https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Jayrenfu&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T12:17:43ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12A_Problems/Problem_13&diff=2067122023 AMC 12A Problems/Problem 132023-12-09T18:27:34Z<p>Jayrenfu: /* Solution 5 (🧀Cheese🧀) */</p>
<hr />
<div>{{duplicate|[[2023 AMC 10A Problems/Problem 16|2023 AMC 10A #16]] and [[2023 AMC 12A Problems/Problem 13|2023 AMC 12A #13]]}}<br />
<br />
==Problem==<br />
<br />
In a table tennis tournament every participant played every other participant exactly once. Although there were twice as many right-handed players as left-handed players, the number of games won by left-handed players was <math>40\%</math> more than the number of games won by right-handed players. (There were no ties and no ambidextrous players.) What is the total number of games played?<br />
<br />
<math>\textbf{(A) }15\qquad\textbf{(B) }36\qquad\textbf{(C) }45\qquad\textbf{(D) }48\qquad\textbf{(E) }66</math><br />
<br />
==Solution 1==<br />
We know that the total amount of games must be the sum of games won by left and right handed players. Then, we can write <math>g = l + r</math>, and since <math>l = 1.4r</math>, <math>g = 2.4r</math>. Given that <math>r</math> and <math>g</math> are both integers, <math>g/2.4</math> also must be an integer. From here we can see that <math>g</math> must be divisible by 12, leaving only answers B and D. Now we know the formula for how many games are played in this tournament is <math>n(n-1)/2</math>, the sum of the first <math>n-1</math> triangular numbers. Now, setting 36 and 48 equal to the equation will show that two consecutive numbers must have a product of 72 or 96. Clearly, <math>72=8*9</math>, so the answer is <math>\boxed{\textbf{(B) }36}</math>.<br />
<br />
~~ Antifreeze5420<br />
<br />
==Solution 2==<br />
First, we know that every player played every other player, so there's a total of <math>\dbinom{n}{2}</math> games since each pair of players forms a bijection to a game. Therefore, that rules out D. Also, if we assume the right-handed players won a total of <math>x</math> games, the left-handed players must have won a total of <math>\dfrac{7}{5}x</math> games, meaning that the total number of games played was <math>\dfrac{12}{5}x</math>. Thus, the total number of games must be divisible by <math>12</math>. Therefore leaving only answer choices B and D. Since answer choice D doesn't satisfy the first condition, the only answer that satisfies both conditions is <math>\boxed{\textbf{(B) }36}</math><br />
<br />
==Solution 3==<br />
Let <math>r</math> be the amount of games the right-handed won. Since the left-handed won <math>1.4r</math> games, the total number of games played can be expressed as <math>(2.4)r</math>, or <math>12/5r</math>, meaning that the answer is divisible by 12. This brings us down to two answer choices, <math>B</math> and <math>D</math>. <br />
We note that the answer is some number <math>x</math> choose <math>2</math>. This means the answer is in the form <math>x(x-1)/2</math>. Since answer choice D gives <math>48 = x(x-1)/2</math>, and <math>96 = x(x-1)</math> has no integer solutions, we know that <math>\boxed{\textbf{(B) }36}</math> is the only possible choice.<br />
<br />
==Solution 4==<br />
Here is the rigid way to prove that 36 is the only result. Let the number of left-handed players be n, so the right-handed player is 2n. The number of games won by the left-handed players comes in two ways: (1) the games played by two left-left pairs, which is <math>\frac{n(n-1)}{2}</math>, and (2) the games played by left-right pairs, which is x. And <math>x\leq 2n^2</math>. so <cmath>\frac{\frac{n(n-1)}{2}+x}{\frac{2n(2n-1)}{2}+2n^2-x}=1.4</cmath> which gives <cmath>x=\frac{17n^2}{8}-\frac{3n}{8}</cmath> By setting up the inequality <math>x\leq 2n^2</math>, it comes <math>n\leq 3</math>. So the total number of players can only be <math>3</math>, <math>6</math>, and <math>9</math>. Then we can rule out all other possible values for the total number of games they played.<br />
<br />
<br />
~ ~ ggao5uiuc<br />
~ ~ yingkai_0_ (minor edits)<br />
<br />
==Solution 5 (🧀Cheese🧀)==<br />
If there are <math>n</math> players, the total number of games played must be <math>\binom{n}{2}</math>, so it has to be a triangular number. The ratio of games won by left-handed to right-handed players is <math>7:5</math>, so the number of games played must also be divisible by <math>12</math>. Finally, we notice that only <math>\boxed{\textbf{(B) }36}</math> satisfies both of these conditions.<br />
<br />
~MathFun1000<br />
<br />
==Video Solution (âš¡Under 2 minâš¡)==<br />
https://youtu.be/wSeSYxDCOZ8<br />
<i> ~Education, the Study of Everything </i><br />
<br />
== Video Solution 1 by OmegaLearn ==<br />
https://youtu.be/BXgQIV2WbOA<br />
<br />
== Video Solution 2 by TheBeautyofMath ==<br />
https://www.youtube.com/watch?v=sLtsF1k9Fx8&t=227s<br />
<br />
==Video Solution==<br />
<br />
https://youtu.be/S9l1C6pjXWI<br />
<br />
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)<br />
<br />
==Video Solution by Math-X (First understand the problem!!!)==<br />
https://youtu.be/N2lyYRMuZuk?si=7IEmYDGBHi5i_XKt&t=1452<br />
~Math-X<br />
<br />
==Video Solution by SpreadTheMathLove==<br />
https://www.youtube.com/watch?v=sypOvNiR3sw<br />
<br />
==See Also==<br />
{{AMC10 box|year=2023|ab=A|num-b=15|num-a=17}}<br />
{{AMC12 box|year=2023|ab=A|num-b=12|num-a=14}}<br />
{{MAA Notice}}</div>Jayrenfuhttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_18&diff=1863902017 AMC 8 Problems/Problem 182023-01-12T04:26:11Z<p>Jayrenfu: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
In the non-convex quadrilateral <math>ABCD</math> shown below, <math>\angle BCD</math> is a right angle, <math>AB=12</math>, <math>BC=4</math>, <math>CD=3</math>, and <math>AD=13</math>. What is the area of quadrilateral <math>ABCD</math>?<br />
<br />
<asy>draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); label("$B$", (0, 0), SW); label("$A$", (12, 0), ESE); label("$C$", (2.4, 3.6), SE); label("$D$", (0, 5), N);</asy><br />
<br />
<math>\textbf{(A) }12 \qquad \textbf{(B) }24 \qquad \textbf{(C) }26 \qquad \textbf{(D) }30 \qquad \textbf{(E) }36</math><br />
<br />
==Solution 1==<br />
We first connect point <math>B</math> with point <math>D</math>. <br />
<br />
<asy>draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); draw((0,0)--(0,5)); label("$B$", (0, 0), SW); label("$A$", (12, 0), ESE); label("$C$", (2.4, 3.6), SE); label("$D$", (0, 5), N);</asy> <br />
<br />
We can see that <math>\triangle BCD</math> is a 3-4-5 right triangle. We can also see that <math>\triangle BDA</math> is a right triangle, by the 5-12-13 Pythagorean triple. With these lengths, we can solve the problem. The area of <math>\triangle BDA</math> is <math>\frac{5\cdot 12}{2}</math>, and the area of <math>\triangle BCD</math> is <math>\frac{3\cdot 4}{2}</math>. Thus, the area of quadrilateral <math>ABCD</math> is <math>30-6 = \boxed{\textbf{(B)}\ 24}.</math><br />
<br />
==Solution 2==<br />
<math>\triangle BCD</math> is a 3-4-5 right triangle. So the area of <math>\triangle BCD</math> is 6. Then we can use Heron's formula to compute the area of <math>\triangle ABD</math> whose sides have lengths 5,12,and 13. The area of <math>\triangle ABD</math> = <math>\sqrt{s(s-5)(s-12)(s-13)}</math> , where s is the semi-perimeter of the triangle, that is <math>s=(5+12+13)/2=15.</math> Thus, the area of <math>\triangle ABD</math> is 30, so the area of <math>ABCD</math> is <math>30-6 = \boxed{\textbf{(B)}\ 24}.</math> ---LarryFlora<br />
<br />
==Video Solution==<br />
http://youtube/DU95-maui9U<br />
<br />
~savannahsolver<br />
<br />
==Video Solution by OmegaLearn==<br />
https://youtu.be/51K3uCzntWs?t=2010<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also==<br />
{{AMC8 box|year=2017|num-b=17|num-a=19}}<br />
<br />
{{MAA Notice}}</div>Jayrenfuhttps://artofproblemsolving.com/wiki/index.php?title=2022_AMC_8_Problems/Problem_18&diff=1860772022 AMC 8 Problems/Problem 182023-01-08T23:32:01Z<p>Jayrenfu: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
<br />
The midpoints of the four sides of a rectangle are <math>(-3,0), (2,0), (5,4),</math> and <math>(0,4).</math> What is the<br />
area of the rectangle? <br />
<br />
<math>\textbf{(A) } 20 \qquad \textbf{(B) } 25 \qquad \textbf{(C) } 40 \qquad \textbf{(D) } 50 \qquad \textbf{(E) } 80</math><br />
<br />
==Solution 1==<br />
<br />
The midpoints of the four sides of every rectangle are the vertices of a rhombus whose area is half the area of the rectangle.<br />
<br />
Let <math>A=(-3,0), B=(2,0), C=(5,4),</math> and <math>D=(0,4).</math> Note that <math>A,B,C,</math> and <math>D</math> are the vertices of a rhombus whose diagonals have lengths <math>AC=4\sqrt{5}</math> and <math>BD=2\sqrt{5}.</math> Since the diagonals of rhombus <math>ABCD</math> are equal to the sides of the rectangle, the rectangle's area is <math>4\sqrt{5}\cdot2\sqrt{5}=40</math>.<br />
<br />
~MRENTHUSIASM<br />
<br />
~Edited by jayrenfu<br />
<br />
==Solution 2==<br />
<br />
If a rectangle has area <math>K,</math> then the area of the quadrilateral formed by its midpoints is <math>\frac{K}{2}.</math><br />
<br />
Define points <math>A,B,C,</math> and <math>D</math> as Solution 1 does. Since <math>A,B,C,</math> and <math>D</math> are the midpoints of the rectangle, the rectangle's area is <math>2[ABCD].</math> Now, note that <math>ABCD</math> is a parallelogram since <math>AB=CD</math> and <math>\overline{AB}\parallel\overline{CD}.</math> As the parallelogram's height from <math>D</math> to <math>\overline{AB}</math> is <math>4</math> and <math>AB=5,</math> its area is <math>4\cdot5=20.</math> Therefore, the area of the rectangle is <math>20\cdot2=\boxed{\textbf{(C) } 40}.</math><br />
<br />
~Fruitz<br />
==Video Solution==<br />
https://youtu.be/Ij9pAy6tQSg?t=1564<br />
<br />
~Interstigation<br />
<br />
==Video Solution by Ismail.maths==<br />
https://www.youtube.com/watch?v=JHBcnevL5_U<br />
<br />
~Ismail.maths93<br />
<br />
==Video Solution==<br />
https://youtu.be/hs6y4PWnoWg?t=188<br />
<br />
~STEMbreezy<br />
<br />
==Video Solution==<br />
https://youtu.be/9-TlEV5SGqM<br />
<br />
~savannahsolver<br />
<br />
