC$ , thus creating an equilateral triangle whose area has a ratio of $18:8$ when compared to $\triangle AEF$ (based on our conclusion from the first paragraph). Since these triangles are both equilateral, they are similar, and since the area ratio $18:8$ reduces to $9:4$ , the ratio of their sides must be $3:2$ . So, because $FC$ and $AF$ represent sides of these triangles, and they add to $840$ $AF$ must equal two-fifths of $840$ , or $\boxed{336}$


Answer: 32.0

Equilateral triangle $T$ is inscribed in circle $A$ , which has radius $10$ . Circle $B$ with radius $3$ is internally tangent to circle $A$ at one vertex of $T$ . Circles $C$ and $D$ , both with radius $2$ , are internally tangent to circle $A$ at the other two vertices of $T$ . Circles $B$ $C$ , and $D$ are all externally tangent to circle $E$ , which has radius $\dfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] unitsize(3mm); defaultpen(linewidth(.8pt)); dotfactor=4;  pair A=(0,0), D=8*dir(330), C=8*dir(210), B=7*dir(90); pair Ep=(0,4-27/5); pair[] dotted={A,B,C,D,Ep};  draw(Circle(A,10)); draw(Circle(B,3)); draw(Circle(C,2)); draw(Circle(D,2)); draw(Circle(Ep,27/5));  dot(dotted); label("$E$",Ep,E); label("$A$",A,W); label("$B$",B,W); label("$C$",C,W); label("$D$",D,E); [/asy]

Solution:
Let $X$ be the intersection of the circles with centers $B$ and $E$ , and $Y$ be the intersection of the circles with centers $C$ and $E$ . Since the radius of $B$ is $3$ $AX =4$ . Assume $AE$ $p$ . Then $EX$ and $EY$ are radii of circle $E$ and have length $4+p$ $AC = 8$ , and angle $CAE = 60$ degrees because we are given that triangle $T$ is equilateral. Using the Law of Cosines on triangle $CAE$ , we obtain
$(6+p)^2 =p^2 + 64 - 2(8)(p) \cos 60$
The $2$ and the $\cos 60$ terms cancel out:
$p^2 + 12p +36 = p^2 + 64 - 8p$
$12p+ 36 = 64 - 8p$
$p =\frac {28}{20} = \frac {7}{5}$ . The radius of circle $E$ is $4 + \frac {7}{5} = \frac {27}{5}$ , so the answer is $27 + 5 = \boxed{032}$


Answer: 378.0

Equilateral triangle $\triangle ABC$ is inscribed in circle $\omega$ with radius $18.$ Circle $\omega_A$ is tangent to sides $\overline{AB}$ and $\overline{AC}$ and is internally tangent to $\omega.$ Circles $\omega_B$ and $\omega_C$ are defined analogously. Circles $\omega_A,$ $\omega_B,$ and $\omega_C$ meet in six points---two points for each pair of circles. The three intersection points closest to the vertices of $\triangle ABC$ are the vertices of a large equilateral triangle in the interior of $\triangle ABC,$ and the other three intersection points are the vertices of a smaller equilateral triangle in the interior of $\triangle ABC.$ The side length of the smaller equilateral triangle can be written as $\sqrt{a} - \sqrt{b},$ where $a$ and $b$ are positive integers. Find $a+b.$

Solution:
We can extend $AB$ and $AC$ to $B'$ and $C'$ respectively such that circle $\omega_A$ is the incircle of $\triangle AB'C'$  Since the diameter of the circle is the height of this triangle, the height of this triangle is $36$ . We can use inradius or equilateral triangle properties to get the inradius of this triangle is $12$ (The incenter is also a centroid in an equilateral triangle, and the distance from a side to the centroid is a third of the height). Therefore, the radius of each of the smaller circles is $12$
Let $O=\omega$ be the center of the largest circle. We will set up a coordinate system with $O$ as the origin. The center of $\omega_A$ will be at $(0,-6)$ because it is directly beneath $O$ and is the length of the larger radius minus the smaller radius, or $18-12 = 6$ . By rotating this point $120^{\circ}$ around $O$ , we get the center of $\omega_B$ . This means that the magnitude of vector $\overrightarrow{O\omega_B}$ is $6$ and is at a $30$ degree angle from the horizontal. Therefore, the coordinates of this point are $(3\sqrt{3},3)$ and by symmetry the coordinates of the center of $\omega_C$ is $(-3\sqrt{3},3)$
The upper left and right circles intersect at two points, the lower of which is $X$ . The equations of these two circles are: \begin{align*} (x+3\sqrt3)^2 + (y-3)^2 &= 12^2, \\ (x-3\sqrt3)^2 + (y-3)^2 &= 12^2. \end{align*} We solve this system by subtracting to get $x = 0$ . Plugging back in to the first equation, we have $(3\sqrt{3})^2 + (y-3)^2 = 144 \implies (y-3)^2 = 117 \implies y-3 = \pm \sqrt{117} \implies y = 3 \pm \sqrt{117}$ . Since we know $X$ is the lower solution, we take the negative value to get $X = (0,3-\sqrt{117})$
We can solve the problem two ways from here. We can find $Y$ by rotation and use the distance formula to find the length, or we can be somewhat more clever. We notice that it is easier to find $OX$ as they lie on the same vertical, $\angle XOY$ is $120$ degrees so we can make use of $30-60-90$ triangles, and $OX = OY$ because $O$ is the center of triangle $XYZ$ . We can draw the diagram as such:  Note that $OX = OY = \sqrt{117} - 3$ . It follows that \begin{align*} XY &= 2 \cdot \frac{OX\cdot\sqrt{3}}{2} \\ &= OX \cdot \sqrt{3} \\ &= (\sqrt{117}-3) \cdot \sqrt{3} \\ &= \sqrt{351}-\sqrt{27}. \end{align*} Finally, the answer is $351+27 = \boxed{378}$

Solution:
 For equilateral triangle with side length $l$ , height $h$ , and circumradius $r$ , there are relationships: $h = \frac{\sqrt{3}}{2} l$ $r = \frac{2}{3} h = \frac{\sqrt{3}}{3} l$ , and $l = \sqrt{3}r$
There is a lot of symmetry in the figure. The radius of the big circle $\odot \omega$ is $R = 18$ , let the radius of the small circles $\odot \omega_A$ $\odot \omega_B$ $\odot \omega_C$ be $r$
We are going to solve this problem in $3$ steps:
$\textbf{Step 1:}$
We have $\triangle A \omega_A D$ is a $30-60-90$ triangle, and $A \omega_A = 2 \cdot \omega_A D$ $A \omega_A = 2R-r$ $\odot \omega$ and $\odot \omega_A$ are tangent), and $\omega_A D = r$ . So, we get $2R-r = 2r$ and $r = \frac{2}{3} \cdot R = 12$
Since $\odot \omega$ and $\odot \omega_A$ are tangent, we get $\omega \omega_A = R - r = \frac{1}{3} \cdot R = 6$
Note that $\triangle \omega_A \omega_B \omega_C$ is an equilateral triangle, and $\omega$ is its center, so $\omega_B \omega_C = \sqrt{3} \cdot \omega \omega_A = 6 \sqrt{3}$
$\textbf{Step 2:}$
Note that $\triangle \omega_C E X$ is an isosceles triangle, so \[EX = 2 \sqrt{(\omega_C E)^2 - \left(\frac{\omega_B \omega_C}{2}\right)^2} = 2 \sqrt{r^2 - \left(\frac{\omega_B \omega_C}{2}\right)^2} = 2 \sqrt{12^2 - (3 \sqrt{3})^2} = 2 \sqrt{117}.\]
$\textbf{Step 3:}$
In $\odot \omega_C$ , Power of a Point gives $\omega X \cdot \omega E = r^2 - (\omega_C \omega)^2$ and $\omega E = EX - \omega X = 2\sqrt{117} - \omega X$
It follows that $\omega X \cdot (2\sqrt{117} - \omega X) = 12^2 - 6^2$ . We solve this quadratic equation: $\omega X = \sqrt{117} - 3$
Since $\omega X$ is the circumradius of equilateral $\triangle XYZ$ , we have $XY = \sqrt{3} \cdot \omega X = \sqrt{3} \cdot (\sqrt{117} - 3) = \sqrt{351}-\sqrt{27}$
Therefore, the answer is $351+27 = \boxed{378}$

Solution:
Let $O$ be the center, $R = 18$ be the radius, and $CC'$ be the diameter of $\omega.$ Let $r$ be the radius, $E,D,F$ are the centers of $\omega_A, \omega_B,\omega_C.$ Let $KGH$ be the desired triangle with side $x.$ We find $r$ using \[CC' = 2R = C'K + KC = r + \frac{r}{\sin 30^\circ} = 3r.\] \[r = \frac{2R}{3} = 12.\] \[OE = R – r = 6.\] Triangles $\triangle DEF$ and $\triangle KGH$ – are equilateral triangles with a common center $O,$ therefore in the triangle $OEH$ $OE = 6, \angle EOH = 120^\circ, OH = \frac{x}{\sqrt3}.$
We apply the Law of Cosines to $\triangle OEH$ and get \[OE^2 + OH^2 + OE \cdot OH = EH^2.\] \[6^2 + \frac{x^2}{3} + \frac{6x}{\sqrt3} = 12^2.\] \[x^2 + 6x \sqrt{3} = 324\] \[x= \sqrt{351} - \sqrt{27}  \implies 351 + 27 = \boxed{378}\]

Solution: Let $O$ be the center of $\omega$ $X$ be the intersection of $\omega_B,\omega_C$ further from $A$ , and $O_A$ be the center of $\omega_A$ . Define $Y, Z, O_B, O_C$ similarly. It is well-known that the $A$ -mixtilinear inradius $R_A$ is $\tfrac{r}{\cos^2\left(\frac{\angle A}{2}\right)} = \tfrac{9}{\cos^2\left(30^{\circ}\right)} = 12$ , so in particular this means that $OO_B = 18 -  R_B = 6 = OO_C$ . Since $\angle O_BOO_C = \angle BOC = 120^\circ$ , it follows by Law of Cosines on $\triangle OO_BO_C$ that $O_BO_C = 6\sqrt{3}$ . Then the Pythagorean theorem gives that the altitude of $O_BO_CX$ is $\sqrt{117}$ , so $OY = OX = \text{dist}(X, YZ) - \text{dist}(O, YZ) = \sqrt{117} - 3$ and $YZ = \tfrac{O_BO_C\cdot OY}{OO_B} = \tfrac{6\sqrt{3}(\sqrt{117} - 3)}{6}=\sqrt{351} - \sqrt{27}$ so the answer is $351 + 27 = \boxed{378}$


