https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Infinity26484&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T08:55:00ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_19&diff=1644572016 AMC 10A Problems/Problem 192021-11-02T01:44:13Z<p>Infinity26484: /* Solution 1 (Similar Triangles) */</p>
<hr />
<div>== Problem ==<br />
<br />
In rectangle <math>ABCD,</math> <math>AB=6</math> and <math>BC=3</math>. Point <math>E</math> between <math>B</math> and <math>C</math>, and point <math>F</math> between <math>E</math> and <math>C</math> are such that <math>BE=EF=FC</math>. Segments <math>\overline{AE}</math> and <math>\overline{AF}</math> intersect <math>\overline{BD}</math> at <math>P</math> and <math>Q</math>, respectively. The ratio <math>BP:PQ:QD</math> can be written as <math>r:s:t</math> where the greatest common factor of <math>r,s,</math> and <math>t</math> is <math>1.</math> What is <math>r+s+t</math>?<br />
<br />
<math>\textbf{(A) } 7 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 20</math><br />
<br />
== Solution 1 (Similar Triangles)==<br />
<br />
<asy><br />
size(6cm);<br />
pair D=(0,0), C=(6,0), B=(6,3), A=(0,3);<br />
draw(A--B--C--D--cycle);<br />
draw(B--D);<br />
draw(A--(6,2));<br />
draw(A--(6,1));<br />
label("$A$", A, dir(135));<br />
label("$B$", B, dir(45));<br />
label("$C$", C, dir(-45));<br />
label("$D$", D, dir(-135));<br />
label("$Q$", extension(A,(6,1),B,D),dir(-90));<br />
label("$P$", extension(A,(6,2),B,D), dir(90));<br />
label("$F$", (6,1), dir(0));<br />
label("$E$", (6,2), dir(0));<br />
</asy><br />
<br />
Use similar triangles. Our goal is to put the ratio in terms of <math>{BD}</math>. Since <math>\triangle APD \sim \triangle EPB,</math> <math>\frac{DP}{PB}=\frac{AD}{BE}=3.</math> Therefore, <math>PB=\frac{BD}{4}</math>. Similarly, <math>\frac{DQ}{QB}=\frac{3}{2}</math>. This means that <math>{DQ}=\frac{3\cdot BD}{5}</math>. Therefore, <math>r:s:t=\frac{1}{4}:\frac{2}{5}-\frac{1}{4}:\frac{3}{5}=5:3:12,</math> so <math>r+s+t=\boxed{\textbf{(E) }20.}</math><br />
<br />
==Solution 2(Coordinate Bash)==<br />
<br />
We can set coordinates for the points. <math>D=(0,0), C=(6,0), B=(6,3),</math> and <math>A=(0,3)</math>. The line <math>BD</math>'s equation is <math>y = \frac{1}{2}x</math>, line <math>AE</math>'s equation is <math>y = -\frac{1}{6}x + 3</math>, and line <math>AF</math>'s equation is <math>y = -\frac{1}{3}x + 3</math>. Adding the equations of lines <math>BD</math> and <math>AE</math>, we find that the coordinates of <math>P</math> are <math>\left(\frac{9}{2},\frac{9}{4}\right)</math>. Furthermore we find that the coordinates of <math>Q</math> are <math>\left(\frac{18}{5}, \frac{9}{5}\right)</math>. Using the [[Pythagorean Theorem]], we get that the length of <math>QD</math> is <math>\sqrt{\left(\frac{18}{5}\right)^2+\left(\frac{9}{5}\right)^2} = \sqrt{\frac{405}{25}} = \frac{\sqrt{405}}{5} = \frac{9\sqrt{5}}{5}</math>, and the length of <math>DP</math> is <math>\sqrt{\left(\frac{9}{2}\right)^2+\left(\frac{9}{4}\right)^2} = \sqrt{\frac{81}{4} + \frac{81}{16}} = \sqrt{\frac{405}{16}} = \frac{\sqrt{405}}{4} = \frac{9\sqrt{5}}{4}.</math> <math>PQ = DP - DQ = \frac{9\sqrt{5}}{4} - \frac{9\sqrt{5}}{5} = \frac{9\sqrt{5}}{20}.</math> The length of <math>DB = \sqrt{6^2 + 3^2} = \sqrt{45} = 3\sqrt{5}</math>. Then <math>BP= 3\sqrt{5} - \frac{9\sqrt{5}}{4} = \frac{3\sqrt{5}}{4}.</math> The ratio <math>BP : PQ : QD = \frac{3\sqrt{5}}{4} : \frac{9\sqrt{5}}{20} : \frac{9\sqrt{5}}{5} = 15\sqrt{5} : 9\sqrt{5} : 36\sqrt{5} = 15 : 9 : 36 = 5 : 3 : 12.</math> Then <math>r, s,</math> and <math>t</math> is <math>5, 3,</math> and <math>12</math>, respectively. The problem tells us to find <math>r + s + t</math>, so <math>5 + 3 + 12 = \boxed{\textbf{(E) }20}</math> ~ minor LaTeX edits by dolphin7<br />
<br />
==Solution 3==<br />
<br />
Extend <math>AF</math> to meet <math>CD</math> at point <math>T</math>. Since <math>FC=1</math> and <math>BF=2</math>, <math>TC=3</math> by similar triangles <math>\triangle TFC</math> and <math>\triangle AFB</math>. It follows that <math>\frac{BQ}{QD}=\frac{BP+PQ}{QD}=\frac{2}{3}</math>. Now, using similar triangles <math>\triangle BEP</math> and <math>\triangle DAP</math>, <math>\frac{BP}{PD}=\frac{BP}{PQ+QD}=\frac{1}{3}</math>. WLOG let <math>BP=1</math>. Solving for <math>PQ, QD</math> gives <math>PQ=\frac{3}{5}</math> and <math>QD=\frac{12}{5}</math>. So our desired ratio is <math>5:3:12</math> and <math>5+3+12=\boxed{\textbf{(E) } 20}</math>.<br />
<br />
==Solution 4 (Mass Points)==<br />
<br />
Draw line segment <math>AC</math>, and call the intersection between <math>AC</math> and <math>BD</math> point <math>K</math>. In <math>\delta ABC</math>, observe that <math>BE:EC=1:2</math> and <math>AK:KC=1:1</math>. Using mass points, find that <math>BP:PK=1:1</math>. Again utilizing <math>\delta ABC</math>, observe that <math>BF:FC=2:1</math> and <math>AK:KC=1:1</math>. Use mass points to find that <math>BQ:QK=4:1</math>. Now, draw a line segment with points <math>B</math>,<math>P</math>,<math>Q</math>, and <math>K</math> ordered from left to right. Set the values <math>BP=x</math>,<math>PK=x</math>,<math>BQ=4y</math> and <math>QK=y</math>. Setting both sides segment <math>BK</math> equal, we get <math>y= \frac{2}{5}x</math>. Plugging in and solving gives <math>QK= \frac{2}{5}x</math>, <math>PQ=\frac{3}{5}x</math>,<math>BP=x</math>. The question asks for <math>BP:PQ:QD</math>, so we add <math>2x</math> to <math>QK</math> and multiply the ratio by <math>5</math> to create integers. This creates <math>5(1:\frac{3}{5}:\frac{12}{5})= 5:3:12</math>. This sums up to <math>3+5+12=\boxed{\textbf{(E) }20}</math><br />
<br />
==Solution 5 (Easy Coord Bash)==<br />
We set coordinates for the points. Let <math>A=(0,3), B=(6,3), C=(6,0)</math> and <math>D=(0,0)</math>. Then the equation of line <math>AE</math> is <math>y = -\frac{1}{6}x + 3,</math> the equation of line <math>AF</math> is <math>y = -\frac{1}{3}x + 3,</math> and the equation of line <math>BD</math> is <math>y = \frac{1}{2}x</math>. We find that the x-coordinate of point <math>P</math> is <math>\frac 9 2</math> by solving <math> -\frac{1}{6}x + 3=\frac{1}{2}x.</math> Similarly we find that the x-coordinate of point <math>Q</math> is <math>\frac {18} 5</math> by solving <math>-\frac{1}{3}x + 3=\frac{1}{2}x.</math> It follows that <math>BP:PQ:QD=6-\frac 9 2 : \frac 9 2 - \frac {18} 5 : \frac {18} 5= \frac 3 2 : \frac 9 {10} : \frac {18} 5 = 5:3:12.</math> Hence <math>r,s,t=5,3,12</math> and <math>r+s+t=5+3+12=\boxed{\textbf{(E) } 20}.</math> ~ Solution by dolphin7<br />
<br />
==Video Solution==<br />
<br />
https://www.youtube.com/watch?v=aG9JiBMd0ag<br />
<br />
==Video Solution 2==<br />
https://youtu.be/us3e2jMjWnw<br />
<br />
~IceMatrix<br />
<br />
== Video Solution 3 ==<br />
https://youtu.be/4_x1sgcQCp4?t=3406<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2016|ab=A|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Infinity26484https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_19&diff=1644562016 AMC 10A Problems/Problem 192021-11-02T01:42:15Z<p>Infinity26484: /* Solution 1 (Similar Triangles) */</p>
<hr />
<div>== Problem ==<br />
<br />
In rectangle <math>ABCD,</math> <math>AB=6</math> and <math>BC=3</math>. Point <math>E</math> between <math>B</math> and <math>C</math>, and point <math>F</math> between <math>E</math> and <math>C</math> are such that <math>BE=EF=FC</math>. Segments <math>\overline{AE}</math> and <math>\overline{AF}</math> intersect <math>\overline{BD}</math> at <math>P</math> and <math>Q</math>, respectively. The ratio <math>BP:PQ:QD</math> can be written as <math>r:s:t</math> where the greatest common factor of <math>r,s,</math> and <math>t</math> is <math>1.</math> What is <math>r+s+t</math>?<br />
<br />
<math>\textbf{(A) } 7 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 20</math><br />
<br />
== Solution 1 (Similar Triangles)==<br />
<br />
<asy><br />
size(6cm);<br />
pair D=(0,0), C=(6,0), B=(6,3), A=(0,3);<br />
draw(A--B--C--D--cycle);<br />
draw(B--D);<br />
draw(A--(6,2));<br />
draw(A--(6,1));<br />
label("$A$", A, dir(135));<br />
label("$B$", B, dir(45));<br />
label("$C$", C, dir(-45));<br />
label("$D$", D, dir(-135));<br />
label("$Q$", extension(A,(6,1),B,D),dir(-90));<br />
label("$P$", extension(A,(6,2),B,D), dir(90));<br />
label("$F$", (6,1), dir(0));<br />
label("$E$", (6,2), dir(0));<br />
</asy><br />
<br />
Use similar triangles. Our goal is to put the ratio in terms of <math>{BD}</math>. Since <math>\triangle APD \sim \triangle EPB,</math> <math>\frac{DP}{PB}=\frac{AD}{BE}=3.</math> Similarly, <math>\frac{DQ}{QB}=\frac{3}{2}</math>. This means that <math>{DQ}=\frac{3\cdot BD}{5}</math>. In addition, <math>PB=\frac{BD}{4}</math>. Therefore, <math>r:s:t=\frac{1}{4}:\frac{2}{5}-\frac{1}{4}:\frac{3}{5}=5:3:12,</math> so <math>r+s+t=\boxed{\textbf{(E) }20.}</math><br />
<br />
==Solution 2(Coordinate Bash)==<br />
<br />
We can set coordinates for the points. <math>D=(0,0), C=(6,0), B=(6,3),</math> and <math>A=(0,3)</math>. The line <math>BD</math>'s equation is <math>y = \frac{1}{2}x</math>, line <math>AE</math>'s equation is <math>y = -\frac{1}{6}x + 3</math>, and line <math>AF</math>'s equation is <math>y = -\frac{1}{3}x + 3</math>. Adding the equations of lines <math>BD</math> and <math>AE</math>, we find that the coordinates of <math>P</math> are <math>\left(\frac{9}{2},\frac{9}{4}\right)</math>. Furthermore we find that the coordinates of <math>Q</math> are <math>\left(\frac{18}{5}, \frac{9}{5}\right)</math>. Using the [[Pythagorean Theorem]], we get that the length of <math>QD</math> is <math>\sqrt{\left(\frac{18}{5}\right)^2+\left(\frac{9}{5}\right)^2} = \sqrt{\frac{405}{25}} = \frac{\sqrt{405}}{5} = \frac{9\sqrt{5}}{5}</math>, and the length of <math>DP</math> is <math>\sqrt{\left(\frac{9}{2}\right)^2+\left(\frac{9}{4}\right)^2} = \sqrt{\frac{81}{4} + \frac{81}{16}} = \sqrt{\frac{405}{16}} = \frac{\sqrt{405}}{4} = \frac{9\sqrt{5}}{4}.</math> <math>PQ = DP - DQ = \frac{9\sqrt{5}}{4} - \frac{9\sqrt{5}}{5} = \frac{9\sqrt{5}}{20}.</math> The length of <math>DB = \sqrt{6^2 + 3^2} = \sqrt{45} = 3\sqrt{5}</math>. Then <math>BP= 3\sqrt{5} - \frac{9\sqrt{5}}{4} = \frac{3\sqrt{5}}{4}.</math> The ratio <math>BP : PQ : QD = \frac{3\sqrt{5}}{4} : \frac{9\sqrt{5}}{20} : \frac{9\sqrt{5}}{5} = 15\sqrt{5} : 9\sqrt{5} : 36\sqrt{5} = 15 : 9 : 36 = 5 : 3 : 12.</math> Then <math>r, s,</math> and <math>t</math> is <math>5, 3,</math> and <math>12</math>, respectively. The problem tells us to find <math>r + s + t</math>, so <math>5 + 3 + 12 = \boxed{\textbf{(E) }20}</math> ~ minor LaTeX edits by dolphin7<br />
