https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Derpguys&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T09:35:30ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_22&diff=724922014 AMC 10B Problems/Problem 222015-10-16T03:30:34Z<p>Derpguys: </p>
<hr />
<div>==Problem==<br />
Eight semicircles line the inside of a square with side length 2 as shown. What is the radius of the circle tangent to all of these semicircles?<br />
<br />
<math>\text{(A) } \dfrac{1+\sqrt2}4 \quad \text{(B) } \dfrac{\sqrt5-1}2 \quad \text{(C) } \dfrac{\sqrt3+1}4 \quad \text{(D) } \dfrac{2\sqrt3}5 \quad \text{(E) } \dfrac{\sqrt5}3</math><br />
<br />
<asy><br />
scale(200);<br />
draw(scale(.5)*((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle));<br />
path p = arc((.25,-.5),.25,0,180)--arc((-.25,-.5),.25,0,180);<br />
draw(p);<br />
p=rotate(90)*p; draw(p);<br />
p=rotate(90)*p; draw(p);<br />
p=rotate(90)*p; draw(p);<br />
draw(scale((sqrt(5)-1)/4)*unitcircle);<br />
</asy><br />
[[Category: Introductory Geometry Problems]]<br />
<br />
==Solution==<br />
<br />
We connect the centers of the circle and one of the semicircles, then draw the perpendicular from the center of the middle circle to that side, as shown.<br />
<br />
<asy><br />
scale(200);<br />
draw(scale(.5)*((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle));<br />
path p = arc((.25,-.5),.25,0,180)--arc((-.25,-.5),.25,0,180);<br />
draw(p);<br />
p=rotate(90)*p; draw(p);<br />
p=rotate(90)*p; draw(p);<br />
p=rotate(90)*p; draw(p);<br />
draw(scale((sqrt(5)-1)/4)*unitcircle);<br />
pair OO=(0,0);<br />
pair XX=(-.25,-.5);<br />
pair YY=(0,-.5);<br />
draw(YY--OO--XX--cycle,black+1bp);<br />
label("$\frac12$",.5*(XX+YY),S);<br />
label("$1$",.5*YY,E);<br />
</asy><br />
<br />
Let's make an equation by the pythagorean theorem.<br />
<br />
<math>\sqrt{1^2 + \left(\frac12\right)^2} = \sqrt{\frac54} = \frac{\sqrt5}{2}</math><br />
<br />
Let's call <math>r</math> as the radius of the circle that we want to find. We see that the hypotenuse of the bold right triangle is <math>\dfrac{1}{2}+r</math>, and thus <math>r</math> is <math>\boxed{\textbf{(B)} \frac{\sqrt{5}-1}{2}}</math><br />
<br />
<br />
==See Also==<br />
{{AMC10 box|year=2014|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Derpguyshttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_22&diff=724912014 AMC 10B Problems/Problem 222015-10-16T03:06:35Z<p>Derpguys: </p>
<hr />
<div>==Problem==<br />
Eight semicircles line the inside of a square with side length 2 as shown. What is the radius of the circle tangent to all of these semicircles?<br />
<br />
<math>\text{(A) } \dfrac{1+\sqrt2}4 \quad \text{(B) } \dfrac{\sqrt5-1}2 \quad \text{(C) } \dfrac{\sqrt3+1}4 \quad \text{(D) } \dfrac{2\sqrt3}5 \quad \text{(E) } \dfrac{\sqrt5}3</math><br />
<br />
<asy><br />
scale(200);<br />
draw(scale(.5)*((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle));<br />
path p = arc((.25,-.5),.25,0,180)--arc((-.25,-.5),.25,0,180);<br />
draw(p);<br />
p=rotate(90)*p; draw(p);<br />
p=rotate(90)*p; draw(p);<br />
p=rotate(90)*p; draw(p);<br />
draw(scale((sqrt(5)-1)/4)*unitcircle);<br />
</asy><br />
[[Category: Introductory Geometry Problems]]<br />
<br />
<br />
==See Also==<br />