Answer: 4.0

Estimate the time it takes to send $60$ blocks of data over a communications channel if each block consists of $512$ "chunks" and the channel can transmit $120$ chunks per second.
$\textbf{(A)}\ 0.04 \text{ seconds}\qquad \textbf{(B)}\ 0.4 \text{ seconds}\qquad \textbf{(C)}\ 4 \text{ seconds}\qquad \textbf{(D)}\ 4\text{ minutes}\qquad \textbf{(E)}\ 4\text{ hours}$

Solution: We want to figure out the number of chunks in $60$ blocks, so we have $60\cdot 512 \approx 30000$ . We divide this by $120$ to determine the number of seconds necessary to transmit. $30000/120 \approx 250$ , which means that it takes approximately $4$ minutes to transmit. Thus, the answer is $\boxed{4}$

Solution: This solution is if you are running out of time and just want to write down an answer. So, this is quite unreliable. You can logic it out. It doesn't make sense for the first three options to be the answer since that is way too quick. The last option is way too long. That just leaves $\boxed{4}$


Answer: 2075.0

Estimate the year in which the population of Nisos will be approximately 6,000.
$\text{(A)}\ 2050 \qquad \text{(B)}\ 2075 \qquad \text{(C)}\ 2100 \qquad \text{(D)}\ 2125 \qquad \text{(E)}\ 2150$

Solution:
We could triple the population every $25$ years and make a chart:
Year: 2000
Population: 200
Year: 2025
Population: 600
Year: 2050
Population: 1800
Year: 2075
Population: 5400
Year: 2100
Population: 16200
The closest year is 2075, or $\boxed{2075}$

Solution:
We could find out how many periods of 25 years we need to triple by dividing our total from our present number.
$\frac{6000}{200}=30$
The power of $3$ that $30$ is closest to is $27=3^3$
Therefore, after $3$ periods, we will be closest to $6000$
$\boxed{2075}$


Answer: 2000.0

Estimate to determine which of the following numbers is closest to $\frac{401}{.205}$
$\text{(A)}\ .2 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 200 \qquad \text{(E)}\ 2000$

Solution: $401$ is around $400$ and $.205$ is around $.2$ so the fraction is approximately \[\frac{400}{.2}=2000\rightarrow \boxed{2000}\]


Answer: 250.0

Euler's formula states that for a convex polyhedron with $V$ vertices $E$ edges , and $F$ faces $V-E+F=2$ . A particular convex polyhedron has 32 faces, each of which is either a triangle or a pentagon . At each of its $V$ vertices, $T$ triangular faces and $P$ pentagonal faces meet. What is the value of $100P+10T+V$

Solution:
The convex polyhedron of the problem can be easily visualized; it corresponds to a dodecahedron (a regular solid with $12$ equilateral pentagons) in which the $20$ vertices have all been truncated to form $20$ equilateral triangles with common vertices. The resulting solid has then $p=12$ smaller equilateral pentagons and $t=20$ equilateral triangles yielding a total of $t+p=F=32$ faces. In each vertex, $T=2$ triangles and $P=2$ pentagons are concurrent. Now, the number of edges $E$ can be obtained if we count the number of sides that each triangle and pentagon contributes: $E=\frac{3t+5p}{2}$ , (the factor $2$ in the denominator is because we are counting twice each edge, since two adjacent faces share one edge). Thus, $E=60$ . Finally, using Euler's formula we have $V=E-30=30$
In summary, the solution to the problem is $100P+10T+V=\boxed{250}$

Solution:
As seen above, $E=V+30$ . Every vertex $V$ , there is a triangle for every $T$ and a pentagon for every $P$ by the given. However, there are three times every triangle will be counted and five times every pentagon will be counted because of their numbers of vertices. From this observation, $\frac{VT}3+\frac{VP}5=32\implies V(5T+3P)=480$ . Also, at every vertex $V$ , there are $T+P$ edges coming out from that vertex (one way to see this is to imagine the leftmost segment of each triangle and pentagon that is connected to the given vertex, and note that it includes every one of the edges exactly once), so $\frac{V(T+P)}2=E\implies V(T+P)=2E\implies V(5T+5P)=10E$ , and subtracting the other equation involving the vertices from this gives $2VP=10E-480\implies VP=5E-240=5(V+30)-240=5V-90$ $\implies V(5-P)=90$ .
Since $V|480$ from the first vertex-related observation and $P>0\implies5-P<5$ , and it quickly follows that $V=30\implies E=60\implies P=2\implies T=2\implies100P+10T+V=\boxed{250}$

Solution:
Notice that at each vertex, we must have the sum of the angles be less than $360$ degrees or we will not be able to fold the polyhedron.  Therefore, we have $60T + 108P < 360.$ Now, let there be $t$ triangles and $p$ pentagons total such that $t+p = 32.$ From the given, we know that $E = V + 30.$ Lastly, we see that $E = \frac{3t+5p}{2}$ and $V = \frac{3t}{T}=\frac{5p}{P}.$
Now, we do casework on what $P$ is.
Case 1: $P = 2$ Notice that we must have $t$ and $p$ integral.  Trying $T = 1, 2$ yields a solution with $t=2.$ Trying other cases of $P$ and $T$ yields no solutions.  Therefore, $T=2, P=2$ and after solving for $t, p$ we get $V=30.$ Finally, we have $100P+10T+V = \boxed{250}$

Solution: We know that $V-E = -30 \implies V = E-30$ based off the problem condition. Furthermore, if we draw out a few pentagons as well as triangles on each of side of the pentagons, it's clear that each vertex has 4 edges connected to it, with two triangles and two pentagons for each vertex. However, each edge is used for two vertices and thus counted twice. Therefore, we have that $E = \frac{4V}{2} = 2V$ . Plugging this in, we have that $V=30$ and so our answer is $200+20+30 = \boxed{250}$


Answer: 256.0

Solution: Plugging in $2$ as $x$ gives $4^4$ , which is merely $\boxed{256}$


Answer: 104.0

Solution:
More generally, let $(x,y,z)=\left(\sqrt5,\sqrt6,\sqrt7\right)$ so that $\left(x^2,y^2,z^2\right)=(5,6,7).$
We rewrite the original expression in terms of $x,y,$ and $z,$ then apply the difference of squares repeatedly: \begin{align*} (x+y+z)(x+y-z)(x-y+z)(-x+y+z) &= \left[((x+y)+z)((x+y)-z)\right]\left[((z+(x-y))(z-(x-y))\right] \\ &= \left[(x+y)^2-z^2\right]\left[z^2 - (x-y)^2\right] \\ &= \left[x^2+2xy+y^2-z^2\right]\left[z^2-x^2+2xy-y^2\right] \\ &= \left[2xy + \left(x^2+y^2-z^2\right)\right]\left[2xy - \left(x^2+y^2-z^2\right)\right] \\ &= \left(2xy\right)^2 - \left(x^2+y^2-z^2\right)^2 \\ &= \left(2\cdot\sqrt5\cdot\sqrt6\right)^2 - \left(5+6-7\right)^2 \\ &= \boxed{104} Remark

Solution: We group the first and last factors as well as the two middle factors, then apply the difference of squares repeatedly: \begin{align*} \left(\left(\sqrt{6} + \sqrt{7}\right)^2 - \sqrt{5}^2\right)\left(\sqrt{5}^2 - \left(\sqrt{6} - \sqrt{7}\right)^2\right) &= \left(13 + 2\sqrt{42} - 5\right)\left(5 - \left(13 - 2\sqrt{42}\right)\right) \\ &= \left(2\sqrt{42} + 8\right)\left(2\sqrt{42} - 8\right) \\ &= \left(2\sqrt{42}\right)^2 - 8^2 \\ &= \boxed{104} ~Azjps (Solution)

Solution:
Notice that in a triangle with side-lengths $2\sqrt5,2\sqrt6,$ and $2\sqrt7,$ by Heron's Formula, the area is the square root of the original expression.
Let $\theta$ be the measure of the angle opposite the $2\sqrt7$ side. By the Law of Cosines, \[\cos\theta=\frac{\left(2\sqrt5\right)^2+\left(2\sqrt{6}\right)^2-\left(2\sqrt7\right)^2}{2\cdot 2\sqrt5\cdot2\sqrt6}=\frac{16}{8\sqrt{30}}=\sqrt{\frac{2}{15}},\] so $\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{\frac{13}{15}}.$
The area of the triangle is then \[\frac{\sin\theta}{2}\cdot 2\sqrt5\cdot 2\sqrt6=\sqrt{\frac{26}{30}}\cdot 2\sqrt{30}=\sqrt{104},\] so our answer is $\left(\sqrt{104}\right)^2=\boxed{104}.$


Answer: 117.0

Every card in a deck has a picture of one shape - circle, square, or triangle, which is painted in one of the three colors - red, blue, or green. Furthermore, each color is applied in one of three shades - light, medium, or dark. The deck has 27 cards, with every shape-color-shade combination represented. A set of three cards from the deck is called complementary if all of the following statements are true:
i. Either each of the three cards has a different shape or all three of the cards have the same shape.
ii. Either each of the three cards has a different color or all three of the cards have the same color.
iii. Either each of the three cards has a different shade or all three of the cards have the same shade.
How many different complementary three-card sets are there?