<br />
==Solution 3==<br />
<br />
Extend <math>AF</math> to meet <math>CD</math> at point <math>T</math>. Since <math>FC=1</math> and <math>BF=2</math>, <math>TC=3</math> by similar triangles <math>\triangle TFC</math> and <math>\triangle AFB</math>. It follows that <math>\frac{BQ}{QD}=\frac{BP+PQ}{QD}=\frac{2}{3}</math>. Now, using similar triangles <math>\triangle BEP</math> and <math>\triangle DAP</math>, <math>\frac{BP}{PD}=\frac{BP}{PQ+QD}=\frac{1}{3}</math>. WLOG let <math>BP=1</math>. Solving for <math>PQ, QD</math> gives <math>PQ=\frac{3}{5}</math> and <math>QD=\frac{12}{5}</math>. So our desired ratio is <math>5:3:12</math> and <math>5+3+12=\boxed{\textbf{(E) } 20}</math>.<br />
<br />
==Solution 4 (Mass Points)==<br />
<br />
Draw line segment <math>AC</math>, and call the intersection between <math>AC</math> and <math>BD</math> point <math>K</math>. In <math>\delta ABC</math>, observe that <math>BE:EC=1:2</math> and <math>AK:KC=1:1</math>. Using mass points, find that <math>BP:PK=1:1</math>. Again utilizing <math>\delta ABC</math>, observe that <math>BF:FC=2:1</math> and <math>AK:KC=1:1</math>. Use mass points to find that <math>BQ:QK=4:1</math>. Now, draw a line segment with points <math>B</math>,<math>P</math>,<math>Q</math>, and <math>K</math> ordered from left to right. Set the values <math>BP=x</math>,<math>PK=x</math>,<math>BQ=4y</math> and <math>QK=y</math>. Setting both sides segment <math>BK</math> equal, we get <math>y= \frac{2}{5}x</math>. Plugging in and solving gives <math>QK= \frac{2}{5}x</math>, <math>PQ=\frac{3}{5}x</math>,<math>BP=x</math>. The question asks for <math>BP:PQ:QD</math>, so we add <math>2x</math> to <math>QK</math> and multiply the ratio by <math>5</math> to create integers. This creates <math>5(1:\frac{3}{5}:\frac{12}{5})= 5:3:12</math>. This sums up to <math>3+5+12=\boxed{\textbf{(E) }20}</math><br />
<br />
==Solution 5 (Easy Coord Bash)==<br />
We set coordinates for the points. Let <math>A=(0,3), B=(6,3), C=(6,0)</math> and <math>D=(0,0)</math>. Then the equation of line <math>AE</math> is <math>y = -\frac{1}{6}x + 3,</math> the equation of line <math>AF</math> is <math>y = -\frac{1}{3}x + 3,</math> and the equation of line <math>BD</math> is <math>y = \frac{1}{2}x</math>. We find that the x-coordinate of point <math>P</math> is <math>\frac 9 2</math> by solving <math> -\frac{1}{6}x + 3=\frac{1}{2}x.</math> Similarly we find that the x-coordinate of point <math>Q</math> is <math>\frac {18} 5</math> by solving <math>-\frac{1}{3}x + 3=\frac{1}{2}x.</math> It follows that <math>BP:PQ:QD=6-\frac 9 2 : \frac 9 2 - \frac {18} 5 : \frac {18} 5= \frac 3 2 : \frac 9 {10} : \frac {18} 5 = 5:3:12.</math> Hence <math>r,s,t=5,3,12</math> and <math>r+s+t=5+3+12=\boxed{\textbf{(E) } 20}.</math> ~ Solution by dolphin7<br />
<br />
==Video Solution==<br />
<br />
https://www.youtube.com/watch?v=aG9JiBMd0ag<br />
<br />
==Video Solution 2==<br />
https://youtu.be/us3e2jMjWnw<br />
<br />
~IceMatrix<br />
<br />
== Video Solution 3 ==<br />
https://youtu.be/4_x1sgcQCp4?t=3406<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2016|ab=A|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Infinity26484https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_19&diff=1644552016 AMC 10A Problems/Problem 192021-11-02T01:40:55Z<p>Infinity26484: /* Solution 1 (Similar Triangles) */</p>
<hr />
<div>== Problem ==<br />
<br />
In rectangle <math>ABCD,</math> <math>AB=6</math> and <math>BC=3</math>. Point <math>E</math> between <math>B</math> and <math>C</math>, and point <math>F</math> between <math>E</math> and <math>C</math> are such that <math>BE=EF=FC</math>. Segments <math>\overline{AE}</math> and <math>\overline{AF}</math> intersect <math>\overline{BD}</math> at <math>P</math> and <math>Q</math>, respectively. The ratio <math>BP:PQ:QD</math> can be written as <math>r:s:t</math> where the greatest common factor of <math>r,s,</math> and <math>t</math> is <math>1.</math> What is <math>r+s+t</math>?<br />
<br />
<math>\textbf{(A) } 7 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 20</math><br />
<br />
== Solution 1 (Similar Triangles)==<br />
<br />
<asy><br />
size(6cm);<br />
pair D=(0,0), C=(6,0), B=(6,3), A=(0,3);<br />
draw(A--B--C--D--cycle);<br />
draw(B--D);<br />
draw(A--(6,2));<br />
draw(A--(6,1));<br />
label("$A$", A, dir(135));<br />
label("$B$", B, dir(45));<br />
label("$C$", C, dir(-45));<br />
label("$D$", D, dir(-135));<br />
label("$Q$", extension(A,(6,1),B,D),dir(-90));<br />
label("$P$", extension(A,(6,2),B,D), dir(90));<br />
label("$F$", (6,1), dir(0));<br />
label("$E$", (6,2), dir(0));<br />
</asy><br />
<br />
Use similar triangles. Our goal is to put the ratio in terms of <math>{BD}</math>. Since <math>\triangle APD \sim \triangle EPB,</math> <math>\frac{DP}{PB}=\frac{AD}{BE}=3.</math> Similarly, <math>\frac{DQ}{QB}=\frac{3}{2}</math>. This means that <math>{DQ}=\frac{3\cdot BD}{5}</math>. Thus <math>PB=\frac{BD}{4}</math>. Therefore, <math>r:s:t=\frac{1}{4}:\frac{2}{5}-\frac{1}{4}:\frac{3}{5}=5:3:12,</math> so <math>r+s+t=\boxed{\textbf{(E) }20.}</math><br />
<br />
==Solution 2(Coordinate Bash)==<br />
<br />
We can set coordinates for the points. <math>D=(0,0), C=(6,0), B=(6,3),</math> and <math>A=(0,3)</math>. The line <math>BD</math>'s equation is <math>y = \frac{1}{2}x</math>, line <math>AE</math>'s equation is <math>y = -\frac{1}{6}x + 3</math>, and line <math>AF</math>'s equation is <math>y = -\frac{1}{3}x + 3</math>. Adding the equations of lines <math>BD</math> and <math>AE</math>, we find that the coordinates of <math>P</math> are <math>\left(\frac{9}{2},\frac{9}{4}\right)</math>. Furthermore we find that the coordinates of <math>Q</math> are <math>\left(\frac{18}{5}, \frac{9}{5}\right)</math>. Using the [[Pythagorean Theorem]], we get that the length of <math>QD</math> is <math>\sqrt{\left(\frac{18}{5}\right)^2+\left(\frac{9}{5}\right)^2} = \sqrt{\frac{405}{25}} = \frac{\sqrt{405}}{5} = \frac{9\sqrt{5}}{5}</math>, and the length of <math>DP</math> is <math>\sqrt{\left(\frac{9}{2}\right)^2+\left(\frac{9}{4}\right)^2} = \sqrt{\frac{81}{4} + \frac{81}{16}} = \sqrt{\frac{405}{16}} = \frac{\sqrt{405}}{4} = \frac{9\sqrt{5}}{4}.</math> <math>PQ = DP - DQ = \frac{9\sqrt{5}}{4} - \frac{9\sqrt{5}}{5} = \frac{9\sqrt{5}}{20}.</math> The length of <math>DB = \sqrt{6^2 + 3^2} = \sqrt{45} = 3\sqrt{5}</math>. Then <math>BP= 3\sqrt{5} - \frac{9\sqrt{5}}{4} = \frac{3\sqrt{5}}{4}.</math> The ratio <math>BP : PQ : QD = \frac{3\sqrt{5}}{4} : \frac{9\sqrt{5}}{20} : \frac{9\sqrt{5}}{5} = 15\sqrt{5} : 9\sqrt{5} : 36\sqrt{5} = 15 : 9 : 36 = 5 : 3 : 12.</math> Then <math>r, s,</math> and <math>t</math> is <math>5, 3,</math> and <math>12</math>, respectively. The problem tells us to find <math>r + s + t</math>, so <math>5 + 3 + 12 = \boxed{\textbf{(E) }20}</math> ~ minor LaTeX edits by dolphin7<br />
<br />
==Solution 3==<br />
<br />
Extend <math>AF</math> to meet <math>CD</math> at point <math>T</math>. Since <math>FC=1</math> and <math>BF=2</math>, <math>TC=3</math> by similar triangles <math>\triangle TFC</math> and <math>\triangle AFB</math>. It follows that <math>\frac{BQ}{QD}=\frac{BP+PQ}{QD}=\frac{2}{3}</math>. Now, using similar triangles <math>\triangle BEP</math> and <math>\triangle DAP</math>, <math>\frac{BP}{PD}=\frac{BP}{PQ+QD}=\frac{1}{3}</math>. WLOG let <math>BP=1</math>. Solving for <math>PQ, QD</math> gives <math>PQ=\frac{3}{5}</math> and <math>QD=\frac{12}{5}</math>. So our desired ratio is <math>5:3:12</math> and <math>5+3+12=\boxed{\textbf{(E) } 20}</math>.<br />
<br />
==Solution 4 (Mass Points)==<br />
<br />
Draw line segment <math>AC</math>, and call the intersection between <math>AC</math> and <math>BD</math> point <math>K</math>. In <math>\delta ABC</math>, observe that <math>BE:EC=1:2</math> and <math>AK:KC=1:1</math>. Using mass points, find that <math>BP:PK=1:1</math>. Again utilizing <math>\delta ABC</math>, observe that <math>BF:FC=2:1</math> and <math>AK:KC=1:1</math>. Use mass points to find that <math>BQ:QK=4:1</math>. Now, draw a line segment with points <math>B</math>,<math>P</math>,<math>Q</math>, and <math>K</math> ordered from left to right. Set the values <math>BP=x</math>,<math>PK=x</math>,<math>BQ=4y</math> and <math>QK=y</math>. Setting both sides segment <math>BK</math> equal, we get <math>y= \frac{2}{5}x</math>. Plugging in and solving gives <math>QK= \frac{2}{5}x</math>, <math>PQ=\frac{3}{5}x</math>,<math>BP=x</math>. The question asks for <math>BP:PQ:QD</math>, so we add <math>2x</math> to <math>QK</math> and multiply the ratio by <math>5</math> to create integers. This creates <math>5(1:\frac{3}{5}:\frac{12}{5})= 5:3:12</math>. This sums up to <math>3+5+12=\boxed{\textbf{(E) }20}</math><br />
<br />
==Solution 5 (Easy Coord Bash)==<br />
We set coordinates for the points. Let <math>A=(0,3), B=(6,3), C=(6,0)</math> and <math>D=(0,0)</math>. Then the equation of line <math>AE</math> is <math>y = -\frac{1}{6}x + 3,</math> the equation of line <math>AF</math> is <math>y = -\frac{1}{3}x + 3,</math> and the equation of line <math>BD</math> is <math>y = \frac{1}{2}x</math>. We find that the x-coordinate of point <math>P</math> is <math>\frac 9 2</math> by solving <math> -\frac{1}{6}x + 3=\frac{1}{2}x.</math> Similarly we find that the x-coordinate of point <math>Q</math> is <math>\frac {18} 5</math> by solving <math>-\frac{1}{3}x + 3=\frac{1}{2}x.</math> It follows that <math>BP:PQ:QD=6-\frac 9 2 : \frac 9 2 - \frac {18} 5 : \frac {18} 5= \frac 3 2 : \frac 9 {10} : \frac {18} 5 = 5:3:12.</math> Hence <math>r,s,t=5,3,12</math> and <math>r+s+t=5+3+12=\boxed{\textbf{(E) } 20}.</math> ~ Solution by dolphin7<br />