{{AMC10 box|year=2014|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Derpguyshttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_21&diff=724902014 AMC 10B Problems/Problem 212015-10-16T03:03:38Z<p>Derpguys: </p>
<hr />
<div>==Problem==<br />
Trapezoid <math> ABCD </math> has parallel sides <math> \overline{AB} </math> of length <math> 33 </math> and <math> \overline {CD} </math> of length <math> 21 </math>. The other two sides are of lengths <math> 10 </math> and <math> 14 </math>. The angles <math> A </math> and <math> B </math> are acute. What is the length of the shorter diagonal of <math> ABCD </math>?<br />
<br />
<math> \textbf{(A) }10\sqrt{6}\qquad\textbf{(B) }25\qquad\textbf{(C) }8\sqrt{10}\qquad\textbf{(D) }18\sqrt{2}\qquad\textbf{(E) }26 </math><br />
[[Category: Introductory Geometry Problems]]<br />
<br />
==Solution==<br />
<br />
<asy><br />
size(7cm);<br />
pair A,B,C,D,CC,DD;<br />
A = (-2,7);<br />
B = (14,7);<br />
C = (10,0);<br />
D = (0,0);<br />
CC = (10,7);<br />
DD = (0,7);<br />
draw(A--B--C--D--cycle);<br />
//label("33",(A+B)/2,N);<br />
label("21",(C+D)/2,S);<br />
label("10",(A+D)/2,W);<br />
label("14",(B+C)/2,E);<br />
label("$A$",A,NW);<br />
label("$B$",B,NE);<br />
label("$C$",C,SE);<br />
label("$D$",D,SW);<br />
draw(C--CC); draw(D--DD);<br />
</asy><br />
<br />
In the diagram, <math>\overline{DE} \perp \overline{AB}, \overline{FC} \perp \overline{AB}</math>. <br />
Denote <math>\overline{AE} = x</math> and <math>\overline{DE} = h</math>. In right triangle <math>AED</math>, we have from the Pythagorean theorem: <math>x^2+h^2=100</math>. Note that since <math>EF = DC</math>, we have <math>BF = 33-DC-x = 12-x</math>. Using the Pythagorean theorem in right triangle <math>BFC</math>, we have <math>(12-x)^2 + h^2 = 196</math>. <br />
<br />
<br />
We isolate the <math>h^2</math> term in both equations, getting <math>h^2= 100-x^2</math> and<br />
<math>h^2 = 196-(12-x)^2</math>.<br />
<br />
Setting these equal, we have <math>100-x^2 = 196 - 144 + 24x -x^2 \implies 24x = 48 \implies x = 2</math>. Now, we can determine that <math>h^2 = 100-4 \implies h = \sqrt{96}</math>.<br />
<br />
<asy><br />
size(7cm);<br />
pair A,B,C,D,CC,DD;<br />
A = (-2,7);<br />
B = (14,7);<br />
C = (10,0);<br />
D = (0,0);<br />
CC = (10,7);<br />
DD = (0,7);<br />
draw(A--B--C--D--cycle);<br />
//label("33",(A+B)/2,N);<br />
label("21",(C+D)/2,S);<br />
label("10",(A+D)/2,W);<br />
label("14",(B+C)/2,E);<br />
label("$A$",A,NW);<br />
label("$B$",B,NE);<br />
label("$C$",C,SE);<br />
label("$D$",D,SW);<br />
draw(C--CC); draw(D--DD);<br />
label("21",(CC+DD)/2,N);<br />
label("$2$",(A+DD)/2,N);<br />
label("$10$",(CC+B)/2,N);<br />
label("$\sqrt{96}$",(C+CC)/2,W);<br />
label("$\sqrt{96}$",(D+DD)/2,E);<br />
pair X = (-2,0);<br />
//draw(X--C--A--cycle,black+2bp);<br />
</asy><br />
<br />
The two diagonals are <math>\overline{AC}</math> and <math>\overline{BD}</math>. Using the Pythagorean theorem again on <math>\bigtriangleup AFC</math> and <math>\bigtriangleup BED</math>, we can find these lengths to be <math>\sqrt{96+529} = 25</math> and <math>\sqrt{96+961} = \sqrt{1057}</math>. Obviously, <math>25</math> is the shorter length, and thus the answer is <math>\boxed{\textbf{(B) }25}</math>.<br />
<br />
<br />
==See Also==<br />