Solution: Adding the cases up, we get $27 + 54 + 36 = \boxed{117}$

Solution: Let's say we have picked two cards. We now compare their attributes to decide how we can pick the third card to make a complement set. For each of the three attributes, should the two values be the same we have one option - choose a card with the same value for that attribute. Furthermore, should the two be different there is only one option- choose the only value that is remaining. In this way, every two card pick corresponds to exactly one set, for a total of $\binom{27}{2} = 27*13 = 351$ possibilities. Note, however, that each set is generated by ${3\choose 2} = 3$ pairs, so we've overcounted by a multiple of 3 and the answer is $\frac{351}{3} = \boxed{117}$

Solution: Treat the sets as ordered. Then for each of the three criterion, there are $3!=6$ choices if the attribute is different and there are $3$ choices is the attribute is the same. Thus all three attributes combine to a total of $(6+3)^3=729$ possibilities. However if all three attributes are the same then the set must be composed of three cards that are the same, which is impossible. This takes out $3^3=27$ possibilities. Notice that we have counted every set $3!=6$ times by treating the set as ordered. The final solution is then $\frac{729-27}{6}=\boxed{117}$


Answer: 23.0

Every high school in the city of Euclid sent a team of $3$ students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed $37$ th and $64$ th , respectively. How many schools are in the city?
$\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 25 \qquad \textbf{(E)}\ 26$

Solution: There are $x$ schools. This means that there are $3x$ people. Because no one's score was the same as another person's score, that means that there could only have been $1$ median score. This implies that $x$ is an odd number. $x$ cannot be less than $23$ , because there wouldn't be a $64$ th place if x was. $x$ cannot be greater than $23$ either, because that would tie Andrea and Beth or Andrea's place would be worse than Beth's. Thus, the only possible answer is $23 \Rightarrow \boxed{23}$


Answer: 23.0

Solution:
Let $a$ be Andrea's place. We know that she was the highest on her team, so $a < 37$
Since $a$ is the median, there are $a-1$ to the left and right of the median, so the total number of people is $2a-1$ and the number of schools is $(2a-1)/3$ . This implies that $2a-1 \equiv 0 \pmod{3} \implies a \equiv 2 \pmod{3}$
Also, since $2a-1$ is the rank of the last-place person, and one of Andrea's teammates already got 64th place, $2a-1 > 64 \implies a \ge 33$
Putting it all together: $33 \le a < 37$ and $a \equiv 2 \pmod{3}$ , so clearly $a = 35$ , and the number of schools as we got before is $(2a-1)/3 = 69/3 = \boxed{23}$


Answer: 23.0

Solution: There are $x$ schools. This means that there are $3x$ people. Because no one's score was the same as another person's score, that means that there could only have been $1$ median score. This implies that $x$ is an odd number. $x$ cannot be less than $23$ , because there wouldn't be a $64$ th place if x was. $x$ cannot be greater than $23$ either, because that would tie Andrea and Beth or Andrea's place would be worse than Beth's. Thus, the only possible answer is $23 \Rightarrow \boxed{23}$


Answer: 23.0

Solution:
Let $a$ be Andrea's place. We know that she was the highest on her team, so $a < 37$
Since $a$ is the median, there are $a-1$ to the left and right of the median, so the total number of people is $2a-1$ and the number of schools is $(2a-1)/3$ . This implies that $2a-1 \equiv 0 \pmod{3} \implies a \equiv 2 \pmod{3}$
Also, since $2a-1$ is the rank of the last-place person, and one of Andrea's teammates already got 64th place, $2a-1 > 64 \implies a \ge 33$
Putting it all together: $33 \le a < 37$ and $a \equiv 2 \pmod{3}$ , so clearly $a = 35$ , and the number of schools as we got before is $(2a-1)/3 = 69/3 = \boxed{23}$


Answer: 495.0

Every positive integer $k$ has a unique factorial base expansion $(f_1,f_2,f_3,\ldots,f_m)$ , meaning that $k=1!\cdot f_1+2!\cdot f_2+3!\cdot f_3+\cdots+m!\cdot f_m$ , where each $f_i$ is an integer, $0\le f_i\le i$ , and $0<f_m$ . Given that $(f_1,f_2,f_3,\ldots,f_j)$ is the factorial base expansion of $16!-32!+48!-64!+\cdots+1968!-1984!+2000!$ , find the value of $f_1-f_2+f_3-f_4+\cdots+(-1)^{j+1}f_j$

Solution:
Note that $1+\sum_{k=1}^{n-1} {k\cdot k!} = 1+\sum_{k=1}^{n-1} {((k+1)\cdot k!- k!)} = 1+\sum_{k=1}^{n-1} {((k+1)!- k!)} = 1 + ((2! - 1!) + (3! - 2!) + \cdots + (n! - (n-1)!)) = n!$
Thus for all $m\in\mathbb{N}$
$(32m+16)!-(32m)! = \left(1+\sum_{k=1}^{32m+15} {k\cdot k!}\right)-\left(1+\sum_{k=1}^{32m-1} {k\cdot k!}\right) = \sum_{k=32m}^{32m+15}k\cdot k!.$
So now,
\begin{align*} 16!-32!+48!-64!+\cdots+1968!-1984!+2000!&=16!+(48!-32!)+(80!-64!)\cdots+(2000!-1984!)\\ &=16! +\sum_{m=1}^{62}(32m+16)!-(32m)!\\ &=16! +\sum_{m=1}^{62}\sum_{k=32m}^{32m+15}k\cdot k! \end{align*}
Therefore we have $f_{16} = 1$ $f_k=k$ if $32m\le k \le 32m+15$ for some $m=1,2,\ldots,62$ , and $f_k = 0$ for all other $k$
Therefore we have:
\begin{align*} f_1-f_2+f_3-f_4+\cdots+(-1)^{j+1}f_j &= (-1)^{17}\cdot 1 + \sum_{m=1}^{62}\sum_{k=32m}^{32m+15}(-1)^{k+1}k\\ &= -1 + \sum_{m=1}^{62}\left[\sum_{j=16m}^{16m+7}(-1)^{2j+1}2j+\sum_{j=16m}^{16m+7}(-1)^{2j+2}(2j+1)\right]\\ &= -1 + \sum_{m=1}^{62}\sum_{j=16m}^{16m+7}[(-1)^{2j+1}2j+(-1)^{2j+2}(2j+1)]\\ &= -1 + \sum_{m=1}^{62}\sum_{j=16m}^{16m+7}[-2j+(2j+1)]\\ &= -1 + \sum_{m=1}^{62}\sum_{j=16m}^{16m+7}1\\ &= -1 + \sum_{m=1}^{62}8\\ &= -1 + 8\cdot 62\\ &= \boxed{495}

Solution:
This is equivalent to Solution 1. I put up this solution merely for learners to see the intuition.
Let us consider a base $n$ number system. It’s a well known fact that when we take the difference of two integral powers of $n$ , (such as $10000_{10} - 100_{10}$ ) the result will be an integer in base $n$ composed only of the digits $n - 1$ and $0$ (in this example, $9900$ ). More specifically, the difference $(n^k)_n - (n^j)_n$ $j<k$ , is an integer $k$ digits long (note that $(n^k)_n$ has $k + 1$ digits). This integer is made up of $(k-j)$ $(n - 1)$ ’s followed by $j$ $0$ ’s.
It should make sense that this fact carries over to the factorial base, albeit with a modification. Whereas in the general base $n$ , the largest digit value is $n - 1$ , in the factorial base, the largest digit value is the argument of the factorial in that place. (for example, $321_!$ is a valid factorial base number, as is $3210_!$ . However, $31_!$ is not, as $3$ is greater than the argument of the second place factorial, $2$ $31_!$ should be represented as $101_!$ , and is $7_{10}$ .) Therefore, for example, $1000000_! - 10000_!$ is not $990000_!$ , but rather is $650000_!$ . Thus, we may add or subtract factorials quite easily by converting each factorial to its factorial base expression, with a $1$ in the argument of the factorial’s place and $0$ ’s everywhere else, and then using a standard carry/borrow system accounting for the place value.
With general intuition about the factorial base system out of the way, we may tackle the problem. We use the associative property of addition to regroup the terms as follows: $(2000! - 1984!) + (1968! - 1952!) + \cdots + (48! - 32!) + 16!$ we now apply our intuition from paragraph 2. $2000!_{10}$ is equivalent to $1$ followed by $1999$ $0$ ’s in the factorial base, and $1984!$ is $1$ followed by $1983$ $0$ ’s, and so on. Therefore, $2000! - 1984! = (1999)(1998)(1997)\cdots(1984)$ followed by $1983$ $0$ ’s in the factorial base. $1968! - 1952! = (1967)(1966)\cdots(1952)$ followed by $1951$ $0$ ’s, and so on for the rest of the terms, except $16!$ , which will merely have a $1$ in the $16!$ place followed by $0$ ’s. To add these numbers, no carrying will be necessary, because there is only one non-zero value for each place value in the sum. Therefore, the factorial base place value $f_k$ is $k$ for all $32m \leq k \leq 32m+15$ if $1\leq m \in\mathbb{Z} \leq 62$ $f_{16} = 1$ , and $f_k = 0$ for all other $k$
Therefore, to answer, we notice that $1999 - 1998 = 1997 - 1996 = 1$ , and this will continue. Therefore, $f_{1999} - f_{1998} + \cdots - f_{1984} = 8$ . We have 62 sets that sum like this, and each contains $8$ pairs of elements that sum to $1$ , so our answer is almost $8 \cdot 62$ . However, we must subtract the $1$ in the $f_{16}$ place, and our answer is $8 \cdot 62 - 1 = \boxed{495}$

Solution:
Let $S = 16!-32!+\cdots-1984!+2000!$ . Note that since $|S - 2000!| << 2000!$ (or $|S - 2000!| = 1984! + \cdots$ is significantly smaller than $2000!$ ), it follows that $1999! < S < 2000!$ . Hence $f_{2000} = 0$ . Then $2000! = 2000 \cdot 1999! = 1999 \cdot 1999! + 1999!$ , and as $S - 2000! << 1999!$ , it follows that $1999 \cdot 1999! < S < 2000 \cdot 1999!$ . Hence $f_{1999} = 1999$ , and we now need to find the factorial base expansion of
\[S_2 = S - 1999 \cdot 1999! = 1999! - 1984! + 1962! - 1946! + \cdots + 16!\]
Since $|S_2 - 1999!| << 1999!$ , we can repeat the above argument recursively to yield $f_{1998} = 1998$ , and so forth down to $f_{1985} = 1985$ . Now $S_{16} = 1985! - 1984! + 1962! + \cdots = 1984 \cdot 1984! + 1962! + \cdots$ , so $f_{1984} = 1984$
The remaining sum is now just $1962! - 1946! + \cdots + 16!$ . We can repeatedly apply the argument from the previous two paragraphs to find that $f_{16} = 1$ , and $f_k=k$ if $32m\le k \le 32m+15$ for some $m=1,2,\ldots,62$ , and $f_k = 0$ for all other $k$
Now for each $m$ , we have $-f_{32m} + f_{32m+1} - f_{32m+2} + \cdots + f_{32m + 31}$ $= -32m + (32m + 1) - (32m + 2) + \cdots - (32m - 30) + (32 m + 31)$ $= 1 + 1 + \cdots + 1 + 1$ $= 8$ . Thus, our answer is $-f_{16} + 8 \cdot 62 = \boxed{495}$


