https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Ngeorge&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T12:11:04ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12A_Problems/Problem_20&diff=805602008 AMC 12A Problems/Problem 202016-10-09T16:13:02Z<p>Ngeorge: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
Triangle <math>ABC</math> has <math>AC=3</math>, <math>BC=4</math>, and <math>AB=5</math>. Point <math>D</math> is on <math>\overline{AB}</math>, and <math>\overline{CD}</math> bisects the right angle. The inscribed circles of <math>\triangle ADC</math> and <math>\triangle BCD</math> have radii <math>r_a</math> and <math>r_b</math>, respectively. What is <math>r_a/r_b</math>?<br />
<br />
<math>\mathrm{(A)}\ \frac{1}{28}\left(10-\sqrt{2}\right)\qquad\mathrm{(B)}\ \frac{3}{56}\left(10-\sqrt{2}\right)\qquad\mathrm{(C)}\ \frac{1}{14}\left(10-\sqrt{2}\right)\qquad\mathrm{(D)}\ \frac{5}{56}\left(10-\sqrt{2}\right)\\\mathrm{(E)}\ \frac{3}{28}\left(10-\sqrt{2}\right)</math><br />
<br />
== Solution 1==<br />
<center><asy><br />
import olympiad;<br />
size(300);<br />
defaultpen(0.8);<br />
pair C=(0,0),A=(0,3),B=(4,0),D=(4-2.28571,1.71429);<br />
pair O=incenter(A,C,D), P=incenter(B,C,D);<br />
picture p = new picture; <br />
draw(p,Circle(C,0.2)); draw(p,Circle(B,0.2));<br />
clip(p,B--C--D--cycle);<br />
add(p);<br />
draw(A--B--C--D--C--cycle);<br />
draw(incircle(A,C,D));<br />
draw(incircle(B,C,D));<br />
dot(O);dot(P);<br />
label("\(A\)",A,W);<br />
label("\(B\)",B,E);<br />
label("\(C\)",C,W);<br />
label("\(D\)",D,NE);<br />
label("\(O_A\)",O,W);<br />
label("\(O_B\)",P,W);<br />
label("\(3\)",(A+C)/2,W);<br />
label("\(4\)",(B+C)/2,S);<br />
label("\(\frac{15}{7}\)",(A+D)/2,NE);<br />
label("\(\frac{20}{7}\)",(B+D)/2,NE);<br />
label("\(45^{\circ}\)",(.2,.1),E);<br />
label("\(\sin \theta = \frac{3}{5}\)",B-(.2,-.1),W);<br />
</asy></center><br />
<br />
By the [[Angle Bisector Theorem]], <br />
<cmath>\frac{BD}{4} = \frac{5-BD}{3} \Longrightarrow BD = \frac{20}7</cmath><br />
By [[Law of Sines]] on <math>\triangle BCD</math>, <br />
<cmath>\frac{BD}{\sin 45^{\circ}} = \frac{CD}{\sin \angle B} \Longrightarrow \frac{20/7}{\sqrt{2}/2} = \frac{CD}{3/5} \Longrightarrow CD=\frac{12\sqrt{2}}{7}</cmath><br />
Since the area of a triangle satisfies <math>[\triangle]=rs</math>, where <math>r = </math> the [[inradius]] and <math>s =</math> the [[semiperimeter]], we have <br />
<cmath>\frac{r_A}{r_B} = \frac{[ACD] \cdot s_B}{[BCD] \cdot s_A}</cmath><br />
<!--Using any of various formulas for triangle area, we find the area <math>[BCD]</math> to be <br />
<cmath>[BCD] = \frac{1}{2} (\sin \angle CBD) \cdot (BD) \cdot (CD) = \frac 12 \cdot \frac 35 \cdot \frac{20}{7} \cdot 4 = \frac{24}{7}</cmath><br />
and <br />
<cmath>[ACD] = [ABC] - [BCD] = \frac 12 (3)(4) - \frac{24}{7} = \frac{18}{7}</cmath>--><br />
Since <math>\triangle ACD</math> and <math>\triangle BCD</math> share the [[altitude]] (to <math>\overline{AB}</math>), their areas are the ratio of their bases, or <cmath>\frac{[ACD]}{[BCD]} = \frac{AD}{BD} = \frac{3}{4}</cmath><br />
The semiperimeters are <math>s_A = \left(3 + \frac{15}{7} + \frac{12\sqrt{2}}{7}\right)\left/\right.2 = \frac{18+6\sqrt{2}}{7}</math> and <math>s_B = \frac{24+ 6\sqrt{2}}{7}</math>. Thus,<br />
<cmath>\begin{align*}<br />
\frac{r_A}{r_B} &= \frac{[ACD] \cdot s_B}{[BCD] \cdot s_A} = \frac{3}{4} \cdot \frac{(24+ 6\sqrt{2})/7}{(18+6\sqrt{2})/7} \\<br />
&= \frac{3(4+\sqrt{2})}{4(3+\sqrt{2})} \cdot \left(\frac{3-\sqrt{2}}{3-\sqrt{2}}\right) = \frac{3}{28}(10-\sqrt{2}) \Rightarrow \mathrm{(E)}\qquad \blacksquare \end{align*}</cmath><br />
