https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Carrot+karen&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T20:04:00ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12B_Problems/Problem_18&diff=1426702020 AMC 12B Problems/Problem 182021-01-18T18:43:09Z<p>Carrot karen: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
In square <math>ABCD</math>, points <math>E</math> and <math>H</math> lie on <math>\overline{AB}</math> and <math>\overline{DA}</math>, respectively, so that <math>AE=AH.</math> Points <math>F</math> and <math>G</math> lie on <math>\overline{BC}</math> and <math>\overline{CD}</math>, respectively, and points <math>I</math> and <math>J</math> lie on <math>\overline{EH}</math> so that <math>\overline{FI} \perp \overline{EH}</math> and <math>\overline{GJ} \perp \overline{EH}</math>. See the figure below. Triangle <math>AEH</math>, quadrilateral <math>BFIE</math>, quadrilateral <math>DHJG</math>, and pentagon <math>FCGJI</math> each has area <math>1.</math> What is <math>FI^2</math>?<br />
<br />
<asy><br />
real x=2sqrt(2);<br />
real y=2sqrt(16-8sqrt(2))-4+2sqrt(2);<br />
real z=2sqrt(8-4sqrt(2));<br />
pair A, B, C, D, E, F, G, H, I, J;<br />
A = (0,0);<br />
B = (4,0);<br />
C = (4,4);<br />
D = (0,4);<br />
E = (x,0);<br />
F = (4,y);<br />
G = (y,4);<br />
H = (0,x);<br />
I = F + z * dir(225);<br />
J = G + z * dir(225);<br />
<br />
draw(A--B--C--D--A);<br />
draw(H--E);<br />
draw(J--G^^F--I);<br />
draw(rightanglemark(G, J, I), linewidth(.5));<br />
draw(rightanglemark(F, I, E), linewidth(.5));<br />
<br />
dot("$A$", A, S);<br />
dot("$B$", B, S);<br />
dot("$C$", C, dir(90));<br />
dot("$D$", D, dir(90));<br />
dot("$E$", E, S);<br />
dot("$F$", F, dir(0));<br />
dot("$G$", G, N);<br />
dot("$H$", H, W);<br />
dot("$I$", I, SW);<br />
dot("$J$", J, SW);<br />
<br />
</asy><br />
<br />
<math>\textbf{(A) } \frac{7}{3} \qquad \textbf{(B) } 8-4\sqrt2 \qquad \textbf{(C) } 1+\sqrt2 \qquad \textbf{(D) } \frac{7}{4}\sqrt2 \qquad \textbf{(E) } 2\sqrt2</math><br />
<br />
== Solution 1 ==<br />
Since the total area is <math>4</math>, the side length of square <math>ABCD</math> is <math>2</math>. We see that since triangle <math>HAE</math> is a right isosceles triangle with area 1, we can determine sides <math>HA</math> and <math>AE</math> both to be <math>\sqrt{2}</math>. Now, consider extending <math>FB</math> and <math>IE</math> until they intersect. Let the point of intersection be <math>K</math>. We note that <math>EBK</math> is also a right isosceles triangle with side <math>2-\sqrt{2}</math> and find it's area to be <math>3-2\sqrt{2}</math>. Now, we notice that <math>FIK</math> is also a right isosceles triangle and find it's area to be <math>\frac{1}{2}</math><math>FI^2</math>. This is also equal to <math>1+3-2\sqrt{2}</math> or <math>4-2\sqrt{2}</math>. Since we are looking for <math>FI^2</math>, we want two times this. That gives <math>\boxed{\textbf{(B)}\ 8-4\sqrt{2}}</math>.~TLiu<br />
<br />
== Solution 2 (Lucky Measuring) ==<br />
Since this is a geometry problem involving sides, and we know that <math>HE</math> is <math>2</math>, we can use our ruler and find the ratio between <math>FI</math> and <math>HE</math>. Measuring(on the booklet), we get that <math>HE</math> is about <math>1.8</math> inches and <math>FI</math> is about <math>1.4</math> inches. Thus, we can then multiply the length of <math>HE</math> by the ratio of <math>\frac{1.4}{1.8},</math> of which we then get <math>FI= \frac{14}{9}.</math> We take the square of that and get <math>\frac{196}{81},</math> and the closest answer to that is <math>\boxed{\textbf{(B)}\ 8-4\sqrt{2}}</math>. ~Celloboy (Note that this is just a strategy I happened to use that worked. Do not press your luck with this strategy, for it was a lucky guess)<br />
<br />
== Solution 3 ==<br />
Draw the auxiliary line <math>AC</math>. Denote by <math>M</math> the point it intersects with <math>HE</math>, and by <math>N</math> the point it intersects with <math>GF</math>. Last, denote by <math>x</math> the segment <math>FN</math>, and by <math>y</math> the segment <math>FI</math>. We will find two equations for <math>x</math> and <math>y</math>, and then solve for <math>y^2</math>. <br />
<br />
Since the overall area of <math>ABCD</math> is <math>4 \;\; \Longrightarrow \;\; AB=2</math>, and <math>AC=2\sqrt{2}</math>. In addition, the area of <math>\bigtriangleup AME = \frac{1}{2} \;\; \Longrightarrow \;\; AM=1</math>.<br />
<br />
The two equations for <math>x</math> and <math>y</math> are then:<br />
<br />
<math>\bullet</math> Length of <math>AC</math>: <math>1+y+x = 2\sqrt{2} \;\; \Longrightarrow \;\; x = (2\sqrt{2}-1) - y</math> <br />
<br />
<math>\bullet</math> Area of CMIF: <math>\frac{1}{2}x^2+xy = \frac{1}{2} \;\; \Longrightarrow \;\; x(x+2y)=1</math>.<br />
<br />
Substituting the first into the second, yields <br />
<math>\left[\left(2\sqrt{2}-1\right)-y\right]\cdot \left[\left(2\sqrt{2}-1\right)+y\right]=1</math><br />
<br />
Solving for <math>y^2</math> gives <math>\boxed{\textbf{(B)}\ 8-4\sqrt{2}}</math> ~DrB<br />
<br />
== Solution 4 ==<br />
Plot a point <math>F'</math> such that <math>F'I</math> and <math>AB</math> are parallel and extend line <math>FB</math> to point <math>B'</math> such that <math>FIB'F'</math> forms a square. Extend line <math>AE</math> to meet line <math>F'B'</math> and point <math>E'</math> is the intersection of the two. The area of this square is equivalent to <math>FI^2</math>. We see that the area of square <math>ABCD</math> is <math>4</math>, meaning each side is of length 2. The area of the pentagon <math>EIFF'E'</math> is <math>2</math>. Length <math>AE=\sqrt{2}</math>, thus <math>EB=2-\sqrt{2}</math>. Triangle <math>EB'E'</math> is isosceles, and the area of this triangle is <math>\frac{1}{2}(4-2\sqrt{2})(2-\sqrt{2})=6-4\sqrt{2}</math>. Adding these two areas, we get <cmath>2+6-4\sqrt{2}=8-4\sqrt{2}\rightarrow \boxed{\mathrm{(B)}}</cmath>. --OGBooger<br />
<br />
== Solution 5 (HARD Calculation) ==<br />
We can easily observe that the area of square <math>ABCD</math> is 4 and its side length is 2 since all four regions that build up the square has area 1. <br />
Extend <math>FI</math> and let the intersection with <math>AB</math> be <math>K</math>. Connect <math>AC</math>, and let the intersection of <math>AC</math> and <math>HE</math> be <math>L</math>.<br />
Notice that since the area of triangle <math>AEH</math> is 1 and <math>AE=AH</math> , <math>AE=AH=\sqrt{2}</math>, therefore <math>BE=HD=2-\sqrt{2}</math>.<br />
Let <math>CG=CF=m</math>, then <math>BF=DG=2-m</math>.<br />
Also notice that <math>KB=2-m</math>, thus <math>KE=KB-BE=2-m-(2-\sqrt{2})=\sqrt{2}-m</math>.<br />
Now use the condition that the area of quadrilateral <math>BFIE</math> is 1, we can set up the following equation: <br />
<math>\frac{1}{2}(2-m)^2-\frac{1}{4}(\sqrt{2}-m)^2=1</math><br />
We solve the equation and yield <math>m=\frac{8-2\sqrt{2}-\sqrt{64-32\sqrt{2}}}{2}</math>.<br />
Now notice that<br />
<math>FI=AC-AL-\frac{m}{\sqrt{2}}=2\sqrt{2}-1-\frac{\sqrt{2}}{2}*\frac{8-2\sqrt{2}-\sqrt{64-32\sqrt{2}}}{2}</math><br />
<math>=2\sqrt{2}-1-\frac{8\sqrt{2}-4-\sqrt{128-64\sqrt2}}{4}</math><br />
<math>=\frac{\sqrt{128-64\sqrt{2}}}{4}</math>.<br />