Answer: 23.0

Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was $A$ dollars. The cost of his movie ticket was $20\%$ of the difference between $A$ and the cost of his soda, while the cost of his soda was $5\%$ of the difference between $A$ and the cost of his movie ticket. To the nearest whole percent, what fraction of $A$ did Roger pay for his movie ticket and soda?
$\textbf{(A) } 9\%\qquad \textbf{(B) } 19\%\qquad \textbf{(C) } 22\%\qquad \textbf{(D) }  23\%\qquad \textbf{(E) } 25\%$

Solution:
Let $m$ = cost of movie ticket Let $s$ = cost of soda
We can create two equations:
\[m = \frac{1}{5}(A - s)\] \[s  = \frac{1}{20}(A - m)\]
Substituting we get:
\[m = \frac{1}{5}(A - \frac{1}{20}(A - m))\] which yields: \[m = \frac{19}{99}A\]
Now we can find s and we get:
\[s = \frac{4}{99}A\]
Since we want to find what fraction of $A$ did Roger pay for his movie ticket and soda, we add $m$ and $s$ to get:
\[\frac{19}{99}A + \frac{4}{99}A \implies \boxed{23}\]

Solution:
We have two equations from the problem: $5M=A-S$ and $20S=A-M$ If we replace $A$ with $100$ we get a system of equations, and the sum of the values of $M$ and $S$ is the percentage of $A$ .
Solving, we get $S=\frac{400}{99}$ and $M=\frac{1900}{99}$ .
Adding, we get $\frac{2300}{99}$ , which is closest to $23$ which is $\boxed{23}$

Solution:
Let $m$ be the price of a movie ticket and $s$ be the price of a soda.
Then,
\[m=\frac{A-s}{5}\] and \[s=\frac{A-m}{20}\] Then, we can turn this into \[5m=A-s\] \[20s=A-m\]
Subtracting and getting rid of A, we have $20s-5m=-m+s \rightarrow 19s=4m$ . Assume WLOG that $s=4$ $m=19$ , thus making a solution for this equation. Substituting this into the 1st equation, we get $A=99$ . Hence, $\frac{m+s}{A} = \frac{19+4}{99} \approx \boxed{23}$


Answer: 24.0

Everyday at school, Jo climbs a flight of $6$ stairs. Jo can take the stairs $1$ $2$ , or $3$ at a time. For example, Jo could climb $3$ , then $1$ , then $2$ . In how many ways can Jo climb the stairs?
$\textbf{(A)}\ 13 \qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$

Solution:
A dynamics programming approach is quick and easy. The number of ways to climb one stair is $1$ . There are $2$ ways to climb two stairs: $1$ $1$ or $2$ . For 3 stairs, there are $4$ ways:
( $1$ $1$ $1$ )
( $1$ $2$ )
( $2$ $1$ )
( $3$
For four stairs, consider what step they came from to land on the fourth stair. They could have hopped straight from the 1st, done a double from #2, or used a single step from #3. The ways to get to each of these steps are $1+2+4=7$ ways to get to step 4. The pattern can then be extended: $4$ steps: $1+2+4=7$ ways. $5$ steps: $2+4+7=13$ ways. $6$ steps: $4+7+13=24$ ways.
Thus, there are $\boxed{24}$ ways to get to step $6.$

Solution:
Complementary counting is also possible. Considering the six steps, Jo has to land on the last step, so there are $2^5=32$ subsets (hit steps) of the other five steps. After that, subtract the number of ways to climb the steps while taking a leap of $4$ $5$ , or $6$ . The eight possible ways for this is ( $4$ $1$ $1$ ), ( $4$ $2$ ), ( $1$ $4$ $1$ ), ( $1$ $1$ $4$ ), ( $2$ $4$ ), ( $1$ $5$ ), ( $5$ $1$ ), and ( $6$
Altogether this makes for $32-8= \boxed{24}$ valid ways for Jo to get to step 6.

Solution:
We can set up a recursion to solve this problem. Suppose $f(n)$ represents the number of valid ways to get to the $n$ th step. $f(0)=1$ because there is 1 way for Jo to get to the " $0$ th" step (i.e. the ground). There is $1$ way to get to the first step (a $1$ -step), so $f(1)=1$ . There are $2$ ways to get to the second step (two $1$ -steps or one $2$ -step). Thus, $f(2) = 2$ . In general, $f(n) = f(n-1) + f(n-2) + f(n-3)$ . This is because from the $n-3$ th step, Jo can take a $3$ -step to get to the $n$ th step, from the $n-2$ th step, Jo can take a $2$ -step to get to the $n$ th step, and from the $n-1$ th step, Jo can take a $1$ -step to get to the $n$ th step. We now iteratively calculate values of $f(n)$
$f(0) = 1$
$f(1) = 1$
$f(2) = 2$
$f(3) = f(2) + f(1) + f(0) = 2 + 1 + 1 = 4$
$f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7$
$f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13$
$f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = \boxed{24}$


Answer: 6.0

Exactly three of the interior angles of a convex polygon are obtuse. What is the maximum number of sides of such a polygon?
$\mathrm{(A)\ } 4 \qquad \mathrm{(B) \ }5 \qquad \mathrm{(C) \  } 6 \qquad \mathrm{(D) \  } 7 \qquad \mathrm{(E) \  }8$

Solution:
Suppose that such a polygon has $n$ sides. Let the three obtuse angle measures, in degrees, be $o_1$ $o_2$ , and $o_3$ and the $(n-3)$ acute angle measures, again in degrees, be $a_1,a_2,a_3, \dotsc a_{n-3}$
Since $90 < o_i < 180$ for each $i$ , we have \[3(90) = 270 < o_1+o_2+o_3 < 3(180) = 540,\] and similarly, since $0 < a_i < 90$ for each $i$ \[0 < a_1+a_2+a_3+\dotsb+a_{n-3} < 90(n-3) = 90n-270.\] It follows that \[270+0 < o_1+o_2+o_3+a_1+a_2+a_3+\dotsb+a_{n-3} < 540+90n-270,\] and recalling that the sum of the interior angle measures of an $n$ -gon is $180(n-2) = 180n-360$ , this reduces to $270 < 180n-360 < 90n+270$ . Hence \[\frac{540}{180} < n < \frac{270+360}{90} \iff 3 < n < 7,\] so an upper bound is $n \leq 6$ , and it is easy to check that this bound can be attained by e.g. a convex hexagon with a right angle, $2$ acute angles, and $3$ obtuse angles, as shown below:
HrKCfF2ETNSQSy3uGlkg hexagonsurvey1.gif
Accordingly, the maximum possible number of sides of such a polygon is $\boxed{6}$


Answer: 618.0

Except for the first two terms, each term of the sequence $1000, x, 1000 - x,\ldots$ is obtained by subtracting the preceding term from the one before that.  The last term of the sequence is the first negative term encounted.  What positive integer $x$ produces a sequence of maximum length?

Solution:
We can start to write out some of the inequalities now:
And in general,
It is apparent that the bounds are slowly closing in on $x$ , so we can just calculate $x$ for some large value of $n$ (randomly, 10, 11):
$x < \frac{F_{7}}{F_{8}} \cdot 1000 = \frac{13}{21} \cdot 1000 = 618.\overline{18}$
$x > \frac{F_{8}}{F_{9}} \cdot 1000 = \frac{21}{34} \cdot 1000 \approx 617.977$
Thus the sequence is maximized when $x = \boxed{618}.$

Solution: It is well known that $\lim_{n\rightarrow\infty} \frac{F_{n-1}}{F_n} = \phi - 1 =\frac{1 + \sqrt{5}}{2} - 1 \approx .61803$ , so $1000 \cdot \frac{F_{n-1}}{F_n}$ approaches $x = \boxed{618}.$


Answer: 166.0

Solution:
Let $0<x_{}^{}<1$ . Then we may write $A_{k}^{}={N\choose k}x^{k}=\frac{N!}{k!(N-k)!}x^{k}=\frac{(N-k+1)!}{k!}x^{k}$ . Taking logarithms in both sides of this last equation and using the well-known fact $\log(a_{}^{}b)=\log a + \log b$ (valid if $a_{}^{},b_{}^{}>0$ ), we have
$\log(A_{k})=\log\left[\frac{(N-k+1)!}{k!}x^{k}\right]=\log\left[\prod_{j=1}^{k}\frac{(N-j+1)x}{j}\right]=\sum_{j=1}^{k}\log\left[\frac{(N-j+1)x}{j}\right]\, .$
Now, $\log(A_{k}^{})$ keeps increasing with $k_{}^{}$ as long as the arguments $\frac{(N-j+1)x}{j}>1$ in each of the $\log\big[\;\big]$ terms (recall that $\log y_{}^{} <0$ if $0<y_{}^{}<1$ ). Therefore, the integer $k_{}^{}$ that we are looking for must satisfy $k=\Big\lfloor\frac{(N+1)x}{1+x}\Big\rfloor$ , where $\lfloor z_{}^{}\rfloor$ denotes the greatest integer less than or equal to $z_{}^{}$
In summary, substituting $N_{}^{}=1000$ and $x_{}^{}=0.2$ we finally find that $k=\boxed{166}$

Solution:
We know that once we have found the largest value of $k$ , all values after $A_k$ are less than $A_k$ . Therefore, we are looking for the smallest possible value such that:
${\frac{1}{5}}^k\cdot {{1000} \choose {k}}>{\frac{1}{5}}^{k+1}\cdot {{1000} \choose {k+1}}$
Dividing by ${\frac{1}{5}}^k$ gives:
${1000\choose k}>{\frac{1}{5}}\cdot{1000\choose {k+1}}$
We can express these binomial coefficients as factorials.
$\frac{1000!}{(1000-k)!\cdot(k)!}>{\frac{1}{5}}\cdot\frac{1000!}{(1000-k-1)!\cdot{(k+1)!}}$
We note that the $1000!$ can cancel. Also, $(1000-k)!=(1000-k)(1000-k-1)!$ . Similarly, $(k+1)!=(k+1)k!$
Canceling these terms yields,
$\frac{1}{(1000-k)}>{\frac{1}{5}}\cdot\frac{1}{(k+1)}$
Cross multiplying gives:
$5k+5>1000-k \Rightarrow k>165.8$
Therefore, since this identity holds for all values of $k>165.8$ , the largest possible value of $k$ is $\boxed{166}$