<br />
==Solution 2==<br />
<center><asy><br />
import olympiad;<br />
import geometry;<br />
size(300);<br />
defaultpen(0.8);<br />
<br />
pair C=(0,0),A=(0,3),B=(4,0),D=(4-2.28571,1.71429);<br />
pair O=incenter(A,C,D), P=incenter(B,C,D);<br />
line cd = line(C, D);<br />
<br />
picture p = new picture;<br />
picture q = new picture; <br />
picture r = new picture;<br />
picture s = new picture;<br />
<br />
draw(p,Circle(C,0.2));<br />
clip(p,P--C--D--cycle);<br />
<br />
draw(q, Circle(C, 0.3));<br />
clip(q, O--C--D--cycle);<br />
<br />
line l1 = perpendicular(O, cd);<br />
draw(r, l1);<br />
clip(r, C--D--O--cycle);<br />
<br />
line l2 = perpendicular(P, cd);<br />
draw(s, l2);<br />
clip(s, C--P--D--cycle); <br />
<br />
add(p);<br />
add(q);<br />
add(r);<br />
add(s);<br />
<br />
draw(A--B--C--D--C--cycle);<br />
draw(incircle(A,C,D));<br />
draw(incircle(B,C,D));<br />
draw(C--O);<br />
draw(C--P);<br />
dot(O);<br />
dot(P);<br />
<br />
point inter1 = intersectionpoint(l1, cd);<br />
point inter2 = intersectionpoint(l2, cd);<br />
dot(inter1);<br />
dot(inter2);<br />
<br />
label("\(A\)",A,W);<br />
label("\(B\)",B,E);<br />
label("\(C\)",C,W);<br />
label("\(D\)",D,NE);<br />
label("\(O_a\)",O,W);<br />
label("\(O_b\)",P,E);<br />
label("\(3\)",(A+C)/2,W);<br />
label("\(4\)",(B+C)/2,S);<br />
label("\(\frac{15}{7}\)",(A+D)/2,NE);<br />
label("\(\frac{20}{7}\)",(B+D)/2,NE);<br />
label("\(M\)", inter1, 2W);<br />
label("\(N\)", inter2, 2E);<br />
</asy></center><br />
<br />
We start by finding the length of <math>AD</math> and <math>BD</math> as in solution 1. Using the angle bisector theorem, we see that <math>AD = \frac{15}{7}</math> and <math>BD = \frac{20}{7}</math>. Using Stewart's Theorem gives us the equation <math>5d^2 + \frac{1500}{49} = \frac{240}{7} + \frac{180}{7}</math>, where <math>d</math> is the length of <math>CD</math>. Solving gives us <math>d = \frac{12\sqrt{2}}{7}</math>, so <math>CD = \frac{12\sqrt{2}}{7}</math>.<br />
<br />
Call the incenters of triangles <math>ACD</math> and <math>BCD</math> <math>O_a</math> and <math>O_b</math> respectively. Since <math>O_a</math> is an incenter, it lies on the angle bisector of <math>\angle ACD</math>. Similarly, <math>O_b</math> lies on the angle bisector of <math>\angle BCD</math>. Call the point on <math>CD</math> tangent to <math>O_a</math> <math>M</math>, and the point tangent to <math>O_b</math> <math>N</math>. Since <math>\triangle CO_aM</math> and <math>\triangle CO_bN</math> are right, and <math>\angle O_aCM = \angle O_bCN</math>, <math>\triangle CO_aM \sim \triangle CO_bN</math>. Then, <math>\frac{r_a}{r_b} = \frac{CM}{CN}</math>. <br />
<br />
We now use common tangents to find the length of <math>CM</math> and <math>CN</math>. Let <math>CM = m</math>, and the length of the other tangents be <math>n</math> and <math>p</math>. Since common tangents are equal, we can write that <math>m + n = \frac{12\sqrt{2}}{7}</math>, <math>n + p = \frac{15}{7}</math> and <math>m + p = 3</math>. Solving gives us that <math>CM = m = \frac{6\sqrt{2} + 3}{7}</math>. Similarly, <math>CN = \frac{6\sqrt{2} + 4}{7}</math>. <br />
<br />
We see now that <math>\frac{r_a}{r_b} = \frac{\frac{6\sqrt{2} + 3}{7}}{\frac{6\sqrt{2} + 4}{7}} = \frac{6\sqrt{2} + 3}{6\sqrt{2} + 4} = \frac{60-6\sqrt{2}}{56} = \frac{3}{28}(10 - \sqrt{2}) \Rightarrow \boxed{E}</math><br />
<br />
<br />
(Thanks to above solution writer for the framework of my diagram)<br />
<br />
==See Also==<br />
{{AMC12 box|year=2008|num-b=19|num-a=21|ab=A}}<br />