==See Also== <br />
{{AMC8 box|year=2022|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Jayrenfuhttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_22&diff=1851792019 AMC 8 Problems/Problem 222022-12-29T21:56:45Z<p>Jayrenfu: /* Video Solution */</p>
<hr />
<div>==Problem 22==<br />
A store increased the original price of a shirt by a certain percent and then lowered the new price by the same amount. Given that the resulting price was <math>84\%</math> of the original price, by what percent was the price increased and decreased<math>?</math><br />
<br />
<math>\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }40</math><br />
<br />
==Solution 1==<br />
Suppose the fraction of discount is <math>x</math>. That means <math>(1-x)(1+x)=0.84</math>; so, <math>1-x^{2}=0.84</math>, and <math>(x^{2})=0.16</math>, obtaining <math>x=0.4</math>. Therefore, the price was increased and decreased by <math>40</math>%, or <math>\boxed{\textbf{(E)}\ 40}</math>.<br />
<br />
==Solution 2 (Answer options)==<br />
We can try out every option and see which one works. By this method, we get <math>\boxed{\textbf{(E)}\ 40}</math>.<br />
<br />
==Solution 3==<br />
Let x be the discount. We can also work in reverse such as (<math>84</math>)<math>(\frac{100}{100-x})</math><math>(\frac{100}{100+x})</math> = <math>100</math>.<br />
<br />
Thus, <math>8400</math> = <math>(100+x)(100-x)</math>. Solving for <math>x</math> gives us <math>x = 40, -40</math>. But <math>x</math> has to be positive. Thus, <math>x</math> = <math>40</math>.<br />
<br />
==Solution 4 ~ using the answer choices==<br />
<br />
Let our original cost be <math>\$ 100.</math> We are looking for a result of <math>\$ 84,</math> then. We try 16% and see it gets us higher than 84. We try 20% and see it gets us lower than 16 but still higher than 84. We know that the higher the percent, the less the value. We try 36, as we are not progressing much, and we are close! We try <math>\boxed{40\%}</math>, and we have the answer; it worked.<br />
<br />
==Solution 5 (A Variation of Solution 4)==<br />
<br />
Let our original cost be <math>\$ 100</math>, so we are looking for a whole number of <math>\$ 84</math>. Also, we can see that (A), (C), and (D) give us answers with decimals while we know that (B) and (E) give us whole numbers. Therefore, we only need to try these two: (B) <math>\$100</math> increased by 20% = <math>\$120</math>, and <math>\$120</math> decreased by 20% = <math>\$96</math>, a whole number, and (E) <math>\$100</math> increased by 40% = <math>\$140</math>, and <math>\$140</math> decreased by 40% = <math>\$84</math>, a whole number.<br />
<br />
Thus, <math>40</math>% or <math>\boxed{\textbf{(E)}\ 40}</math> is the answer.<br />
<br />
~ SaxStreak<br />
<br />
==Video Solution== <br />
<br />
https://www.youtube.com/watch?v=_TheVi-6LWE <br />
<br />
- Happytwin<br />
<br />
Associated video - https://www.youtube.com/watch?v=aJX27Cxvwlc<br />
<br />
https://youtu.be/gX_l0PGsQao<br />
<br />
https://www.youtube.com/watch?v=RcBDdB35Whk&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=4 <br />
<br />
~ MathEx<br />
<br />
https://youtu.be/0MoZFiPp8LA<br />
<br />
~savannahsolver<br />
<br />
https://www.youtube.com/watch?v=DaF8uD8V8u0&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=24<br />
<br />
https://www.youtube.com/watch?v=i1x2b3_hmzA<br />
<br />
https://youtu.be/2Y294ssDjVQ (The Fastest Way!)<br />
<br />
~ SaxStreak<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=21|num-a=23}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Jayrenfuhttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_9&diff=1851782019 AMC 8 Problems/Problem 92022-12-29T21:53:27Z<p>Jayrenfu: /* Problem 9 */</p>
<hr />
<div>==Problem 9==<br />
Alex and Felicia each have cats as pets. Alex buys cat food in cylindrical cans that are <math>6</math> cm in diameter and <math>12</math> cm high. Felicia buys cat food in cylindrical cans that are <math>12</math> cm in diameter and <math>6</math> cm high. What is the ratio of the volume of one of Alex's cans to the volume of one of Felicia's cans?<br />
<br />
<math>\textbf{(A) }1:4\qquad\textbf{(B) }1:2\qquad\textbf{(C) }1:1\qquad\textbf{(D) }(B)2:1\qquad\textbf{(E) }4:1</math><br />
<br />
==Solution 1==<br />
<br />
Using the formula for the volume of a cylinder, we get Alex, <math>\pi108</math>, and Felicia, <math>\pi216</math>. We can quickly notice that <math>\pi</math> cancels out on both sides, and that Alex's volume is <math>1/2</math> of Felicia's leaving <math>1/2 = \boxed{1:2}</math> as the answer. <br />
<br />
~aopsav<br />
<br />
==Solution 2==<br />
<br />
Using the formula for the volume of a cylinder, we get that the volume of Alex's can is <math>3^2\cdot12\cdot\pi</math>, and that the volume of Felicia's can is <math>6^2\cdot6\cdot\pi</math>. Now, we divide the volume of Alex's can by the volume of Felicia's can, so we get <math>\frac{1}{2}</math>, which is <math>\boxed{\textbf{(B)}\ 1:2}</math>. <br />
<br />
lol, this is something no one should be able to do.-(Algebruh123)2020<br />
<br />
==Solution 3==<br />
<br />
The ratio of the numbers is <math>1/2</math>. Looking closely at the formula <math>r^2 * h * \pi</math>, we see that the <math>r * h * \pi</math> will cancel, meaning that the ratio of them will be <math>\frac{1(2)}{2(2)}</math> = <math>\boxed{\textbf{(B)}\ 1:2}</math>.<br />
<br />
-Lcz<br />
<br />
== Video Solution ==<br />
The Learning Royal : https://youtu.be/8njQzoztDGc<br />
== Video Solution 2 ==<br />
https://youtu.be/FDgcLW4frg8?t=2440<br />
<br />
~ pi_is_3.14<br />
<br />
== Video Solution ==<br />
<br />
Solution detailing how to solve the problem: https://www.youtube.com/watch?v=G-gEdWP0S9M&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=10<br />
<br />
==Video Solution==<br />
https://youtu.be/FLT3iOKBC8c<br />
<br />
~savannahsolver<br />
<br />
==See also==<br />
{{AMC8 box|year=2019|num-b=8|num-a=10}}<br />
<br />
{{MAA Notice}}</div>Jayrenfuhttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_21&diff=1850422017 AMC 8 Problems/Problem 212022-12-27T23:21:03Z<p>Jayrenfu: /* Video Solutions */</p>
<hr />
<div>==Problem==<br />
<br />
Suppose <math>a</math>, <math>b</math>, and <math>c</math> are nonzero real numbers, and <math>a+b+c=0</math>. What are the possible value(s) for <math>\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}</math>?<br />
<br />
<math>\text{(A) }0\qquad\text{(B) }1\text{ and }-1\qquad\text{(C) }2\text{ and }-2\qquad\text{(D) }0,2,\text{ and }-2\qquad\text{(E) }0,1,\text{ and }-1</math><br />
<br />
==Solution 1==<br />
<br />
There are <math>2</math> cases to consider:<br />
<br />
Case <math>1</math>: <math>2</math> of <math>a</math>, <math>b</math>, and <math>c</math> are positive and the other is negative. WLOG, we can assume that <math>a</math> and <math>b</math> are positive and <math>c</math> is negative. In this case, we have that <cmath>\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}=1+1-1-1=0.</cmath><br />
<br />
Case <math>2</math>: <math>2</math> of <math>a</math>, <math>b</math>, and <math>c</math> are negative and the other is positive. Without loss of generality, we can assume that <math>a</math> and <math>b</math> are negative and <math>c</math> is positive. In this case, we have that <cmath>\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}=-1-1+1+1=0.</cmath><br />
<br />
Note these are the only valid cases, for neither <math>3</math> negatives nor <math>3</math> positives would work, as they cannot sum up to <math>0</math>. In both cases, we get that the given expression equals <math>\boxed{\textbf{(A)}\ 0}</math>.<br />
<br />
==Video Solutions==<br />
https://youtu.be/FUEHirfk-tw<br />
<br />
https://youtu.be/V9wCBTwvIZo - Happytwin<br />
<br />
https://youtu.be/7an5wU9Q5hk?t=2362<br />
<br />
https://youtu.be/xN0dnJC1hv8<br />
<br />
~savannahsolver<br />
<br />
==See Also==<br />
{{AMC8 box|year=2017|num-b=20|num-a=22}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Jayrenfuhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_23&diff=1849292015 AMC 8 Problems/Problem 232022-12-26T18:46:08Z<p>Jayrenfu: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
Tom has twelve slips of paper which he wants to put into five cups labeled <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, <math>E</math>. He wants the sum of the numbers on the slips in each cup to be an integer. Furthermore, he wants the five integers to be consecutive and increasing from <math>A</math> to <math>E</math>. The numbers on the papers are <math>2, 2, 2, 2.5, 2.5, 3, 3, 3, 3, 3.5, 4,</math> and <math>4.5</math>. If a slip with <math>2</math> goes into cup <math>E</math> and a slip with <math>3</math> goes into cup <math>B</math>, then the slip with <math>3.5</math> must go into what cup?<br />
<br />
<math><br />
\textbf{(A) } A \qquad<br />
\textbf{(B) } B \qquad<br />
\textbf{(C) } C \qquad<br />
\textbf{(D) } D \qquad<br />
\textbf{(E) } E<br />
</math><br />
<br />
==Solution==<br />
<br />
The numbers have a sum of <math>6+5+12+4+8=35</math>, which averages to <math>7</math>, which means <math>A, B, C, D, <br />
E</math> will have the values <math>5, 6, 7, 8</math> and <math>9</math>, respectively. Now it's the process of elimination: Cup <math>A</math><br />
will have a sum of <math>5</math>, so putting a <math>3.5</math> slip in the cup will leave <math>5-3.5=1.5</math>; However,<br />
all of our slips are bigger than <math>1.5</math>, so this is impossible. Cup <math>B</math> has a sum of <math>6</math>, but we are told that it already has a <math>3</math> slip, leaving <math>6-3=3</math>, which is too small for the<br />
<math>3.5</math> slip. Cup <math>C</math> is a little bit trickier but still manageable. It must have a value of <math>7</math>, so adding the <math>3.5</math> slip leaves room for <math>7-3.5=3.5</math>. This looks good at first, as we have slips smaller than that, but upon closer inspection, we see that no slip fits <br />
exactly, and the smallest sum of two slips is <math>2+2=4</math>, which is too big, so this case is also impossible. Cup <math>E</math> has a sum of <math>9</math>, but we are told it already has a <math>2</math> slip, so we<br />
are left with <math>9-2=7</math>, which is identical to Cup C and thus also impossible.<br />