<br />
==Video Solution==<br />
<br />
https://www.youtube.com/watch?v=aG9JiBMd0ag<br />
<br />
==Video Solution 2==<br />
https://youtu.be/us3e2jMjWnw<br />
<br />
~IceMatrix<br />
<br />
== Video Solution 3 ==<br />
https://youtu.be/4_x1sgcQCp4?t=3406<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2016|ab=A|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Infinity26484https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12A_Problems/Problem_15&diff=1639442017 AMC 12A Problems/Problem 152021-10-23T18:12:06Z<p>Infinity26484: /* Solution 3(More Detailed Answer of Solution 1) */</p>
<hr />
<div>==Problem==<br />
<br />
Let <math>f(x) = \sin{x} + 2\cos{x} + 3\tan{x}</math>, using radian measure for the variable <math>x</math>. In what interval does the smallest positive value of <math>x</math> for which <math>f(x) = 0</math> lie?<br />
<br />
<math> \textbf{(A)}\ (0,1) <br />
\qquad \textbf{(B)}\ (1, 2)<br />
\qquad\textbf{(C)}\ (2, 3)<br />
\qquad\textbf{(D)}\ (3, 4)<br />
\qquad\textbf{(E)}\ (4,5) </math><br />
<br />
==Solution 1==<br />
We must first get an idea of what <math>f(x)</math> looks like:<br />
<br />
Between <math>0</math> and <math>1</math>, <math>f(x)</math> starts at <math>2</math> and increases; clearly there is no zero here.<br />
<br />
Between <math>1</math> and <math>\frac{\pi}{2}</math>, <math>f(x)</math> starts at a positive number and increases to <math>\infty</math>; there is no zero here either.<br />
<br />
Between <math>\frac{\pi}{2}</math> and 3, <math>f(x)</math> starts at <math>-\infty</math> and increases to some negative number; there is no zero here either.<br />
<br />
Between <math>3</math> and <math>\pi</math>, <math>f(x)</math> starts at some negative number and increases to -2; there is no zero here either.<br />
<br />
Between <math>\pi</math> and <math>\pi+\frac{\pi}{4} < 4</math>, <math>f(x)</math> starts at -2 and increases to <math>-\frac{\sqrt2}{2} + 2\left(-\frac{\sqrt2}{2}\right) + 3\left(1\right)=3\left(1-\frac{\sqrt2}{2}\right)>0</math>. There is a zero here by the Intermediate Value Theorem. Therefore, the answer is <math>\boxed{\textbf{(D)}}</math>.<br />
<br />
==Solution 2 (Graphing)==<br />
If you quickly take a moment to sketch the graphs of the three functions, you will see that between <math>0</math> and <math>\frac{\pi}{2}</math> everything is positive, while the positive number created by the sin does not outweigh the negative by the cos and tan function. Upon further examination, it is clear that the positive the tan function creates will balance the other two functions, and thus the first solution is a little bit after <math>\pi</math>, which is around <math>3.14</math>. Hence the answer is <math>\boxed{\textbf{(D)}}</math>.<br />
<br />
Solution by roadchicken~<br />
<br />
(Not original author) Here is the graph:<br />
<asy><br />
Label f; <br />
f.p=fontsize(6); <br />
xaxis(-5,5,Ticks(f, 1.0)); <br />
yaxis(-8,8,Ticks(f, 2.0));<br />
real f(real x)<br />
{<br />
return sin(x);<br />
}<br />
draw(graph(f, -5,5));<br />
real g(real x)<br />
{<br />
return 2*cos(x);<br />
}<br />
draw(graph(g, -5,5));<br />
real h(real x)<br />
{<br />
return 3*tan(x);<br />
}<br />
draw(graph(h, -1.2,1.2));<br />
draw(graph(h, 1.94, 4.34));<br />
draw(graph(h, -4.34, -1.94));<br />
</asy><br />
<br />
==Solution 3(More Detailed Answer of Solution 1)==<br />
Denote <math>D_{f}</math> as the domain of <math>f(x).</math> Obviously <math>D_{f}=\left \{x|x\neq \frac{(2k+1)\pi}{2} \right \} .</math><br />
<br />
<math>f^{'}(x)=\cos x-2\sin x+3\sec^{2}x=\sqrt{5}\cos(x+\phi)+3\sec^{2}x > 0 </math> for all <math>x</math> in sub-intervals of <math> D_{f}.</math><br />
<br />
When <math>x\in (0,\frac{\pi}{2}), f(0)=2, \lim_{x \to \frac{\pi}{2}^{-} }f(x)=+\infty .</math> Hence <math>f(x)>0</math> in this interval.<br />
<br />
When <math>x\in (\frac{\pi}{2},\pi),\lim_{x \to \frac{\pi}{2}^{+} }f(x)=-\infty, f(\pi)=-2.</math> Hence <math>f(x)<0</math> in this interval.<br />
<br />
When <math>x\in (\pi,\frac{3\pi}{2}),f(\pi)=-2<0.</math> <br />
<br />
Notice that <math>\frac{5\pi}{4}<\frac{3.15*5}{4}<4<\frac{3\pi}{2}, f(\frac{5\pi}{4})=3-\frac{3\sqrt{2}}{2}>0 .</math> Hence there must be a root located in the interval <math>(3,4).</math> Choose <math>\boxed{\textbf{(D)}}</math>.<br />
<br />
<br />
~PythZhou<br />
<br />
==See Also==<br />
{{AMC12 box|year=2017|ab=A|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Infinity26484https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_21&diff=1417502016 AMC 10A Problems/Problem 212021-01-09T03:09:31Z<p>Infinity26484: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
Circles with centers <math>P, Q</math> and <math>R</math>, having radii <math>1, 2</math> and <math>3</math>, respectively, lie on the same side of line <math>l</math> and are tangent to <math>l</math> at <math>P', Q'</math> and <math>R'</math>, respectively, with <math>Q'</math> between <math>P'</math> and <math>R'</math>. The circle with center <math>Q</math> is externally tangent to each of the other two circles. What is the area of triangle <math>PQR</math>?<br />
<br />
<math>\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{6}/3\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{6}/2</math><br />
<br />
==Solution 1== <br />
<asy><br />
size(250);<br />
defaultpen(linewidth(0.4));<br />
//Variable Declarations<br />
pair P,Q,R,Pp,Qp,Rp;<br />
pair A,B;<br />
<br />
//Variable Definitions<br />
A=(-5, 0);<br />
B=(8, 0);<br />
P=(-2.828,1);<br />
Q=(0,2);<br />
R=(4.899,3);<br />
Pp=foot(P,A,B);<br />
Qp=foot(Q,A,B);<br />
Rp=foot(R,A,B);<br />
path PQR = P--Q--R--cycle;<br />
//Initial Diagram<br />
dot(P);<br />
dot(Q);<br />
dot(R);<br />
dot(Pp);<br />
dot(Qp);<br />
dot(Rp);<br />
draw(Circle(P, 1), linewidth(0.8));<br />
draw(Circle(Q, 2), linewidth(0.8));<br />
draw(Circle(R, 3), linewidth(0.8));<br />
draw(A--B,Arrows);<br />
label("$P$",P,N);<br />
label("$Q$",Q,N);<br />
label("$R$",R,N);<br />
label("$P'$",Pp,S);<br />
label("$Q'$",Qp,S);<br />
label("$R'$",Rp,S);<br />
label("$l$",B,E);<br />
<br />
//Added lines<br />
draw(PQR);<br />
draw(P--Pp);<br />
draw(Q--Qp);<br />
draw(R--Rp);<br />
<br />
//Angle marks<br />
draw(rightanglemark(P,Pp,B));<br />
draw(rightanglemark(Q,Qp,B));<br />
draw(rightanglemark(R,Rp,B));<br />
</asy><br />
Notice that we can find <math>[P'PQRR']</math> in two different ways: <math>[P'PQQ']+[Q'QRR']</math> and <math>[PQR]+[P'PRR']</math>, so <math>[P'PQQ']+[Q'QRR']=[PQR]+[P'PRR']</math> <br />
<math>\break</math><br />
<br />
<math>P'Q'=\sqrt{PQ^2-(QQ'-PP')^2}=\sqrt{9-1}=\sqrt{8}=2\sqrt{2}</math>. Additionally, <math>Q'R'=\sqrt{QR^2-(RR'-QQ')^2}=\sqrt{5^2-1^2}=\sqrt{24}=2\sqrt{6}</math>. Therefore, <math>[P'PQQ']=\frac{P'P+Q'Q}{2}*2\sqrt{2}=\frac{1+2}{2}*2\sqrt{2}=3\sqrt{2}</math>. Similarly, <math>[Q'QRR']=5\sqrt6</math>. We can calculate <math>[P'PRR']</math> easily because <math>P'R'=P'Q'+Q'R'=2\sqrt{2}+2\sqrt{6}</math>. <math>[P'PRR']=4\sqrt{2}+4\sqrt{6}</math>. <math>\newline</math><br />
<br />
Plugging into first equation, the two sums of areas, <math>3\sqrt{2}+5\sqrt{6}=4\sqrt{2}+4\sqrt{6}+[PQR]</math>. <math>\newline</math><br />
<br />
<math>[PQR]=\sqrt{6}-\sqrt{2}\rightarrow \fbox{D}</math>.<br />
<br />
==Solution 2== <br />
<br />
Use the [[Shoelace Theorem]].<br />
<br />
Let the center of the first circle of radius 1 be at <math>(0, 1)</math>. <br />
<br />
Draw the trapezoid <math>PQQ'P'</math> and using the Pythagorean Theorem, we get that <math>P'Q' = 2\sqrt{2}</math> so the center of the second circle of radius 2 is at <math>(2\sqrt{2}, 2)</math>.<br />
<br />
Draw the trapezoid <math>QRR'Q'</math> and using the Pythagorean Theorem, we get that <math>Q'R' = 2\sqrt{6}</math> so the center of the third circle of radius 3 is at <math>(2\sqrt{2}+2\sqrt{6}, 3)</math>.<br />
<br />
Now, we may use the Shoelace Theorem!<br />
<br />
<math>(0,1)</math><br />
<br />
<math>(2\sqrt{2}, 2)</math><br />
<br />
<math>(2\sqrt{2}+2\sqrt{6}, 3)</math><br />
<br />
<math>\frac{1}{2}|(2\sqrt{2}+4\sqrt{2}+4\sqrt{6})-(6\sqrt{2}+2\sqrt{2}+2\sqrt{6})|</math><br />
<br />
<math>= \sqrt{6}-\sqrt{2}</math> <math>\fbox{D}</math>.<br />
<br />
==Solution 3==<br />
<math>PQ = 3</math> and <math>QR = 5</math> because they are the sum of two radii. <math>QQ' - PP' = 1</math> and <math>RR' - QQ' = 1</math>, the difference of the radii. Using pythagorean theorem, we find that <math>P'Q'</math> and <math>Q'R'</math> are <math>\sqrt{8}</math> and <math>\sqrt{24}</math>, <math>P'R' = \sqrt{8} + \sqrt{24}</math>. <br />
<br />
Draw a perpendicular from <math>P</math> to line <math>RR'</math>, then we can use the Pythagorean theorem to find <math>PR</math>. <math>RR' - PP' = 2</math>. We get <cmath>PR^2 = (\sqrt{8} + \sqrt{24})^2 + 4 = 36 + 16\sqrt{3} \Rightarrow PR = \sqrt{36 + 16\sqrt{3}} = 2\sqrt{9 + 4\sqrt{3}}</cmath><br />
<br />
To make our calculations easier, let <math>\sqrt{9 + 4\sqrt{3}} = a</math>. The semi-perimeter of our triangle is <math>\frac{3 + 5 + 2a}{2} = 4 + a</math>. Symbolize the area of the triangle with <math>A</math>. Using Heron's formula, we have <cmath>A^2 = (4 + a)(4 + a - 2a)(4 + a - 3)(4 + a - 5) = (4 + a)(4 - a)(a + 1)(a - 1) = (16 - a^2)(a^2 - 1)</cmath> We can remove the outer root of a. <cmath>A^2 = (16 - 9 - 4\sqrt{3})(9 + 4\sqrt{3} - 1) = (7 - 4\sqrt{3})(8 + 4\sqrt{3}) = 8 - 4\sqrt{3} \rightarrow A = \sqrt{8 - 4\sqrt{3}}</cmath><br />
<br />
We solve the nested root. We want to turn <math>8 - 4\sqrt{3}</math> into the square of something. If we have <math>(a - b) ^ 2 = 8 - 4\sqrt{3}</math>, then we get <cmath>\begin{cases} a^2 + b^2 = 8 \\ ab = 2\sqrt{3} \end{cases}</cmath> Solving the system of equations, we get <math>a = \sqrt{6}</math> and <math>b = \sqrt{2}</math>. Alternatively, you can square all the possible solutions until you find one that is equal to <math>8 - 4\sqrt{3}</math>. <cmath>A = \sqrt{8 - 4\sqrt{3}} = \sqrt{(\sqrt{6} - \sqrt{2})^2} = \sqrt{6} - \sqrt{2} \rightarrow \fbox{D}</cmath><br />