Hence <math>FI^2=\frac{128-64\sqrt{2}}{16}=8-4\sqrt{2}</math>. -HarryW<br />
<br />
-edit: annabelle0913<br />
<br />
== Solution 6 ==<br />
<br />
<asy><br />
real x=2sqrt(2);<br />
real y=2sqrt(16-8sqrt(2))-4+2sqrt(2);<br />
real z=2sqrt(8-4sqrt(2));<br />
real k= 8-2sqrt(2);<br />
real l= 2sqrt(2)-4;<br />
pair A, B, C, D, E, F, G, H, I, J, L, M, K;<br />
A = (0,0);<br />
B = (4,0);<br />
C = (4,4);<br />
D = (0,4);<br />
E = (x,0);<br />
F = (4,y);<br />
G = (y,4);<br />
H = (0,x);<br />
I = F + z * dir(225);<br />
J = G + z * dir(225);<br />
L = (k,0);<br />
M = F + z * dir(315);<br />
K = (4,l);<br />
<br />
draw(A--B--C--D--A);<br />
draw(H--E);<br />
draw(J--G^^F--I);<br />
draw(F--M);<br />
draw(M--L);<br />
draw(E--K,dashed+linewidth(.5));<br />
draw(K--L,dashed+linewidth(.5));<br />
draw(B--L);<br />
draw(rightanglemark(G, J, I), linewidth(.5));<br />
draw(rightanglemark(F, I, E), linewidth(.5));<br />
draw(rightanglemark(F, M, L), linewidth(.5));<br />
fill((4,0)--(k,0)--M--(4,y)--cycle, gray);<br />
dot("$A$", A, S);<br />
dot("$C$", C, dir(90));<br />
dot("$D$", D, dir(90));<br />
dot("$E$", E, S);<br />
dot("$G$", G, N);<br />
dot("$H$", H, W);<br />
dot("$I$", I, SW);<br />
dot("$J$", J, SW);<br />
dot("$K$", K, S);<br />
dot("$F(G)$", F, E);<br />
dot("$J'$", M, dir(90));<br />
dot("$H'$", L, S);<br />
dot("$B(D)$", B, S);<br />
<br />
<br />
</asy><br />
Easily, we can find that: quadrilateral <math>BFIE</math> and <math>DHJG</math> are congruent with each other, so we can move <math>DHJG</math> to the shaded area (<math>F</math> and <math>G</math>, <math>B</math> and <math>D</math> overlapping) to form a square <math>FIKJ'</math> (<math>DG</math> = <math>FB</math>, <math>CG</math> = <math>FC</math>, <math>{\angle} CGF</math> = <math>{\angle}CFG</math> = <math>45^{\circ}</math> so <math>{\angle} IFJ'= 90^{\circ}</math>). Then we can solve <math>AH</math> = <math>AE</math> = <math>\sqrt{2}</math>, <math>EB</math> = <math>2-\sqrt{2}</math>, <math>EK</math> = <math>2\sqrt{2}-2</math>. <br />
<br />
<math>FI^2</math> = <math>area</math> of <math>BFIE</math> <math>+</math> <math>area</math> of <math>FJ'H'B</math> <math>+</math> <math>area</math> of <math>EH'K</math> = <math>1 + 1 + \frac{1}{2}(2\sqrt{2}-2)^2=8-4\sqrt{2}\rightarrow \boxed{\mathrm{(B)}}</math><br />
<br />
--Ryan Zhang @BRS<br />
<br />
==Video Solution 1==<br />
https://www.youtube.com/watch?v=AKJXB07Sat0&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=7 ~ MathEx<br />
<br />
==Video Solution 2 by the Beauty of Math==<br />
Solution starts at 3:09: https://youtu.be/VZYe3Hu88OA<br />
<br />
==See Also==<br />
{{AMC10 box|year=2020|ab=B|num-b=20|num-a=22}}<br />
{{AMC12 box|year=2020|ab=B|num-b=17|num-a=19}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Carrot karenhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_11&diff=1425032020 AMC 10B Problems/Problem 112021-01-18T01:06:56Z<p>Carrot karen: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Ms. Carr asks her students to read any 5 of the 10 books on a reading list. Harold randomly selects 5 books from this list, and Betty does the same. What is the probability that there are exactly 2 books that they both select?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{8} \qquad\textbf{(B)}\ \frac{5}{36} \qquad\textbf{(C)}\ \frac{14}{45} \qquad\textbf{(D)}\ \frac{25}{63} \qquad\textbf{(E)}\ \frac{1}{2}</math><br />
<br />
==Solution 1==<br />
<br />
We don't care about which books Harold selects. We just care that Betty picks <math>2</math> books from Harold's list and <math>3</math> that aren't on Harold's list.<br />
<br />
The total amount of combinations of books that Betty can select is <math>\binom{10}{5}=252</math>. <br />
<br />
There are <math>\binom{5}{2}=10</math> ways for Betty to choose <math>2</math> of the books that are on Harold's list.<br />
<br />
From the remaining <math>5</math> books that aren't on Harold's list, there are <math>\binom{5}{3}=10</math> ways to choose <math>3</math> of them.<br />
<br />
<math>\frac{10\cdot10}{252}=\boxed{\textbf{(D) }\frac{25}{63}}</math> ~quacker88<br />
<br />
==Solution 2==<br />
<br />
We can analyze this as two containers with <math>10</math> balls each, with the two people grabbing <math>5</math> balls each. First, we need to find the probability of two of the balls being the same among five: <math>\frac{1}{3}\frac{4}{9}\frac{3}{8}\frac{5}{7}\frac{2}{4}</math>. After that we must, multiply this probability by <math>{5 \choose 2}</math>, for the 2 balls that are the same are chosen among 5 balls. The answer will be <math>\frac{5}{126}*10 = \frac{25}{63}</math>. <math>\boxed{(D) \frac{25}{63}}</math><br />
<br />
<br />
==Video Solution==<br />
https://youtu.be/t6yjfKXpwDs<br />
<br />
~IceMatrix<br />
<br />
https://youtu.be/RbKdVmZRxkk<br />
<br />
~savannahsolver<br />
<br />
==See Also==<br />
{{AMC10 box|year=2020|ab=B|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Carrot karenhttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_18&diff=1413282019 AMC 10B Problems/Problem 182021-01-02T18:52:57Z<p>Carrot karen: /* Solution 3 (not rigorous; similar to 2) */</p>
<hr />
<div>==Problem==<br />
<br />
Henry decides one morning to do a workout, and he walks <math>\tfrac{3}{4}</math> of the way from his home to his gym. The gym is <math>2</math> kilometers away from Henry's home. At that point, he changes his mind and walks <math>\tfrac{3}{4}</math> of the way from where he is back toward home. When he reaches that point, he changes his mind again and walks <math>\tfrac{3}{4}</math> of the distance from there back toward the gym. If Henry keeps changing his mind when he has walked <math>\tfrac{3}{4}</math> of the distance toward either the gym or home from the point where he last changed his mind, he will get very close to walking back and forth between a point <math>A</math> kilometers from home and a point <math>B</math> kilometers from home. What is <math>|A-B|</math>?<br />
<br />
<math>\textbf{(A) } \frac{2}{3} \qquad \textbf{(B) } 1 \qquad \textbf{(C) } \frac{6}{5} \qquad \textbf{(D) } \frac{5}{4} \qquad \textbf{(E) } \frac{3}{2}</math><br />
<br />
==Solution 1==<br />
Let the two points that Henry walks in between be <math>P</math> and <math>Q</math>, with <math>P</math> being closer to home. As given in the problem statement, the distances of the points <math>P</math> and <math>Q</math> from his home are <math>A</math> and <math>B</math> respectively. By symmetry, the distance of point <math>Q</math> from the gym is the same as the distance from home to point <math>P</math>. Thus, <math>A = 2 - B</math>. In addition, when he walks from point <math>Q</math> to home, he walks <math>\frac{3}{4}</math> of the distance, ending at point <math>P</math>. Therefore, we know that <math>B - A = \frac{3}{4}B</math>. By substituting, we get <math>B - A = \frac{3}{4}(2 - A)</math>. Adding these equations now gives <math>2(B - A) = \frac{3}{4}(2 + B - A)</math>. Multiplying by <math>4</math>, we get <math>8(B - A) = 6 + 3(B - A)</math>, so <math>B - A = \frac{6}{5} = \boxed{\textbf{(C) } \frac{6}{5}}</math>.<br />
<br />
==Solution 2 (not rigorous)==<br />