{{MAA Notice}}</div>Ngeorgehttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_17&diff=783032003 AMC 12A Problems/Problem 172016-04-28T00:03:30Z<p>Ngeorge: /* Solution 4 */</p>
<hr />
<div>== Problem ==<br />
Square <math>ABCD</math> has sides of length <math>4</math>, and <math>M</math> is the midpoint of <math>\overline{CD}</math>. A circle with radius <math>2</math> and center <math>M</math> intersects a circle with radius <math>4</math> and center <math>A</math> at points <math>P</math> and <math>D</math>. What is the distance from <math>P</math> to <math>\overline{AD}</math>?<br />
<br />
[[Image:5d50417537c6cddfb70810403c62787b889cdcb1.png]]<br />
<br />
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \frac {16}{5} \qquad \textbf{(C)}\ \frac {13}{4} \qquad \textbf{(D)}\ 2\sqrt {3} \qquad \textbf{(E)}\ \frac {7}{2}</math><br />
<br />
== Solution 1==<br />
<br />
Let <math>D</math> be the origin. <math>A</math> is the point <math>(0,4)</math> and <math>M</math> is the point <math>(2,0)</math>. We are given the radius of the quarter circle and semicircle as <math>4</math> and <math>2</math>, respectively, so their equations, respectively, are:<br />
<br />
<math>x^2 + (y-4)^2 = 4^2</math><br />
<br />
<math>(x-2)^2 + y^2 = 2^2</math><br />
<br />
Subtract the second equation from the first:<br />
<br />
<math>x^2 + (y - 4)^2 - (x - 2)^2 - y^2 = 12</math><br />
<br />
<math>4x - 8y + 12 = 12</math><br />
<br />
<math>x = 2y.</math><br />
<br />
Then substitute:<br />
<br />
<math>(2y)^2 + (y - 4)^2 = 16</math><br />
<br />
<math>4y^2 + y^2 - 8y + 16 = 16</math><br />
<br />
<math>5y^2 - 8y = 0</math><br />
<br />
<math>y(5y - 8) = 0.</math><br />
<br />
Thus <math>y = 0</math> and <math>y = \frac{8}{5}</math> making <math>x = 0</math> and <math>x = \frac{16}{5}</math>.<br />
<br />
The first value of <math>0</math> is obviously referring to the x-coordinate of the point where the circles intersect at the origin, <math>D</math>, so the second value must be referring to the x coordinate of <math>P</math>. Since <math>\overline{AD}</math> is the y-axis, the distance to it from <math>P</math> is the same as the x-value of the coordinate of <math>P</math>, so the distance from <math>P</math> to <math>\overline{AD}</math> is <math>\frac{16}{5} \Rightarrow B</math><br />
<br />
==Solution 2==<br />
<br />
Note that <math>P</math> is merely a reflection of <math>D</math> over <math>AM</math>. Call the intersection of <math>AM</math> and <math>DP</math> <math>X</math>. Drop perpendiculars from <math>X</math> and <math>P</math> to <math>AD</math>, and denote their respective points of intersection by <math>J</math> and <math>K</math>. We then have <math>\triangle DXJ\sim\triangle DPK</math>, with a scale factor of 2. Thus, we can find <math>XJ</math> and double it to get our answer. With some analytical geometry, we find that <math>XJ=\frac{8}{5}</math>, implying that <math>PK=\frac{16}{5}</math>.<br />
<br />
==Solution 3==<br />
As in Solution 2, draw in <math>DP</math> and <math>AM</math> and denote their intersection point <math>X</math>. Next, drop a perpendicular from <math>P</math> to <math>AD</math> and denote the foot as <math>Z</math>. <math>AP \cong AD</math> as they are both radii and similarly <math>DM \cong MP</math> so <math>APMD</math> is a kite and <math>DX \perp XM</math> by a well-known theorem. <br />
<br />
Pythagorean theorem gives us <math>AM=2 \sqrt{5}</math>. Clearly <math>\triangle XMD \sim \triangle XDA \sim \triangle DMA \sim \triangle ZDP</math> by angle-angle and <math>\triangle XMD \cong \triangle XMP</math> by Hypotenuse Leg.<br />
Manipulating similar triangles gives us <math>PZ=\frac{16}{5}</math><br />