~ZericH<br />
<br />
==Solution 4==<br />
<asy><br />
// Initial Pen Sizing<br />
size(250);<br />
defaultpen(linewidth(0.4));<br />
defaultpen(fontsize(10pt));<br />
<br />
// Variable Declarations<br />
pair P,Q,R,Pp,Qp,Rp,X,Y,Z,A,B;<br />
<br />
// Variable Definitions<br />
A=(-5, 0);<br />
B=(8, 0);<br />
P=(-2.828,1);<br />
Q=(0,2);<br />
R=(4.899,3);<br />
X=(0,1);<br />
Y=(4.899,1);<br />
Z=(4.899,2);<br />
Pp=foot(P,A,B);<br />
Qp=foot(Q,A,B);<br />
Rp=foot(R,A,B);<br />
path PQR = P--Q--R--cycle;<br />
<br />
//Initial Diagram<br />
dot(P);<br />
dot(Q);<br />
dot(R);<br />
dot(Pp);<br />
dot(Qp);<br />
dot(Rp);<br />
dot(X);<br />
dot(Y);<br />
dot(Z);<br />
draw(Circle(P, 1), linewidth(0.3));<br />
draw(Circle(Q, 2), linewidth(0.3));<br />
draw(Circle(R, 3), linewidth(0.3));<br />
draw(A--B,Arrows);<br />
label("$P$",P,N);<br />
label("$Q$",Q,N);<br />
label("$R$",R,N);<br />
label("$P'$",Pp,S);<br />
label("$Q'$",Qp,S);<br />
label("$R'$",Rp,S);<br />
label("$l$",B,E);<br />
label("$X$",X,NE);<br />
label("$Y$",Y,E);<br />
label("$Z$",Z,E);<br />
<br />
//Added lines<br />
filldraw(PQR,gray(0.8));<br />
draw(P--Pp,linetype("8 8"));<br />
draw(Q--Qp,linetype("8 8"));<br />
draw(R--Rp,linetype("8 8"));<br />
draw(P--Y,linetype("8 8"));<br />
draw(Q--Z,linetype("8 8"));<br />
<br />
//Angle marks<br />
draw(rightanglemark(R,Y,P));<br />
draw(rightanglemark(Q,X,P));<br />
draw(rightanglemark(R,Z,Q));<br />
<br />
//Length labeling<br />
label("$2\sqrt{2}$",P--X,fontsize(8pt));<br />
label("$2\sqrt{6}$",X--Y,fontsize(8pt));<br />
label("$2\sqrt{6}$",Q--Z,fontsize(8pt));<br />
label("$1$",R--Z,E,fontsize(8pt));<br />
label("$1$",Z--Y,E,fontsize(8pt));<br />
label("$3$",(-2.828,1.3)--Q,W,fontsize(8pt));<br />
label("$5$",Q--R,N,fontsize(8pt));<br />
</asy><br />
The above diagram can be achieved relatively simply using basic knowledge of the Pythagorean theorem and the fact that the radius from the center to the point of tangency is perpendicular to the tangent line. From there, observe that <math>[PQRY]</math> can be calculated in two ways: <math>[\triangle PQX] + [QZYX] + [\triangle QRZ]</math> and <math>[\triangle PRY] + [\triangle PQR]</math>. Solving, we get:<br />
<cmath><br />
\begin{align*}<br />
[\triangle PQR] &= [PQRY] - [\triangle PRY] \\ &= [\triangle PQX] + [QZYX] + [\triangle QRZ] - [\triangle PRY] \\ &= \sqrt{2} + 2\sqrt{6} + \sqrt{6}- 2\sqrt{2}-2\sqrt{6} \\ &= \boxed{\textbf{(D)} \sqrt{6}-\sqrt{2}}<br />
\end{align*}<br />
</cmath><br />
<br />
- ColtsFan10, diagram partially borrowed from Solution 1<br />
<br />
==Solution 5 (Heron’s)==<br />
<br />
We can use the Pythagorean theorem to find that the lengths are <math>3,5,2\sqrt3</math>. If we apply Heron’s, we know that it must be the sum (or difference) of two or more square roots, by instinct. This means that <math>\boxed{\textbf{(D)} \sqrt{6}-\sqrt{2}}</math> is the answer.<br />
<br />
<br />
==Solution 6 (Educated Guess)==<br />
<br />
Like Solution 1, we can use the Pythagorean theorem to find <math>P'Q'</math> and <math>Q'R'</math>, which are <math>2\sqrt2</math> and <math>2\sqrt6</math> respectively. Since the only answer choice that has <math>\sqrt2</math> and <math>\sqrt6</math> is <math>D</math>, we can make an educated guess that <math>\boxed{\textbf{(D)} \sqrt{6}-\sqrt{2}}</math> is the answer.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2016|ab=A|num-b=20|num-a=22}}<br />
{{AMC12 box|year=2016|ab=A|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Infinity26484https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_4&diff=1412252018 AIME I Problems/Problem 42021-01-01T01:13:45Z<p>Infinity26484: /* Solution 1*/</p>
<hr />
<div>==Problem 4==<br />
In <math>\triangle ABC, AB = AC = 10</math> and <math>BC = 12</math>. Point <math>D</math> lies strictly between <math>A</math> and <math>B</math> on <math>\overline{AB}</math> and point <math>E</math> lies strictly between <math>A</math> and <math>C</math> on <math>\overline{AC}</math> so that <math>AD = DE = EC</math>. Then <math>AD</math> can be expressed in the form <math>\dfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.<br />
<br />
==Solution 1==<br />
<math>\cos(A) = \frac{5^2+5^2-6^2}{2*5*5} = \frac{7}{25}</math>. Let <math>M</math> be midpoint of <math>AE</math>, then <math>\frac{7}{25} = \frac{10-x}{2x} \iff x =\frac{250}{39}</math>. So, our answer is <math>250+39=\boxed{289}</math>.<br />
<br />
==Solution 1 (No Trig)==<br />
<center><br />
<asy><br />
import cse5;<br />
unitsize(10mm);<br />
pathpen=black;<br />
dotfactor=3;<br />
<br />
pair B = (0,0), A = (6,8), C = (12,0), D = (2.154,2.872), E = (8.153, 5.128), F=(7.68,5.76), G=(7.077,6.564), H=(5.6,4.3), I=(4.5,6), J=(10,2.66);<br />
pair[] dotted = {A,B,C,D,E,F,G};<br />
<br />
D(A--B);<br />
D(C--B);<br />
D(A--C);<br />
D(D--E);<br />
pathpen=dashed;<br />
D(B--F);<br />
D(D--G);<br />
<br />
dot(dotted);<br />
label("$A$",A,N);<br />
label("$B$",B,SW);<br />
label("$C$",C,SE);<br />
label("$D$",D,NW);<br />
label("$E$",E,NE);<br />
label("$F$",F,NE);<br />
label("$G$",G,NE);<br />
label("$x$",H,NW);<br />
label("$x$",I,NW);<br />
label("$x$",J,NE);<br />
</asy><br />
</center><br />
<br />
We draw the altitude from <math>B</math> to <math>\overline{AC}</math> to get point <math>F</math>. We notice that the triangle's height from <math>A</math> to <math>\overline{BC}</math> is 8 because it is a <math>3-4-5</math> Right Triangle. To find the length of <math>\overline{BF}</math>, we let <math>h</math> represent <math>\overline{BF}</math> and set up an equation by finding two ways to express the area. The equation is <math>(8)(12)=(10)(h)</math>, which leaves us with <math>h=9.6</math>. We then solve for the length <math>\overline{AF}</math>, which is done through pythagorean theorm and get <math>\overline{AF}</math> = <math>2.8</math>. We can now see that <math>\triangle ABF</math> is a <math>7-24-25</math> Right Triangle. Thus, we set <math>\overline{AG}</math> as <math>5-</math><math>\tfrac{x}{2}</math>, and yield that <math>\overline{AD}</math> <math>=</math> <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Now, we can see <math>x</math> = <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Solving this equation, we yield <math>39x=250</math>, or <math>x=</math> <math>\tfrac{250}{39}</math>. Thus, our final answer is <math>250+39=\boxed{289}</math>.<br />
~bluebacon008<br />
<br />
==Solution 2 (Easy Similar Triangles)==<br />
We start by adding a few points to the diagram. Call <math>F</math> the midpoint of <math>AE</math>, and <math>G</math> the midpoint of <math>BC</math>. (Note that <math>DF</math> and <math>AG</math> are altitudes of their respective triangles). We also call <math>\angle BAC = \theta</math>. Since triangle <math>ADE</math> is isosceles, <math>\angle AED = \theta</math>, and <math>\angle ADF = \angle EDF = 90 - \theta</math>. Since <math>\angle DEA = \theta</math>, <math>\angle DEC = 180 - \theta</math> and <math>\angle EDC = \angle ECD = \frac{\theta}{2}</math>. Since <math>FDC</math> is a right triangle, <math>\angle FDC = 180 - 90 - \frac{\theta}{2} = \frac{180-m}{2}</math>. <br />
<br />
Since <math>\angle BAG = \frac{\theta}{2}</math> and <math>\angle ABG = \frac{180-m}{2}</math>, triangles <math>ABG</math> and <math>CDF</math> are similar by Angle-Angle similarity. Using similar triangle ratios, we have <math>\frac{AG}{BG} = \frac{CF}{DF}</math>. <math>AG = 8</math> and <math>BG = 6</math> because there are <math>2</math> <math>6-8-10</math> triangles in the problem. Call <math>AD = x</math>. Then <math>CE = x</math>, <math>AE = 10-x</math>, and <math>EF = \frac{10-x}{2}</math>. Thus <math>CF = x + \frac{10-x}{2}</math>. Our ratio now becomes <math>\frac{8}{6} = \frac{x+ \frac{10-x}{2}}{DF}</math>. Solving for <math>DF</math> gives us <math>DF = \frac{30+3x}{8}</math>. Since <math>DF</math> is a height of the triangle <math>ADE</math>, <math>FE^2 + DF^2 = x^2</math>, or <math>DF = \sqrt{x^2 - (\frac{10-x}{2})^2}</math>. Solving the equation <math>\frac{30+3x}{8} = \sqrt{x^2 - (\frac{10-x}{2})^2}</math> gives us <math>x = \frac{250}{39}</math>, so our answer is <math>250+39 = \boxed{289}</math>.<br />
<br />
==Solution 3 (Algebra w/ Law of Cosines)==<br />
As in the diagram, let <math>DE = x</math>. Consider point <math>G</math> on the diagram shown above. Our goal is to be able to perform Pythagorean Theorem on <math>DG, GC</math>, and <math>DC</math>. Let <math>GE = 10-x</math>. Therefore, it is trivial to see that <math>GC^2 = \Big(x + \frac{10-x}{2}\Big)^2</math> (leave everything squared so that it cancels nicely at the end). By Pythagorean Theorem on Triangle <math>DGE</math>, we know that <math>DG^2 = x^2 - \Big(\frac{10-x}{2}\Big)^2</math>. Finally, we apply Law of Cosines on Triangle <math>DBC</math>. We know that <math>\cos(\angle DBC) = \frac{3}{5}</math>. Therefore, we get that <math>DC^2 = (10-x)^2 + 12^2 - 2(12)(10-x)\frac{3}{5}</math>. We can now do our final calculation:<br />
<cmath><br />
DG^2 + GC^2 = DC^2 \implies x^2 - \Big(\frac{10-x}{2}\Big)^2 + \Big(x + \frac{10-x}{2}\Big)^2 = (10-x)^2 + 12^2 - 2(12)(10-x)\frac{3}{5}<br />
</cmath><br />
After some quick cleaning up, we get<br />
<cmath><br />
30x = \frac{72}{5} + 100 \implies x = \frac{250}{39}<br />
</cmath><br />
Therefore, our answer is <math>250+39=\boxed{289}</math>.<br />
<br />
~awesome1st<br />
<br />
<br />
==Solution 4 (Coordinates)==<br />
Let <math>B = (0, 0)</math>, <math>C = (12, 0)</math>, and <math>A = (6, 8)</math>. Then, let <math>x</math> be in the interval <math>0<x<2</math> and parametrically define <math>D</math> and <math>E</math> as <math>(6-3x, 8-4x)</math> and <math>(12-3x, 4x)</math> respectively. Note that <math>AD = 5x</math>, so <math>DE = 5x</math>. This means that<br />
<cmath>\begin{align*}<br />
\sqrt{36+(8x-8)^2} &= 5x\\<br />
36+(8x-8)^2 &= 25x^2\\<br />
64x^2-128x+100 &= 25x^2\\<br />
39x^2-128x+100 &= 0\\<br />
x &= \dfrac{128\pm\sqrt{16384-15600}}{78}\\<br />
x &= \dfrac{100}{78}, 2\\<br />
\end{align*}</cmath><br />