We assume that Henry is walking back and forth exactly between points <math>P</math> and <math>Q</math>, with <math>P</math> closer to Henry's home than <math>Q</math>. Denote Henry's home as a point <math>H</math> and the gym as a point <math>G</math>. Then <math>HP:PQ = 1:3</math> and <math>PQ:QG = 3:1</math>, so <math>HP:PQ:QG = 1:3:1</math>. Therefore, <math>|A-B| = PQ = \frac{3}{1+3+1} \cdot 2 = \frac{6}{5} = \boxed{\textbf{(C) } \frac{6}{5}}</math>.<br />
<br />
==Solution 3 (not rigorous; similar to 2)==<br />
Since Henry is very close to walking back and forth between two points, let us denote <math>A</math> closer to his house, and <math>B</math> closer to the gym. Then, let us denote the distance from <math>A</math> to <math>B</math> as <math>x</math>. If Henry was at <math>B</math> and walked <math>\frac{3}{4}</math> of the way, he would end up at <math>A</math>, vice versa. Thus we can say that the distance from <math>A</math> to the gym is <math>\frac{1}{4}</math> the distance from <math>B</math> to his house. That means it is <math>\frac{1}{3}x</math>. This also applies to the other side. Furthermore, we can say <math>\frac{1}{3}x</math> + <math>x</math> + <math>\frac{1}{3}x</math> = <math>2</math>. We solve for <math>x</math> and get <math>x=\frac{6}{5}</math>. Therefore, the answer is <math>\boxed{\textbf{(C) } \frac{6}{5}}</math>.<br />
<br />
~aryam<br />
<br />
==Solution 4 ==<br />
<br />
Let <math>A</math> have a distance of <math>x</math> from the home. Then, the distance to the gym is <math>2-x</math>. This means point <math>B</math> and point <math>A</math> are <math>\frac{3}{4} \cdot (2-x)</math> away from one another. It also means that Point <math>B</math> is located at <math>\frac{3}{4} (2-x) + x.</math> So, the distance between the home and point <math>B</math> is also <math>\frac{3}{4} (2-x) + x.</math><br />
<br />
It follows that point <math>A</math> must be at a distance of <math>\frac{3}{4} \left( \frac{3}{4} (2-x) + x \right)</math> from point <math>B</math>. However, we also said that this distance has length <math>\frac{3}{4} (2-x)</math>. So, we can set those equal, which gives the equation: <cmath>\frac{3}{4} \left( \frac{3}{4} (2-x) + x \right) = \frac{3}{4} (2-x).</cmath><br />
<br />
Solving, we get <math>x = \frac{2}{5}</math>. This means <math>A</math> is at point <math>\frac{2}{5}</math> and <math>B</math> is at point <math>\frac{3}{4} \cdot \frac{8}{5} + \frac{2}{5} = \frac{8}{5}.</math><br />
<br />
So, <math>|A - B| = \boxed{\frac{6}{5}}.</math><br />
<br />
==Video Solution==<br />
For those who want a video solution: https://youtu.be/45kdBy3htOg<br />
<br />
==Video Solution 2==<br />
https://youtu.be/U5PjjZ-5MIQ<br />
<br />
~IceMatrix<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2019|ab=B|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Carrot karenhttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_15&diff=1407522019 AMC 10A Problems/Problem 152020-12-27T19:40:11Z<p>Carrot karen: /* Solution 4 (Not rigorous at all) */</p>
<hr />
<div>{{duplicate|[[2019 AMC 10A Problems|2019 AMC 10A #15]] and [[2019 AMC 12A Problems|2019 AMC 12A #9]]}}<br />
<br />
==Problem==<br />
<br />
A sequence of numbers is defined recursively by <math>a_1 = 1</math>, <math>a_2 = \frac{3}{7}</math>, and<br />
<cmath>a_n=\frac{a_{n-2} \cdot a_{n-1}}{2a_{n-2} - a_{n-1}}</cmath>for all <math>n \geq 3</math> Then <math>a_{2019}</math> can be written as <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. What is <math>p+q ?</math><br />
<br />
<math>\textbf{(A) } 2020 \qquad\textbf{(B) } 4039 \qquad\textbf{(C) } 6057 \qquad\textbf{(D) } 6061 \qquad\textbf{(E) } 8078</math><br />
<br />
==Video Solution==<br />
https://youtu.be/h9rgKc_YVQ0<br />
<br />
Education, the Study of Everything<br />
<br />
<br />
<br />
==Solution 1 (Induction)==<br />
<br />
Using the recursive formula, we find <math>a_3=\frac{3}{11}</math>, <math>a_4=\frac{3}{15}</math>, and so on. It appears that <math>a_n=\frac{3}{4n-1}</math>, for all <math>n</math>. Setting <math>n=2019</math>, we find <math>a_{2019}=\frac{3}{8075}</math>, so the answer is <math>\boxed{\textbf{(E) }8078}</math>.<br />
<br />
To prove this formula, we use induction. We are given that <math>a_1=1</math> and <math>a_2=\frac{3}{7}</math>, which satisfy our formula. Now assume the formula holds true for all <math>n\le m</math> for some positive integer <math>m</math>. By our assumption, <math>a_{m-1}=\frac{3}{4m-5}</math> and <math>a_m=\frac{3}{4m-1}</math>. Using the recursive formula, <cmath>a_{m+1}=\frac{a_{m-1}\cdot a_m}{2a_{m-1}-a_m}=\frac{\frac{3}{4m-5}\cdot\frac{3}{4m-1}}{2\cdot\frac{3}{4m-5}-\frac{3}{4m-1}}=\frac{\left(\frac{3}{4m-5}\cdot\frac{3}{4m-1}\right)(4m-5)(4m-1)}{\left(2\cdot\frac{3}{4m-5}-\frac{3}{4m-1}\right)(4m-5)(4m-1)}=\frac{9}{6(4m-1)-3(4m-5)}=\frac{3}{4(m+1)-1},</cmath><br />
so our induction is complete.<br />
<br />
==Solution 2==<br />
<br />
Since we are interested in finding the sum of the numerator and the denominator, consider the sequence defined by <math>b_n = \frac{1}{a_n}</math>.<br />
<br />
We have <math>\frac{1}{a_n} = \frac{2a_{n-2}-a_{n-1}}{a_{n-2} \cdot a_{n-1}}=\frac{2}{a_{n-1}}-\frac{1}{a_{n-2}}</math>, so in other words, <math>b_n = 2b_{n-1}-b_{n-2}=3b_{n-2}-2b_{n-3}=4b_{n-3}-3b_{n-4}=\dots</math>.<br />
<br />
By recursively following this pattern, we can see that <math>b_n=(n-1) \cdot b_2 - (n-2) \cdot b_1</math>.<br />
<br />
By plugging in 2019, we thus find <math>b_{2019} = 2018 \cdot \frac{7}{3}-2017 = \frac{8075}{3}</math>. Since the numerator and the denominator are relatively prime, the answer is <math>\boxed{\textbf{(E) } 8078}</math>.<br />
<br />
-eric2020<br />
<br />
==Solution 3==<br />
<br />
It seems reasonable to transform the equation into something else. Let <math>a_{n}=x</math>, <math>a_{n-1}=y</math>, and <math>a_{n-2}=z</math>. Therefore, we have <cmath>x=\frac{zy}{2z-y}</cmath><br />
<cmath>2xz-xy=zy</cmath><br />
<cmath>2xz=y(x+z)</cmath><br />
<cmath>y=\frac{2xz}{x+z}</cmath><br />
Thus, <math>y</math> is the harmonic mean of <math>x</math> and <math>z</math>. This implies <math>a_{n}</math> is a harmonic sequence or equivalently <math>b_{n}=\frac{1}{a_{n}}</math> is arithmetic. Now, we have <math>b_{1}=1</math>, <math>b_{2}=\frac{7}{3}</math>, <math>b_{3}=\frac{11}{3}</math>, and so on. Since the common difference is <math>\frac{4}{3}</math>, we can express <math>b_{n}</math> explicitly as <math>b_{n}=\frac{4}{3}(n-1)+1</math>. This gives <math>b_{2019}=\frac{4}{3}(2019-1)+1=\frac{8075}{3}</math> which implies <math>a_{2019}=\frac{3}{8075}=\frac{p}{q}</math>. <math>p+q=\boxed{\textbf{(E) } 8078}</math><br />
~jakeg314<br />
<br />
==Solution 4 (Not rigorous at all)==<br />
Noticing that there is clearly a pattern, but the formula for it is hideous, we first find the first few terms of the sequence to see if there is any pattern: <math>1, \frac{3}{7}, \frac{3}{11}, \frac{1}{5}, \frac{3}{19}, \frac{3}{23} ...</math><br />
<br />
Now, we notice that the numerator seems to be in a pattern: <math>1, 3, 3, 1, 3, 3, 1, 3, 3...</math> Then, we notice that the only time the numerator is <math>1</math> is when the index is a multiple of <math>4</math>. Clearly, <math>2019</math> is NOT a multiple of <math>4</math>, so the numerator is <math>3</math>. Then, using the positions of each term, we can come up with a simple formula for the denominator with <math>n</math> as the position or index (This only applies for the numbers with numerator <math>3</math>): <math>3n + (n - 1)</math>.<br />