{{AMC10 box|year=2014|ab=B|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Derpguyshttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_21&diff=724892014 AMC 10B Problems/Problem 212015-10-16T02:58:22Z<p>Derpguys: </p>
<hr />
<div>==Problem==<br />
Trapezoid <math> ABCD </math> has parallel sides <math> \overline{AB} </math> of length <math> 33 </math> and <math> \overline {CD} </math> of length <math> 21 </math>. The other two sides are of lengths <math> 10 </math> and <math> 14 </math>. The angles <math> A </math> and <math> B </math> are acute. What is the length of the shorter diagonal of <math> ABCD </math>?<br />
<br />
<math> \textbf{(A) }10\sqrt{6}\qquad\textbf{(B) }25\qquad\textbf{(C) }8\sqrt{10}\qquad\textbf{(D) }18\sqrt{2}\qquad\textbf{(E) }26 </math><br />
[[Category: Introductory Geometry Problems]]<br />
<br />
<br />
==See Also==<br />
{{AMC10 box|year=2014|ab=B|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Derpguyshttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_20&diff=724882014 AMC 10B Problems/Problem 202015-10-16T02:45:33Z<p>Derpguys: </p>
<hr />
<div>==Problem==<br />
For how many integers <math>x</math> is the number <math>x^4-51x^2+50</math> negative?<br />
<br />
<math> \textbf {(A) } 8 \qquad \textbf {(B) } 10 \qquad \textbf {(C) } 12 \qquad \textbf {(D) } 14 \qquad \textbf {(E) } 16</math><br />
<br />
==Solution 1==<br />
First, note that <math>50+1=51</math>, which motivates us to factor the polynomial as <math>(x^2-50)(x^2-1)</math>. Since this expression is negative, one term must be negative and the other positive. Also, the first term is obviously smaller than the second, so <math>x^2-50<0<x^2-1</math>. Solving this inequality, we find <math>1<x^2<50</math>. There are exactly 12 integers <math>x</math> that satisfy this inequality, <math>\pm 2,3,4,5,6,7</math>.<br />
<br />
Thus our answer is <math>\boxed{\textbf {(C) } 12}</math><br />
<br />
==Solution 2==<br />
Since the <math>x^4-51x^2</math> part of <math>x^4-51x^2+50</math> has to be less than <math>-50</math> (because we want <math>x^4-51x^2+50</math> to be negative), we have the inequality <math>x^4-51x^2<-50</math> --> <math>x^2(x^2-51) <-50</math>. <math>x^2</math> has to be positive, so <math>(x^2-51)</math> is negative. Then we have <math>x^2<51</math>. Try answers to find <math> \boxed{\textbf{(C) }12} </math>.<br />
<br />
<br />
==See Also==<br />
{{AMC10 box|year=2014|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Derpguyshttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_20&diff=724872014 AMC 10B Problems/Problem 202015-10-16T02:43:06Z<p>Derpguys: </p>
<hr />
<div>==Problem==<br />
For how many integers <math>x</math> is the number <math>x^4-51x^2+50</math> negative?<br />
<br />
<math> \textbf {(A) } 8 \qquad \textbf {(B) } 10 \qquad \textbf {(C) } 12 \qquad \textbf {(D) } 14 \qquad \textbf {(E) } 16</math><br />
<br />
<br />
<br />
<br />
==See Also==<br />
{{AMC10 box|year=2014|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Derpguyshttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_25&diff=724862014 AMC 10A Problems/Problem 252015-10-16T02:42:05Z<p>Derpguys: </p>
<hr />
<div>{{duplicate|[[2014 AMC 12A Problems|2014 AMC 12A #22]] and [[2014 AMC 10A Problems|2014 AMC 10A #25]]}}<br />
==ProbIem==<br />
<br />