<br />
==Solution 4==<br />
Using the double-angle formula for sine, what we need to find is <math>AP\cdot \sin(DAP) = AP\cdot 2\sin( DAM) \cos(DAM) = 4\cdot 2\cdot \frac{2}{\sqrt{20}}\cdot\frac{4}{\sqrt{20}} = \frac{16}{5}</math>.<br />
<br />
== See Also ==<br />
*[[2003 AMC 12A Problems]]<br />
*[[2003 AMC 12A Problems/Problem 16|Previous Problem]]<br />
*[[2003 AMC 12A Problems/Problem 18|Next Problem]]<br />
{{MAA Notice}}</div>Ngeorgehttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_17&diff=783022003 AMC 12A Problems/Problem 172016-04-28T00:03:11Z<p>Ngeorge: /* Solution 4 */</p>
<hr />
<div>== Problem ==<br />
Square <math>ABCD</math> has sides of length <math>4</math>, and <math>M</math> is the midpoint of <math>\overline{CD}</math>. A circle with radius <math>2</math> and center <math>M</math> intersects a circle with radius <math>4</math> and center <math>A</math> at points <math>P</math> and <math>D</math>. What is the distance from <math>P</math> to <math>\overline{AD}</math>?<br />
<br />
[[Image:5d50417537c6cddfb70810403c62787b889cdcb1.png]]<br />
<br />
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \frac {16}{5} \qquad \textbf{(C)}\ \frac {13}{4} \qquad \textbf{(D)}\ 2\sqrt {3} \qquad \textbf{(E)}\ \frac {7}{2}</math><br />
<br />
== Solution 1==<br />
<br />
Let <math>D</math> be the origin. <math>A</math> is the point <math>(0,4)</math> and <math>M</math> is the point <math>(2,0)</math>. We are given the radius of the quarter circle and semicircle as <math>4</math> and <math>2</math>, respectively, so their equations, respectively, are:<br />
<br />
<math>x^2 + (y-4)^2 = 4^2</math><br />
<br />
<math>(x-2)^2 + y^2 = 2^2</math><br />
<br />
Subtract the second equation from the first:<br />
<br />
<math>x^2 + (y - 4)^2 - (x - 2)^2 - y^2 = 12</math><br />
<br />
<math>4x - 8y + 12 = 12</math><br />
<br />
<math>x = 2y.</math><br />
<br />
Then substitute:<br />
<br />
<math>(2y)^2 + (y - 4)^2 = 16</math><br />
<br />
<math>4y^2 + y^2 - 8y + 16 = 16</math><br />
<br />
<math>5y^2 - 8y = 0</math><br />
<br />
<math>y(5y - 8) = 0.</math><br />
<br />
Thus <math>y = 0</math> and <math>y = \frac{8}{5}</math> making <math>x = 0</math> and <math>x = \frac{16}{5}</math>.<br />
<br />
The first value of <math>0</math> is obviously referring to the x-coordinate of the point where the circles intersect at the origin, <math>D</math>, so the second value must be referring to the x coordinate of <math>P</math>. Since <math>\overline{AD}</math> is the y-axis, the distance to it from <math>P</math> is the same as the x-value of the coordinate of <math>P</math>, so the distance from <math>P</math> to <math>\overline{AD}</math> is <math>\frac{16}{5} \Rightarrow B</math><br />
<br />
==Solution 2==<br />
<br />
Note that <math>P</math> is merely a reflection of <math>D</math> over <math>AM</math>. Call the intersection of <math>AM</math> and <math>DP</math> <math>X</math>. Drop perpendiculars from <math>X</math> and <math>P</math> to <math>AD</math>, and denote their respective points of intersection by <math>J</math> and <math>K</math>. We then have <math>\triangle DXJ\sim\triangle DPK</math>, with a scale factor of 2. Thus, we can find <math>XJ</math> and double it to get our answer. With some analytical geometry, we find that <math>XJ=\frac{8}{5}</math>, implying that <math>PK=\frac{16}{5}</math>.<br />
<br />
==Solution 3==<br />