<br />
Plugging <math>n</math> in for <math>2019</math>, we get <math>8075</math> for the denominator. Adding <math>3</math> (The numerator) gives <math>\boxed{\textbf{(E) }8078}</math><br />
<br />
~EricShi1685<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2019|ab=A|num-b=14|num-a=16}}<br />
{{AMC12 box|year=2019|ab=A|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Carrot karenhttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_17&diff=1404122017 AMC 10A Problems/Problem 172020-12-23T23:59:23Z<p>Carrot karen: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
Distinct points <math>P</math>, <math>Q</math>, <math>R</math>, <math>S</math> lie on the circle <math>x^{2}+y^{2}=25</math> and have integer coordinates. The distances <math>PQ</math> and <math>RS</math> are irrational numbers. What is the greatest possible value of the ratio <math>\frac{PQ}{RS}</math>?<br />
<br />
<math>\mathrm{\textbf{(A)}}\ 3\qquad\mathrm{\textbf{(B)}}\ 5\qquad\mathrm{\textbf{(C)}}\ 3\sqrt{5}\qquad\mathrm{\textbf{(D)}}\ 7\qquad\mathrm{\textbf{(E)}}\ 5\sqrt{2}</math><br />
<br />
==Solution 1==<br />
<br />
Because <math>P</math>, <math>Q</math>, <math>R</math>, and <math>S</math> are lattice points, there are only a few coordinates that actually satisfy the equation. The coordinates are <math>(\pm 3,\pm 4), (\pm 4, \pm 3), (0,\pm 5),</math> and <math>(\pm 5,0).</math> We want to maximize <math>PQ</math> and minimize <math>RS.</math> They also have to be the square root of something, because they are both irrational. The greatest value of <math>PQ</math> happens when <math>P</math> and <math>Q</math> are almost directly across from each other and are in different quadrants. For example, the endpoints of the segment could be <math>(-4,3)</math> and <math>(3,-4)</math> because the two points are almost across from each other. Another possible pair could be <math>(-4,3)</math> and <math>(5,0)</math>. To find out which segment is longer, we have to compare the distances from their endpoints to a diameter (which must be the longest possible segment). The closest diameter would be from <math>(-4,3)</math> to <math>(4,-3)</math>. The distance between <math>(3,-4)</math> and <math>(4,-3)</math> is shorter than the distance between <math>(5,0)</math> and <math>(4,-3)</math>. Therefore, the segment from <math>(-4,3)</math> to <math>(3,-4)</math> is the longest attainable. The least value of <math>RS</math> is when the two endpoints are in the same quadrant and are very close to each other. This can occur when, for example, <math>R</math> is <math>(3,4)</math> and <math>S</math> is <math>(4,3).</math> They are in the same quadrant and no other point on the circle with integer coordinates is closer to the point <math>(3,4)</math> than <math>(4,3).</math> Using the distance formula, we get that <math>PQ</math> is <math>\sqrt{98}</math> and that <math>RS</math> is <math>\sqrt{2}.</math> <math>\frac{\sqrt{98}}{\sqrt{2}}=\sqrt{49}=\boxed{\mathrm{\textbf{(D)}}\ 7}</math><br />
<br />
==Solution 2==<br />
So what we can do is look at the option choices. Since we are aiming for the highest possible ratio, let's try using <math>7</math>. Now, looking at the problem alone, we know that to have the largest ratio possible, we have to let <math>RS</math> be the minimum possible value while at the same time using integer coordinates. Thus, the smallest possible value of <math>RS</math> is <math>\sqrt{1^{2}+1^{2}} = \sqrt{2}</math>. Assuming that <math>\frac{PQ}{RS} = 7</math>, we plug in <math>RS = \sqrt{2}</math> and solve for PQ: <math>PQ=7\sqrt{2}</math>. Remember, we don't know if this is possible yet, we are only trying to figure out if it is. But for what values of <math>x</math> and <math>y</math> does <math>\sqrt{x^{2}+y^{2}}=7\sqrt2</math>? We see that this can easily be made into a <math>45-45-90</math> triangle. But, instead of substituting <math>y=x</math> into the equation and then using a whole lot of algebra, we can save time and use the little trick, that if in a <math>45-45-90</math> triangle, the two <math>45</math> degree sides have side length <math>s</math>, then the hypotenuse is <math>s\sqrt2</math>. Using this, we can see that <math>s=7</math>, and since our equation does in fact yield a sensible solution, we can be assured that our answer is <math>\boxed{\mathrm{\textbf{(D)}}\ 7}</math>.<br />
<br />
Quality Control by fasterthanlight<br />
<br />
(Note by Carrot_Karen: We tried 7, but some might be confused why we concluded that it was the answer after verifying without trying the others, like why wasn't option <math>\textbf{(E)}</math> tried? This is because the problem can only have one correct answer, so if we have an option that already works, we can conclude that none of the others work and <math>\textbf{(D)}</math> is the answer.<br />
<br />
==Video Solution==<br />
https://youtu.be/umr2Aj9ViOA<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=A|num-b=16|num-a=18}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Geometry Problems]]</div>Carrot karenhttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_19&diff=1404112017 AMC 10A Problems/Problem 192020-12-23T23:58:42Z<p>Carrot karen: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of <math>5</math> chairs under these conditions?<br />
<br />
<math> \textbf{(A)}\ 12\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 40</math><br />
<br />
==Solution 1: Casework==<br />
<br />
For notation purposes, let Alice be A, Bob be B, Carla be C, Derek be D, and Eric be E.<br />
We can split this problem up into two cases:<br />
<br />
<math>\textbf{Case 1: }</math> A sits on an edge seat.<br />
<br />
Then, since B and C can't sit next to A, that must mean either D or E sits next to A. After we pick either D or E, then either B or C must sit next to D/E. Then, we can arrange the two remaining people in two ways. Since there are two different edge seats that A can sit in, there are a total of <math>2 \cdot 2 \cdot 2 \cdot 2 = 16</math>.<br />
<br />
<math>\textbf{Case 2: }</math> A does not sit in an edge seat.<br />
<br />
In this case, then only two people that can sit next to A are D and E, and there are two ways to permute them, and this also handles the restriction that D can't sit next to E. Then, there are two ways to arrange B and C, the remaining people. However, there are three initial seats that A can sit in, so there are <math>3 \cdot 2 \cdot 2 = 12</math> seatings in this case.<br />
<br />
Adding up all the cases, we have <math>16+12 = \boxed{\textbf{(C) } 28}</math>.<br />
<br />
==Solution 2==<br />
<br />
Label the seats (from left to right) <math>1</math> through <math>5</math>. The number of ways to seat Derek and Eric in the five seats with no restrictions is <math>5 \cdot 4=20</math>. The number of ways to seat Derek and Eric such that they sit next to each other is <math>8</math> (we can treat Derek and Eric as a "block". There are four ways to seat this "block", and two ways to permute Derek and Eric, for a total of <math>4\cdot 2=8</math>), so the number of ways such that Derek and Eric don't sit next to each other is <math>20-8=12</math>. Note that once Derek and Eric are seated, there are three cases. <br />