With all other choices removed, we are left with the answer: Cup <math>\boxed{\textbf{(D)}~D}</math><br />
<br />
==Video Solution==<br />
https://youtu.be/SgEXCLfk_AU<br />
<br />
~savannahsolver<br />
<br />
==See Also==<br />
<br />
{{AMC8 box|year=2015|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Jayrenfuhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_23&diff=1849282015 AMC 8 Problems/Problem 232022-12-26T18:45:22Z<p>Jayrenfu: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
Tom has twelve slips of paper which he wants to put into five cups labeled <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, <math>E</math>. He wants the sum of the numbers on the slips in each cup to be an integer. Furthermore, he wants the five integers to be consecutive and increasing from <math>A</math> to <math>E</math>. The numbers on the papers are <math>2, 2, 2, 2.5, 2.5, 3, 3, 3, 3, 3.5, 4,</math> and <math>4.5</math>. If a slip with <math>2</math> goes into cup <math>E</math> and a slip with <math>3</math> goes into cup <math>B</math>, then the slip with <math>3.5</math> must go into what cup?<br />
<br />
<math><br />
\textbf{(A) } A \qquad<br />
\textbf{(B) } B \qquad<br />
\textbf{(C) } C \qquad<br />
\textbf{(D) } D \qquad<br />
\textbf{(E) } E<br />
</math><br />
<br />
==Solution==<br />
<br />
The numbers have a sum of <math>6+5+12+4+8=35</math>, which averages to <math>7</math>, which means <math>A, B, C, D, <br />
E</math> will have the values <math>5, 6, 7, 8, and 9</math>, respectively. Now it's the process of elimination: Cup <math>A</math><br />
will have a sum of <math>5</math>, so putting a <math>3.5</math> slip in the cup will leave <math>5-3.5=1.5</math>; However,<br />
all of our slips are bigger than <math>1.5</math>, so this is impossible. Cup <math>B</math> has a sum of <math>6</math>, but we are told that it already has a <math>3</math> slip, leaving <math>6-3=3</math>, which is too small for the<br />
<math>3.5</math> slip. Cup <math>C</math> is a little bit trickier but still manageable. It must have a value of <math>7</math>, so adding the <math>3.5</math> slip leaves room for <math>7-3.5=3.5</math>. This looks good at first, as we have slips smaller than that, but upon closer inspection, we see that no slip fits <br />
exactly, and the smallest sum of two slips is <math>2+2=4</math>, which is too big, so this case is also impossible. Cup <math>E</math> has a sum of <math>9</math>, but we are told it already has a <math>2</math> slip, so we<br />
are left with <math>9-2=7</math>, which is identical to Cup C and thus also impossible.<br />
With all other choices removed, we are left with the answer: Cup <math>\boxed{\textbf{(D)}~D}</math><br />
<br />
==Video Solution==<br />
https://youtu.be/SgEXCLfk_AU<br />
<br />
~savannahsolver<br />
<br />
==See Also==<br />
<br />
{{AMC8 box|year=2015|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Jayrenfuhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_19&diff=1849262015 AMC 8 Problems/Problem 192022-12-26T18:40:21Z<p>Jayrenfu: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
A triangle with vertices as <math>A=(1,3)</math>, <math>B=(5,1)</math>, and <math>C=(4,4)</math> is plotted on a <math>6\times5</math> grid. What fraction of the grid is covered by the triangle?<br />
<br />
<math>\textbf{(A) }\frac{1}{6} \qquad \textbf{(B) }\frac{1}{5} \qquad \textbf{(C) }\frac{1}{4} \qquad \textbf{(D) }\frac{1}{3} \qquad \textbf{(E) }\frac{1}{2}</math><br />
<br />
<asy><br />
draw((1,0)--(1,5),linewidth(.5));<br />
draw((2,0)--(2,5),linewidth(.5));<br />
draw((3,0)--(3,5),linewidth(.5));<br />
draw((4,0)--(4,5),linewidth(.5));<br />
draw((5,0)--(5,5),linewidth(.5));<br />
draw((6,0)--(6,5),linewidth(.5));<br />
draw((0,1)--(6,1),linewidth(.5));<br />
draw((0,2)--(6,2),linewidth(.5));<br />
draw((0,3)--(6,3),linewidth(.5));<br />
draw((0,4)--(6,4),linewidth(.5));<br />
draw((0,5)--(6,5),linewidth(.5)); <br />
draw((0,0)--(0,6),EndArrow);<br />
draw((0,0)--(7,0),EndArrow);<br />
draw((1,3)--(4,4)--(5,1)--cycle);<br />
label("$y$",(0,6),W); label("$x$",(7,0),S);<br />
label("$A$",(1,3),dir(210)); label("$B$",(5,1),SE); label("$C$",(4,4),dir(100));<br />
</asy><br />
<br />
==Solutions==<br />
<br />
===Solution 1===<br />
<br />
The area of <math>\triangle ABC</math> is equal to half the product of its base and height. By the Pythagorean Theorem, we find its height is <math>\sqrt{1^2+2^2}=\sqrt{5}</math>, and its base is <math>\sqrt{2^2+4^2}=\sqrt{20}</math>. We multiply these and divide by <math>2</math> to find the area of the triangle is <math>\frac{\sqrt{5 \cdot 20}}2=\frac{\sqrt{100}}2=\frac{10}2=5</math>. Since the grid has an area of <math>30</math>, the fraction of the grid covered by the triangle is <math>\frac 5{30}=\boxed{\textbf{(A) }\frac{1}{6}}</math>.<br />
<br />
===Solution 2===<br />
Note angle <math>\angle ACB</math> is right, thus the area is <math>\sqrt{1^2+3^2} \times \sqrt{1^2+3^2}\times \dfrac{1}{2}=10 \times \dfrac{1}{2}=5</math> thus the fraction of the total is <math>\dfrac{5}{30}=\boxed{\textbf{(A)}~\dfrac{1}{6}}</math><br />
<br />
===Solution 3===<br />
By the [[Shoelace Theorem]], the area of <math>\triangle ABC=|\dfrac{1}{2}(15+4+4-1-20-12)|=|\dfrac{1}{2}(-10)|=5</math>.<br />
<br />
This means the fraction of the total area is <math>\dfrac{5}{30}=\boxed{\textbf{(A)}~\dfrac{1}{6}}</math><br />
<br />
===Solution 4===<br />
<br />
The smallest rectangle that follows the grid lines and completely encloses <math>\triangle ABC</math> has an area of <math>12</math>, where <math>\triangle ABC</math> splits the rectangle into four triangles. The area of <math>\triangle ABC</math> is therefore <math>12 - (\frac{4 \cdot 2}{2}+\frac{3 \cdot 1}{2}+\frac{3 \cdot 1}{2}) = 12 - (4 + \frac{3}{2} + \frac{3}{2}) = 12 - 7 = 5</math>. That means that <math>\triangle ABC</math> takes up <math>\frac{5}{30} = \boxed{\textbf{(A)}~\frac{1}{6}}</math> of the grid.<br />
<br />
===Solution 5===<br />
<br />
Using [[Pick's Theorem]], the area of the triangle is <math>4 + \dfrac{4}{2} - 1=5</math>. Therefore, the triangle takes up <math>\dfrac{5}{30}=\boxed{\textbf{(A)}~\frac{1}{6}}</math> of the grid.<br />
<br />
===Solution 6 (Heron's Formula, Not Recommended)===<br />
<br />
We can find the lengths of the sides by using the [[Pythagorean Theorem]]. Then, we apply [[Heron's Formula]] to find the area. <br />
<cmath> \sqrt{\left(\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}\right)\left(\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}-\sqrt{10}\right)\left(\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}-\sqrt{10}\right)\left(\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}-2\sqrt{5}\right)}. </cmath><br />
This simplifies to<br />
<cmath> \sqrt{\left(\sqrt{10}+\sqrt{5}\right)\left(\sqrt{10}+\sqrt{5}-\sqrt{10}\right)\left(\sqrt{10}+\sqrt{5}-\sqrt{10}\right)\left(\sqrt{10}+\sqrt{5}-2\sqrt{5}\right)}. </cmath><br />
Again, we simplify to get<br />
<cmath> \sqrt{\left(\sqrt{10}+\sqrt{5}\right)\left(\sqrt{5}\right)\left(\sqrt{5}\right)\left(\sqrt{10}-\sqrt{5}\right)}. </cmath><br />
The middle two terms inside the square root multiply to <math> 5 </math>, and the first and last terms inside the square root multiply to <math> \sqrt{10}^2-\sqrt{5}^2=10-5=5. </math> This means that the area of the triangle is<br />
<cmath> \sqrt{5\cdot 5}=5. </cmath><br />
The area of the grid is <math> 6\cdot 5=30. </math> Thus, the answer is <math> \frac{5}{30}=\boxed{\textbf{(A) }\frac{1}{6}} </math>. -BorealBear<br />
<br />
==Video Solution==<br />
https://youtu.be/EyDGtLc6xGE<br />
<br />
~savannahsolver<br />
<br />
==See Also==<br />
<br />
{{AMC8 box|year=2015|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Jayrenfuhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_19&diff=1849252015 AMC 8 Problems/Problem 192022-12-26T18:39:53Z<p>Jayrenfu: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
A triangle with vertices as <math>A=(1,3)</math>, <math>B=(5,1)</math>, and <math>C=(4,4)</math> is plotted on a <math>6\times5</math> grid. What fraction of the grid is covered by the triangle?<br />
<br />
<math>\textbf{(A) }\frac{1}{6} \qquad \textbf{(B) }\frac{1}{5} \qquad \textbf{(C) }\frac{1}{4} \qquad \textbf{(D) }\frac{1}{3} \qquad \textbf{(E) }\frac{1}{2}</math><br />
<br />
<asy><br />
draw((1,0)--(1,5),linewidth(.5));<br />
draw((2,0)--(2,5),linewidth(.5));<br />
draw((3,0)--(3,5),linewidth(.5));<br />
draw((4,0)--(4,5),linewidth(.5));<br />
draw((5,0)--(5,5),linewidth(.5));<br />
draw((6,0)--(6,5),linewidth(.5));<br />
draw((0,1)--(6,1),linewidth(.5));<br />
draw((0,2)--(6,2),linewidth(.5));<br />
draw((0,3)--(6,3),linewidth(.5));<br />
draw((0,4)--(6,4),linewidth(.5));<br />
draw((0,5)--(6,5),linewidth(.5)); <br />
draw((0,0)--(0,6),EndArrow);<br />
draw((0,0)--(7,0),EndArrow);<br />
draw((1,3)--(4,4)--(5,1)--cycle);<br />
label("$y$",(0,6),W); label("$x$",(7,0),S);<br />
label("$A$",(1,3),dir(210)); label("$B$",(5,1),SE); label("$C$",(4,4),dir(100));<br />
</asy><br />
<br />
==Solutions==<br />
<br />
===Solution 1===<br />
<br />
The area of <math>\triangle ABC</math> is equal to half the product of its base and height. By the Pythagorean Theorem, we find its height is <math>\sqrt{1^2+2^2}=\sqrt{5}</math>, and its base is <math>\sqrt{2^2+4^2}=\sqrt{20}</math>. We multiply these and divide by <math>2</math> to find the of the triangle is <math>\frac{\sqrt{5 \cdot 20}}2=\frac{\sqrt{100}}2=\frac{10}2=5</math>. Since the grid has an area of <math>30</math>, the fraction of the grid covered by the triangle is <math>\frac 5{30}=\boxed{\textbf{(A) }\frac{1}{6}}</math>.<br />
<br />
===Solution 2===<br />
Note angle <math>\angle ACB</math> is right, thus the area is <math>\sqrt{1^2+3^2} \times \sqrt{1^2+3^2}\times \dfrac{1}{2}=10 \times \dfrac{1}{2}=5</math> thus the fraction of the total is <math>\dfrac{5}{30}=\boxed{\textbf{(A)}~\dfrac{1}{6}}</math><br />
<br />
===Solution 3===<br />
By the [[Shoelace Theorem]], the area of <math>\triangle ABC=|\dfrac{1}{2}(15+4+4-1-20-12)|=|\dfrac{1}{2}(-10)|=5</math>.<br />
<br />
This means the fraction of the total area is <math>\dfrac{5}{30}=\boxed{\textbf{(A)}~\dfrac{1}{6}}</math><br />
<br />
===Solution 4===<br />
<br />