As in Solution 2, draw in <math>DP</math> and <math>AM</math> and denote their intersection point <math>X</math>. Next, drop a perpendicular from <math>P</math> to <math>AD</math> and denote the foot as <math>Z</math>. <math>AP \cong AD</math> as they are both radii and similarly <math>DM \cong MP</math> so <math>APMD</math> is a kite and <math>DX \perp XM</math> by a well-known theorem. <br />
<br />
Pythagorean theorem gives us <math>AM=2 \sqrt{5}</math>. Clearly <math>\triangle XMD \sim \triangle XDA \sim \triangle DMA \sim \triangle ZDP</math> by angle-angle and <math>\triangle XMD \cong \triangle XMP</math> by Hypotenuse Leg.<br />
Manipulating similar triangles gives us <math>PZ=\frac{16}{5}</math><br />
<br />
==Solution 4==<br />
Using the double-angle formula for sine, what we need to find is <math>AP\cdot \sin(DAP) = AD\cdot 2\sin( DAM) \cos(DAM) = 4\cdot 2\cdot \frac{2}{\sqrt{20}}\cdot\frac{4}{\sqrt{20}} = \frac{16}{5}</math>.<br />
<br />
== See Also ==<br />
*[[2003 AMC 12A Problems]]<br />
*[[2003 AMC 12A Problems/Problem 16|Previous Problem]]<br />
*[[2003 AMC 12A Problems/Problem 18|Next Problem]]<br />
{{MAA Notice}}</div>Ngeorgehttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_19&diff=765662016 AMC 10B Problems/Problem 192016-02-21T14:29:50Z<p>Ngeorge: /* Problem */</p>
<hr />
<div>==Problem==<br />
Rectangle <math>ABCD</math> has <math>AB=5</math> and <math>BC=4</math>. Point <math>E</math> lies on <math>\overline{AB}</math> so that <math>EB=1</math>, point <math>G</math> lies on <math>\overline{BC}</math> so that <math>CG=1</math>. and point <math>F</math> lies on <math>\overline{CD}</math> so that <math>DF=2</math>. Segments <math>\overline{AG}</math> and <math>\overline{AC}</math> intersect <math>\overline{EF}</math> at <math>Q</math> and <math>P</math>, respectively. What is the value of <math>\frac{PQ}{EF}</math>?<br />
<br />
<br />
<asy>pair A1=(2,0),A2=(4,4);<br />
pair B1=(0,4),B2=(5,1);<br />
pair C1=(5,0),C2=(0,4); <br />
draw(A1--A2);<br />
draw(B1--B2);<br />
draw(C1--C2);<br />
draw((0,0)--B1--(5,4)--C1--cycle);<br />
dot((20/7,12/7));<br />
dot((3.07692307692,2.15384615384));<br />
label("$Q$",(3.07692307692,2.15384615384),N);<br />
label("$P$",(20/7,12/7),W);<br />
label("$A$",(0,4), NW);<br />
label("$B$",(5,4), NE);<br />
label("$C$",(5,0),SE);<br />
label("$D$",(0,0),SW);<br />
label("$F$",(2,0),S); label("$G$",(5,1),E);<br />
label("$E$",(4,4),N);</asy><br />
<br />
<math>\textbf{(A)}~\frac{\sqrt{13}}{16} \qquad<br />
\textbf{(B)}~\frac{\sqrt{2}}{13} \qquad<br />
\textbf{(C)}~\frac{9}{82} \qquad<br />
\textbf{(D)}~\frac{10}{91}\qquad<br />
\textbf{(E)}~\frac19</math><br />
<br />
<br />
==Solution==<br />
<math>\textbf{(D)}~\frac{10}{91}</math><br />
<br />
solution by ngeorge</div>Ngeorgehttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_23&diff=765652016 AMC 10B Problems/Problem 232016-02-21T14:27:19Z<p>Ngeorge: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
In regular hexagon <math>ABCDEF</math>, points <math>W</math>, <math>X</math>, <math>Y</math>, and <math>Z</math> are chosen on sides <math>\overline{BC}</math>, <math>\overline{CD}</math>, <math>\overline{EF}</math>, and <math>\overline{FA}</math> respectively, so lines <math>AB</math>, <math>ZW</math>, <math>YX</math>, and <math>ED</math> are parallel and equally spaced. What is the ratio of the area of hexagon <math>WCXYFZ</math> to the area of hexagon <math>ABCDEF</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{3}\qquad\textbf{(B)}\ \frac{10}{27}\qquad\textbf{(C)}\ \frac{11}{27}\qquad\textbf{(D)}\ \frac{4}{9}\qquad\textbf{(E)}\ \frac{13}{27}</math><br />
<br />
<br />
==Solution==<br />
<math>\textbf{(C)}\ \frac{11}{27}</math><br />
<br />