Solution:
We know that $A_k$ will increase as $k$ increases until certain $k=m$ , where $A_0 < A_1 < A_2 < \dots < A_{m-2} < A_{m-1} < A_m$ and
$A_m > A_{m+1} > A_{m+2} > \dots > A_{1000}.$
Next, to change $A_{k-1}$ to $A_k$ , we multiply $A_{k-1}$ by $\frac{1000-k+1}{5k}$ . It follows that the numerator must be greater than the
denominator or
\[1000-k+1>5k\] \[k<166.8\]
We conclude that the answer is $\boxed{166}$

Solution:
Notice that the expansion is the largest the moment BEFORE the $nC_p < 5$ (this reasoning can probably be found in the other solutions; basically, if we have a number k and then k+1, the value is the largest when k+1 is larger than k, or in other words $nC_p*\frac{1}{5} > 1$
Say we have ${1000 \choose 5}$ ... this equals $\frac{1000*999*998*997*996}{5*4*3*2*1}$ , and if we have k+1, then it's just going to be equivalent to multiplying this fraction by $\frac{995}{6}$ . Notice that this fraction's numerator plus denominator is equal to $1001$ . Calling the numerator x and the denominator y, we get that \[x+y = 1001\] \[\frac{x}{y} > 5\] \[x>5y\] \[6y<1001\] \[y<166.83333\] and since the denominator is what we are specifically choosing, or in other words what k is, we conclude k = $\boxed{166}$

Solution:
Notice the relation between $A_m$ and $A_{m+1}$ . We have that: $A_{m+1}  = A_m \cdot \frac{1}{5} \cdot \frac{1000-m}{m+1}$ . This is true because from $A_m$ to $A_{m+1}$ we have to multiply by $\frac{1}{5}=0.2$ once,and then we must resolve the factorial issue. To do this, we must realize that
\[{1000 \choose m+1} = \frac{1000!}{(m+1)!(1000-m-1)!} = \frac{1000 \cdots (1001-m)(1000-m)}{(m+1)!}\] and that \[{1000 \choose m} = \frac{1000!}{m!(1000-m)!} = \frac{1000 \cdots (1001-m)}{m!}\]
\[\Longrightarrow {1000 \choose m} \cdot \frac{1000-m}{m+1}\]
So now, we must find out for which $m$ is $1000-m< 5 \cdot (m+1)$ (when this happens the numerator is less than the denominator for the fraction $\frac{1000-m}{m+1}$ so then we will have $A_m > A_{m+1}$ ) Then we find that for all $m > 165.833333$ , the above inequality ( $1000-m< 5 \cdot (m+1)$ ) holds, so then $m=165$ so $k=m+1= \boxed{166}$


Answer: 12.0

Externally tangent circles with centers at points $A$ and $B$ have radii of lengths $5$ and $3$ , respectively. A line externally tangent to both circles intersects ray $AB$ at point $C$ . What is $BC$
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 4.8\qquad\textbf{(C)}\ 10.2\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 14.4$

Solution:
Let $D$ and $E$ be the points of tangency on circles $A$ and $B$ with line $CD$ $AB=8$ . Also, let $BC=x$ . As $\angle ADC$ and $\angle BEC$ are right angles (a radius is perpendicular to a tangent line at the point of tangency) and both triangles share $\angle ACD$ $\triangle ADC \sim \triangle BEC$ . From this we can get a proportion.
$\frac{BC}{AC}=\frac{BE}{AD} \rightarrow \frac{x}{x+8}=\frac{3}{5} \rightarrow 5x=3x+24 \rightarrow x=\boxed{12}$


Answer: 320.0

Faces $ABC^{}_{}$ and $BCD^{}_{}$ of tetrahedron $ABCD^{}_{}$ meet at an angle of $30^\circ$ . The area of face $ABC^{}_{}$ is $120^{}_{}$ , the area of face $BCD^{}_{}$ is $80^{}_{}$ , and $BC=10^{}_{}$ . Find the volume of the tetrahedron.

Solution:
Since the area $BCD=80=\frac{1}{2}\cdot10\cdot16$ , the perpendicular from $D$ to $BC$ has length $16$
The perpendicular from $D$ to $ABC$ is $16 \cdot \sin 30^\circ=8$ . Therefore, the volume is $\frac{8\cdot120}{3}=\boxed{320}$


Answer: 390.0

Fifteen distinct points are designated on $\triangle ABC$ : the 3 vertices $A$ $B$ , and $C$ $3$ other points on side $\overline{AB}$ $4$ other points on side $\overline{BC}$ ; and $5$ other points on side $\overline{CA}$ . Find the number of triangles with positive area whose vertices are among these $15$ points.

Solution: Every triangle is uniquely determined by 3 points. There are $\binom{15}{3}=455$ ways to choose 3 points, but that counts the degenerate triangles formed by choosing three points on a line. There are $\binom{5}{3}$ invalid cases on segment $AB$ $\binom{6}{3}$ invalid cases on segment $BC$ , and $\binom{7}{3}$ invalid cases on segment $CA$ for a total of $65$ invalid cases. The answer is thus $455-65=\boxed{390}$


Answer: 8.0

Fifteen integers $a_1, a_2, a_3, \dots, a_{15}$ are arranged in order on a number line. The integers are equally spaced and have the property that \[1 \le a_1 \le 10, \thickspace 13 \le a_2 \le 20, \thickspace \text{ and } \thickspace 241 \le a_{15}\le 250.\] What is the sum of digits of $a_{14}?$
$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 11 \qquad \textbf{(E)}\ 12$

Solution:
We can find the possible values of the common difference by finding the numbers which satisfy the conditions. To do this, we find the minimum of the last two: $241-20=221$ , and the maximum– $250-13=237$ . There is a difference of $13$ between them, so only $17$ and $18$ work, as $17\cdot13=221$ , so $17$ satisfies $221\leq 13x\leq237$ . The number $18$ is similarly found. $19$ , however, is too much.
Now, we check with the first and last equations using the same method. We know $241-10\leq 14x\leq250-1$ . Therefore, $231\leq 14x\leq249$ . We test both values we just got, and we can realize that $18$ is too large to satisfy this inequality. On the other hand, we can now find that the difference will be $17$ , which satisfies this inequality.
The last step is to find the first term. We know that the first term can only be from $1$ to $3$ since any larger value would render the second inequality invalid. Testing these three, we find that only $a_1=3$ will satisfy all the inequalities. Therefore, $a_{14}=13\cdot17+3=224$ . The sum of the digits is therefore $\boxed{8}$

Solution:
Let the common difference between consecutive $a_i$ be $d$ .
Since $a_{15} - a_1 = 14d$ , we find from the first and last inequalities that $231 \le 14d \le 249$ . As $d$ must be an integer, this means $d = 17$ . Substituting this into all of the given inequalities so we may extract information about $a_1$ gives \[1 \le a_1 \le 10, \thickspace 13 \le a_1 + 17 \le 20, \thickspace 241 \le a_1 + 238 \le 250.\] The second inequality tells us that $1 \le a_1 \le 3$ while the last inequality tells us $3 \le a_1 \le 12$ , so we must have $a_1 = 3$ . Finally, to solve for $a_{14}$ , we simply have $a_{14} = a_1 + 13d = 3 + 13(17) = 3 + 221 = 224$ , so our answer is $2 + 2 + 4 = \boxed{8}$


Answer: 7.0

Solution:
The side of the large square is $1 + 3 + 1 = 5$ , so the area of the large square is $5^2 = 25$
The area of the middle square is $3^2$ , and the sum of the areas of the two smaller squares is $2 * 1^2 = 2$
Thus, the big square minus the three smaller squares is $25 - 9 - 2 = 14$ .  This is the area of the two congruent L-shaped regions.
So the area of one L-shaped region is $\frac{14}{2} = 7$ , and the answer is $\boxed{7}$

Solution: The shaded area can be divided into two regions:  one rectangle that is 1 by 3, and one rectangle that is 4 by 1.  (Or the reverse, depending on which rectangle the 1 by 1 square is "joined" to.)  Either way, the total area of these two regions is $3 + 4 = 7$ , and the answer is $\boxed{7}$

Solution: Chop the entire 5 by 5 region into $25$ squares like a piece of graph paper.  When you draw all the lines, you can count that only $7$ of the small 1 by 1 squares will be shaded, giving $\boxed{7}$ as the answer.