The smallest rectangle that follows the grid lines and completely encloses <math>\triangle ABC</math> has an area of <math>12</math>, where <math>\triangle ABC</math> splits the rectangle into four triangles. The area of <math>\triangle ABC</math> is therefore <math>12 - (\frac{4 \cdot 2}{2}+\frac{3 \cdot 1}{2}+\frac{3 \cdot 1}{2}) = 12 - (4 + \frac{3}{2} + \frac{3}{2}) = 12 - 7 = 5</math>. That means that <math>\triangle ABC</math> takes up <math>\frac{5}{30} = \boxed{\textbf{(A)}~\frac{1}{6}}</math> of the grid.<br />
<br />
===Solution 5===<br />
<br />
Using [[Pick's Theorem]], the area of the triangle is <math>4 + \dfrac{4}{2} - 1=5</math>. Therefore, the triangle takes up <math>\dfrac{5}{30}=\boxed{\textbf{(A)}~\frac{1}{6}}</math> of the grid.<br />
<br />
===Solution 6 (Heron's Formula, Not Recommended)===<br />
<br />
We can find the lengths of the sides by using the [[Pythagorean Theorem]]. Then, we apply [[Heron's Formula]] to find the area. <br />
<cmath> \sqrt{\left(\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}\right)\left(\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}-\sqrt{10}\right)\left(\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}-\sqrt{10}\right)\left(\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}-2\sqrt{5}\right)}. </cmath><br />
This simplifies to<br />
<cmath> \sqrt{\left(\sqrt{10}+\sqrt{5}\right)\left(\sqrt{10}+\sqrt{5}-\sqrt{10}\right)\left(\sqrt{10}+\sqrt{5}-\sqrt{10}\right)\left(\sqrt{10}+\sqrt{5}-2\sqrt{5}\right)}. </cmath><br />
Again, we simplify to get<br />
<cmath> \sqrt{\left(\sqrt{10}+\sqrt{5}\right)\left(\sqrt{5}\right)\left(\sqrt{5}\right)\left(\sqrt{10}-\sqrt{5}\right)}. </cmath><br />
The middle two terms inside the square root multiply to <math> 5 </math>, and the first and last terms inside the square root multiply to <math> \sqrt{10}^2-\sqrt{5}^2=10-5=5. </math> This means that the area of the triangle is<br />
<cmath> \sqrt{5\cdot 5}=5. </cmath><br />
The area of the grid is <math> 6\cdot 5=30. </math> Thus, the answer is <math> \frac{5}{30}=\boxed{\textbf{(A) }\frac{1}{6}} </math>. -BorealBear<br />
<br />
==Video Solution==<br />
https://youtu.be/EyDGtLc6xGE<br />
<br />
~savannahsolver<br />
<br />
==See Also==<br />
<br />
{{AMC8 box|year=2015|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Jayrenfuhttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_19&diff=1849242015 AMC 8 Problems/Problem 192022-12-26T18:39:31Z<p>Jayrenfu: /* Solutions */</p>
<hr />
<div>== Problem ==<br />
A triangle with vertices as <math>A=(1,3)</math>, <math>B=(5,1)</math>, and <math>C=(4,4)</math> is plotted on a <math>6\times5</math> grid. What fraction of the grid is covered by the triangle?<br />
<br />
<math>\textbf{(A) }\frac{1}{6} \qquad \textbf{(B) }\frac{1}{5} \qquad \textbf{(C) }\frac{1}{4} \qquad \textbf{(D) }\frac{1}{3} \qquad \textbf{(E) }\frac{1}{2}</math><br />
<br />
<asy><br />
draw((1,0)--(1,5),linewidth(.5));<br />
draw((2,0)--(2,5),linewidth(.5));<br />
draw((3,0)--(3,5),linewidth(.5));<br />
draw((4,0)--(4,5),linewidth(.5));<br />
draw((5,0)--(5,5),linewidth(.5));<br />
draw((6,0)--(6,5),linewidth(.5));<br />
draw((0,1)--(6,1),linewidth(.5));<br />
draw((0,2)--(6,2),linewidth(.5));<br />
draw((0,3)--(6,3),linewidth(.5));<br />
draw((0,4)--(6,4),linewidth(.5));<br />
draw((0,5)--(6,5),linewidth(.5)); <br />
draw((0,0)--(0,6),EndArrow);<br />
draw((0,0)--(7,0),EndArrow);<br />
draw((1,3)--(4,4)--(5,1)--cycle);<br />
label("$y$",(0,6),W); label("$x$",(7,0),S);<br />
label("$A$",(1,3),dir(210)); label("$B$",(5,1),SE); label("$C$",(4,4),dir(100));<br />
</asy><br />
<br />
==Solutions==<br />
<br />
===Solution 1===<br />
<br />
The area of <math>\triangle ABC</math> is equal to half the product of its base and height. By the Pythagorean Theorem, we find its height is <math>\sqrt{1^2+2^2}=\sqrt{5}</math>, and its base is <math>\sqrt{2^2+4^2}=\sqrt{20}</math>. We multiply these and divide by <math>2</math> to find the base of the triangle is <math>\frac{\sqrt{5 \cdot 20}}2=\frac{\sqrt{100}}2=\frac{10}2=5</math>. Since the grid has an area of <math>30</math>, the fraction of the grid covered by the triangle is <math>\frac 5{30}=\boxed{\textbf{(A) }\frac{1}{6}}</math>.<br />
<br />
===Solution 2===<br />
Note angle <math>\angle ACB</math> is right, thus the area is <math>\sqrt{1^2+3^2} \times \sqrt{1^2+3^2}\times \dfrac{1}{2}=10 \times \dfrac{1}{2}=5</math> thus the fraction of the total is <math>\dfrac{5}{30}=\boxed{\textbf{(A)}~\dfrac{1}{6}}</math><br />
<br />
===Solution 3===<br />
By the [[Shoelace Theorem]], the area of <math>\triangle ABC=|\dfrac{1}{2}(15+4+4-1-20-12)|=|\dfrac{1}{2}(-10)|=5</math>.<br />
<br />
This means the fraction of the total area is <math>\dfrac{5}{30}=\boxed{\textbf{(A)}~\dfrac{1}{6}}</math><br />
<br />
===Solution 4===<br />
<br />
The smallest rectangle that follows the grid lines and completely encloses <math>\triangle ABC</math> has an area of <math>12</math>, where <math>\triangle ABC</math> splits the rectangle into four triangles. The area of <math>\triangle ABC</math> is therefore <math>12 - (\frac{4 \cdot 2}{2}+\frac{3 \cdot 1}{2}+\frac{3 \cdot 1}{2}) = 12 - (4 + \frac{3}{2} + \frac{3}{2}) = 12 - 7 = 5</math>. That means that <math>\triangle ABC</math> takes up <math>\frac{5}{30} = \boxed{\textbf{(A)}~\frac{1}{6}}</math> of the grid.<br />
<br />
===Solution 5===<br />
<br />
Using [[Pick's Theorem]], the area of the triangle is <math>4 + \dfrac{4}{2} - 1=5</math>. Therefore, the triangle takes up <math>\dfrac{5}{30}=\boxed{\textbf{(A)}~\frac{1}{6}}</math> of the grid.<br />
<br />
===Solution 6 (Heron's Formula, Not Recommended)===<br />
<br />
We can find the lengths of the sides by using the [[Pythagorean Theorem]]. Then, we apply [[Heron's Formula]] to find the area. <br />
<cmath> \sqrt{\left(\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}\right)\left(\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}-\sqrt{10}\right)\left(\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}-\sqrt{10}\right)\left(\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}-2\sqrt{5}\right)}. </cmath><br />
This simplifies to<br />
<cmath> \sqrt{\left(\sqrt{10}+\sqrt{5}\right)\left(\sqrt{10}+\sqrt{5}-\sqrt{10}\right)\left(\sqrt{10}+\sqrt{5}-\sqrt{10}\right)\left(\sqrt{10}+\sqrt{5}-2\sqrt{5}\right)}. </cmath><br />
Again, we simplify to get<br />
<cmath> \sqrt{\left(\sqrt{10}+\sqrt{5}\right)\left(\sqrt{5}\right)\left(\sqrt{5}\right)\left(\sqrt{10}-\sqrt{5}\right)}. </cmath><br />
The middle two terms inside the square root multiply to <math> 5 </math>, and the first and last terms inside the square root multiply to <math> \sqrt{10}^2-\sqrt{5}^2=10-5=5. </math> This means that the area of the triangle is<br />
<cmath> \sqrt{5\cdot 5}=5. </cmath><br />
The area of the grid is <math> 6\cdot 5=30. </math> Thus, the answer is <math> \frac{5}{30}=\boxed{\textbf{(A) }\frac{1}{6}} </math>. -BorealBear<br />
<br />
==Video Solution==<br />
https://youtu.be/EyDGtLc6xGE<br />
<br />
~savannahsolver<br />
<br />
==See Also==<br />
<br />
{{AMC8 box|year=2015|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Jayrenfuhttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_24&diff=1675152019 AMC 8 Problems/Problem 242021-12-08T23:13:25Z<p>Jayrenfu: /* Solution 15 (Straightforward Solution) */</p>
<hr />
<div>==Problem 24==<br />
In triangle <math>ABC</math>, point <math>D</math> divides side <math>\overline{AC}</math> so that <math>AD:DC=1:2</math>. Let <math>E</math> be the midpoint of <math>\overline{BD}</math> and let <math>F</math> be the point of intersection of line <math>BC</math> and line <math>AE</math>. Given that the area of <math>\triangle ABC</math> is <math>360</math>, what is the area of <math>\triangle EBF</math>?<br />
<br />
<asy><br />
unitsize(2cm);<br />
pair A,B,C,DD,EE,FF;<br />
B = (0,0); C = (3,0); <br />
A = (1.2,1.7);<br />
DD = (2/3)*A+(1/3)*C;<br />
EE = (B+DD)/2;<br />
FF = intersectionpoint(B--C,A--A+2*(EE-A));<br />
draw(A--B--C--cycle);<br />
draw(A--FF); <br />
draw(B--DD);dot(A); <br />
label("$A$",A,N);<br />
dot(B); <br />
label("$B$",<br />
B,SW);dot(C); <br />
label("$C$",C,SE);<br />
dot(DD); <br />
label("$D$",DD,NE);<br />
dot(EE); <br />
label("$E$",EE,NW);<br />
dot(FF); <br />
label("$F$",FF,S);<br />
</asy><br />
<br />
<br />
<math>\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40</math><br />
<br />
==Solution 1==<br />
<br />
Draw <math>X</math> on <math>\overline{AF}</math> such that <math>XD</math> is parallel to <math>BC</math>. That makes triangles <math>BEF</math> and <math>EXD</math> congruent since <math>BE = ED</math>. <math>FC=3XD</math> so <math>BC=4BF</math>. Since <math>AF=3EF</math> (<math>XE=EF</math> and <math>AX=\frac13 AF</math>, so <math>XE=EF=\frac13 AF</math>), the altitude of triangle <math>BEF</math> is equal to <math>\frac{1}{3}</math> of the altitude of <math>ABC</math>. The area of <math>ABC</math> is <math>360</math>, so the area of <math>BEF=\frac{1}{3} \cdot \frac{1}{4} \cdot 360=\boxed{\textbf{(B) }30}</math><br />
<br />
==Solution 2 (Mass Points)==<br />
<asy><br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
import graph; size(7cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73; /* image dimensions */<br />
pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
/* draw figures */<br />
draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr); <br />
draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr); <br />
draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr); <br />
draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr); <br />
draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr); <br />
draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr); <br />
draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr); <br />
draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr); <br />
/* dots and labels */<br />
dot((0.28,2.39),dotstyle); <br />
label("$A$", (0.36,2.59), NE * labelscalefactor); <br />
dot((-2.8,-1.17),dotstyle); <br />
label("$B$", (-2.72,-0.97), NE * labelscalefactor); <br />
dot((3.78,-1.05),dotstyle); <br />
label("$C$", (3.86,-0.85), NE * labelscalefactor); <br />
dot((1.2887445398528459,1.3985482236874887),dotstyle); <br />
label("$D$", (1.36,1.59), NE * labelscalefactor); <br />
dot((-0.7199623188673492,-1.1320661821070033),dotstyle); <br />