The number <math>5^{867}</math> is between <math>2^{2013}</math> and <math>2^{2014}</math>. How many pairs of integers <math>(m,n)</math> are there such that <math>1\leq m\leq 2012</math> and <cmath>5^n<2^m<2^{m+2}<5^{n+1}?</cmath><br />
<math>\textbf{(A) }278\qquad<br />
\textbf{(B) }279\qquad<br />
\textbf{(C) }280\qquad<br />
\textbf{(D) }281\qquad<br />
\textbf{(E) }282\qquad</math><br />
<br />
==Solution==<br />
<br />
Between any two consecutive powers of <math>5</math> there are either <math>2</math> or <math>3</math> powers of <math>2</math> (because <math>2^2<5^1<2^3</math>). Consider the intervals <math>(5^0,5^1),(5^1,5^2),\dots (5^{866},5^{867})</math>. We want the number of intervals with <math>3</math> powers of <math>2</math>.<br />
<br />
From the given that <math>2^{2013}<5^{867}<2^{2014}</math>, we know that these <math>867</math> intervals together have <math>2013</math> powers of <math>2</math>. Let <math>x</math> of them have <math>2</math> powers of <math>2</math> and <math>y</math> of them have <math>3</math> powers of <math>2</math>. Thus we have the system<br />
<cmath>x+y=867</cmath><cmath>2x+3y=2013</cmath><br />
from which we get <math>y=279</math>, so the answer is <math>\boxed{\textbf{(B)}}</math>.<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2014|ab=A|num-b=24|after=Last Problem}}<br />
{{AMC12 box|year=2014|ab=A|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Derpguyshttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_25&diff=724852014 AMC 10A Problems/Problem 252015-10-16T02:41:42Z<p>Derpguys: /* Solution */</p>
<hr />
<div>{{duplicate|[[2014 AMC 12A Problems|2014 AMC 12A #22]] and [[2014 AMC 10A Problems|2014 AMC 10A #25]]}}<br />
==ProbIem==<br />
<br />
The number <math>5^{867}</math> is between <math>2^{2013}</math> and <math>2^{2014}</math>. How many pairs of integers <math>(m,n)</math> are there such that <math>1\leq m\leq 2012</math> and <cmath>5^n<2^m<2^{m+2}<5^{n+1}?</cmath><br />
<math>\textbf{(A) }278\qquad<br />
\textbf{(B) }279\qquad<br />
\textbf{(C) }280\qquad<br />
\textbf{(D) }281\qquad<br />
\textbf{(E) }282\qquad</math><br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2014|ab=A|num-b=24|after=Last Problem}}<br />
{{AMC12 box|year=2014|ab=A|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Derpguyshttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_25&diff=724842014 AMC 10A Problems/Problem 252015-10-16T02:41:22Z<p>Derpguys: </p>
<hr />
<div>{{duplicate|[[2014 AMC 12A Problems|2014 AMC 12A #22]] and [[2014 AMC 10A Problems|2014 AMC 10A #25]]}}<br />
==ProbIem==<br />
<br />
The number <math>5^{867}</math> is between <math>2^{2013}</math> and <math>2^{2014}</math>. How many pairs of integers <math>(m,n)</math> are there such that <math>1\leq m\leq 2012</math> and <cmath>5^n<2^m<2^{m+2}<5^{n+1}?</cmath><br />
<math>\textbf{(A) }278\qquad<br />
\textbf{(B) }279\qquad<br />
\textbf{(C) }280\qquad<br />
\textbf{(D) }281\qquad<br />
\textbf{(E) }282\qquad</math><br />
<br />
==Solution==<br />
<br />
Between any two consecutive powers of <math>5</math> there are either <math>2</math> or <math>3</math> powers of <math>2</math> (because <math>2^2<5^1<2^3</math>). Consider the intervals <math>(5^0,5^1),(5^1,5^2),\dots (5^{866},5^{867})</math>. We want the number of intervals with <math>3</math> powers of <math>2</math>.<br />