However, since <math>2</math> is extraneous by definition, <math>x=\dfrac{50}{39}\implies AD = \dfrac{250}{39}\implies\boxed{289}</math> ~ mathwiz0803<br />
<br />
==Solution 5 (Law of Cosines)==<br />
As shown in the diagram, let <math>x</math> denote <math>\overline{AD}</math>. Let us denote the foot of the altitude of <math>A</math> to <math>\overline{BC}</math> as <math>F</math>. Note that <math>\overline{AE}</math> can be expressed as <math>10-x</math> and <math>\triangle{ABF}</math> is a <math>6-8-10</math> triangle . Therefore, <math>\sin(\angle{BAF})=\frac{3}{5}</math> and <math>\cos(\angle{BAF})=\frac{4}{5}</math>. Before we can proceed with the Law of Cosines, we must determine <math>\cos(\angle{BAC})=\cos(2\cdot \angle{BAF})=\cos^2(\angle{BAF})-\sin^2(\angle{BAF})=\frac{7}{25}</math>. Using LOC, we can write the following statement:<br />
<cmath>(\overline{DE})^2=(\overline{AD})^2+\overline{AE}^2-2(\overline{AD})(\overline{AE})\cos(\angle{BAC})\implies</cmath><br />
<cmath>x^2=x^2+(10-x)^2-2(x)(10-x)\left(\frac{7}{25}\right)\implies</cmath><br />
<cmath>0=(10-x)^2-\frac{14x}{25}(10-x)\implies</cmath><br />
<cmath>0=10-x-\frac{14x}{25}\implies</cmath><br />
<cmath>10=\frac{39x}{25}\implies x=\frac{250}{39}</cmath><br />
Thus, the desired answer is <math>\boxed{289}</math> ~ blitzkrieg21<br />
<br />
==Solution 6==<br />
In isosceles triangle, draw the altitude from <math>D</math> onto <math>\overline{AD}</math>. Let the point of intersection be <math>X</math>. Clearly, <math>AE=10-AD</math>, and hence <math>AX=\frac{10-AD}{2}</math>.<br />
<br />
Now, we recognise that the perpendicular from <math>A</math> onto <math>\overline{AD}</math> gives us two <math>6</math>-<math>8</math>-<math>10</math> triangles. So, we calculate <math>\sin \angle ABC=\frac{4}{5}</math> and <math>\cos \angle ABC=\frac{3}{5}</math><br />
<br />
<math>\angle BAC = 180-2\cdot\angle ABC</math>. And hence,<br />
<br />
<math>\cos \angle BAC = \cos \angle (180-2\cdot\angle ABC)<br />
= -\cos (2\cdot\angle ABC)<br />
= \sin^2 \angle ABC - \cos^2 \angle ABC<br />
= 2\cos^2 \angle ABC - 1<br />
= \frac{32}{25}-\frac{25}{25}=\frac{7}{25}</math><br />
<br />
Inspecting <math>\triangle ADX</math> gives us <math>\cos \angle BAC = \frac{\frac{10-x}{2}}{x} = \frac{10-x}{2x}</math><br />
Solving the equation <math>\frac{10-x}{2x}=\frac{7}{25}</math> gives <math>x= \frac{250}{39} \implies\boxed{289}</math><br />
<br />
~novus677<br />
<br />
==Solution 7 (Fastest via Law of Cosines)==<br />
We can have 2 Law of Cosines applied on <math>\angle A</math> (one from <math>\triangle ADE</math> and one from <math>\triangle ABC</math>),<br />
<br />
<math>x^2=x^2+(10-x)^2-2(x)(10-x)\cdot \cos{A}</math> and <math>12^2=10^2+10^2-2(10)(10)\cdot \cos{A}</math><br />
<br />
Solving for <math>\cos{A}</math> in both equations, we get<br />
<br />
<math>\cos{A} = \frac{(10-x)^2}{(2)(10-x)(x)}</math> and <math>cos A = \frac{7}{25} \implies \frac{(10-x)^2}{(2)(10-x)(x)} = \frac{7}{25} \implies x = \frac{250}{39}</math>, so the answer is <math>\boxed {289}</math> <br />
<br />
'''-RootThreeOverTwo'''<br />
<br />
==Solution 8 (Easiest way- Coordinates without bash)==<br />
Let <math>B=(0, 0)</math>, and <math>C=(12, 0)</math>. From there, we know that <math>A=(6, 8)</math>, so line <math>AB</math> is <math>y=\frac{4}{3}x</math>. Hence, <math>D=(a, \frac{4}{3}a)</math> for some <math>a</math>, and <math>BD=\frac{5}{3}a</math> so <math>AD=10-\frac{5}{3}a</math>. Now, notice that by symmetry, <math>E=(6+a, 8-\frac{4}{3}a)</math>, so <math>ED^2=6^2+(8-\frac{8}{3}a)^2</math>. Because <math>AD=ED</math>, we now have <math>(10-\frac{5}{3})^2=6^2+(8-\frac{8}{3}a)^2</math>, which simplifies to <math>\frac{64}{9}a^2-\frac{128}{3}a+100=\frac{25}{9}a^2-\frac{100}{3}a+100</math>, so <math>\frac{39}{9}a=\frac{13}{3}a=\frac{28}{3}</math>, and <math>a=\frac{28}{13}</math>.<br />
It follows that <math>AD=10-\frac{5}{3}\times\frac{28}{13}=10-\frac{140}{39}=\frac{390-140}{39}=\frac{250}{39}</math>, and our answer is <math>250+39=\boxed{289}</math>.<br />
<br />
-Stormersyle<br />
<br />
== Solution 9 Even Faster Law of Cosines(1 variable equation)==<br />
<br />
Doing law of cosines we know that <math>\cos A</math> is <math>\frac{7}{25}.</math>* Dropping the perpendicular from <math>D</math> to <math>AE</math> we get that <cmath>\frac{10-x}{2}=\frac{7x}{25}.</cmath> <br />
Solving for <math>x</math> we get <math>\frac{250}{39}</math> so our answer is <math>289</math>.<br />
<br />
-harsha12345<br />
<br />
* It is good to remember that doubling the smallest angle of a 3-4-5 triangle gives the larger (not right) angle in a 7-24-25 triangle.<br />
<br />
== Solution 10 (Law of Sines)==<br />
<br />
Let's label <math>\angle A = \theta</math> and <math>\angle ECD = \alpha</math>. Using isosceles triangle properties and the triangle angle sum equation, we get <cmath>180-(180-2\theta+\alpha) + \frac{180-\theta}{2} + \left(\frac{180-\theta}{2} - \alpha\right) = 180.</cmath> Solving, we find <math>\theta = 2 \alpha</math>. <br />
<br />
<br />
Relabelling our triangle, we get <math>\angle ABC = 90 - \alpha</math>. Dropping an altitude from <math>A</math> to <math>BC</math> and using the Pythagorean theorem, we find <math>[ABC] = 48</math>. Using the sine area formula, we see <math>\frac12 \cdot 10 \cdot 12 \cdot \sin(90-\alpha) = 48</math>. Plugging in our sine angle cofunction identity, <math>\sin(90-\alpha) = \cos(\alpha)</math>, we get <math>\alpha = \cos{^{-1}}{\frac45}</math>. <br />
<br />
<br />
Now, using the Law of Sines on <math>\triangle ADE</math>, we get <cmath>\frac{\sin{2\alpha}}{\frac{p}{q}} = \frac{\sin{(180-4\alpha)}}{10-\frac{p}{q}}.</cmath> After applying numerous trigonometric and algebraic tricks, identities, and simplifications, such as <math>\sin{(180-4\alpha)}=\sin{4\alpha}</math> and <math>\sin{\left(\cos{^{-1}}{\frac45}\right)} = \frac35</math>, we find <math>\frac{p}{q} = \frac{10\sin{2\alpha}}{\sin{4\alpha}+\sin{2\alpha}} = \frac{250}{39}</math>. <br />
<br />
<br />
<br />
Therefore, our answer is <math>250 + 39 = \boxed{289}</math>.<br />
<br />
<br />
~Tiblis<br />
<br />
== Solution 11 (Trigonometry)==<br />
We start by labelling a few angles (all of them in degrees). Let <math>\angle{BAC}=2\alpha = \angle{AED}, \angle{EDC}=\angle{ECD}=\alpha, \angle{DEC}=180-2\alpha, \angle{BDC}=3\alpha, \angle{DCB}=90-2\alpha, \angle{DBC}=90-\alpha</math>. Also let <math>AD=a</math>. By sine rule in <math>\triangle{ADE},</math> we get <math>\frac{a}{\sin{2\alpha}}=\frac{10-a}{\sin{4\alpha}} \implies \cos{2\alpha}=\frac{5}{a}-\frac{1}{2}</math><br />
Using sine rule in <math>\triangle{ABC}</math>, we get <math>\sin{\alpha}=\frac{3}{5}</math>. Hence we get <math>\cos{2\alpha}=1-2\sin^2{\alpha}=1-\frac{18}{25}=\frac{7}{25}</math>. Hence <math>\frac{5}{a}=\frac{1}{2}+\frac{7}{25}=\frac{39}{50} \implies a=\frac{250}{39}</math>. Therefore, our answer is <math>\boxed{289}</math><br />
<br />
Alternatively, use sine rule in <math>\triangle{BDC}</math>. (It’s easier)<br />
<br />
~Prabh1512<br />
<br />
==Video Solution==<br />
<br />
https://www.youtube.com/watch?v=iE8paW_ICxw<br />
<br />
<br />
https://youtu.be/dI6uZ67Ae2s ~yofro<br />
<br />
==See Also==<br />
{{AIME box|year=2018|n=I|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Infinity26484https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_2&diff=1412132018 AIME I Problems/Problem 22021-01-01T00:22:38Z<p>Infinity26484: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
<br />
The number <math>n</math> can be written in base <math>14</math> as <math>\underline{a}\text{ }\underline{b}\text{ }\underline{c}</math>, can be written in base <math>15</math> as <math>\underline{a}\text{ }\underline{c}\text{ }\underline{b}</math>, and can be written in base <math>6</math> as <math>\underline{a}\text{ }\underline{c}\text{ }\underline{a}\text{ }\underline{c}\text{ }</math>, where <math>a > 0</math>. Find the base-<math>10</math> representation of <math>n</math>.<br />
==Solution 1==<br />
<br />
We have these equations:<br />
<math>196a+14b+c=225a+15c+b=222a+37c</math>.<br />
Taking the last two we get <math>3a+b=22c</math>. Because <math>c \neq 0</math> otherwise <math>a \ngtr 0</math>, and <math>a \leq 5</math>, <math>c=1</math>.<br />
<br />
Then we know <math>3a+b=22</math>.<br />
Taking the first two equations we see that <math>29a+14c=13b</math>. Combining the two gives <math>a=4, b=10, c=1</math>. Then we see that <math>222 \times 4+37 \times1=\boxed{925}</math>.<br />
<br />
==Solution 2==<br />
<br />
We know that <math>196a+14b+c=225a+15c+b=222a+37c</math>. Combining the first and third equations give that <math>196a+14b+c=222a+37c</math>, or <cmath>7b=13a+18c</cmath><br />
The second and third gives <math>222a+37c=225a+15c+b</math>, or <cmath>22c-3a=b </cmath><cmath> 154c-21a=7b=13a+18c </cmath><cmath> 4c=a</cmath><br />
We can have <math>a=4,8,12</math>, but only <math>a=4</math> falls within the possible digits of base <math>6</math>. Thus <math>a=4</math>, <math>c=1</math>, and thus you can find <math>b</math> which equals <math>10</math>. Thus, our answer is <math>4\cdot225+1\cdot15+10=\boxed{925}</math>.<br />
<br />
==Video Solution==<br />
<br />
https://www.youtube.com/watch?v=WVtbD8x9fCM<br />
~Shreyas S<br />
<br />
==See Also==<br />
{{AIME box|year=2018|n=I|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Infinity26484https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_2&diff=1412122018 AIME I Problems/Problem 22021-01-01T00:21:57Z<p>Infinity26484: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
<br />
The number <math>n</math> can be written in base <math>14</math> as <math>\underline{a}\text{ }\underline{b}\text{ }\underline{c}</math>, can be written in base <math>15</math> as <math>\underline{a}\text{ }\underline{c}\text{ }\underline{b}</math>, and can be written in base <math>6</math> as <math>\underline{a}\text{ }\underline{c}\text{ }\underline{a}\text{ }\underline{c}\text{ }</math>, where <math>a > 0</math>. Find the base-<math>10</math> representation of <math>n</math>.<br />
==Solution 1==<br />