label("$F$", (-0.64,-0.93), NE * labelscalefactor); <br />
dot((-0.2815567696989588,0.41208536204620183),linewidth(4pt) + dotstyle); <br />
label("$E$", (-0.2,0.57), NE * labelscalefactor); <br />
label("2", (-0.18,2.81), NE * labelscalefactor,wrwrwr); <br />
label("1", (4.4,-1.39), NE * labelscalefactor,wrwrwr); <br />
label("3", (1.9,1.45), NE * labelscalefactor,wrwrwr); <br />
label("3", (-3.44,-1.73), NE * labelscalefactor,wrwrwr); <br />
label("6", (0.08,0.03), NE * labelscalefactor,wrwrwr); <br />
label("4", (-0.82,-1.93), NE * labelscalefactor,wrwrwr); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
/* end of picture */<br />
</asy><br />
<br />
First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us. <br />
<br />
First, we assign a mass of <math>2</math> to point <math>A</math>. We figure out that <math>C</math> has a mass of <math>1</math> since <math>2\times1 = 1\times2</math>. Then, by adding <math>1+2 = 3</math>, we get that point <math>D</math> has a mass of <math>3</math>. By equality, point <math>B</math> has a mass of <math>3</math> also. <br />
<br />
Now, we add <math>3+3 = 6</math> for point <math>E</math> and <math>3+1 = 4</math> for point <math>F</math>.<br />
<br />
Now, <math>BF</math> is a common base for triangles <math>ABF</math> and <math>EBF</math>, so we figure out that the ratios of the areas is the ratios of the heights which is <math>\frac{AE}{EF} = 2:1</math>. So, <math>EBF</math>'s area is one third the area of <math>ABF</math>, and we know the area of <math>ABF</math> is <math>\frac{1}{4}</math> the area of <math>ABC</math> since they have the same heights but different bases.<br />
<br />
So we get the area of <math>EBF</math> as <math>\frac{1}{3}\times\frac{1}{4}\times360 = \boxed{\textbf{(B) }30}</math><br />
-Brudder<br />
<br />
Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of <math>EBF</math> over the product of the mass points of <math>ABC</math> which is <math>\frac{2\times3\times1}{3\times6\times4}\times360</math> which also yields <math>\boxed{\textbf{(B) }30}</math><br />
-Brudder<br />
<br />
==Solution 3==<br />
<math>\frac{BF}{FC}</math> is equal to <math>\frac{\textrm{The area of triangle ABE}}{\textrm{The area of triangle ACE}}</math>. The area of triangle <math>ABE</math> is equal to <math>60</math> because it is equal to on half of the area of triangle <math>ABD</math>, which is equal to one third of the area of triangle <math>ABC</math>, which is <math>360</math>. The area of triangle <math>ACE</math> is the sum of the areas of triangles <math>AED</math> and <math>CED</math>, which is respectively <math>60</math> and <math>120</math>. So, <math>\frac{BF}{FC}</math> is equal to <math>\frac{60}{180}</math>=<math>\frac{1}{3}</math>, so the area of triangle <math>ABF</math> is <math>90</math>. That minus the area of triangle <math>ABE</math> is <math>\boxed{\textbf{(B) }30}</math>. ~~SmileKat32<br />
<br />
==Solution 4 (Similar Triangles)==<br />
Extend <math>\overline{BD}</math> to <math>G</math> such that <math>\overline{AG} \parallel \overline{BC}</math> as shown:<br />
<asy><br />
size(8cm);<br />
pair A, B, C, D, E, F, G;<br />
B = (0,0);<br />
A = (2, 3);<br />
C = (5, 0);<br />
D = (3, 2);<br />
E = (1.5, 1);<br />
F = (1.25, 0);<br />
G = (4.5, 3);<br />
<br />
draw(A--B--C--A--G--B);<br />
draw(A--F);<br />
label("$A$", A, N);<br />
label("$B$", B, WSW);<br />
label("$C$", C, ESE);<br />
label("$D$", D, dir(0)*1.5);<br />
label("$E$", E, SE);<br />
label("$F$", F, S);<br />
label("$G$", G, ENE);<br />
</asy><br />
Then <math>\triangle ADG \sim \triangle CDB</math> and <math>\triangle AEG \sim \triangle FEB</math>. Since <math>CD = 2AD</math>, triangle <math>CDB</math> has four times the area of triangle <math>ADG</math>. Since <math>[CDB] = 240</math>, we get <math>[ADG] = 60</math>.<br />
<br />
Since <math>[AED]</math> is also <math>60</math>, we have <math>ED = DG</math> because triangles <math>AED</math> and <math>ADG</math> have the same height and same areas and so their bases must be the congruent. Thus triangle <math>AEG</math> has twice the side lengths and therefore four times the area of triangle <math>BEF</math>, giving <math>[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}</math>.<br />
<br />
<asy><br />
size(8cm);<br />
pair A, B, C, D, E, F, G;<br />
B = (0,0);<br />
A = (2, 3);<br />
C = (5, 0);<br />
D = (3, 2);<br />
E = (1.5, 1);<br />
F = (1.25, 0);<br />
G = (4.5, 3);<br />
<br />
draw(A--B--C--A--G--B);<br />
draw(A--F);<br />
label("$A$", A, N);<br />
label("$B$", B, WSW);<br />
label("$C$", C, ESE);<br />
label("$D$", D, dir(0)*1.5);<br />
label("$E$", E, SE);<br />
label("$F$", F, S);<br />
label("$G$", G, ENE);<br />
label("$60$", (A+E+D)/3);<br />
label("$60$", (A+E+B)/3);<br />
label("$60$", (A+G+D)/3);<br />
label("$30$", (B+E+F)/3);<br />
</asy><br />
(Credit to MP8148 for the idea)<br />
<br />
==Solution 5 (Area Ratios)==<br />
<asy><br />
size(8cm);<br />
pair A, B, C, D, E, F;<br />
B = (0,0);<br />
A = (2, 3);<br />
C = (5, 0);<br />
D = (3, 2);<br />
E = (1.5, 1);<br />
F = (1.25, 0);<br />
<br />
draw(A--B--C--A--D--B);<br />
draw(A--F);<br />
draw(E--C);<br />
label("$A$", A, N);<br />
label("$B$", B, WSW);<br />
label("$C$", C, ESE);<br />
label("$D$", D, dir(0)*1.5);<br />
label("$E$", E, SSE);<br />
label("$F$", F, S);<br />
label("$60$", (A+E+D)/3);<br />
label("$60$", (A+E+B)/3);<br />
label("$120$", (D+E+C)/3);<br />
label("$x$", (B+E+F)/3);<br />
label("$120-x$", (F+E+C)/3);<br />
</asy><br />
As before we figure out the areas labeled in the diagram. Then we note that <cmath>\dfrac{EF}{AE} = \dfrac{x}{60} = \dfrac{120-x}{180}.</cmath>Solving gives <math>x = \boxed{\textbf{(B) }30}</math>. <br />
(Credit to scrabbler94 for the idea)<br />
<br />
==Solution 6 (Coordinate Bashing)==<br />
Let <math>ADB</math> be a right triangle, and <math>BD=CD</math><br />
<br />
Let <math>A=(-2\sqrt{30}, 0)</math><br />
<br />
<math>B=(0, 4\sqrt{30})</math><br />
<br />
<math>C=(4\sqrt{30}, 0)</math><br />
<br />
<math>D=(0, 0)</math><br />
<br />
<math>E=(0, 2\sqrt{30})</math><br />
<br />
<math>F=(\sqrt{30}, 3\sqrt{30})</math><br />
<br />
The line <math>\overleftrightarrow{AE}</math> can be described with the equation <math>y=x-2\sqrt{30}</math><br />
<br />
The line <math>\overleftrightarrow{BC}</math> can be described with <math>x+y=4\sqrt{30}</math><br />
<br />
Solving, we get <math>x=3\sqrt{30}</math> and <math>y=\sqrt{30}</math><br />
<br />
Now we can find <math>EF=BF=2\sqrt{15}</math><br />
<br />
<math>[\bigtriangleup EBF]=\frac{(2\sqrt{15})^2}{2}=\boxed{\textbf{(B) }30}\blacksquare</math><br />
<br />
-Trex4days<br />
<br />
== Solution 7 ==<br />
<asy><br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
import graph; size(15cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -6.61, xmax = 16.13, ymin = -6.4, ymax = 6.42; /* image dimensions */<br />
<br />
/* draw figures */<br />
draw(circle((0,0), 5), linewidth(2)); <br />
draw((-4,-3)--(4,3), linewidth(2)); <br />
draw((-4,-3)--(0,5), linewidth(2)); <br />
draw((0,5)--(4,3), linewidth(2)); <br />
draw((12,-1)--(-4,-3), linewidth(2)); <br />
draw((0,5)--(0,-5), linewidth(2)); <br />
draw((-4,-3)--(0,-5), linewidth(2)); <br />
draw((4,3)--(0,2.48), linewidth(2)); <br />
draw((4,3)--(12,-1), linewidth(2)); <br />
draw((-4,-3)--(4,3), linewidth(2)); <br />
/* dots and labels */<br />
dot((0,0),dotstyle); <br />
label("E", (0.27,-0.24), NE * labelscalefactor); <br />
dot((-5,0),dotstyle); <br />
dot((-4,-3),dotstyle); <br />
label("B", (-4.45,-3.38), NE * labelscalefactor); <br />
dot((4,3),dotstyle); <br />
label("$D$", (4.15,3.2), NE * labelscalefactor); <br />
dot((0,5),dotstyle); <br />
label("A", (-0.09,5.26), NE * labelscalefactor); <br />
dot((12,-1),dotstyle); <br />
label("C", (12.23,-1.24), NE * labelscalefactor); <br />
dot((0,-5),dotstyle); <br />
label("$G$", (0.19,-4.82), NE * labelscalefactor); <br />
dot((0,2.48),dotstyle); <br />
label("I", (-0.33,2.2), NE * labelscalefactor); <br />
dot((0,0),dotstyle); <br />
label("E", (0.27,-0.24), NE * labelscalefactor); <br />
dot((0,-2.5),dotstyle); <br />
label("F", (0.23,-2.2), NE * labelscalefactor); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
/* end of picture */<br />
</asy><br />
<br />
Let <math>A[\Delta XYZ]</math> = <math>\text{Area of Triangle XYZ}</math> <br />
<br />
<br />
<math>A[\Delta ABD]: A[\Delta DBC] :: 1:2 :: 120:240</math><br />
<br />
<br />
<math>A[\Delta ABE] = A[\Delta AED] = 60</math> (the median divides the area of the triangle into two equal parts)<br />
<br />
<br />
Construction: Draw a circumcircle around <math>\Delta ABD</math> with <math>BD</math> as is diameter. Extend <math>AF</math> to <math>G</math> such that it meets the circle at <math>G</math>. Draw line <math>BG</math>.<br />
<br />
<br />
<math>A[\Delta ABD] = A[\Delta ABG] = 120</math> (Since <math>\square ABGD</math> is cyclic)<br />
<br />
<br />
But <math>A[\Delta ABE]</math> is common in both with an area of 60. So, <math>A[\Delta AED] = A[\Delta BEG]</math>.<br />
<br />
\therefore <math>A[\Delta AED] \cong A[\Delta BEG]</math> (SAS Congruency Theorem).<br />
<br />
In <math>\Delta AED</math>, let <math>DI</math> be the median of <math>\Delta AED</math>.<br />
<br />
Which means <math>A[\Delta AID] = 30 = A[\Delta EID]</math><br />
<br />
<br />
Rotate <math>\Delta DEA</math> to meet <math>D</math> at <math>B</math> and <math>A</math> at <math>G</math>. <math>DE</math> will fit exactly in <math>BE</math> (both are radii of the circle). From the above solutions, <math>\frac{AE}{EF} = 2:1</math>.<br />
<br />
<math>AE</math> is a radius and <math>EF</math> is half of it implies <math>EF</math> = <math>\frac{radius}{2}</math>.<br />
<br />
Which means <math>A[\Delta BEF] \cong A[\Delta DEI]</math><br />
<br />
Thus <math>A[\Delta BEF] = \boxed{\textbf{(B) }30}</math><br />
<br />
<br />
~phoenixfire & flamewavelight<br />
<br />
== Solution 8 ==<br />
<asy><br />
import geometry;<br />
unitsize(2cm);<br />
pair A,B,C,DD,EE,FF, M;<br />
B = (0,0); C = (3,0); M = (1.45,0);<br />
A = (1.2,1.7);<br />
DD = (2/3)*A+(1/3)*C;<br />
EE = (B+DD)/2;<br />
FF = intersectionpoint(B--C,A--A+2*(EE-A));<br />
draw(A--B--C--cycle);<br />
draw(A--FF); <br />
draw(B--DD);dot(A); <br />
label("$A$",A,N);<br />