<br />
From the given that <math>2^{2013}<5^{867}<2^{2014}</math>, we know that these <math>867</math> intervals together have <math>2013</math> powers of <math>2</math>. Let <math>x</math> of them have <math>2</math> powers of <math>2</math> and <math>y</math> of them have <math>3</math> powers of <math>2</math>. Thus we have the system<br />
<cmath>x+y=867</cmath><cmath>2x+3y=2013</cmath><br />
from which we get <math>y=279</math>, so the answer is <math>\boxed{\textbf{(B)}}</math>.<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2014|ab=A|num-b=24|after=Last Problem}}<br />
{{AMC12 box|year=2014|ab=A|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Derpguyshttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_25&diff=724832014 AMC 10A Problems/Problem 252015-10-16T02:37:33Z<p>Derpguys: /* Solution */</p>
<hr />
<div>{{duplicate|[[2014 AMC 12A Problems|2014 AMC 12A #22]] and [[2014 AMC 10A Problems|2014 AMC 10A #25]]}}<br />
==ProbIem==<br />
<br />
The number <math>5^{867}</math> is between <math>2^{2013}</math> and <math>2^{2014}</math>. How many pairs of integers <math>(m,n)</math> are there such that <math>1\leq m\leq 2012</math> and <cmath>5^n<2^m<2^{m+2}<5^{n+1}?</cmath><br />
<math>\textbf{(A) }278\qquad<br />
\textbf{(B) }279\qquad<br />
\textbf{(C) }280\qquad<br />
\textbf{(D) }281\qquad<br />
\textbf{(E) }282\qquad</math><br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2014|ab=A|num-b=24|after=Last Problem}}<br />
{{AMC12 box|year=2014|ab=A|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Derpguyshttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_25&diff=724822014 AMC 10A Problems/Problem 252015-10-16T02:30:53Z<p>Derpguys: /* ProbIem */</p>
<hr />
<div>{{duplicate|[[2014 AMC 12A Problems|2014 AMC 12A #22]] and [[2014 AMC 10A Problems|2014 AMC 10A #25]]}}<br />
==ProbIem==<br />
<br />
The number <math>5^{867}</math> is between <math>2^{2013}</math> and <math>2^{2014}</math>. How many pairs of integers <math>(m,n)</math> are there such that <math>1\leq m\leq 2012</math> and <cmath>5^n<2^m<2^{m+2}<5^{n+1}?</cmath><br />
<math>\textbf{(A) }278\qquad<br />
\textbf{(B) }279\qquad<br />
\textbf{(C) }280\qquad<br />
\textbf{(D) }281\qquad<br />
\textbf{(E) }282\qquad</math><br />
<br />
==Solution==<br />
<br />
Between any two consecutive powers of <math>5</math> there are either <math>2</math> or <math>3</math> powers of <math>2</math> (because <math>2^2<5^1<2^3</math>). Consider the intervals <math>(5^0,5^1),(5^1,5^2),\dots (5^{866},5^{867})</math>. We want the number of intervals with <math>3</math> powers of <math>2</math>.<br />
<br />
From the given that <math>2^{2013}<5^{867}<2^{2014}</math>, we know that these <math>867</math> intervals together have <math>2013</math> powers of <math>2</math>. Let <math>x</math> of them have <math>2</math> powers of <math>2</math> and <math>y</math> of them have <math>3</math> powers of <math>2</math>. Thus we have the system<br />
<cmath>x+y=867</cmath><cmath>2x+3y=2013</cmath><br />
from which we get <math>y=279</math>, so the answer is <math>\boxed{\textbf{(B)}}</math>.<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2014|ab=A|num-b=24|after=Last Problem}}<br />
{{AMC12 box|year=2014|ab=A|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Derpguys