<br />
The first case is that they sit at each end. There are two ways to seat Derek and Eric. But this is impossible because then Alice, Bob, and Carla would have to sit in some order in the middle three seats which would lead to Alice sitting next to Bob or Carla, a contradiction. So this case gives us <math>0</math> ways.<br />
<br />
Another possible case is if Derek and Eric seat in seats <math>2</math> and <math>4</math> in some order. There are <math>2</math> possible ways to seat Derek and Eric like this. This leaves Alice, Bob, and Carla to sit in any order in the remaining three seats. Since no two of these three seats are consecutive, there are <math>3!=6</math> ways to do this. So the second case gives us <math>2 \cdot 6=12</math> total ways for the second case.<br />
<br />
The last case is if once Derek and Eric are seated, exactly one pair of consecutive seats are available. There are <math>12-2-2=8</math> ways to seat Derek and Eric like this. Once they are seated like this, Alice cannot sit in one of the two consecutive available seats without sitting next to Bob or Carla. So Alice has to sit in the other remaining chair. Then, there are two ways to seat Bob and Carla in the remaining two seats (which are consecutive). So this case gives us <math>8 \cdot 2=16</math> ways.<br />
<br />
So in total there are <math>12+16=28</math>. So our answer is <math>\boxed{\textbf{(C)}\ 28}</math>.<br />
<br><br />
Minor LaTeX edits by fasterthanlight<br />
<br />
Minor clarity edits by Carrot_Karen<br />
<br />
==Video Solution==<br />
https://youtu.be/umr2Aj9ViOA<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=A|num-b=18|num-a=20}}<br />
{{AMC12 box|year=2017|ab=A|num-b=13|num-a=15}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]</div>Carrot karenhttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_19&diff=1404102017 AMC 10A Problems/Problem 192020-12-23T23:58:24Z<p>Carrot karen: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of <math>5</math> chairs under these conditions?<br />
<br />
<math> \textbf{(A)}\ 12\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 40</math><br />
<br />
==Solution 1: Casework==<br />
<br />
For notation purposes, let Alice be A, Bob be B, Carla be C, Derek be D, and Eric be E.<br />
We can split this problem up into two cases:<br />
<br />
<math>\textbf{Case 1: }</math> A sits on an edge seat.<br />
<br />
Then, since B and C can't sit next to A, that must mean either D or E sits next to A. After we pick either D or E, then either B or C must sit next to D/E. Then, we can arrange the two remaining people in two ways. Since there are two different edge seats that A can sit in, there are a total of <math>2 \cdot 2 \cdot 2 \cdot 2 = 16</math>.<br />
<br />
<math>\textbf{Case 2: }</math> A does not sit in an edge seat.<br />
<br />
In this case, then only two people that can sit next to A are D and E, and there are two ways to permute them, and this also handles the restriction that D can't sit next to E. Then, there are two ways to arrange B and C, the remaining people. However, there are three initial seats that A can sit in, so there are <math>3 \cdot 2 \cdot 2 = 12</math> seatings in this case.<br />
<br />
Adding up all the cases, we have <math>16+12 = \boxed{\textbf{(C) } 28}</math>.<br />
<br />
==Solution 2==<br />
<br />
Label the seats (from left to right) <math>1</math> through <math>5</math>. The number of ways to seat Derek and Eric in the five seats with no restrictions is <math>5 \cdot 4=20</math>. The number of ways to seat Derek and Eric such that they sit next to each other is <math>8</math> (we can treat Derek and Eric as a "block". There are four ways to seat this "block", and two ways to permute Derek and Eric, for a total of <math>4\cdot 2=8</math>), so the number of ways such that Derek and Eric don't sit next to each other is <math>20-8=12</math>. Note that once Derek and Eric are seated, there are three cases. <br />
<br />
The first case is that they sit at each end. There are two ways to seat Derek and Eric. But this is impossible because then Alice, Bob, and Carla would have to sit in some order in the middle three seats which would lead to Alice sitting next to Bob or Carla, a contradiction. So this case gives us <math>0</math> ways.<br />
<br />
Another possible case is if Derek and Eric seat in seats <math>2</math> and <math>4</math> in some order. There are <math>2</math> possible ways to seat Derek and Eric like this. This leaves Alice, Bob, and Carla to sit in any order in the remaining three seats. Since no two of these three seats are consecutive, there are <math>3!=6</math> ways to do this. So the second case gives us <math>2 \cdot 6=12</math> total ways for the second case.<br />
<br />
The last case is if once Derek and Eric are seated, exactly one pair of consecutive seats are available. There are <math>12-2-2=8</math> ways to seat Derek and Eric like this. Once they are seated like this, Alice cannot sit in one of the two consecutive available seats without sitting next to Bob or Carla. So Alice has to sit in the other remaining chair. Then, there are two ways to seat Bob and Carla in the remaining two seats (which are consecutive). So this case gives us <math>8 \cdot 2=16</math> ways.<br />
<br />
So in total there are <math>12+16=28</math>. So our answer is <math>\boxed{\textbf{(C)}\ 28}</math>.<br />
<br><br />
Minor LaTeX edits by fasterthanlight<br />
Minor clarity edits by Carrot_Karen<br />
<br />
==Video Solution==<br />
https://youtu.be/umr2Aj9ViOA<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=A|num-b=18|num-a=20}}<br />
{{AMC12 box|year=2017|ab=A|num-b=13|num-a=15}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]</div>Carrot karenhttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_19&diff=1404092017 AMC 10A Problems/Problem 192020-12-23T23:55:52Z<p>Carrot karen: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of <math>5</math> chairs under these conditions?<br />
<br />
<math> \textbf{(A)}\ 12\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 40</math><br />
<br />
==Solution 1: Casework==<br />
<br />
For notation purposes, let Alice be A, Bob be B, Carla be C, Derek be D, and Eric be E.<br />
We can split this problem up into two cases:<br />
<br />
<math>\textbf{Case 1: }</math> A sits on an edge seat.<br />
<br />
Then, since B and C can't sit next to A, that must mean either D or E sits next to A. After we pick either D or E, then either B or C must sit next to D/E. Then, we can arrange the two remaining people in two ways. Since there are two different edge seats that A can sit in, there are a total of <math>2 \cdot 2 \cdot 2 \cdot 2 = 16</math>.<br />
<br />
<math>\textbf{Case 2: }</math> A does not sit in an edge seat.<br />
<br />
In this case, then only two people that can sit next to A are D and E, and there are two ways to permute them, and this also handles the restriction that D can't sit next to E. Then, there are two ways to arrange B and C, the remaining people. However, there are three initial seats that A can sit in, so there are <math>3 \cdot 2 \cdot 2 = 12</math> seatings in this case.<br />
<br />