<br />
We have these equations:<br />
<math>196a+14b+c=225a+15c+b=222a+37c</math>.<br />
Taking the last two we get <math>3a+b=22c</math>. Because <math>c \neq 0</math> otherwise <math>a \ngtr 0</math>, and <math>a \leq 5</math>, <math>c=1</math>.<br />
<br />
Then we know <math>3a+b=22</math>.<br />
Taking the first two equations we see that <math>29a+14c=13b</math>. Combining the two gives <math>a=4, c=10</math>. Then we see that <math>222 \times 4+37 \times1=\boxed{925}</math>.<br />
<br />
==Solution 2==<br />
<br />
We know that <math>196a+14b+c=225a+15c+b=222a+37c</math>. Combining the first and third equations give that <math>196a+14b+c=222a+37c</math>, or <cmath>7b=13a+18c</cmath><br />
The second and third gives <math>222a+37c=225a+15c+b</math>, or <cmath>22c-3a=b </cmath><cmath> 154c-21a=7b=13a+18c </cmath><cmath> 4c=a</cmath><br />
We can have <math>a=4,8,12</math>, but only <math>a=4</math> falls within the possible digits of base <math>6</math>. Thus <math>a=4</math>, <math>c=1</math>, and thus you can find <math>b</math> which equals <math>10</math>. Thus, our answer is <math>4\cdot225+1\cdot15+10=\boxed{925}</math>.<br />
<br />
==Video Solution==<br />
<br />
https://www.youtube.com/watch?v=WVtbD8x9fCM<br />
~Shreyas S<br />
<br />
==See Also==<br />
{{AIME box|year=2018|n=I|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Infinity26484https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_22&diff=1411002017 AMC 10B Problems/Problem 222020-12-31T00:35:10Z<p>Infinity26484: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
The diameter <math>\overline{AB}</math> of a circle of radius <math>2</math> is extended to a point <math>D</math> outside the circle so that <math>BD=3</math>. Point <math>E</math> is chosen so that <math>ED=5</math> and line <math>ED</math> is perpendicular to line <math>AD</math>. Segment <math>\overline{AE}</math> intersects the circle at a point <math>C</math> between <math>A</math> and <math>E</math>. What is the area of <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}</math><br />
<br />
==Solutions==<br />
<center><asy><br />
size(10cm);<br />
pair A, B, C, D, E, O;<br />
A = (-2,0);<br />
B = (2,0);<br />
C = (2*cos(1.24),2*sin(1.24));<br />
D = (5,0);<br />
E = (5,5);<br />
O = (A+B)/2;<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(D);<br />
dot(E);<br />
dot(O);<br />
draw(Circle((A+B)/2,2));<br />
draw(A--D--E--C--A);<br />
draw(C--B);<br />
draw(rightanglemark(A,C,B,5));<br />
draw(rightanglemark(A,D,E,5));<br />
label("$A$",A,W);<br />
label("$B$",B,SE);<br />
label("$D$",D,SE);<br />
label("$E$",E,NE);<br />
label("$C$",C,N);<br />
label("$2$",(O+B)/2,S);<br />
label("$3$",(B+D)/2,S);<br />
label("$5$",(D+E)/2,NE);<br />
</asy></center><br />
===Solution 1===<br />
Notice that <math>ADE</math> and <math>ABC</math> are right triangles. Then <math>AE = \sqrt{7^2+5^2} = \sqrt{74}</math>. <math>\sin{DAE} = \frac{5}{\sqrt{74}} = \sin{BAE} = \sin{BAC} = \frac{BC}{4}</math>, so <math>BC = \frac{20}{\sqrt{74}}</math>. We also find that <math>AC = \frac{28}{\sqrt{74}}</math>, and thus the area of <math>ABC</math> is <math>\frac{\frac{20}{\sqrt{74}}\cdot\frac{28}{\sqrt{74}}}{2} = \frac{\frac{560}{74}}{2} = \boxed{\textbf{(D) } \frac{140}{37}}</math>.<br />
<br />
===Solution 2===<br />
We note that <math>\triangle ACB \sim \triangle ADE</math> by <math>AA</math> similarity. Also, since the area of <math>\triangle ADE = \frac{7 \cdot 5}2 = \frac{35}2</math> and <math>AE = \sqrt{74}</math>, <math>\frac{[ABC]}{[ADE]} = \frac{[ABC]}{\frac{35}2} = \left(\frac{4}{\sqrt{74}}\right)^2</math>, so the area of <math>\triangle ABC = \boxed{\textbf{(D) } \frac{140}{37}}</math>.<br />
<br />
===Solution 3===<br />
As stated before, note that <math>\triangle ACB</math> is similar to <math>\triangle ADE</math>. By similarity, we note that <math>\frac{\overline{AC}}{\overline{BC}}</math> is equivalent to <math>\frac{7}{5}</math>. We set <math>\overline{AC}</math> to <math>7x</math> and <math>\overline{BC}</math> to <math>5x</math>. By the Pythagorean Theorem, <math>(7x)^2+(5x)^2 = 4^2</math>. Combining, <math>49x^2+25x^2=16</math>. We can add and divide to get <math>x^2=\frac{8}{37}</math>. We square root and rearrange to get <math>x=\frac{2\sqrt{74}}{37}</math>. We know that the legs of the triangle are <math>7x</math> and <math>5x</math>. Multiplying <math>x</math> by <math>7</math> and <math>5</math> eventually gives us <math>\frac {14\sqrt{74}}{37}</math> <math>\frac {10\sqrt{74}}{37}</math>. We divide this by <math>2</math>, since <math>\frac{1}{2}bh</math> is the formula for a triangle. This gives us <math>\boxed{\textbf{(D) } \frac{140}{37}}</math>.<br />
<br />
===Solution 4===<br />
Let's call the center of the circle that segment <math>AB</math> is the diameter of, <math>O</math>. Note that <math>\triangle ODE</math> is an isosceles right triangle. Solving for side <math>OE</math>, using the Pythagorean theorem, we find it to be <math>5\sqrt{2}</math>. Calling the point where segment <math>OE</math> intersects circle <math>O</math>, the point <math>I</math>, segment <math>IE</math> would be <math>5\sqrt{2}-2</math>. Also, noting that <math>\triangle ADE</math> is a right triangle, we solve for side <math>AE</math>, using the Pythagorean Theorem, and get <math>\sqrt{74}</math>. Using Power of Point on point <math>E</math>, we can solve for <math>CE</math>. We can subtract <math>CE</math> from <math>AE</math> to find <math>AC</math> and then solve for <math>CB</math> using Pythagorean theorem once more.<br />
<br />
<math>(AE)(CE)</math> = (Diameter of circle <math>O</math> + <math>IE</math>)<math>(IE)</math> <math>\rightarrow</math> <math>{\sqrt{74}}(CE)</math> = <math>(5\sqrt{2}+2)(5\sqrt{2}-2)</math> <math>\Rightarrow</math> <math>CE</math> = <math>\frac{23\sqrt{74}}{37}</math><br />
<br />
<math>AC = AE - CE</math> <math>\rightarrow</math> <math>AC</math> = <math>{\sqrt74}</math> - <math>\frac{23\sqrt{74}}{37}</math> <math>\Rightarrow</math> <math>AC</math> = <math>\frac{14\sqrt{74}}{37}</math><br />
<br />
Now to solve for <math>CB</math>:<br />
<br />
<math>AB^2</math> - <math>AC^2</math> = <math>CB^2</math> <math>\rightarrow</math> <math>4^2</math> + <math>\frac{14\sqrt{74}}{37}^2</math> = <math>CB^2</math> <math>\Rightarrow</math> <math>CB</math> = <math>\frac{10\sqrt{74}}{37}</math><br />
<br />
Note that <math>\triangle ABC</math> is a right triangle because the hypotenuse is the diameter of the circle. Solving for area using the bases <math>AC</math> and <math>BC</math>, we get the area of triangle <math>ABC</math> to be <math>\boxed{\textbf{(D) } \frac{140}{37}}</math>.<br />
== Coordinate Geo==<br />
<br />
Drawing the picture, we realize that the equation for the line from A to E is <math>y=\frac{5x}{7}</math>, and the equation for the circle is <math>(x-2)^2+y^2=4</math><br />
plugging in <math>\frac{5x}{7}</math> for y we get <math>x(74x-196)=0</math> so <math>x=\frac{98}{37}</math>, that means <math>y = \frac{98}{37} \cdot \frac{5}{7} = \frac{70}{37}</math><br />
<br />
the height is <math>\frac{70}{37}</math> and the base is <math>4</math>, so the area is <math>\boxed{\textbf{(D) } \frac{140}{37}}</math><br />
<br />
-harsha12345<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Infinity26484https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_1&diff=1406682019 AIME II Problems/Problem 12020-12-26T22:08:47Z<p>Infinity26484: /* Solution */</p>
<hr />
<div>==Problem==<br />
Two different points, <math>C</math> and <math>D</math>, lie on the same side of line <math>AB</math> so that <math>\triangle ABC</math> and <math>\triangle BAD</math> are congruent with <math>AB = 9</math>, <math>BC=AD=10</math>, and <math>CA=DB=17</math>. The intersection of these two triangular regions has area <math>\tfrac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
==Solution==<br />
<asy><br />
unitsize(10);<br />
pair A = (0,0);<br />
pair B = (9,0);<br />
pair C = (15,8);<br />
pair D = (-6,8);<br />
pair E = (-6,0);<br />
draw(A--B--C--cycle);<br />
draw(B--D--A);<br />
label("$A$",A,dir(-120));<br />
label("$B$",B,dir(-60));<br />
label("$C$",C,dir(60));<br />
label("$D$",D,dir(120));<br />
label("$E$",E,dir(-135));<br />
label("$9$",(A+B)/2,dir(-90));<br />
label("$10$",(D+A)/2,dir(-150));<br />
label("$10$",(C+B)/2,dir(-30));<br />
label("$17$",(D+B)/2,dir(60));<br />
label("$17$",(A+C)/2,dir(120));<br />
<br />
draw(D--E--A,dotted);<br />
label("$8$",(D+E)/2,dir(180));<br />
label("$6$",(A+E)/2,dir(-90));<br />
</asy><br />
- Diagram by Brendanb4321<br />
<br />
<br />
Extend <math>AB</math> to form a right triangle with legs <math>6</math> and <math>8</math> such that <math>AD</math> is the hypotenuse and connect the points <math>CD</math> so <br />
that you have a rectangle. (We know that <math>\triangle ADE</math> is a <math>6-8-10</math>, since <math>\triangle DEB</math> is an <math>8-15-17</math>.) The base <math>CD</math> of the rectangle will be <math>9+6+6=21</math>. Now, let <math>O</math> be the intersection of <math>BD</math> and <math>AC</math>. This means that <math>\triangle ABO</math> and <math>\triangle DCO</math> are with ratio <math>\frac{21}{9}=\frac73</math>. Set up a proportion, knowing that the two heights add up to 8. We will let <math>y</math> be the height from <math>O</math> to <math>DC</math>, and <math>x</math> be the height of <math>\triangle ABO</math>.<br />
<cmath>\frac{7}{3}=\frac{y}{x}</cmath><br />
<cmath>\frac{7}{3}=\frac{8-x}{x}</cmath><br />
<cmath>7x=24-3x</cmath><br />
<cmath>10x=24</cmath><br />
<cmath>x=\frac{12}{5}</cmath><br />
<br />
This means that the area is <math>A=\tfrac{1}{2}(9)(\tfrac{12}{5})=\tfrac{54}{5}</math>. This gets us <math>54+5=\boxed{059}.</math><br />
<br />
-Solution by the Math Wizard, Number Magician of the Second Order, Head of the Council of the Geometers<br />
<br />
==Solution 2==<br />
<br />
Using the diagram in Solution 1, let <math>E</math> be the intersection of <math>BD</math> and <math>AC</math>. We can see that angle <math>C</math> is in both <br />
