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1991 AHSME Problems/Problem 22 - Revision history
2024-03-29T15:26:43Z
Revision history for this page on the wiki
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https://artofproblemsolving.com/wiki/index.php?title=1991_AHSME_Problems/Problem_22&diff=80596&oldid=prev
Fwbrmath2: /* Solution */
2016-10-11T23:09:05Z
<p><span dir="auto"><span class="autocomment">Solution</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 23:09, 11 October 2016</td>
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<td colspan="2" class="diff-lineno">Line 19:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Solution ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Solution ==</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Using the tangent-tangent theorem, <math>PA=AB=PA'=A'B'=4</math>. We can then drop perpendiculars from the centers of the circles to the points of tangency and use similar triangles. Let us let the center of the smaller circle be point <math>S</math> and the center of the larger circle be point <math>L</math>. If we let the radius of the larger circle be <math>x</math> and the radius of the smaller circle be <math>y</math>, we can see that, using similar triangle, <math>x=2y</math>. In addition, the total hypotenuse of the larger right triangles equals <math>2(x+y)</math> since half of it is <math>x+y</math>, so <math>y^2+4^2=(3y)^2</math>. If we simplify, we get <math>y^2+16=9y^2</math>, so <math>8y^2=16</math>, so <math>y=\sqrt2</math>. This means that the smaller circle has area <math><del class="diffchange diffchange-inline">\fbox{</del>2\pi<del class="diffchange diffchange-inline">}</del></math>, which is answer choice <math>\fbox{B}</math>.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Using the tangent-tangent theorem, <math>PA=AB=PA'=A'B'=4</math>. We can then drop perpendiculars from the centers of the circles to the points of tangency and use similar triangles. Let us let the center of the smaller circle be point <math>S</math> and the center of the larger circle be point <math>L</math>. If we let the radius of the larger circle be <math>x</math> and the radius of the smaller circle be <math>y</math>, we can see that, using similar triangle, <math>x=2y</math>. In addition, the total hypotenuse of the larger right triangles equals <math>2(x+y)</math> since half of it is <math>x+y</math>, so <math>y^2+4^2=(3y)^2</math>. If we simplify, we get <math>y^2+16=9y^2</math>, so <math>8y^2=16</math>, so <math>y=\sqrt2</math>. This means that the smaller circle has area <math>2\pi</math>, which is answer choice <math>\fbox{B}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td></tr>
</table>
Fwbrmath2
https://artofproblemsolving.com/wiki/index.php?title=1991_AHSME_Problems/Problem_22&diff=80595&oldid=prev
Fwbrmath2: /* Solution */
2016-10-11T23:07:32Z
<p><span dir="auto"><span class="autocomment">Solution</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 23:07, 11 October 2016</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l19" >Line 19:</td>
<td colspan="2" class="diff-lineno">Line 19:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Solution ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Solution ==</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><math>\fbox{B}</math></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">Using the tangent-tangent theorem, <math>PA=AB=PA'=A'B'=4</math>. We can then drop perpendiculars from the centers of the circles to the points of tangency and use similar triangles. Let us let the center of the smaller circle be point <math>S</math> and the center of the larger circle be point <math>L</math>. If we let the radius of the larger circle be <math>x</math> and the radius of the smaller circle be <math>y</math>, we can see that, using similar triangle, <math>x=2y</math>. In addition, the total hypotenuse of the larger right triangles equals <math>2(x+y)</math> since half of it is <math>x+y</math>, so <math>y^2+4^2=(3y)^2</math>. If we simplify, we get <math>y^2+16=9y^2</math>, so <math>8y^2=16</math>, so <math>y=\sqrt2</math>. This means that the smaller circle has area <math>\fbox{2\pi}</math>, which is answer choice </ins><math>\fbox{B}</math><ins class="diffchange diffchange-inline">.</ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td></tr>
</table>
Fwbrmath2
https://artofproblemsolving.com/wiki/index.php?title=1991_AHSME_Problems/Problem_22&diff=63690&oldid=prev