Adding up all the cases, we have <math>16+12 = \boxed{\textbf{(C) } 28}</math>.<br />
<br />
==Solution 2==<br />
<br />
Label the seats (from left to right) <math>1</math> through <math>5</math>. The number of ways to seat Derek and Eric in the five seats with no restrictions is <math>5 \cdot 4=20</math>. The number of ways to seat Derek and Eric such that they sit next to each other is <math>8</math> (we can treat Derek and Eric as a "block". There are four ways to seat this "block", and two ways to permute Derek and Eric, for a total of <math>4\cdot 2=8</math>), so the number of ways such that Derek and Eric don't sit next to each other is <math>20-8=12</math>. Note that once Derek and Eric are seated, there are three cases. <br />
<br />
The first case is that they sit at each end. There are two ways to seat Derek and Eric. But this is impossible because then Alice, Bob, and Carla would have to sit in some order in the middle three seats which would lead to Alice sitting next to Bob or Carla, a contradiction. So this case gives us <math>0</math> ways.<br />
<br />
Another possible case is if Derek and Eric seat in seats <math>2</math> and <math>4</math> in some order. There are <math>2</math> possible ways to seat Derek and Eric like this. This leaves Alice, Bob, and Carla to sit in any order in the remaining three seats. Since no two of these three seats are consecutive, there are <math>3!=6</math> ways to do this. So the second case gives us <math>2 \cdot 6=12</math> total ways for the second case.<br />
<br />
The last case is if once Derek and Eric are seated, exactly one pair of consecutive seats are available. There are <math>12-2-2=8</math> ways to seat Derek and Eric like this. Once they are seated like this, Alice cannot not sit in one of the two consecutive available seats without sitting next to Bob and Carla. So Alice has to sit in the other remaining chair. Then, there are two ways to seat Bob and Carla in the remaining two seats (which are consecutive). So this case gives us <math>8 \cdot 2=16</math> ways.<br />
<br />
So in total there are <math>12+16=28</math>. So our answer is <math>\boxed{\textbf{(C)}\ 28}</math>.<br />
<br><br />
Minor LaTeX edits by fasterthanlight<br />
<br />
==Video Solution==<br />
https://youtu.be/umr2Aj9ViOA<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=A|num-b=18|num-a=20}}<br />
{{AMC12 box|year=2017|ab=A|num-b=13|num-a=15}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]</div>Carrot karenhttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_17&diff=1404082017 AMC 10A Problems/Problem 172020-12-23T23:49:10Z<p>Carrot karen: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
Distinct points <math>P</math>, <math>Q</math>, <math>R</math>, <math>S</math> lie on the circle <math>x^{2}+y^{2}=25</math> and have integer coordinates. The distances <math>PQ</math> and <math>RS</math> are irrational numbers. What is the greatest possible value of the ratio <math>\frac{PQ}{RS}</math>?<br />
<br />
<math>\mathrm{\textbf{(A)}}\ 3\qquad\mathrm{\textbf{(B)}}\ 5\qquad\mathrm{\textbf{(C)}}\ 3\sqrt{5}\qquad\mathrm{\textbf{(D)}}\ 7\qquad\mathrm{\textbf{(E)}}\ 5\sqrt{2}</math><br />
<br />
==Solution 1==<br />
<br />
Because <math>P</math>, <math>Q</math>, <math>R</math>, and <math>S</math> are lattice points, there are only a few coordinates that actually satisfy the equation. The coordinates are <math>(\pm 3,\pm 4), (\pm 4, \pm 3), (0,\pm 5),</math> and <math>(\pm 5,0).</math> We want to maximize <math>PQ</math> and minimize <math>RS.</math> They also have to be the square root of something, because they are both irrational. The greatest value of <math>PQ</math> happens when <math>P</math> and <math>Q</math> are almost directly across from each other and are in different quadrants. For example, the endpoints of the segment could be <math>(-4,3)</math> and <math>(3,-4)</math> because the two points are almost across from each other. Another possible pair could be <math>(-4,3)</math> and <math>(5,0)</math>. To find out which segment is longer, we have to compare the distances from their endpoints to a diameter (which must be the longest possible segment). The closest diameter would be from <math>(-4,3)</math> to <math>(4,-3)</math>. The distance between <math>(3,-4)</math> and <math>(4,-3)</math> is shorter than the distance between <math>(5,0)</math> and <math>(4,-3)</math>. Therefore, the segment from <math>(-4,3)</math> to <math>(3,-4)</math> is the longest attainable. The least value of <math>RS</math> is when the two endpoints are in the same quadrant and are very close to each other. This can occur when, for example, <math>R</math> is <math>(3,4)</math> and <math>S</math> is <math>(4,3).</math> They are in the same quadrant and no other point on the circle with integer coordinates is closer to the point <math>(3,4)</math> than <math>(4,3).</math> Using the distance formula, we get that <math>PQ</math> is <math>\sqrt{98}</math> and that <math>RS</math> is <math>\sqrt{2}.</math> <math>\frac{\sqrt{98}}{\sqrt{2}}=\sqrt{49}=\boxed{\mathrm{\textbf{(D)}}\ 7}</math><br />
<br />
==Solution 2==<br />
So what we can do is look at the option choices. Since we are aiming for the highest possible ratio, let's try using <math>7</math>. Now, looking at the problem alone, we know that to have the largest ratio possible, we have to let <math>RS</math> be the minimum possible value while at the same time using integer coordinates. Thus, the smallest possible value of <math>RS</math> is <math>\sqrt{1^{2}+1^{2}} = \sqrt{2}</math>. Assuming that <math>\frac{PQ}{RS} = 7</math>, we plug in <math>RS = \sqrt{2}</math> and solve for PQ: <math>PQ=7\sqrt{2}</math>. Remember, we don't know if this is possible yet, we are only trying to figure out if it is. But for what values of <math>x</math> and <math>y</math> does <math>\sqrt{x^{2}+y^{2}}=7\sqrt2</math>? We see that this can easily be made into a <math>45-45-90</math> triangle. But, instead of substituting <math>y=x</math> into the equation and then using a whole lot of algebra, we can save time and use the little trick, that if in a <math>45-45-90</math> triangle, the two <math>45</math> degree sides have side length <math>s</math>, then the hypotenuse is <math>s\sqrt2</math>. Using this, we can see that <math>s=7</math>, and since our equation does in fact yield a sensible solution, we can be assured that our answer is <math>\boxed{\mathrm{\textbf{(D)}}\ 7}</math>.<br />
<br />
Quality Control by fasterthanlight<br />
<br />
(Note by CarrotKaren: We tried 7, but some might be confused why we concluded that it was the answer after verifying without trying the others, like why wasn't option <math>\textbf{(E)}</math> tried? This is because the problem can only have one correct answer, so if we have an option that already works, we can conclude that none of the others work and <math>\textbf{(D)}</math> is the answer.<br />
<br />
==Video Solution==<br />
https://youtu.be/umr2Aj9ViOA<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=A|num-b=16|num-a=18}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Geometry Problems]]</div>Carrot karenhttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_17&diff=1404042017 AMC 10A Problems/Problem 172020-12-23T23:07:19Z<p>Carrot karen: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