Solution: In the bottpom left corner of the 5 by 5 square there is a 4 by 4 square which has an area of $4\cdot4=16$ . In the top right of that 4 by 4 square is a 3 by 3 square with an area of $3\cdot3=9$ . When we remove the 3 by 3 square from the 4 by 4 square we get the L-shaped figure so our answer is $16-9=\boxed{7}$


Answer: 20.0

Solution:
The area of $\square OPQR$ is $2^2 = 4$
The area of $\triangle PQT$ is $\frac{1}{2}bh = \frac{1}{2}\cdot PT\cdot PQ$
If we set the areas equal, the area of $\triangle PQT$ is $4$ .  Also, note that $PQ=2$ .  Plugging those in, we get:
$4 = \frac{1}{2} \cdot PT \cdot 2$
$PT = 4$
If $PT = 4$ , and $PO = 2$ , then $OT = 2$ , and $T$ must be $2$ units to the left of the origin.  This would be $(-2,0)$ , giving answer $\boxed{2,0}$


Answer: 20201.0

Solution:
Using the recursion from solution 1, we see that the first differences of $4, 8, 12, ...$ form an arithmetic progression, and consequently that the second differences are constant and all equal to $4$ .  Thus, the original sequence can be generated from a quadratic function.
If $f(n) = an^2 + bn + c$ , and $f(0) = 1$ $f(1) = 5$ , and $f(2) = 13$ , we get a system of three equations in three variables:
$f(0) = 1$ gives $c = 1$
$f(1) = 5$ gives $a + b + c = 5$
$f(2) = 13$ gives $4a + 2b + c = 13$
Plugging in $c=1$ into the last two equations gives
$a + b = 4$
$4a + 2b = 12$
Dividing the second equation by 2 gives the system:
$a + b = 4$
$2a + b = 6$
Subtracting the first equation from the second gives $a = 2$ , and hence $b = 2$ .  Thus, our quadratic function is:
$f(n) = 2n^2 + 2n + 1$
Calculating the answer to our problem, $f(100) = 20000 + 200 + 1 = 20201$ , which is choice $\boxed{20201}$

Solution: We can see that each figure $n$ has a central box and 4 columns of $n$ boxes on each side of each square. Therefore, at figure 100, there is a central box with 100 boxes on the top, right, left, and bottom. Knowing that each quarter of each figure has a pyramid structure, we know that for each quarter there are $\sum_{n=1}^{100} n = 5050$ squares. $4 \cdot 5050 = 20200$ . Adding in the original center box we have $20200 + 1 = \boxed{20201}$

Solution:
Let $a_n$ be the number of squares in figure $n$ .  We can easily see that \[a_0=4\cdot 0+1\] \[a_1=4\cdot 1+1\] \[a_2=4\cdot 3+1\] \[a_3=4\cdot 6+1.\] See that we multiply the number we are on to the next consecutive number.
Note that in $a_n$ , the number multiplied by the 4 is the $n$ th triangular number.  Hence, $a_{100}=4\cdot \frac{100\cdot 101}{2}+1=\boxed{20201}$ .
~ edited by mathlover66

Solution:
Let $f_n$ denote the number of unit cubes in a figure. We have \[f_0=1\] \[f_1=5\] \[f_2=13\] \[f_3=25\] \[f_4=41\] \[...\]
Computing the difference between the number of cubes in each figure yields \[4,8,12,16,...\] It is easy to notice that this is an arithmetic sequence, with the first term being $4$ and the difference being $4$ . Let this sequence be $a_n$
From $f_0$ to $f_{100}$ , the sequence will have $100$ terms. Using the arithmetic sum formula yields
\[S_{100}=\frac{100[2\cdot 4+(100-1)4]}{2}\] \[=50(2\cdot 4+99\cdot 4)\] \[=50(101\cdot 4)\] \[=200\cdot 101\] \[=20200\]
So $f_{100}=1+20200=\boxed{20201}$ unit cubes.

Solution:
Newton’s little formula states that $a_n = A \binom{n-1}{0} + B \binom{n-1}{1} + C \binom{n-1}{2} + \cdots + K \binom{n-1}{m}$ if first term is $a_1$ and $A =$ first difference, $B =$ second difference, and so on. Hence we apply the formula (because we start at term 0, term 100 is $a_{101}$ ): $a_{101} = 5 \binom{100}{0} + 8 \binom{100}{1} + 4 \binom{100}{2} = \boxed{20201}.$ ~Peelybonehead
~clarification by LeonidasTheConquerer


Answer: 20201.0

Solution:
Using the recursion from solution 1, we see that the first differences of $4, 8, 12, ...$ form an arithmetic progression, and consequently that the second differences are constant and all equal to $4$ .  Thus, the original sequence can be generated from a quadratic function.
If $f(n) = an^2 + bn + c$ , and $f(0) = 1$ $f(1) = 5$ , and $f(2) = 13$ , we get a system of three equations in three variables:
$f(0) = 1$ gives $c = 1$
$f(1) = 5$ gives $a + b + c = 5$
$f(2) = 13$ gives $4a + 2b + c = 13$
Plugging in $c=1$ into the last two equations gives
$a + b = 4$
$4a + 2b = 12$
Dividing the second equation by 2 gives the system:
$a + b = 4$
$2a + b = 6$
Subtracting the first equation from the second gives $a = 2$ , and hence $b = 2$ .  Thus, our quadratic function is:
$f(n) = 2n^2 + 2n + 1$
Calculating the answer to our problem, $f(100) = 20000 + 200 + 1 = 20201$ , which is choice $\boxed{20201}$

Solution: We can see that each figure $n$ has a central box and 4 columns of $n$ boxes on each side of each square. Therefore, at figure 100, there is a central box with 100 boxes on the top, right, left, and bottom. Knowing that each quarter of each figure has a pyramid structure, we know that for each quarter there are $\sum_{n=1}^{100} n = 5050$ squares. $4 \cdot 5050 = 20200$ . Adding in the original center box we have $20200 + 1 = \boxed{20201}$

Solution:
Let $a_n$ be the number of squares in figure $n$ .  We can easily see that \[a_0=4\cdot 0+1\] \[a_1=4\cdot 1+1\] \[a_2=4\cdot 3+1\] \[a_3=4\cdot 6+1.\] See that we multiply the number we are on to the next consecutive number.
Note that in $a_n$ , the number multiplied by the 4 is the $n$ th triangular number.  Hence, $a_{100}=4\cdot \frac{100\cdot 101}{2}+1=\boxed{20201}$ .
~ edited by mathlover66

Solution:
Let $f_n$ denote the number of unit cubes in a figure. We have \[f_0=1\] \[f_1=5\] \[f_2=13\] \[f_3=25\] \[f_4=41\] \[...\]
Computing the difference between the number of cubes in each figure yields \[4,8,12,16,...\] It is easy to notice that this is an arithmetic sequence, with the first term being $4$ and the difference being $4$ . Let this sequence be $a_n$
From $f_0$ to $f_{100}$ , the sequence will have $100$ terms. Using the arithmetic sum formula yields
\[S_{100}=\frac{100[2\cdot 4+(100-1)4]}{2}\] \[=50(2\cdot 4+99\cdot 4)\] \[=50(101\cdot 4)\] \[=200\cdot 101\] \[=20200\]
So $f_{100}=1+20200=\boxed{20201}$ unit cubes.

Solution:
Newton’s little formula states that $a_n = A \binom{n-1}{0} + B \binom{n-1}{1} + C \binom{n-1}{2} + \cdots + K \binom{n-1}{m}$ if first term is $a_1$ and $A =$ first difference, $B =$ second difference, and so on. Hence we apply the formula (because we start at term 0, term 100 is $a_{101}$ ): $a_{101} = 5 \binom{100}{0} + 8 \binom{100}{1} + 4 \binom{100}{2} = \boxed{20201}.$ ~Peelybonehead
~clarification by LeonidasTheConquerer


Answer: 36.0

Solution:
Since the perimeter of $I$ $12$ , each side is $\frac{12}{4} = 3$
Since the perimeter of $II$ is $24$ , each side is $\frac{24}{4} = 6$
The side of $III$ is equal to the sum of the sides of $I$ and $II$ .  Therefore, the side of $III$ is $3 + 6 = 9$
Since $III$ is also a square, it has an perimeter of $9\cdot 4 = 36$ , and the answer is $\boxed{36}$

Solution: Let a side of $I$ equal $x$ , and let a side of $II$ equal $y$ . The perimeter of $I$ is $4x$ , and the perimeter of $II$ is $4y$ . One side of $III$ has length $x+y$ , so the perimeter is $4x+4y$ , which just so happens to be the sum of the perimeters of $I$ and $II$ , giving us $12+24=36$ , or answer $\boxed{36}$


Answer: 27.0

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}

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. $\boxed{27}$

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 $y=\log_2x=3\sqrt{3}\Longrightarrow(\log_2x)^2=(3\sqrt{3})^2=\boxed{27}$

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: 588.0

Find $3x^2 y^2$ if $x$ and $y$ are integers such that $y^2 + 3x^2 y^2 = 30x^2 + 517$

Solution:
If we move the $x^2$ term to the left side, it is factorable with Simon's Favorite Factoring Trick
\[(3x^2 + 1)(y^2 - 10) = 517 - 10\]
$507$ is equal to $3 \cdot 13^2$ . Since $x$ and $y$ are integers, $3x^2 + 1$ cannot equal a multiple of three. $169$ doesn't work either, so $3x^2 + 1 = 13$ , and $x^2 = 4$ . This leaves $y^2 - 10 = 39$ , so $y^2 = 49$ . Thus, $3x^2 y^2 = 3 \times 4 \times 49 = \boxed{588}$


Answer: 383.0

Find $A^2_{}$ , where $A^{}_{}$ is the sum of the absolute values of all roots of the following equation:

Solution:
$x=\sqrt{19}+\underbrace{\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{x}}}}}}_{x}$
$x=\sqrt{19}+\frac{91}{x}$
$x^2=x\sqrt{19}+91$
$x^2-x\sqrt{19}-91 = 0$
$\left.\begin{array}{l}x_1=\frac{\sqrt{19}+\sqrt{383}}{2}\\\\x_2=\frac{\sqrt{19}-\sqrt{383}}{2}\end{array}\right\}A=|x_1|+|x_2|\Rightarrow\sqrt{383}$
$A^2=\boxed{383}$

Solution:
Let $f(x) = \sqrt{19} + \frac{91}{x}$ . Then $x = f(f(f(f(f(x)))))$ , from which we realize that $f(x) = x$ . This is because if we expand the entire expression, we will get a fraction of the form $\frac{ax + b}{cx + d}$ on the right hand side, which makes the equation simplify to a quadratic. As this quadratic will have two roots, they must be the same roots as the quadratic $f(x)=x$
The given finite expansion can then be easily seen to reduce to the quadratic equation $x_{}^{2}-\sqrt{19}x-91=0$ . The solutions are $x_{\pm}^{}=$ $\frac{\sqrt{19}\pm\sqrt{383}}{2}$ . Therefore, $A_{}^{}=\vert x_{+}\vert+\vert x_{-}\vert=\sqrt{383}$ . We conclude that $A_{}^{2}=\boxed{383}$


Answer: 987.0

Find $a$ if $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^{17} + bx^{16} + 1$