dot(B); <br />
label("$B$",<br />
B,SW);dot(C); <br />
label("$C$",C,SE);<br />
dot(DD); <br />
label("$D$",DD,NE);<br />
dot(EE); <br />
label("$E$",EE,NW);<br />
dot(FF); <br />
label("$F$",FF,S);<br />
draw(EE--M,StickIntervalMarker(1,1));<br />
label("$M$",M,S);<br />
draw(A--DD,invisible,StickIntervalMarker(1,1));<br />
dot((DD+C)/2);<br />
draw(DD--C,invisible,StickIntervalMarker(2,1));<br />
</asy><br />
Using the ratio of <math>\overline{AD}</math> and <math>\overline{CD}</math>, we find the area of <math>\triangle ADB</math> is <math>120</math> and the area of <math>\triangle BDC</math> is <math>240</math>. Also using the fact that <math>E</math> is the midpoint of <math>\overline{BD}</math>, we know <math>\triangle ADE = \triangle ABE = 60</math>.<br />
Let <math>M</math> be a point such <math>\overline{EM}</math> is parellel to <math>\overline{CD}</math>. We immediatley know that <math>\triangle BEM \sim BDC</math> by <math>2</math>. Using that we can conclude <math>EM</math> has ratio <math>1</math>. Using <math>\triangle EFM \sim \triangle AFC</math>, we get <math>EF:AE = 1:2</math>. Therefore using the fact that <math>\triangle EBF</math> is in <math>\triangle ABF</math>, the area has ratio <math>\triangle BEF : \triangle ABE=1:2</math> and we know <math>\triangle ABE</math> has area <math>60</math> so <math>\triangle BEF</math> is <math>\boxed{\textbf{(B) }30}</math>. - fath2012<br />
<br />
==Solution 9 (Menelaus's Theorem)==<br />
<asy><br />
unitsize(2cm);<br />
pair A,B,C,DD,EE,FF;<br />
B = (0,0); C = (3,0); <br />
A = (1.2,1.7);<br />
DD = (2/3)*A+(1/3)*C;<br />
EE = (B+DD)/2;<br />
FF = intersectionpoint(B--C,A--A+2*(EE-A));<br />
draw(A--B--C--cycle);<br />
draw(A--FF); <br />
draw(B--DD);dot(A); <br />
label("$A$",A,N);<br />
dot(B); <br />
label("$B$",<br />
B,SW);dot(C); <br />
label("$C$",C,SE);<br />
dot(DD); <br />
label("$D$",DD,NE);<br />
dot(EE); <br />
label("$E$",EE,NW);<br />
dot(FF); <br />
label("$F$",FF,S);<br />
</asy><br />
By Menelaus's Theorem on triangle <math>BCD</math>, we have <cmath>\dfrac{BF}{FC} \cdot \dfrac{CA}{DA} \cdot \dfrac{DE}{BE} = 3\dfrac{BF}{FC} = 1 \implies \dfrac{BF}{FC} = \dfrac13 \implies \dfrac{BF}{BC} = \dfrac14.</cmath> Therefore, <cmath>[EBF] = \dfrac{BE}{BD}\cdot\dfrac{BF}{BC}\cdot [BCD] = \dfrac12 \cdot \dfrac 14 \cdot \left( \dfrac23 \cdot [ABC]\right) = \boxed{\textbf{(B) }30}.</cmath><br />
<br />
==Solution 10 (Graph Paper)==<br />
<asy><br />
unitsize(2cm);<br />
pair A,B,C,D,E,F,a,b,c,d,e,f;<br />
A = (2,3);<br />
B = (0,2); <br />
C = (2,0);<br />
D = (2/3)*A+(1/3)*C;<br />
E = (B+D)/2;<br />
F = intersectionpoint(B--C,A--A+2*(E-A));<br />
a = (0,0);<br />
b = (1,0);<br />
c = (2,1);<br />
d = (1,3);<br />
e = (0,3);<br />
f = (0,1);<br />
draw(a--C,dashed);<br />
draw(f--c,dashed);<br />
draw(e--A,dashed);<br />
draw(a--e,dashed);<br />
draw(b--d,dashed);<br />
draw(A--B--C--cycle);<br />
draw(A--F); <br />
draw(B--D);<br />
dot(A); <br />
label("$A$",A,NE);<br />
dot(B); <br />
label("$B$",B,dir(180));<br />
dot(C); <br />
label("$C$",C,SE);<br />
dot(D); <br />
label("$D$",D,dir(0));<br />
dot(E); <br />
label("$E$",E,SE);<br />
dot(F); <br />
label("$F$",F,SW);<br />
</asy><br />
<b>Note:</b> If graph paper is unavailable, this solution can still be used by constructing a small grid on a sheet of blank paper.<br><br />
<br><br />
As triangle <math>ABC</math> is loosely defined, we can arrange its points such that the diagram fits nicely on a coordinate plane. By doing so, we can construct it on graph paper and be able to visually determine the relative sizes of the triangles.<br><br />
<br><br />
As point <math>D</math> splits line segment <math>\overline{AC}</math> in a <math>1:2</math> ratio, we draw <math>\overline{AC}</math> as a vertical line segment <math>3</math> units long. Point <math>D</math> is thus <math>1</math> unit below point <math>A</math> and <math>2</math> units above point <math>C</math>. By definition, Point <math>E</math> splits line segment <math>\overline{BD}</math> in a <math>1:1</math> ratio, so we draw <math>\overline{BD}</math> <math>2</math> units long directly left of <math>D</math> and draw <math>E</math> directly between <math>B</math> and <math>D</math>, <math>1</math> unit away from both.<br><br />
<br><br />
We then draw line segments <math>\overline{AB}</math> and <math>\overline{BC}</math>. We can easily tell that triangle <math>ABC</math> occupies <math>3</math> square units of space. Constructing line <math>AE</math> and drawing <math>F</math> at the intersection of <math>AE</math> and <math>BC</math>, we can easily see that triangle <math>EBF</math> forms a right triangle occupying <math>\frac{1}{4}</math> of a square unit of space.<br><br />
<br><br />
The ratio of the areas of triangle <math>EBF</math> and triangle <math>ABC</math> is thus <math>\frac{1}{4}\div3=\frac{1}{12}</math>, and since the area of triangle <math>ABC</math> is <math>360</math>, this means that the area of triangle <math>EBF</math> is <math>\frac{1}{12}\times360=\boxed{\textbf{(B) }30}</math>. ~[[User:emerald_block|emerald_block]]<br><br />
<br><br />
<b>Additional note:</b> There are many subtle variations of this triangle; this method is one of the more compact ones. ~[[User:i_equal_tan_90|i_equal_tan_90]]<br />
<br />
==Solution 11==<br />
<asy><br />
unitsize(2cm);<br />
pair A,B,C,DD,EE,FF,G;<br />
B = (0,0); C = (3,0); <br />
A = (1.2,1.7);<br />
DD = (2/3)*A+(1/3)*C;<br />
EE = (B+DD)/2;<br />
FF = intersectionpoint(B--C,A--A+2*(EE-A));<br />
G = (1.5,0);<br />
draw(A--B--C--cycle);<br />
draw(A--FF); <br />
draw(B--DD); <br />
draw(G--DD);<br />
label("$A$",A,N);<br />
label("$B$",<br />
B,SW); <br />
label("$C$",C,SE);<br />
label("$D$",DD,NE); <br />
label("$E$",EE,NW);<br />
label("$F$",FF,S);<br />
label("$G$",G,S);<br />
</asy><br />
We know that <math>AD = \dfrac{1}{3} AC</math>, so <math>[ABD] = \dfrac{1}{3} [ABC] = 120</math>. Using the same method, since <math>BE = \dfrac{1}{2} BD</math>, <math>[ABE] = \dfrac{1}{2} [ABD] = 60</math>. Next, we draw <math>G</math> on <math>\overline{BC}</math> such that <math>\overline{DG}</math> is parallel to <math>\overline{AF}</math> and create segment <math>DG</math>. We then observe that <math>\triangle AFC \sim \triangle DGC</math>, and since <math>AD:DC = 1:2</math>, <math>FG:GC</math> is also equal to <math>1:2</math>. Similarly (no pun intended), <math>\triangle DBG \sim \triangle EBF</math>, and since <math>BE:ED = 1:1</math>, <math>BF:FG</math> is also equal to <math>1:1</math>. Combining the information in these two ratios, we find that <math>BF:FG:GC = 1:1:2</math>, or equivalently, <math>BF = \dfrac{1}{4} BC</math>. Thus, <math>[BFA] = \dfrac{1}{4} [BCA] = 90</math>. We already know that <math>[ABE] = 60</math>, so the area of <math>\triangle EBF</math> is <math>[BFA] - [ABE] = \boxed{\textbf{(B) }30}</math>. ~[[User:i_equal_tan_90|i_equal_tan_90]]<br />
<br />
==Solution 12 (Fastest Solution if you have no time)==<br />
The picture is misleading. Assume that the triangle ABC is right. <br />
<br />
Then find two factors of <math>720</math> that are the closest together so that the picture becomes easier in your mind. Quickly searching for squares near <math>720</math> to use difference of squares, we find <math>24</math> and <math>30</math> as our numbers. Then the coordinates of D are <math>(10,16)</math>(note, A=0,0). E is then <math>(5,8)</math>. Then the equation of the line AE is <math>-16x/5+24=y</math>. Plugging in <math>y=0</math>, we have <math>x=\dfrac{15}{2}</math>. Now notice that we have both the height and the base of EBF. <br />
<br />
Solving for the area, we have <math>(8)(15/2)(1/2)=30</math>.<br />
<br />
== Solution 13 ==<br />
<math>AD : DC = 1:2</math>, so <math>ADB</math> has area <math>120</math> and <math>CDB</math> has area <math>240</math>. <math>BE = ED</math> so the area of <math>ABE</math> is equal to the area of <math>ADE = 60</math>.<br />
Draw <math>\overline{DG}</math> parallel to <math>\overline{AF}</math>.<br> <br />
Set area of BEF = <math>x</math>. BEF is similar to BDG in ratio of 1:2<br> <br />
so area of BDG = <math>4x</math>, area of EFDG=<math>3x</math>, and area of CDG<math>=240-4x</math>.<br><br />
CDG is similar to CAF in ratio of 2:3 so area CDG = <math>4/9</math> area CAF, and area AFDG=<math>5/4</math> area CDG.<br> <br />
Thus <math>60+3x=5/4(240-4x)</math> and <math>x=30</math>.<br />
~EFrame<br />
<br />
== Solution 14 - Geometry & Algebra==<br />
<asy><br />
unitsize(2cm);<br />
pair A,B,C,DD,EE,FF;<br />
B = (0,0); C = (3,0); <br />
A = (1.2,1.7);<br />
DD = (2/3)*A+(1/3)*C;<br />
EE = (B+DD)/2;<br />
FF = intersectionpoint(B--C,A--A+2*(EE-A));<br />
draw(A--B--C--cycle);<br />
draw(A--FF); <br />
draw(DD--FF,blue);<br />
draw(B--DD);dot(A); <br />
label("$A$",A,N);<br />
dot(B); <br />
label("$B$",<br />
B,SW);dot(C); <br />
label("$C$",C,SE);<br />
dot(DD); <br />
label("$D$",DD,NE);<br />
dot(EE); <br />
label("$E$",EE,NW);<br />
dot(FF); <br />
label("$F$",FF,S);<br />
</asy><br />
<br />
We draw line <math>FD</math> so that we can define a variable <math>x</math> for the area of <math> \triangle BEF = \triangle DEF</math>. Knowing that <math> \triangle ABE</math> and <math> \triangle ADE</math> share both their height and base, we get that <math>ABE = ADE = 60</math>.<br />
<br />
Since we have a rule where 2 triangles, (<math>\triangle A</math> which has base <math>a</math> and vertex <math>c</math>), and (<math>\triangle B</math> which has Base <math>b</math> and vertex <math>c</math>)who share the same vertex (which is vertex <math>c</math> in this case), and share a common height, their relationship is : Area of <math>A : B = a : b</math> (the length of the two bases), we can list the equation where <math>\frac{ \triangle ABF}{\triangle ACF} = \frac{\triangle DBF}{\triangle DCF}</math>. Substituting <math>x</math> into the equation we get: <br />
<br />
<cmath>\frac{x+60}{300-x} = \frac{2x}{240-2x}</cmath>. <cmath>(2x)(300-x) = (60+x)(240-x).</cmath> <cmath>600-2x^2 = 14400 - 120x + 240x - 2x^2.</cmath> <cmath>480x = 14400.</cmath> and we now have that <math> \triangle BEF=30.</math> <br />
~<math>\bold{\color{blue}{onionheadjr}}</math><br />
<br />
==Video Solutions==<br />
Associated video - https://www.youtube.com/watch?v=DMNbExrK2oo<br />