<math>\triangle BCE</math> and <math>\triangle ABC</math>. Since <math>\triangle BCE</math> and <math>\triangle ADE</math> are congruent by AAS, we can then state <math>AE=BE</math> and <math>DE=CE</math>. It follows that <math>BE=AE</math> and <math>CE=17-BE</math>. We can now state that the area of <math>\triangle ABE</math> is the area of <math>\triangle ABC-</math> the area of <math>\triangle BCE</math>. Using Heron's formula, we compute the area of <math>\triangle ABC=36</math>. Using the Law of Cosines on angle <math>C</math>, we obtain <br />
<br />
<cmath>9^2=17^2+10^2-2(17)(10)cosC</cmath><br />
<cmath>-308=-340cosC</cmath><br />
<cmath>cosC=\frac{308}{340}</cmath><br />
(For convenience, we're not going to simplify.) <br />
<br />
Applying the Law of Cosines on <math>\triangle BCE</math> yields <br />
<cmath>BE^2=10^2+(17-BE)^2-2(10)(17-BE)cosC</cmath><br />
<cmath>BE^2=389-34BE+BE^2-20(17-BE)(\frac{308}{340})</cmath><br />
<cmath>0=389-34BE-(340-20BE)(\frac{308}{340})</cmath><br />
<cmath>0=389-34BE+\frac{308BE}{17}</cmath><br />
<cmath>0=81-\frac{270BE}{17}</cmath><br />
<cmath>81=\frac{270BE}{17}</cmath><br />
<cmath>BE=\frac{51}{10}</cmath> <br />
This means <math>CE=17-BE=17-\frac{51}{10}=\frac{119}{10}</math>. Next, apply Heron's formula to get the area of <math>\triangle BCE</math>, which equals <math>\frac{126}{5}</math> after simplifying. Subtracting the area of <math>\triangle BCE</math> from the area of <math>\triangle ABC</math> yields the area of <math>\triangle ABE</math>, which is <math>\frac{54}{5}</math>, giving us our answer, which is <math>54+5=\boxed{059}.</math><br />
-Solution by flobszemathguy<br />
<br />
==Solution 3 (Very quick)==<br />
<asy><br />
unitsize(10);<br />
pair A = (0,0);<br />
pair B = (9,0);<br />
pair C = (15,8);<br />
pair D = (-6,8);<br />
draw(A--B--C--cycle);<br />
draw(B--D--A);<br />
label("$A$",A,dir(-120));<br />
label("$B$",B,dir(-60));<br />
label("$C$",C,dir(60));<br />
label("$D$",D,dir(120));<br />
label("$9$",(A+B)/2,dir(-90));<br />
label("$10$",(D+A)/2,dir(-150));<br />
label("$10$",(C+B)/2,dir(-30));<br />
label("$17$",(D+B)/2,dir(60));<br />
label("$17$",(A+C)/2,dir(120));<br />
<br />
draw(D--(-6,0)--A,dotted);<br />
label("$8$",(D+(-6,0))/2,dir(180));<br />
label("$6$",(A+(-6,0))/2,dir(-90));<br />
<br />
draw((4.5,0)--(4.5,2.4),dotted);<br />
label("$h$", (4.5,1.2), dir(180));<br />
label("$4.5$", (6,0), dir(90));<br />
<br />
</asy><br />
- Diagram by Brendanb4321 extended by Duoquinquagintillion<br />
<br />
Begin with the first step of solution 1, seeing <math>AD</math> is the hypotenuse of a <math>6-8-10</math> triangle and calling the intersection of <math>DB</math> and <math>AC</math> point <math>E</math>. Next, notice <math>DB</math> is the hypotenuse of an <math>8-15-17</math> triangle. Drop an altitude from <math>E</math> with length <math>h</math>, so the other leg of the new triangle formed has length <math>4.5</math>. Notice we have formed similar triangles, and we can solve for <math>h</math>.<br />
<br />
<cmath>\frac{h}{4.5} = \frac{8}{15}</cmath><br />
<cmath>h = \frac{36}{15} = \frac{12}{5}</cmath><br />
<br />
So <math>\triangle ABE</math> has area <cmath>\frac{ \frac{12}{5} \cdot 9}{2} = \frac{54}{5}</cmath><br />
And <math>54+5=\boxed{059}.</math><br />
- Solution by Duoquinquagintillion<br />
<br />
== Solution 4 ==<br />
Let <math>a = \angle{CAB}</math>. By Law of Cosines,<br />
<cmath>\cos a = \frac{17^2+9^2-10^2}{2*9*17} = \frac{15}{17}</cmath><br />
<cmath>\sin a = \sqrt{1-\cos^2 a} = \frac{8}{17}</cmath><br />
<cmath>\tan a = \frac{8}{15}</cmath><br />
<cmath>A = \frac{1}{2}* 9*\frac{9}{2}\tan a = \frac{54}{5}</cmath><br />
And <math>54+5=\boxed{059}.</math><br />
<br />
- by Mathdummy<br />
<br />
== Solution 5 ==<br />
Because <math>AD = BC</math> and <math>\angle BAD = \angle ABC</math>, quadrilateral <math>ABCD</math> is cyclic. So, Ptolemy's theorem tells us that<br />
<cmath>AB \cdot CD + BC \cdot AD = AC \cdot BD \implies 9 \cdot CD + 10^2 = 17^2 \implies CD = 21.</cmath><br />
<br />
From here, there are many ways to finish which have been listed above. If we let <math>AB \cap CD = P</math>, then <br />
<cmath>\triangle APB \sim \triangle CPD \implies \frac{AP}{AB} = \frac{CP}{CD} \implies \frac{AP}{9} = \frac{17-AP}{21} \implies AP = 5.1.</cmath><br />
<br />
Using Heron's formula on <math>\triangle ABP</math>, we see that <br />
<cmath>[ABC] = \sqrt{9.6(9.6-5.1)(9.6-5.1)(9.6-9)} = 10.8 = \frac{54}{5}.</cmath><br />
<br />
Thus, our answer is <math>059</math>. ~a.y.711<br />
<br />
==See Also==<br />
{{AIME box|year=2019|n=II|before=First Problem|num-a=2}}<br />
[[Category: Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Infinity26484https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1386632020 AMC 10B Problems/Problem 222020-11-28T22:40:44Z<p>Infinity26484: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the denominator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2==<br />
Similar to Solution 1, let <math>x=2^{50}</math>. It suffices to find remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>. Dividing polynomials results in a remainder of <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==MAA Original Solution==<br />
<br />
<cmath>2^{202} + 202 = (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201</cmath><cmath>= (2^{101} + 1)^2 - 2^{102} + 201</cmath> <cmath>= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
So<br />
<br />
<br />
==Solution 3==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 4==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Infinity26484https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10A_Problems/Problem_19&diff=1306122006 AMC 10A Problems/Problem 192020-08-04T21:40:10Z<p>Infinity26484: /* Solution 2(Stars and Bars) */</p>
<hr />
<div>== Problem ==<br />
How many non-[[similar]] [[triangle]]s have [[angle]]s whose [[degree]] measures are distinct positive integers in [[arithmetic progression]]? <br />
<br />
<math>\mathrm{(A) \ } 0\qquad\mathrm{(B) \ } 1\qquad\mathrm{(C) \ } 59\qquad\mathrm{(D) \ } 89\qquad\mathrm{(E) \ } 178\qquad</math><br />
<br />
== Solution ==<br />
The sum of the angles of a triangle is <math>180</math> degrees. For an arithmetic progression with an odd number of terms, the middle term is equal to the average of the sum of all of the terms, making it <math>\frac{180}{3} = 60</math> degrees. The minimum possible value for the smallest angle is <math>1</math> and the highest possible is <math>59</math> (since the numbers are distinct), so there are <math>59</math> possibilities <math>\Longrightarrow \mathrm{C}</math>.<br />
<br />
==Solution 2(Stars and Bars)==<br />
Let the first angle be <math>x</math>, and the common difference be <math>d</math>. The arithmetic progression can now be expressed as <math>x + (x + d) + (x + 2d) = 180</math>. Simplifiying, <math>x + d = 60</math>. Now, using stars and bars, we have <math>61_{c_1} = 61</math>. <br />
However, we must subtract the two cases in which either <math>a</math> or <math>d</math> equal <math>0</math>, so we have <math>61 - 2</math> = C.<br />
<br />
== See also ==<br />
{{AMC10 box|year=2006|ab=A|num-b=18|num-a=20}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Infinity26484https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10A_Problems/Problem_19&diff=1306112006 AMC 10A Problems/Problem 192020-08-04T21:39:52Z<p>Infinity26484: /* Solution 2(Stars and Bars) */</p>
<hr />
<div>== Problem ==<br />
How many non-[[similar]] [[triangle]]s have [[angle]]s whose [[degree]] measures are distinct positive integers in [[arithmetic progression]]? <br />
<br />
<math>\mathrm{(A) \ } 0\qquad\mathrm{(B) \ } 1\qquad\mathrm{(C) \ } 59\qquad\mathrm{(D) \ } 89\qquad\mathrm{(E) \ } 178\qquad</math><br />
<br />
== Solution ==<br />
The sum of the angles of a triangle is <math>180</math> degrees. For an arithmetic progression with an odd number of terms, the middle term is equal to the average of the sum of all of the terms, making it <math>\frac{180}{3} = 60</math> degrees. The minimum possible value for the smallest angle is <math>1</math> and the highest possible is <math>59</math> (since the numbers are distinct), so there are <math>59</math> possibilities <math>\Longrightarrow \mathrm{C}</math>.<br />
<br />
==Solution 2(Stars and Bars)==<br />
Let the first angle be <math>x</math>, and the common difference be <math>d</math>. The arithmetic progression can now be expressed as <math>x + (x + d) + (x + 2d) = 180</math>. Simplifiying, <math>x + d = 60</math>. Now, using stars and bars, we have <math>61_{c1} = 61</math>. <br />
However, we must subtract the two cases in which either <math>a</math> or <math>d</math> equal <math>0</math>, so we have <math>61 - 2</math> = C.<br />
<br />
== See also ==<br />
{{AMC10 box|year=2006|ab=A|num-b=18|num-a=20}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Infinity26484https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_24&diff=1143262017 AMC 10B Problems/Problem 242020-01-05T17:39:19Z<p>Infinity26484: /* Solution 4 (5-second sol) */</p>
<hr />
<div>==Problem 24==<br />
The vertices of an equilateral triangle lie on the hyperbola <math>xy=1</math>, and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?<br />
<br />
<math>\textbf{(A)}\ 48\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 108\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 169</math><br />
<br />
==Solution 1==<br />
WLOG, let the centroid of <math>\triangle ABC</math> be <math>I = (-1,-1)</math>. The centroid of an equilateral triangle is the same as the circumcenter. It follows that the circumcircle must intersect the graph exactly three times. Therefore, <math>A = (1,1)</math>, so <math>AI = BI = CI = 2\sqrt{2}</math>, so since <math>\triangle AIB</math> is isosceles and <math>\angle AIB = 120^{\circ}</math>, then by Law of Cosines, <math>AB = 2\sqrt{6}</math>. Alternatively, we can use the fact that the circumradius of an equilateral triangle is equal to <math>\frac {s}{\sqrt{3}}</math>. Therefore, the area of the triangle is <math>\frac{(2\sqrt{6})^2\sqrt{3}}4 = 6\sqrt{3}</math>, so the square of the area of the triangle is <math>\boxed{\textbf{(C) } 108}</math>.<br />