Solution: Let $F_n$ represent the $n$ th number in the Fibonacci sequence. Therefore, \begin{align*} x^2 - x - 1 = 0&\Longrightarrow x^n = F_n(x), \ n\in N \\ &\Longrightarrow x^{n + 2} = F_{n + 1}\cdot x + F_n,\ n\in N. \end{align*} The above uses the similarity between the Fibonacci recursion|recursive definition, $F_{n+2} - F_{n+1} - F_n = 0$ , and the polynomial $x^2 - x - 1 = 0$ \begin{align*} 0 = ax^{17} + bx^{16} + 1 = a(F_{17}\cdot x + F_{16}) + b(F_{16}\cdot x + F_{15}) + 1 &\Longrightarrow (aF_{17} + bF_{16})\cdot x + (aF_{16} + bF_{15} + 1) = 0,\ x\not\in Q \\ &\Longrightarrow aF_{17} + bF_{16} = 0 \text{ and } aF_{16} + bF_{15} + 1 = 0 \\ &\Longrightarrow a = F_{16},\ b = - F_{17} \\ &\Longrightarrow a = \boxed{987}

Solution: We can long divide and search for a pattern; then the remainder would be set to zero to solve for $a$ . Writing out a few examples quickly shows us that the remainders after each subtraction follow the Fibonacci sequence. Carrying out this pattern, we find that the remainder is \[(F_{16}b + F_{17}a)x + F_{15}b + F_{16}a + 1 = 0.\] Since the coefficient of $x$ must be zero, this gives us two equations, $F_{16}b + F_{17}a = 0$ and $F_{15}b + F_{16}a + 1 = 0$ . Solving these two as above, we get that $a = \boxed{987}$

Solution:
Trying to divide $ax^{17} + bx^{16} + 1$ by $x^2-x-1$ would be very tough, so let's try to divide using smaller degrees of $x$ . Doing $\frac{ax^3+bx^2+1}{x^2-x-1}$ , we get the following systems of equations: \begin{align*} a+b &= -1, \\ 2a+b &= 0. \end{align*} Continuing with $\frac{ax^4+bx^3+1}{x^2-x-1}$ \begin{align*} 2a+b &= -1, \\ 3a+2b &= 0. \end{align*} There is somewhat of a pattern showing up, so let's try $\frac{ax^5+bx^4+1}{x^2-x-1}$ to verify. We get: \begin{align*} 3a+2b &= -1, \\ 5a+3b &= 0. \end{align*} Now we begin to see that our pattern is actually the Fibonacci Numbers! Using the previous equations, we can make a general statement about $\frac{ax^n+bx^{n-1}+1}{x^2-x-1}$ \begin{align*} af_{n-1}+bf_{n-2} &= -1, \\ af_n+bf_{n-1} &= 0. \end{align*} Also, noticing our solutions from the previous systems of equations, we can create the following statement:
If $ax^n+bx^{n-1}+1$ has $x^2-x-1$ as a factor, then $a=f_{n-1}$ and $b = f_n.$
Thus, if $ax^{17}+bx^{16}+1$ has $x^2-x-1$ as a factor, we get that $a = 987$ and $b = -1597,$ so $a = \boxed{987}$

Solution:
Let's work backwards! Let $F(x) = ax^{17} + bx^{16} + 1$ and let $P(x)$ be the polynomial such that $P(x)(x^2 - x - 1) = F(x)$
Clearly, the constant term of $P(x)$ must be $- 1$ . Now, we have \[(x^2 - x - 1)(c_1x^{15} + c_2x^{14} + \cdots + c_{15}x - 1),\] where $c_{i}$ is some coefficient. However, since $F(x)$ has no $x$ term, it must be true that $c_{15} = 1$
Let's find $c_{14}$ now. Notice that all we care about in finding $c_{14}$ is that $(x^2 - x - 1)(\cdots + c_{14}x^2 + x - 1) = \text{something} + 0x^2 + \text{something}$ . Therefore, $c_{14} = - 2$ . Undergoing a similar process, $c_{13} = 3$ $c_{12} = - 5$ $c_{11} = 8$ , and we see a nice pattern. The coefficients of $P(x)$ are just the Fibonacci sequence with alternating signs! Therefore, $a = c_1 = F_{16}$ , where $F_{16}$ denotes the 16th Fibonnaci number and $a = \boxed{987}$

Solution: The roots of $x^2-x-1$ are $\phi$ (the Golden Ratio) and $1-\phi$ . These two must also be roots of $ax^{17}+bx^{16}+1$ . Thus, we have two equations: \begin{align*} a\phi^{17}+b\phi^{16}+1=0, \\ a(1-\phi)^{17}+b(1-\phi)^{16}+1=0. \end{align*} Subtract these two and divide by $\sqrt{5}$ to get \[\frac{a(\phi^{17}-(1-\phi)^{17})}{\sqrt{5}}+\frac{b(\phi^{16}-(1-\phi)^{16})}{\sqrt{5}}=0.\] Noting that the formula for the $n$ th Fibonacci number is $\frac{\phi^n-(1-\phi)^n}{\sqrt{5}}$ , we have $1597a+987b=0$ . Since $1597$ and $987$ are coprime, the solutions to this equation under the integers are of the form $a=987k$ and $b=-1597k$ , of which the only integral solutions for $a$ on $[0,999]$ are $0$ and $987$ $(a,b)=(0,0)$ cannot work since $x^2-x-1$ does not divide $1$ , so the answer must be $\boxed{987}$ . (Note that this solution would not be valid on an Olympiad or any test that does not restrict answers as integers between $000$ and $999$ ).

Solution:
We are given that $x^2 - x - 1$ is a factor of $ax^{17} + bx^{16} + 1,$ so the roots of $x^2 - x - 1$ must also be roots of $ax^{17} + bx^{16} + 1.$
Let $x=r$ be a root of $x^2 - x - 1$ so that $r^2 - r - 1 = 0,$ or $r^2 = r + 1.$ It follows that \[ar^{17} + br^{16} + 1 = 0. \hspace{20mm} (\bigstar)\] Note that \begin{align*} r^4 &= (r+1)^2 \\ &= r^2 + 2r + 1 \\ &= (r+1) + 2r + 1 \\ &= 3r + 2, \\ r^8 &= (3r+2)^2 \\ &= 9r^2 + 12r + 4 \\ &= 9(r+1) + 12r + 4 \\ &= 21r + 13, \\ r^{16} &= (21r + 13)^2 \\ &= 441r^2 + 546r + 169 \\ &= 441(r+1) +546r + 169 \\ &= 987r + 610. \end{align*} We rewrite the left side of $(\bigstar)$ as a linear expression of $r:$ \begin{align*} (ar+b)r^{16} + 1 &= 0 \\ (ar+b)(987r + 610) + 1 &= 0 \\ 987ar^2 + (610a+987b)r + 610b + 1 &= 0 \\ 987a(r+1) + (610a+987b)r + 610b + 1 &= 0 \\ (1597a+987b)r + (987a + 610b + 1) &= 0.  \end{align*} Since this linear equation has two solutions of $r,$ it must be an identity. Therefore, we have the following system of equations: \begin{align*} 1597a+987b &= 0, \\ 987a+610b &= -1. \end{align*} To eliminate $b,$ we multiply the first equation by $610$ and multiply the second equation by $987,$ then subtract the resulting equations: \begin{align*} 610(1597a)+610(987b) &= 0, \\ 987(987a)+987(610b) &= -987, \end{align*} from which \begin{align*} (610\cdot1597-987\cdot987)a&=987 \\ (974170-974169)a&=987 \\ a&=\boxed{987} ~MRENTHUSIASM

Solution: For simplicity, let $f(x) =ax^{17} + bx^{16} + 1$ and $g(x) = x^2-x-1$ . Notice that the roots of $g(x)$ are also roots of $f(x)$ . Let these roots be $u,v$ . We get the system \begin{align*} au^{17} + bu^{16} + 1 &= 0, \\ av^{17} + bv^{16} + 1 &= 0. \end{align*} If we multiply the first equation by $v^{16}$ and the second by $u^{16}$ we get \begin{align*} u^{17} v^{16} a + u^{16} v^{16} b + v^{16} &= 0, \\ u^{16} v^{17} a + u^{16} v^{16} b + u^{16} &= 0. \end{align*} Now subtracting, we get \[a(u^{17}v^{16} -u^{16} v^{17}) = u^{16}-v^{16} \implies a = \frac{u^{16} - v^{16}}{u^{17}v^{16} -u^{16} v^{17}}.\] By Vieta's, $uv=-1$ so the denominator becomes $u-v$ . By difference of squares and dividing out $u-v$ we get \[a= (u^8+v^8)(u^4+v^4)(u^2+v^2)(u+v).\] A simple exercise of Vieta's gets us $a= \boxed{987}.$

Solution:
We see that $ax^{17} + bx^{16} + 1 = (x^2-x-1)P(x)$ for some polynomial $P(x)$ . Working forwards, we notice that the constant term of $P(x)$ must equal $-1$ , to multiply the constant term of $(x^2-x-1)$ into $1$ . We can similarly continue to use this logic, by repeatedly cancelling out the middle term, and obtain the process: $(x^2-x-1)(-1) = -x^2 + x + 1$ $(x^2-x-1)(-1 + x) = x^3 - 2x^2 + 1$ $(x^2-x-1)(-1 + x - 2x^2) = -2x^4 + 3x^3 + 1$ $(x^2-x-1)(-1 + x - 2x^2 + 3x^3) = 3x^5 - 5x^4 + 1$ .
By this time, you can hopefully notice that the coefficient of the $x^n$ term in $P(x)$ is equal to $(-1)^{n-1} * F_{n}$ , where $F_n$ equals the $n$ th number in the Fibonacci sequence. From here, we just need to find the coefficient of the $x^15$ term in $P(x)$ , which happens to be $F_15 = \boxed{987}$ .
Again, try to only use Engineer's Induction when you have no other options. A rigorous proof is usually not needed, but when you have extra time, checking a solution with a rigorous method is better than worrying about your Engineer's Induction solution.