Distinct points <math>P</math>, <math>Q</math>, <math>R</math>, <math>S</math> lie on the circle <math>x^{2}+y^{2}=25</math> and have integer coordinates. The distances <math>PQ</math> and <math>RS</math> are irrational numbers. What is the greatest possible value of the ratio <math>\frac{PQ}{RS}</math>?<br />
<br />
<math>\mathrm{\textbf{(A)}}\ 3\qquad\mathrm{\textbf{(B)}}\ 5\qquad\mathrm{\textbf{(C)}}\ 3\sqrt{5}\qquad\mathrm{\textbf{(D)}}\ 7\qquad\mathrm{\textbf{(E)}}\ 5\sqrt{2}</math><br />
<br />
==Solution 1==<br />
<br />
Because <math>P</math>, <math>Q</math>, <math>R</math>, and <math>S</math> are lattice points, there are only a few coordinates that actually satisfy the equation. The coordinates are <math>(\pm 3,\pm 4), (\pm 4, \pm 3), (0,\pm 5),</math> and <math>(\pm 5,0).</math> We want to maximize <math>PQ</math> and minimize <math>RS.</math> They also have to be the square root of something, because they are both irrational. The greatest value of <math>PQ</math> happens when <math>P</math> and <math>Q</math> are almost directly across from each other and are in different quadrants. For example, the endpoints of the segment could be <math>(-4,3)</math> and <math>(3,-4)</math> because the two points are almost across from each other. Another possible pair could be <math>(-4,3)</math> and <math>(5,0)</math>. To find out which segment is longer, we have to compare the distances from their endpoints to a diameter (which must be the longest possible segment). The closest diameter would be from <math>(-4,3)</math> to <math>(4,-3)</math>. The distance between <math>(3,-4)</math> and <math>(4,-3)</math> is shorter than the distance between <math>(5,0)</math> and <math>(4,-3)</math>. Therefore, the segment from <math>(-4,3)</math> to <math>(3,-4)</math> is the longest attainable. The least value of <math>RS</math> is when the two endpoints are in the same quadrant and are very close to each other. This can occur when, for example, <math>R</math> is <math>(3,4)</math> and <math>S</math> is <math>(4,3).</math> They are in the same quadrant and no other point on the circle with integer coordinates is closer to the point <math>(3,4)</math> than <math>(4,3).</math> Using the distance formula, we get that <math>PQ</math> is <math>\sqrt{98}</math> and that <math>RS</math> is <math>\sqrt{2}.</math> <math>\frac{\sqrt{98}}{\sqrt{2}}=\sqrt{49}=\boxed{\mathrm{\textbf{(D)}}\ 7}</math><br />
<br />
==Solution 2==<br />
So what we can do is look at the option choices. Since we are aiming for the highest possible ratio, let's try using <math>7</math>. Now, looking at the problem alone, we know that to have the largest ratio possible, we have to let <math>RS</math> be the minimum possible value while at the same time using integer coordinates. Thus, the smallest possible value of <math>RS</math> is <math>\sqrt{1^{2}+1^{2}} = \sqrt{2}</math>. Assuming that <math>\frac{PQ}{RS} = 7</math>, we plug in <math>RS = \sqrt{2}</math> and solve for PQ: <math>PQ=7\sqrt{2}</math>. Remember, we don't know if this is possible yet, we are only trying to figure out if it is. But for what values of <math>x</math> and <math>y</math> does <math>\sqrt{x^{2}+y^{2}}=7\sqrt2</math>? We see that this can easily be made into a <math>45-45-90</math> triangle. But, instead of substituting <math>y=x</math> into the equation and then using a whole lot of algebra, we can save time and use the little trick, that if in a <math>45-45-90</math> triangle, the two <math>45</math> degree sides have side length <math>s</math>, then the hypotenuse is <math>s\sqrt2</math>. Using this, we can see that <math>s=7</math>, and since our equation does in fact yield a sensible solution, we can be assured that our answer is <math>\boxed{\mathrm{\textbf{(D)}}\ 7}</math>.<br />
<br />
Quality Control by fasterthanlight<br />
<br />
(Note by CarrotKaren: We tried 7, but some might be confused why we concluded that it was the answer after verifying without trying the others, like why wasn't option <math>\textbf{(E)}</math> tried? This is because there can only be one answer in the problem that works, so if we have an option that already works, we can conclude that none of the others work and <math>\textbf{(D)}</math> is the answer.<br />
<br />
==Video Solution==<br />
https://youtu.be/umr2Aj9ViOA<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=A|num-b=16|num-a=18}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Geometry Problems]]</div>Carrot karenhttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_17&diff=1404032017 AMC 10A Problems/Problem 172020-12-23T23:05:59Z<p>Carrot karen: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
Distinct points <math>P</math>, <math>Q</math>, <math>R</math>, <math>S</math> lie on the circle <math>x^{2}+y^{2}=25</math> and have integer coordinates. The distances <math>PQ</math> and <math>RS</math> are irrational numbers. What is the greatest possible value of the ratio <math>\frac{PQ}{RS}</math>?<br />
<br />
<math>\mathrm{\textbf{(A)}}\ 3\qquad\mathrm{\textbf{(B)}}\ 5\qquad\mathrm{\textbf{(C)}}\ 3\sqrt{5}\qquad\mathrm{\textbf{(D)}}\ 7\qquad\mathrm{\textbf{(E)}}\ 5\sqrt{2}</math><br />
<br />
==Solution 1==<br />
<br />
Because <math>P</math>, <math>Q</math>, <math>R</math>, and <math>S</math> are lattice points, there are only a few coordinates that actually satisfy the equation. The coordinates are <math>(\pm 3,\pm 4), (\pm 4, \pm 3), (0,\pm 5),</math> and <math>(\pm 5,0).</math> We want to maximize <math>PQ</math> and minimize <math>RS.</math> They also have to be the square root of something, because they are both irrational. The greatest value of <math>PQ</math> happens when <math>P</math> and <math>Q</math> are almost directly across from each other and are in different quadrants. For example, the endpoints of the segment could be <math>(-4,3)</math> and <math>(3,-4)</math> because the two points are almost across from each other. Another possible pair could be <math>(-4,3)</math> and <math>(5,0)</math>. To find out which segment is longer, we have to compare the distances from their endpoints to a diameter (which must be the longest possible segment). The closest diameter would be from <math>(-4,3)</math> to <math>(4,-3)</math>. The distance between <math>(3,-4)</math> and <math>(4,-3)</math> is shorter than the distance between <math>(5,0)</math> and <math>(4,-3)</math>. Therefore, the segment from <math>(-4,3)</math> to <math>(3,-4)</math> is the longest attainable. The least value of <math>RS</math> is when the two endpoints are in the same quadrant and are very close to each other. This can occur when, for example, <math>R</math> is <math>(3,4)</math> and <math>S</math> is <math>(4,3).</math> They are in the same quadrant and no other point on the circle with integer coordinates is closer to the point <math>(3,4)</math> than <math>(4,3).</math> Using the distance formula, we get that <math>PQ</math> is <math>\sqrt{98}</math> and that <math>RS</math> is <math>\sqrt{2}.</math> <math>\frac{\sqrt{98}}{\sqrt{2}}=\sqrt{49}=\boxed{\mathrm{\textbf{(D)}}\ 7}</math><br />
<br />
==Solution 2==<br />