Timneh: /* Problem */
2014-09-28T19:48:09Z
<p><span dir="auto"><span class="autocomment">Problem</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
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<col class="diff-marker" />
<col class="diff-content" />
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 19:48, 28 September 2014</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l1" >Line 1:</td>
<td colspan="2" class="diff-lineno">Line 1:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Problem ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Problem ==</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><asy></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><asy></div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>draw(circle((0,0),<del class="diffchange diffchange-inline">10</del>),black+linewidth(.75));</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>draw(circle((0,<ins class="diffchange diffchange-inline">6sqrt(2)),2sqrt(2)),black+linewidth(.75));</ins></div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>MP(")",(0,0),S);</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">draw(circle((</ins>0<ins class="diffchange diffchange-inline">,3sqrt(2)</ins>),<ins class="diffchange diffchange-inline">sqrt(2)</ins>),black+linewidth(.75)<ins class="diffchange diffchange-inline">);</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">draw((-8/3,16sqrt(2)/3)--(-4/3,8sqrt(2)/3)--(0,0)--(4/3,8sqrt(2)/3)--(8/3,16sqrt(2)/3),dot);</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">MP("B",(-8/3,16*sqrt(2)/3),W);MP("B'",(8/3,16*sqrt(2)/3),E</ins>);</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>MP("<ins class="diffchange diffchange-inline">A",(-4/3,8*sqrt(2)/3),W</ins>)<ins class="diffchange diffchange-inline">;MP("A'",(4/3,8*sqrt(2)/3),E);</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">MP("P</ins>",(0,0),S);</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div></asy></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div></asy></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Two circles are externally tangent. Lines <math>\overline{PAB}</math> and <math>\overline{PA'B'}</math> are common tangents with <math>A</math> and <math>A'</math> on the smaller circle <math>B</math> and <math>B'</math> on the larger circle. If <math>PA=AB=4</math>, then the <del class="diffchange diffchange-inline">are </del>of the smaller circle is</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Two circles are externally tangent. Lines <math>\overline{PAB}</math> and <math>\overline{PA'B'}</math> are common tangents with <math>A</math> and <math>A'</math> on the smaller circle <math>B</math> and <math>B'</math> on the larger circle. If <math>PA=AB=4</math>, then the <ins class="diffchange diffchange-inline">area </ins>of the smaller circle is</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><math>\text{(A) } 1.44\pi\quad</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><math>\text{(A) } 1.44\pi\quad</div></td></tr>
</table>
Timneh
https://artofproblemsolving.com/wiki/index.php?title=1991_AHSME_Problems/Problem_22&diff=63689&oldid=prev
Timneh: Created page with "== Problem == <asy> draw(circle((0,0),10),black+linewidth(.75)); MP(")",(0,0),S); </asy> Two circles are externally tangent. Lines <math>\overline{PAB}</math> and <math>\overli..."
2014-09-28T19:37:42Z
<p>Created page with "== Problem == <asy> draw(circle((0,0),10),black+linewidth(.75)); MP(")",(0,0),S); </asy> Two circles are externally tangent. Lines <math>\overline{PAB}</math> and <math>\overli..."</p>
<p><b>New page</b></p><div>== Problem ==<br />
<asy><br />
draw(circle((0,0),10),black+linewidth(.75));<br />
MP(")",(0,0),S);<br />
</asy><br />
<br />
<br />
Two circles are externally tangent. Lines <math>\overline{PAB}</math> and <math>\overline{PA'B'}</math> are common tangents with <math>A</math> and <math>A'</math> on the smaller circle <math>B</math> and <math>B'</math> on the larger circle. If <math>PA=AB=4</math>, then the are of the smaller circle is<br />
<br />
<math>\text{(A) } 1.44\pi\quad<br />
\text{(B) } 2\pi\quad<br />
\text{(C) } 2.56\pi\quad<br />
\text{(D) } \sqrt{8}\pi\quad<br />
\text{(E) } 4\pi</math><br />
<br />
== Solution ==<br />
<math>\fbox{B}</math><br />
<br />
== See also ==<br />
{{AHSME box|year=1991|num-b=21|num-a=23}} <br />
<br />
[[Category: Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>
Timneh