<br />
https://youtu.be/Ns34Jiq9ofc —DSA_Catachu<br />
<br />
https://www.youtube.com/watch?v=nm-Vj_fsXt4 - Happytwin (Another video solution) <br />
<br />
https://www.youtube.com/watch?v=nyevg9w-CCI&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=6 ~ MathEx <br />
<br />
https://www.youtube.com/watch?v=m04K0Q2SNXY&t=1s<br />
<br />
==Solution 15 (Straightforward Solution)==<br />
<asy><br />
size(8cm);<br />
pair A, B, C, D, E, F;<br />
B = (0,0);<br />
A = (2, 3);<br />
C = (5, 0);<br />
D = (3, 2);<br />
E = (1.5, 1);<br />
F = (1.25, 0);<br />
<br />
draw(A--B--C--A--D--B);<br />
draw(A--F);<br />
draw(E--C);<br />
label("$A$", A, N);<br />
label("$B$", B, WSW);<br />
label("$C$", C, ESE);<br />
label("$D$", D, dir(0)*1.5);<br />
label("$E$", E, SSE);<br />
label("$F$", F, S);<br />
label("$60$", (A+E+D)/3);<br />
label("$60$", (A+E+B)/3);<br />
label("$120$", (D+E+C)/3);<br />
label("$x$", (B+E+F)/3);<br />
label("$120-x$", (F+E+C)/3);<br />
</asy><br />
Since <math>AD:DC=1:2</math> thus <math>\triangle ABD=\frac{1}{3} \cdot 360 = 120.</math><br />
<br />
Similarly, <math>\triangle DBC = \frac{2}{3} \cdot 360 = 240.</math><br />
<br />
Now, since <math>E</math> is a midpoint of <math>BD</math>, <math>\triangle ABE = \triangle AED = 120 \div 2 = 60.</math><br />
<br />
We can use the fact that <math>E</math> is a midpoint of <math>BD</math> even further. Connect lines <math>E</math> and <math>C</math> so that <math>\triangle BEC</math> and <math>\triangle DEC</math> share 2 sides.<br />
<br />
We know that <math>\triangle BEC=\triangle DEC=240 \div 2 = 120</math> since <math>E</math> is a midpoint of <math>BD.</math><br />
<br />
Let's label <math>\triangle BEF</math> <math>x</math>. We know that <math>\triangle EFC</math> is <math>120-x</math> since <math>\triangle BEC = 120.</math><br />
<br />
Note that with this information now, we can deduct more things that are needed to finish the solution.<br />
<br />
Note that <math>\frac{EF}{AE} = \frac{120-x}{180} = \frac{x}{60}.</math> because of triangles <math>EBF, ABE, AEC,</math> and <math>EFC.</math><br />
<br />
We want to find <math>x.</math><br />
<br />
This is a simple equation, and solving we get <math>x=\boxed{\textbf{(B)}30}.</math><br />
<br />
~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea.<br />
<br />
==Note==<br />
This question is extremely similar to 1971 AHSME Problem 26.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=23|num-a=25}}<br />
<br />
[[Category: Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Jayrenfuhttps://artofproblemsolving.com/wiki/index.php?title=1990_AJHSME_Problems/Problem_20&diff=1533401990 AJHSME Problems/Problem 202021-05-07T21:27:02Z<p>Jayrenfu: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
The annual incomes of <math>1,000</math> families range from <math>8200</math> dollars to <math>98,000</math> dollars. In error, the largest income was entered on the computer as <math>980,000</math> dollars. The difference between the mean of the incorrect data and the mean of the actual data is<br />
<br />
<math>\text{(A)}\ \text{882 dollars} \qquad \text{(B)}\ \text{980 dollars} \qquad \text{(C)}\ \text{1078 dollars} \qquad \text{(D)}\ \text{482,000 dollars} \qquad \text{(E)}\ \text{882,000 dollars}</math><br />
<br />
==Solution==<br />
<br />
Let <math>S</math> be the sum of all the incomes but the largest one. For the actual data, the mean is <math>\frac{S+98000}{1000}</math>, and for the incorrect data the mean is <math>\frac{S+980000}{1000}</math>. The difference is <math>882, or \rightarrow \boxed{\text{A}}</math><br />
<br />
==See Also==<br />
<br />
{{AJHSME box|year=1990|num-b=19|num-a=21}}<br />
[[Category:Introductory Algebra Problems]]</div>Jayrenfuhttps://artofproblemsolving.com/wiki/index.php?title=1991_AJHSME_Problems/Problem_25&diff=1532841991 AJHSME Problems/Problem 252021-05-05T21:25:25Z<p>Jayrenfu: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
An equilateral triangle is originally painted black. Each time the triangle is changed, the middle fourth of each black triangle turns white. After five changes, what fractional part of the original area of the black triangle remains black?<br />
<br />
<asy><br />
unitsize(36);<br />
fill((0,0)--(2,0)--(1,sqrt(3))--cycle,gray); draw((0,0)--(2,0)--(1,sqrt(3))--cycle,linewidth(1)); <br />
fill((4,0)--(6,0)--(5,sqrt(3))--cycle,gray); fill((5,0)--(9/2,sqrt(3)/2)--(11/2,sqrt(3)/2)--cycle,white);<br />
draw((5,sqrt(3))--(4,0)--(5,0)--(9/2,sqrt(3)/2)--(11/2,sqrt(3)/2)--(5,0)--(6,0)--cycle,linewidth(1));<br />
fill((8,0)--(10,0)--(9,sqrt(3))--cycle,gray); fill((9,0)--(17/2,sqrt(3)/2)--(19/2,sqrt(3)/2)--cycle,white);<br />
fill((17/2,0)--(33/4,sqrt(3)/4)--(35/4,sqrt(3)/4)--cycle,white);<br />
fill((9,sqrt(3)/2)--(35/4,3*sqrt(3)/4)--(37/4,3*sqrt(3)/4)--cycle,white);<br />
fill((19/2,0)--(37/4,sqrt(3)/4)--(39/4,sqrt(3)/4)--cycle,white);<br />
draw((9,sqrt(3))--(35/4,3*sqrt(3)/4)--(37/4,3*sqrt(3)/4)--(9,sqrt(3)/2)--(35/4,3*sqrt(3)/4)--(33/4,sqrt(3)/4)--(35/4,sqrt(3)/4)--(17/2,0)--(33/4,sqrt(3)/4)--(8,0)--(9,0)--(17/2,sqrt(3)/2)--(19/2,sqrt(3)/2)--(9,0)--(19/2,0)--(37/4,sqrt(3)/4)--(39/4,sqrt(3)/4)--(19/2,0)--(10,0)--cycle,linewidth(1));<br />
label("Change 1",(3,3*sqrt(3)/4),N); label("$\Longrightarrow $",(3,5*sqrt(3)/8),S);<br />
label("Change 2",(7,3*sqrt(3)/4),N); label("$\Longrightarrow $",(7,5*sqrt(3)/8),S);<br />
</asy><br />
<br />
<math>\text{(A)}\ \frac{1}{1024} \qquad \text{(B)}\ \frac{15}{64} \qquad \text{(C)}\ \frac{243}{1024} \qquad \text{(D)}\ \frac{1}{4} \qquad \text{(E)}\ \frac{81}{256}</math><br />
<br />
==Solution==<br />
<br />
With each change, <math>3/4</math> of the black space from the previous stage remains. Since there are <math>5</math> changes, the fractional part of the triangle that remains black is <math>(\frac{3}{4})^5=\frac{243}{1024}\rightarrow \boxed{\text{C}}</math>.<br />
<br />
<br />
edited by stjwyl<br />
<br />
==See Also==<br />
<br />
{{AJHSME box|year=1991|num-b=24|after=Last<br />Problem}}<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Jayrenfuhttps://artofproblemsolving.com/wiki/index.php?title=1991_AJHSME_Problems/Problem_25&diff=1532831991 AJHSME Problems/Problem 252021-05-05T21:25:13Z<p>Jayrenfu: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
An equilateral triangle is originally painted black. Each time the triangle is changed, the middle fourth of each black triangle turns white. After five changes, what fractional part of the original area of the black triangle remains black?<br />
<br />
<asy><br />
unitsize(36);<br />
fill((0,0)--(2,0)--(1,sqrt(3))--cycle,gray); draw((0,0)--(2,0)--(1,sqrt(3))--cycle,linewidth(1)); <br />
fill((4,0)--(6,0)--(5,sqrt(3))--cycle,gray); fill((5,0)--(9/2,sqrt(3)/2)--(11/2,sqrt(3)/2)--cycle,white);<br />
draw((5,sqrt(3))--(4,0)--(5,0)--(9/2,sqrt(3)/2)--(11/2,sqrt(3)/2)--(5,0)--(6,0)--cycle,linewidth(1));<br />
fill((8,0)--(10,0)--(9,sqrt(3))--cycle,gray); fill((9,0)--(17/2,sqrt(3)/2)--(19/2,sqrt(3)/2)--cycle,white);<br />
fill((17/2,0)--(33/4,sqrt(3)/4)--(35/4,sqrt(3)/4)--cycle,white);<br />
fill((9,sqrt(3)/2)--(35/4,3*sqrt(3)/4)--(37/4,3*sqrt(3)/4)--cycle,white);<br />
fill((19/2,0)--(37/4,sqrt(3)/4)--(39/4,sqrt(3)/4)--cycle,white);<br />
draw((9,sqrt(3))--(35/4,3*sqrt(3)/4)--(37/4,3*sqrt(3)/4)--(9,sqrt(3)/2)--(35/4,3*sqrt(3)/4)--(33/4,sqrt(3)/4)--(35/4,sqrt(3)/4)--(17/2,0)--(33/4,sqrt(3)/4)--(8,0)--(9,0)--(17/2,sqrt(3)/2)--(19/2,sqrt(3)/2)--(9,0)--(19/2,0)--(37/4,sqrt(3)/4)--(39/4,sqrt(3)/4)--(19/2,0)--(10,0)--cycle,linewidth(1));<br />
label("Change 1",(3,3*sqrt(3)/4),N); label("$\Longrightarrow $",(3,5*sqrt(3)/8),S);<br />
label("Change 2",(7,3*sqrt(3)/4),N); label("$\Longrightarrow $",(7,5*sqrt(3)/8),S);<br />
</asy><br />
<br />
<math>\text{(A)}\ \frac{1}{1024} \qquad \text{(B)}\ \frac{15}{64} \qquad \text{(C)}\ \frac{243}{1024} \qquad \text{(D)}\ \frac{1}{4} \qquad \text{(E)}\ \frac{81}{256}</math><br />
<br />
==Solution==<br />
<br />
With each change, <math>3/4</math>of the black space from the previous stage remains. Since there are <math>5</math> changes, the fractional part of the triangle that remains black is <math>(\frac{3}{4})^5=\frac{243}{1024}\rightarrow \boxed{\text{C}}</math>.<br />
<br />
<br />
edited by stjwyl<br />
<br />
==See Also==<br />
<br />
{{AJHSME box|year=1991|num-b=24|after=Last<br />Problem}}<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Jayrenfuhttps://artofproblemsolving.com/wiki/index.php?title=1991_AJHSME_Problems/Problem_25&diff=1532821991 AJHSME Problems/Problem 252021-05-05T21:24:51Z<p>Jayrenfu: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
An equilateral triangle is originally painted black. Each time the triangle is changed, the middle fourth of each black triangle turns white. After five changes, what fractional part of the original area of the black triangle remains black?<br />
<br />
<asy><br />
unitsize(36);<br />
fill((0,0)--(2,0)--(1,sqrt(3))--cycle,gray); draw((0,0)--(2,0)--(1,sqrt(3))--cycle,linewidth(1)); <br />
fill((4,0)--(6,0)--(5,sqrt(3))--cycle,gray); fill((5,0)--(9/2,sqrt(3)/2)--(11/2,sqrt(3)/2)--cycle,white);<br />
draw((5,sqrt(3))--(4,0)--(5,0)--(9/2,sqrt(3)/2)--(11/2,sqrt(3)/2)--(5,0)--(6,0)--cycle,linewidth(1));<br />
fill((8,0)--(10,0)--(9,sqrt(3))--cycle,gray); fill((9,0)--(17/2,sqrt(3)/2)--(19/2,sqrt(3)/2)--cycle,white);<br />
fill((17/2,0)--(33/4,sqrt(3)/4)--(35/4,sqrt(3)/4)--cycle,white);<br />
fill((9,sqrt(3)/2)--(35/4,3*sqrt(3)/4)--(37/4,3*sqrt(3)/4)--cycle,white);<br />
fill((19/2,0)--(37/4,sqrt(3)/4)--(39/4,sqrt(3)/4)--cycle,white);<br />
draw((9,sqrt(3))--(35/4,3*sqrt(3)/4)--(37/4,3*sqrt(3)/4)--(9,sqrt(3)/2)--(35/4,3*sqrt(3)/4)--(33/4,sqrt(3)/4)--(35/4,sqrt(3)/4)--(17/2,0)--(33/4,sqrt(3)/4)--(8,0)--(9,0)--(17/2,sqrt(3)/2)--(19/2,sqrt(3)/2)--(9,0)--(19/2,0)--(37/4,sqrt(3)/4)--(39/4,sqrt(3)/4)--(19/2,0)--(10,0)--cycle,linewidth(1));<br />
label("Change 1",(3,3*sqrt(3)/4),N); label("$\Longrightarrow $",(3,5*sqrt(3)/8),S);<br />
label("Change 2",(7,3*sqrt(3)/4),N); label("$\Longrightarrow $",(7,5*sqrt(3)/8),S);<br />
</asy><br />
<br />
<math>\text{(A)}\ \frac{1}{1024} \qquad \text{(B)}\ \frac{15}{64} \qquad \text{(C)}\ \frac{243}{1024} \qquad \text{(D)}\ \frac{1}{4} \qquad \text{(E)}\ \frac{81}{256}</math><br />
<br />
==Solution==<br />
<br />