Answer: 20.0

Find $ax^5 + by^5$ if the real numbers $a,b,x,$ and $y$ satisfy the equations \begin{align*} ax + by &= 3, \\ ax^2 + by^2 &= 7, \\ ax^3 + by^3 &= 16, \\ ax^4 + by^4 &= 42. \end{align*}

Solution:
Set $S = (x + y)$ and $P = xy$ . Then the relationship
\[(ax^n + by^n)(x + y) = (ax^{n + 1} + by^{n + 1}) + (xy)(ax^{n - 1} + by^{n - 1})\]
can be exploited:
\begin{eqnarray*}(ax^2 + by^2)(x + y) & = & (ax^3 + by^3) + (xy)(ax + by) \\ (ax^3 + by^3)(x + y) & = & (ax^4 + by^4) + (xy)(ax^2 + by^2)\end{eqnarray*}
Therefore:
\begin{eqnarray*}7S & = & 16 + 3P \\ 16S & = & 42 + 7P\end{eqnarray*}
Consequently, $S = - 14$ and $P = - 38$ . Finally:
\begin{eqnarray*}(ax^4 + by^4)(x + y) & = & (ax^5 + by^5) + (xy)(ax^3 + by^3) \\ (42)(S) & = & (ax^5 + by^5) + (P)(16) \\ (42)( - 14) & = & (ax^5 + by^5) + ( - 38)(16) \\ ax^5 + by^5 & = & \boxed{020}

Solution:
recurrence of the form $T_n=AT_{n-1}+BT_{n-2}$ will have the closed form $T_n=ax^n+by^n$ , where $x,y$ are the values of the starting term that make the sequence geometric, and $a,b$ are the appropriately chosen constants such that those special starting terms linearly combine to form the actual starting terms.
Suppose we have such a recurrence with $T_1=3$ and $T_2=7$ . Then $T_3=ax^3+by^3=16=7A+3B$ , and $T_4=ax^4+by^4=42=16A+7B$
Solving these simultaneous equations for $A$ and $B$ , we see that $A=-14$ and $B=38$ . So, $ax^5+by^5=T_5=-14(42)+38(16)= \boxed{020}$

Solution:
We first let the answer to this problem be $k.$ Multiplying the first equation by $x$ gives $ax^2 + bxy=3x$
Subtracting this equation from the second equation gives $by^2-bxy=7-3x$ . Similarly, doing the same for the other equations, we obtain: $by^2-bxy=7-3x$ $by^3-bxy^2=16-7x$ $by^4-bxy^3=42-16x$ , and $by^5-bxy^4=k-42x$
Now lets take the first equation. Multiplying this by $y$ and subtracting this from the second gives us $by^3-bxy^2=(7-3x)y$ . We can also obtain $by^4-bxy^3=(16-7x)y$
Now we can solve for $x$ and $y$ $(7-3x)y = 16-7x$ and $(16-7x)y=42-16x$ . Solving for $x$ and $y$ gives us $(-7+\sqrt{87},-7-\sqrt{87})$ (It can be switched, but since the given equations are symmetric, it doesn't matter). $k-42x= (42-16x)y$ , and solving for $k$ gives us $k= \boxed{020}$


Answer: 198.0

Find $c$ if $a$ $b$ , and $c$ are positive integers which satisfy $c=(a + bi)^3 - 107i$ , where $i^2 = -1$

Solution: Expanding out both sides of the given equation we have $c + 107i = (a^3 - 3ab^2) + (3a^2b - b^3)i$ .  Two complex numbers are equal if and only if their real parts and imaginary parts are equal, so $c = a^3 - 3ab^2$ and $107 = 3a^2b - b^3 = (3a^2 - b^2)b$ .  Since $a, b$ are integers , this means $b$ is a divisor of 107, which is a prime number .  Thus either $b = 1$ or $b = 107$ .  If $b = 107$ $3a^2 - 107^2 = 1$ so $3a^2 = 107^2 + 1$ , but $107^2 + 1$ is not divisible by 3, a contradiction.  Thus we must have $b = 1$ $3a^2 = 108$ so $a^2 = 36$ and $a = 6$ (since we know $a$ is positive).  Thus $c = 6^3 - 3\cdot 6 = \boxed{198}$


Answer: 146.0

Find $x^2+y^2_{}$ if $x_{}^{}$ and $y_{}^{}$ are positive integers such that \begin{align*} xy+x+y&=71, \\ x^2y+xy^2&=880. \end{align*}

Solution: Define $a = x + y$ and $b = xy$ . Then $a + b = 71$ and $ab = 880$ . Solving these two equations yields a quadratic $a^2 - 71a + 880 = 0$ , which factors to $(a - 16)(a - 55) = 0$ . Either $a = 16$ and $b = 55$ or $a = 55$ and $b = 16$ . For the first case, it is easy to see that $(x,y)$ can be $(5,11)$ (or vice versa). In the second case, since all factors of $16$ must be $\le 16$ , no two factors of $16$ can sum greater than $32$ , and so there are no integral solutions for $(x,y)$ . The solution is $5^2 + 11^2 = \boxed{146}$

Solution:
Let $a=x+y$ $b=xy$ then we get the equations \begin{align*} a+b&=71\\ ab&=880 \end{align*} After finding the prime factorization of $880=2^4\cdot5\cdot11$ , it's easy to obtain the solution $(a,b)=(16,55)$ . Thus \[x^2+y^2=(x+y)^2-2xy=a^2-2b=16^2-2\cdot55=\boxed{146}\] Note that if $(a,b)=(55,16)$ , the answer would exceed $999$ which is invalid for an AIME answer.
~ Nafer

Solution: First, notice that you can factor $x^2y + xy^2$ as $xy(x + y)$ . From this, we notice that $xy$ and $x + y$ is a common occurrence, so that lends itself to a simple solution by substitution. Let $xy = b$ and $x + y = a$ . From this substitution, we get the following system: \[a + b = 71\] \[ab = 880\] Solving that system gives us the following two pairs $(a, b)$ $(16, 55)$ and $(55, 16)$ . The second one is obviously too big as $55^2$ is clearly out of bounds for an AIME problem (More on that later). Therefore, we proceed with substituting back the pair $(16, 55)$ . This means that $x + y = 16$ and $xy = 55$ Then, instead of solving the system, we can do a clever manipulation by squaring $x + y$ . Doing so, we get: \[(x + y)^2 = (x^2 + y^2) + 2xy\] We see that in this form, we can substitute everything in except for $(x^2 + y^2)$ , which is the desired answer. Substituting, we get: \[256 = (x^2 + y^2) + 110\] so $x^2 + y^2 = \boxed{146}$ . (If we were to go with the pair $(55, 16)$ , then the $(x + y)^2$ would be absurdly out of bounds)


Answer: 115.0

Find a positive integral solution to the equation $\frac{1+3+5+\dots+(2n-1)}{2+4+6+\dots+2n}=\frac{115}{116}$
$\textbf{(A) }110\qquad \textbf{(B) }115\qquad \textbf{(C) }116\qquad \textbf{(D) }231\qquad\\ \textbf{(E) }\text{The equation has no positive integral solutions.}$

Solution:
Solution by e_power_pi_times_i
Notice that the numerator and denominator are the sum of the first $n$ odd and even numbers, respectively. Then the numerator is $n^2$ , and the denominator is $n(n+1)$ .  Then $\frac{n}{n+1} = \frac{115}{116}$ , so $n = \boxed{115}$


Answer: 400.0

Find the area of rhombus $ABCD$ given that the circumradii of triangles $ABD$ and $ACD$ are $12.5$ and $25$ , respectively.

Solution:
The diagonals of the rhombus perpendicularly bisect each other. Call half of diagonal BD $a$ and half of diagonal AC $b$ . The length of the four sides of the rhombus is $\sqrt{a^2+b^2}$
The area of any triangle can be expressed as $\frac{a\cdot b\cdot c}{4R}$ , where $a$ $b$ , and $c$ are the sides and $R$ is the circumradius. Thus, the area of $\triangle ABD$ is $ab=2a(a^2+b^2)/(4\cdot12.5)$ . Also, the area of $\triangle ABC$ is $ab=2b(a^2+b^2)/(4\cdot25)$ . Setting these two expressions equal to each other and simplifying gives $b=2a$ . Substitution yields $a=10$ and $b=20$ , so the area of the rhombus is $20\cdot40/2=\boxed{400}$

Solution: Let $\theta=\angle BDA$ . Let $AB=BC=CD=x$ . By the extended law of sines, \[\frac{x}{\sin\theta}=25\] Since $AC\perp BD$ $\angle CAD=90-\theta$ , so \[\frac{x}{\sin(90-\theta)=\cos\theta}=50\] Hence $x=25\sin\theta=50\cos\theta$ . Solving $\tan\theta=2$ $\sin\theta=\frac{2}{\sqrt{5}}, \cos\theta=\frac{1}{\sqrt{5}}$ . Thus \[x=25\frac{2}{\sqrt{5}}\implies x^2=500\] The height of the rhombus is $x\sin(2\theta)=2x\sin\theta\cos\theta$ , so we want \[2x^2\sin\theta\cos\theta=\boxed{400}\]


Answer: 480.0

Find the area of the region enclosed by the graph of $|x-60|+|y|=\left|\frac{x}{4}\right|.$

Solution:
1987 AIME-4.png
Since $|y|$ is nonnegative $\left|\frac{x}{4}\right| \ge |x - 60|$ . Solving this gives us two equations: $\frac{x}{4} \ge x - 60\ \mathrm{and} \ -\frac{x}{4} \le x - 60$ . Thus, $48 \le x \le 80$ . The maximum and minimum y value is when $|x - 60| = 0$ , which is when $x = 60$ and $y = \pm 15$ . Since the graph is symmetric about the y-axis, we just need casework upon $x$ $\frac{x}{4} > 0$ , so we break up the condition $|x-60|$
The area of the region enclosed by the graph is that of the quadrilateral defined by the points $(48,0),\ (60,15),\ (80,0), \ (60,-15)$ . Breaking it up into triangles and solving or using the Shoelace Theorem , we get $2 \cdot \frac{1}{2}(80 - 48)(15) = \boxed{480}$

Solution:
Since $|y|$ is the only present $y$ "term" in this equation, we know that the area must be symmetrical about the x-axis.
We'll consider the area when $y>0$ and we only consider the portion enclosed with $y=0$ . Then, we'll double that area since the graph is symmetrical.
Now, let us remove the absolute values:
When $x\ge 60$ $x-60+y=0.25x$ . This rearranges to $y=-0.75x+60$
When $0\le x<60$ $60-x+y=0.25$ . So $y=1.25x-60$
When $x<0$ $60-x+y=-0.25x$ . So $y=0.75x-60$
By simple sketching, we see the shape that looks like the image in Solution 1 (graph it out and you'll see). We see that the partial area we seek in this part is the triangle with the vertices $(60,15)$ $(48,0)$ $(80,0)$ . This triangle has an area of $(80-48)*15*0.5=240$
Simply double the area and we get $\boxed{480}$ as our final answer.
~hastapasta