solution by ngeorge</div>Ngeorgehttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_2&diff=765642016 AMC 10B Problems/Problem 22016-02-21T14:25:38Z<p>Ngeorge: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
If <math>n\heartsuit m=n^3m^2</math>, what is <math>\frac{2\heartsuit 4}{4\heartsuit 2}</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4</math><br />
<br />
<br />
<br />
==Solution==<br />
<math>\textbf{(B)}\ \frac{1}{2}</math><br />
<br />
Solution by ngeorge</div>Ngeorgehttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_2&diff=765632016 AMC 10B Problems/Problem 22016-02-21T14:25:13Z<p>Ngeorge: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
If <math>n\heartsuit m=n^3m^2</math>, what is <math>\frac{2\heartsuit 4}{4\heartsuit 2}</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4</math><br />
<br />
<br />
<br />
==Solution==<br />
<math>\textbf{(B)}\ \frac{1}{2}</math></div>Ngeorgehttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_1&diff=765612016 AMC 10B Problems/Problem 12016-02-21T14:20:58Z<p>Ngeorge: /* Solution */</p>
<hr />
<div>==Problem==<br />
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What is the value of <math>\frac{2a^{-1}+\frac{a^{-1}}{2}}{a}</math> when <math>a= \frac{1}{2}</math>?<br />
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<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20</math><br />
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==Solution==<br />
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<math>\frac{\frac{1}{a}*(2+\frac{1}{2})}{a}</math> then becomes <math>\frac{\frac{5}{2}}{a^{2}}</math> which is equal to <math>\frac{5}{2}*2^{2}</math> = <math>{(D)}</math><br />
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Solution by ngeorge</div>Ngeorgehttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_1&diff=765592016 AMC 10B Problems/Problem 12016-02-21T14:19:32Z<p>Ngeorge: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
What is the value of <math>\frac{2a^{-1}+\frac{a^{-1}}{2}}{a}</math> when <math>a= \frac{1}{2}</math>?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20</math><br />
<br />
<br />
<br />
<br />
==Solution==<br />
<br />
<math>\frac{frac{1}{a}*(2+frac{1}{2})}{a}</math> then becomes <math>frac{frac{5}{2}}{a^{2}}</math> which is equal to <math>frac{5}{2}*2^{2}</math> = <math>{(D)}</math><br />
<br />
Solution by ngeorge</div>Ngeorgehttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_1&diff=765582016 AMC 10B Problems/Problem 12016-02-21T14:18:57Z<p>Ngeorge: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
What is the value of <math>\frac{2a^{-1}+\frac{a^{-1}}{2}}{a}</math> when <math>a= \frac{1}{2}</math>?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20</math><br />
<br />
<br />
<br />
<br />
==Solution==<br />
<br />
<math>\frac{frac{1}{a}*(2+frac{1}{2})}{a}</math> then becomes <math>frac{frac{5}{2}}{a^{2}}</math> which is equal to <math>frac{5}{2}*2^{2}</math> = <math>{(D)}\ 10\qquad\textbf{(E)}\ 20</math><br />
<br />
Solution by ngeorge</div>Ngeorgehttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_1&diff=765572016 AMC 10B Problems/Problem 12016-02-21T14:17:01Z<p>Ngeorge: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
What is the value of <math>\frac{2a^{-1}+\frac{a^{-1}}{2}}{a}</math> when <math>a= \frac{1}{2}</math>?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20</math><br />
<br />
<br />
<br />
<br />
==Solution==<br />
<br />
<math>\frac{frac{1}{a}*(2+frac{1}{2})}{a}</math> then becomes <math>frac{frac{5}{2}}{a^{2}}</math> which is equal to <math>frac{5}{2]*2^{2}</math> = <math>{(D)}\ 10\qquad\textbf{(E)}\ 20</math><br />
<br />
Solution by ngeorge</div>Ngeorge