With each change, \frac{3}{4}of the black space from the previous stage remains. Since there are <math>5</math> changes, the fractional part of the triangle that remains black is <math>(\frac{3}{4})^5=\frac{243}{1024}\rightarrow \boxed{\text{C}}</math>.<br />
<br />
<br />
edited by stjwyl<br />
<br />
==See Also==<br />
<br />
{{AJHSME box|year=1991|num-b=24|after=Last<br />Problem}}<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Jayrenfuhttps://artofproblemsolving.com/wiki/index.php?title=1994_AJHSME_Problems/Problem_22&diff=1525891994 AJHSME Problems/Problem 222021-04-24T21:38:34Z<p>Jayrenfu: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
The two wheels shown below are spun and the two resulting numbers are added. The probability that the sum is even is<br />
<br />
<asy><br />
draw(circle((0,0),3));<br />
draw(circle((7,0),3));<br />
draw((0,0)--(3,0));<br />
draw((0,-3)--(0,3));<br />
draw((7,3)--(7,0)--(7+3*sqrt(3)/2,-3/2));<br />
draw((7,0)--(7-3*sqrt(3)/2,-3/2));<br />
draw((0,5)--(0,3.5)--(-0.5,4));<br />
draw((0,3.5)--(0.5,4));<br />
draw((7,5)--(7,3.5)--(6.5,4));<br />
draw((7,3.5)--(7.5,4));<br />
label("$3$",(-0.75,0),W);<br />
label("$1$",(0.75,0.75),NE);<br />
label("$2$",(0.75,-0.75),SE);<br />
label("$6$",(6,0.5),NNW);<br />
label("$5$",(7,-1),S);<br />
label("$4$",(8,0.5),NNE);<br />
</asy><br />
<br />
<math>\text{(A)}\ \dfrac{1}{6} \qquad \text{(B)}\ \dfrac{1}{4} \qquad \text{(C)}\ \dfrac{1}{3} \qquad \text{(D)}\ \dfrac{5}{12} \qquad \text{(E)}\ \dfrac{4}{9}</math><br />
<br />
==Solution==<br />
An even sum occurs when an even is added to an even or an odd is added to an odd. Looking at the areas of the regions, the chance of getting an even in the first wheel is <math>\frac14</math> and the chance of getting an odd is <math>\frac34</math>. On the second wheel, the chance of getting an even is <math>\frac23</math> and an odd is <math>\frac13</math> , resulting in an answer of D.<br />
<br />
<cmath>\frac14 \cdot \frac23 + \frac34 \cdot \frac13 = \frac16 + \frac14 = \boxed{\text{(D)}\ \frac{5}{12}}</cmath><br />
<br />
==See Also==<br />
{{AJHSME box|year=1994|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Jayrenfuhttps://artofproblemsolving.com/wiki/index.php?title=1994_AJHSME_Problems/Problem_22&diff=1525881994 AJHSME Problems/Problem 222021-04-24T21:38:19Z<p>Jayrenfu: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
The two wheels shown below are spun and the two resulting numbers are added. The probability that the sum is even is<br />
<br />
<asy><br />
draw(circle((0,0),3));<br />
draw(circle((7,0),3));<br />
draw((0,0)--(3,0));<br />
draw((0,-3)--(0,3));<br />
draw((7,3)--(7,0)--(7+3*sqrt(3)/2,-3/2));<br />
draw((7,0)--(7-3*sqrt(3)/2,-3/2));<br />
draw((0,5)--(0,3.5)--(-0.5,4));<br />
draw((0,3.5)--(0.5,4));<br />
draw((7,5)--(7,3.5)--(6.5,4));<br />
draw((7,3.5)--(7.5,4));<br />
label("$3$",(-0.75,0),W);<br />
label("$1$",(0.75,0.75),NE);<br />
label("$2$",(0.75,-0.75),SE);<br />
label("$6$",(6,0.5),NNW);<br />
label("$5$",(7,-1),S);<br />
label("$4$",(8,0.5),NNE);<br />
</asy><br />
<br />
<math>\text{(A)}\ \dfrac{1}{6} \qquad \text{(B)}\ \dfrac{1}{4} \qquad \text{(C)}\ \dfrac{1}{3} \qquad \text{(D)}\ \dfrac{5}{12} \qquad \text{(E)}\ \dfrac{4}{9}</math><br />
<br />
==Solution==<br />
An even sum occurs when an even is added to an even or an odd is added to an odd. Looking at the areas of the regions, the chance of getting an even in the first wheel is <math>\frac14</math> and the chance of getting an odd is <math>\frac34</math>. On the second wheel, the chance of getting an even is <math>\frac23</math> and an odd is <math>\frac13</math>, resulting in an answer of D.<br />
<br />
<cmath>\frac14 \cdot \frac23 + \frac34 \cdot \frac13 = \frac16 + \frac14 = \boxed{\text{(D)}\ \frac{5}{12}}</cmath><br />
<br />
==See Also==<br />
{{AJHSME box|year=1994|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Jayrenfuhttps://artofproblemsolving.com/wiki/index.php?title=1994_AJHSME_Problems/Problem_25&diff=1525871994 AJHSME Problems/Problem 252021-04-24T21:35:00Z<p>Jayrenfu: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Find the sum of the digits in the answer to<br />
<br />
<math>\underbrace{9999\cdots 99}_{94\text{ nines}} \times \underbrace{4444\cdots 44}_{94\text{ fours}}</math><br />
<br />
where a string of <math>94</math> nines is multiplied by a string of <math>94</math> fours.<br />
<br />
<math>\text{(A)}\ 846 \qquad \text{(B)}\ 855 \qquad \text{(C)}\ 945 \qquad \text{(D)}\ 954 \qquad \text{(E)}\ 1072</math><br />
<br />
==Solution==<br />
<br />
Notice that:<br />
<br />
<math>9 \cdot 4 = 36</math> and <math>3+6 = 9 = 9 \cdot 1</math><br />
<br />
<math>99 \cdot 44 = 4356</math> and <math>4+5+3+6 = 18 = 9 \cdot 2</math><br />
<br />
So the sum of the digits of <math>x</math> 9s times <math>x</math> 4s is simply <math>x \cdot 9</math> (Try to find the proof that it works for all values of <math>x</math> ~MATHWIZARD10).<br />
<br />
Therefore the answer is <math>94 \cdot 9 = \boxed{\text{(A)}\ 846.}</math><br />
<br />
==See Also==<br />
{{AJHSME box|year=1994|num-b=24|after=Last <br /> Problem}}<br />
{{MAA Notice}}</div>Jayrenfuhttps://artofproblemsolving.com/wiki/index.php?title=1997_AJHSME_Problems/Problem_22&diff=1510631997 AJHSME Problems/Problem 222021-04-05T20:28:26Z<p>Jayrenfu: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
A two-inch cube <math>(2\times 2\times 2)</math> of silver weighs 3 pounds and is worth 200. How much is a three-inch cube of silver worth?<br />
<br />
<math>\text{(A)}\ 300\text{ dollars} \qquad \text{(B)}\ 375\text{ dollars} \qquad \text{(C)}\ 450\text{ dollars} \qquad \text{(D)}\ 560\text{ dollars} \qquad \text{(E)}\ 675\text{ dollars}</math><br />
<br />
==Solution 1==<br />
<br />
The 2x2x2 cube of silver can be divided into <math>8</math> equal cubes that are 1x1x1. Each smaller cube is worth <math>\frac{200}{8} = 25</math> dollars.<br />
<br />
To create a 3x3x3 cube of silver, you need <math>27</math> of those 1x1x1 cubes. The cost of those <math>27</math> cubes is <math>27 \cdot 25 = 675</math> dollars, which is answer <math>\boxed{E}</math><br />
<br />
==Solution 2==<br />
<br />
Since price is proportional to the amount (or volume) of silver, and volume is proportional to the cube of the side, the price ought to be proportional to the cube of the side.<br />
<br />
Setting up a proportion:<br />
<br />
<math>\frac{200}{2^3} = \frac{x}{3^3}</math><br />
<br />
<math>x = 200 \cdot \frac{3^3}{2^3} = 675</math>, which is answer <math>\boxed{E}</math><br />
<br />
== See also ==<br />
{{AJHSME box|year=1997|num-b=21|num-a=23}}<br />
* [[AJHSME]]<br />
* [[AJHSME Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
{{MAA Notice}}</div>Jayrenfuhttps://artofproblemsolving.com/wiki/index.php?title=1997_AJHSME_Problems/Problem_22&diff=1510621997 AJHSME Problems/Problem 222021-04-05T20:28:09Z<p>Jayrenfu: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
A two-inch cube <math>(2\times 2\times 2)</math> of silver weighs 3 pounds and is worth <math>200. How much is a three-inch cube of silver worth?<br />
<br />
</math>\text{(A)}\ 300\text{ dollars} \qquad \text{(B)}\ 375\text{ dollars} \qquad \text{(C)}\ 450\text{ dollars} \qquad \text{(D)}\ 560\text{ dollars} \qquad \text{(E)}\ 675\text{ dollars}$<br />
<br />
==Solution 1==<br />
<br />
The 2x2x2 cube of silver can be divided into <math>8</math> equal cubes that are 1x1x1. Each smaller cube is worth <math>\frac{200}{8} = 25</math> dollars.<br />
<br />
To create a 3x3x3 cube of silver, you need <math>27</math> of those 1x1x1 cubes. The cost of those <math>27</math> cubes is <math>27 \cdot 25 = 675</math> dollars, which is answer <math>\boxed{E}</math><br />
<br />
==Solution 2==<br />
<br />
Since price is proportional to the amount (or volume) of silver, and volume is proportional to the cube of the side, the price ought to be proportional to the cube of the side.<br />
<br />
Setting up a proportion:<br />
<br />
<math>\frac{200}{2^3} = \frac{x}{3^3}</math><br />
<br />
<math>x = 200 \cdot \frac{3^3}{2^3} = 675</math>, which is answer <math>\boxed{E}</math><br />
<br />
== See also ==<br />
{{AJHSME box|year=1997|num-b=21|num-a=23}}<br />
* [[AJHSME]]<br />
* [[AJHSME Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
{{MAA Notice}}</div>Jayrenfuhttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_24&diff=1488662010 AMC 8 Problems/Problem 242021-03-07T23:37:39Z<p>Jayrenfu: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
What is the correct ordering of the three numbers, <math>10^8</math>, <math>5^{12}</math>, and <math>2^{24}</math>?<br />
<br />
<math> \textbf{(A)}\ 2^{24}<10^8<5^{12}\\<br />
\textbf{(B)}\ 2^{24}<5^{12}<10^8 \\<br />
\textbf{(C)}\ 5^{12}<2^{24}<10^8 \\<br />
\textbf{(D)}\ 10^8<5^{12}<2^{24} \\<br />
\textbf{(E)}\ 10^8<2^{24}<5^{12} </math><br />
<br />
==Video Solution==<br />
https://youtu.be/rQUwNC0gqdg?t=381<br />
<br />
==Solution 1==<br />
Use brute force.<br />
<math>10^8=100,000,000</math>, <br />
<math>5^{12}=244,140,625</math>, and<br />
<math>2^{24}=16,777,216</math>.<br />
Therefore, <math>\boxed{\text{(A)}2^{24}<10^8<5^{12}}</math> is the answer. (Not recommended for the contest and will take forever)<br />
<br />
== Solution 2==<br />
Since all of the exponents are multiples of four, we can simplify the problem by taking the fourth root of each number. Evaluating we get <math>10^2=100</math>, <math>5^3=125</math>, and <math>2^6=64</math>. Since <math>64<100<125</math>, it follows that <math>\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}</math> is the correct answer.<br />
<br />
== Solution 3==<br />
First, let us make all exponents equal to 8. Then, it will be easy to order the numbers without doing any computations.<br />
<math>10^8</math> is as fine as it is.<br />
We can rewrite <math>2^{24}</math> as <math>(2^3)^8=8^8</math>.<br />
Then we can rewrite <math>5^{12}</math> as <math>(5^{\frac{3}{2}})^8=(\sqrt{125})^8)</math>.<br />
We take the eighth root of all of these to get <math>{10, 8, \sqrt{125}}</math>.<br />
Obviously, <math>8<10<\sqrt{125}</math>, so the answer is <math>\textbf{(A)}\ 2^{24}<10^8<5^{12}</math>.<br />
Solution by MathHayden<br />
<br />
==See Also==<br />
{{AMC8 box|year=2010|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Jayrenfu