So what we can do is look at the option choices. Since we are aiming for the highest possible ratio, let's try using <math>7</math>. Now, looking at the problem alone, we know that to have the largest ratio possible, we have to let <math>RS</math> be the minimum possible value while at the same time using integer coordinates. Thus, the smallest possible value of <math>RS</math> is <math>\sqrt{1^{2}+1^{2}} = \sqrt{2}</math>. Assuming that <math>\frac{PQ}{RS} = 7</math>, we plug in <math>RS = \sqrt{2}</math> and solve for PQ: <math>PQ=7\sqrt{2}</math>. Remember, we don't know if this is possible yet, we are only trying to figure out if it is. But for what values of <math>x</math> and <math>y</math> does <math>\sqrt{x^{2}+y^{2}}=7\sqrt2</math>? We see that this can easily be made into a <math>45-45-90</math> triangle. But, instead of substituting <math>y=x</math> into the equation and then using a whole lot of algebra, we can save time and use the little trick, that if in a <math>45-45-90</math> triangle, the two <math>45</math> degree sides have side length <math>s</math>, then the hypotenuse is <math>s\sqrt2</math>. Using this, we can see that <math>s=7</math>, and since our equation does in fact yield a sensible solution, we can be assured that our answer is <math>\boxed{\mathrm{\textbf{(D)}}\ 7}</math>.<br />
<br />
Quality Control by fasterthanlight<br />
<br />
(Note by CarrotKaren: We tried 7, but some might be confused why we concluded that it was the answer after verifying without trying the others, like why wasn't option <math>E</math> tried? This is because there can only be one answer in the problem that works, so if we have an option that already works, we can conclude that none of the others work and <math>D</math> is the answer.<br />
<br />
==Video Solution==<br />
https://youtu.be/umr2Aj9ViOA<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=A|num-b=16|num-a=18}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Geometry Problems]]</div>Carrot karenhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_23&diff=1402932020 AMC 8 Problems/Problem 232020-12-22T22:40:30Z<p>Carrot karen: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
Five different awards are to be given to three students. Each student will receive at least one<br />
award. In how many different ways can the awards be distributed?<br />
<br />
<math>\textbf{(A) }120 \qquad \textbf{(B) }150 \qquad \textbf{(C) }180 \qquad \textbf{(D) }210 \qquad \textbf{(E) }240</math><br />
<br />
==Solution 1 (complementary counting)==<br />
Without the restriction that each student receives at least one award, we could simply take each of the <math>5</math> awards and choose one of the <math>3</math> students to give it to, so that there would be <math>3^5=243</math> ways to distribute the awards. We now need to subtract the cases where at least one student doesn't receive an award. If a student doesn't receive an award, there are <math>3</math> choices for which student that is, then <math>2^5 = 32</math> ways of choosing a student to receive each of the awards, for a total of <math>3 \cdot 32 = 96</math>. However, if <math>2</math> students both don't receive an award, then such a case would be counted twice among our <math>96</math>, so we need to add back in these cases. Of course, <math>2</math> students both not receiving an award is equivalent to only <math>1</math> student receiving all <math>5</math> awards, so there are simply <math>3</math> choices for which student that would be. It follows that the total number of ways of distributing the awards is <math>243-96+3=\boxed{\textbf{(B) }150}</math>.<br />
<br />
==Solution 2==<br />
Firstly, observe that it is not possible for a single student to receive <math>4</math> or <math>5</math> awards because this would mean that one of the other students receives no awards. Thus, each student must receive either <math>1</math>, <math>2</math>, or <math>3</math> awards. If a student receives <math>3</math> awards, then the other two students must each receive <math>1</math> award; if a student receives <math>2</math> awards, then another student must also receive <math>2</math> awards and the remaining student must receive <math>1</math> award. We consider each of these two cases in turn. If a student receives three awards, there are <math>3</math> ways to choose which student this is, and <math>\binom{5}{3}</math> ways to give that student <math>3</math> out of the <math>5</math> awards. Next, there are <math>2</math> students left and <math>2</math> awards to give out, with each student getting one award. There are clearly just <math>2</math> ways to distribute these two awards out, giving <math>3\cdot\binom{5}{3}\cdot 2=60</math> ways to distribute the awards in this case.<br />
<br />
In the other case, a student receives <math>2</math> awards. We first have to choose which of the three students we will select to give two awards each to. There are <math>\binom{3}{2}</math> ways to do this, after which there are <math>\binom{5}{2}</math> ways to give the first student his two awards, leaving <math>3</math> awards yet to distribute. There are then <math>\binom{3}{2}</math> ways to give the second student his <math>2</math> awards. Finally, there is only <math>1</math> student and <math>1</math> award left, so there is only <math>1</math> way to distribute this award. This results in <math>\binom{3}{2}\cdot\binom{5}{2}\cdot\binom{3}{2}\cdot 1=90</math> ways to distribute the awards in this case. Adding the results of these two cases, we get <math>60+90=\boxed{\textbf{(B) }150}</math>.<br />
<br />
==Solution 3 (variation of Solution 2)==<br />
If each student must receive at least one award, then, as in Solution 2, we deduce that the only possible ways to split up the <math>5</math> awards are <math>3,1,1</math> and <math>2,2,1</math> (i.e. one student gets three awards and the others get one each, or two students each get two awards and the other student is left with the last one). In the first case, there are <math>3</math> choices for which student gets <math>3</math> awards, and <math>\binom{5}{3} = 10</math> choices for which <math>3</math> awards they get. We are then left with <math>2</math> awards, and there are exactly <math>2</math> choices depending on which remaining student gets which. This yields a total for this case of <math>3 \cdot 10 \cdot 2 = 60</math>. For the second case, there are similarly <math>3</math> choices for which student gets only <math>1</math> award, and <math>5</math> choices for which award he gets. There are then <math>4</math> remaining awards, from which we choose <math>2</math> to give to one student and <math>2</math> to give to the other, which can be done in <math>\binom{4}{2} = 6</math> ways (and we can say that e.g. the <math>2</math> chosen this way go to the first remaining student and the other <math>2</math> go to the second remaining student, which counts all possibilities). This means the total for the second case is <math>3 \cdot 5 \cdot 6 = 90</math>, and the answer is <math>60 + 90 = \boxed{\textbf{(B) }150}</math>.<br />
<br />
==Video Solutions==<br />
https://youtu.be/SPNobOd4t1c (Includes all the problems and has a free class update)<br />
https://youtu.be/tDChKU0pVN4 <br /><br />
https://youtu.be/RUg6QfV5yg4<br />
<br />
==See also== <br />
{{AMC8 box|year=2020|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Carrot karen