<br />
==Solution 2==<br />
WLOG, let the centroid of <math>\triangle ABC</math> be <math>G = (-1,-1)</math>. Then, one of the vertices must be the other curve of the hyperbola. WLOG, let <math>A = (1,1)</math>. Then, point <math>B</math> must be the reflection of <math>C</math> across the line <math>y=x</math>, so let <math>B = (a,\frac{1}{a})</math> and <math>C=(\frac{1}{a},a)</math>, where <math>a <-1</math>. Because <math>G</math> is the centroid, the average of the <math>x</math>-coordinates of the vertices of the triangle is <math>-1</math>. So we know that <math>a + 1/a+ 1 = -3</math>. Multiplying by <math>a</math> and solving gives us <math>a=-2-\sqrt{3}</math>. So <math>B=(-2-\sqrt{3},-2+\sqrt{3})</math> and <math>C=(-2+\sqrt{3},-2-\sqrt{3})</math>. So <math>BC=2\sqrt{6}</math>, and finding the square of the area gives us <math>\boxed{\textbf{(C) } 108}</math>.<br />
<br />
==Solution 3==<br />
WLOG, let the centroid of <math>\triangle ABC</math> be <math>G = (1, 1)</math> and let point <math>A</math> be <math>(-1, -1)</math>. It is known that the centroid is equidistant from the three vertices of <math>\triangle ABC</math>. Because we have the coordinates of both <math>A</math> and <math>G</math>, we know that the distance from <math>G</math> to any vertice of <math>\triangle ABC</math> is <math>\sqrt{(1-(-1))^2+(1-(-1))^2} = 2\sqrt{2}</math>. Therefore, <math>AG=BG=CG=2\sqrt{2}</math>. It follows that from <math>\triangle ABG</math>, where <math>AG=BG=2\sqrt{2}</math> and <math>\angle AGB = \dfrac{360^{\circ}}{3} = 120^{\circ}</math>, <math>[\triangle ABG]= \dfrac{(2\sqrt{2})^2 \cdot \sin(120)}{2} = 4 \cdot \dfrac{\sqrt{3}}{2} = 2\sqrt{3}</math> using the formula for the area of a triangle with sine <math>\left([\triangle ABC]= \dfrac{1}{2} AB \cdot BC \sin(\angle ABC)\right)</math>. Because <math>\triangle ACG</math> and <math>\triangle BCG</math> are congruent to <math>\triangle ABG</math>, they also have an area of <math>2\sqrt{3}</math>. Therefore, <math>[\triangle ABC] = 3(2\sqrt{3}) = 6\sqrt{3}</math>. Squaring that gives us the answer of <math>\boxed{\textbf{(C) }108}</math>.<br />
<br />
==Solution 4 (5-second solution)==<br />
WLOG, let the centroid of the triangle be <math>(1, 1)</math>. By symmetry, the other vertex is <math>(-1, -1)</math>. The distance between these two points is <math>2\sqrt2</math>, so the height of the triangle is <math>3\sqrt 2</math>, the side length is <math>2\sqrt6</math>, and the area is <math>6\sqrt3</math>, yielding an answer of <math>\boxed{\textbf{(C) }108}</math>.<br />
-Stormersyle<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=B|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Infinity26484https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_13&diff=995122017 AMC 10A Problems/Problem 132018-12-17T22:56:56Z<p>Infinity26484: /* Solution */</p>
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<div>==Problem==<br />
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Define a sequence recursively by <math>F_{0}=0,~F_{1}=1,</math> and <math>F_{n}=</math> the remainder when <math>F_{n-1}+F_{n-2}</math> is divided by <math>3,</math> for all <math>n\geq 2.</math> Thus the sequence starts <math>0,1,1,2,0,2,\ldots</math> What is <math>F_{2017}+F_{2018}+F_{2019}+F_{2020}+F_{2021}+F_{2022}+F_{2023}+F_{2024}?</math><br />
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<math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10</math><br />
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==Solution==<br />
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A pattern starts to emerge as the function is continued. The repeating pattern is <math>0,1,1,2,0,2,2,1\ldots</math> The problem asks for the sum of eight consecutive terms in the sequence. Because there are eight numbers in the repeating sequence, we just need to find the sum of the numbers in the sequence, which is <math>\boxed{\textbf{(D)}\ 9}</math><br />
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==See Also==<br />
{{AMC10 box|year=2017|ab=A|num-b=12|num-a=14}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Number Theory Problems]]</div>Infinity26484https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_13&diff=995112017 AMC 10A Problems/Problem 132018-12-17T22:33:14Z<p>Infinity26484: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Define a sequence recursively by <math>F_{0}=0,~F_{1}=1,</math> and <math>F_{n}=</math> the remainder when <math>F_{n-1}+F_{n-2}</math> is divided by <math>3,</math> for all <math>n\geq 2.</math> Thus the sequence starts <math>0,1,1,2,0,2,\ldots</math> What is <math>F_{2017}+F_{2018}+F_{2019}+F_{2020}+F_{2021}+F_{2022}+F_{2023}+F_{2024}?</math><br />
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<math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10</math><br />
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==Solution==<br />
<br />
A pattern starts to emerge as the function is continued. The repeating pattern is <math>0,1,1,2,0,2,2,1\ldots</math> The problem asks for the sum of eight consecutive terms in the sequence. Because there are eight numbers in the repeating sequence, we just need to find the sum of the numbers in the sequence, which is <math>\boxed{\textbf{(D)}\ 9}</math><br />
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NOTE: THIS SOLUTION IS CREATED WITH THE ASSUMPTION THAT 2-2=1 (mod 3) AND 1-2=0 (mod 3)<br />
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==See Also==<br />
{{AMC10 box|year=2017|ab=A|num-b=12|num-a=14}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Number Theory Problems]]</div>Infinity26484https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_16&diff=985732017 AMC 8 Problems/Problem 162018-11-09T23:27:56Z<p>Infinity26484: /* Solution 1 */</p>
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<div>==Problem 16==<br />
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In the figure below, choose point <math>D</math> on <math>\overline{BC}</math> so that <math>\triangle ACD</math> and <math>\triangle ABD</math> have equal perimeters. What is the area of <math>\triangle ABD</math>?<br />
<asy>draw((0,0)--(4,0)--(0,3)--(0,0));<br />
label("$A$", (0,0), SW);<br />
label("$B$", (4,0), ESE);<br />
label("$C$", (0, 3), N);<br />
label("$3$", (0, 1.5), W);<br />
label("$4$", (2, 0), S);<br />
label("$5$", (2, 1.5), NE);</asy><br />
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<math>\textbf{(A) }\frac{3}{4}\qquad\textbf{(B) }\frac{3}{2}\qquad\textbf{(C) }2\qquad\textbf{(D) }\frac{12}{5}\qquad\textbf{(E) }\frac{5}{2}</math><br />
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==Solution 1==<br />
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Essentially, we see that if we draw a line from point A to imaginary point D, that line would apply to both triangles. Let us say that <math>x</math> is the length of the line from B to D. So, the perimeter of <math>\triangle{ABD}</math> would be <math>\overline{AD} + 4 + x</math>, while the perimeter of <math>\triangle{ACD}</math> would be <math>\overline{AD} + 3 + (5 - x)</math>. Notice that we can find out <math>x</math> from these two equations. We can find out that <math>x = 2</math>, so that means that the area of <math>\triangle{ABD} = \frac{2 \cdot 6}{5} = \boxed{\textbf{(D) } \frac{12}{5}}</math><br />
THIS SOLUTION MAKES NO SENSE PLEASE MOVE ON<br />
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==Solution 2==<br />
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We know that the perimeters of the two small triangles are <math>3+CD+AD</math> and <math>4+BD+AD</math>. Setting both equal and using <math>BD+CD = 5</math>, we have <math>BD = 2</math> and <math>CD = 3</math>. Now, we simply have to find the area of <math>\triangle ABD</math>. Since <math>\frac{BD}{CD} = \frac{2}{3}</math>, we must have <math>\frac{[ABD]}{[ACD]} = 2/3</math>. Combining this with the fact that <math>[ABC] = [ABD] + [ACD] = \frac{3\cdot4}{2} = 6</math>, we get <math>[ABD] = \frac{2}{5}[ABC] = \frac{2}{5} \cdot 6 = \boxed{\textbf{(D) } \frac{12}{5}}</math><br />
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==Solution 3== <br />
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Since point <math>D</math> is on line <math>BC</math>, it will split it into <math>CD</math> and <math>DB</math>. Let <math>CD = 5 - x</math> and <math>DB = x</math>. Triangle <math>CAD</math> has side lengths <math>3, 5 - x, AD</math> and triangle <math>DAB</math> has side lengths <math>x, 4, AD</math>. Since both perimeters are equal, we have the equation <math>3 + 5 - x + AD = 4 + x + AD</math>. Eliminating <math>AD</math> and solving the resulting linear equation gives <math>x = 2</math>. Draw a perpendicular from point <math>D</math> to <math>AB</math>. Call the point of intersection <math>F</math>. Because angle <math>ABC</math> is common to both triangles <math>DBF</math> and <math>ABC</math>, and both are right triangles, both are similar. The hypotenuse of triangle <math>DBF</math> is 2, so the altitude must be <math>6/5</math> Because <math>DBF</math> and <math>ABD</math> share the same altitude, the height of <math>ABD</math> therefore must be <math>6/5</math>. The base of <math>ABD</math> is 4, so <math>[ABD] = \frac{1}{2} \cdot 4 \cdot \frac{6}{5} = \frac{12}{5} \implies \boxed{\textbf{(D) } \frac{12}{5}}</math><br />
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==Solution 4==<br />
Using any preferred method, realize <math>BD = 2</math>. Since we are given a 3-4-5 right triangle, we know the value of <math>\sin(\angle ABC) = \frac{3}{5}</math>. Since we are given <math>AB = 4</math>, apply the Sine Area Formula to get <math>\frac{1}{2} \cdot 4 \cdot 2 \cdot \frac{3}{5} = \boxed{\textbf{(D) } \frac{12}{5}}</math>.<br />
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==See Also==<br />
{{AMC8 box|year=2017|num-b=15|num-a=17}}<br />
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{{MAA Notice}}</div>Infinity26484