https://artofproblemsolving.com/wiki/index.php?title=1980_AHSME_Problems/Problem_5&feed=atom&action=history
1980 AHSME Problems/Problem 5 - Revision history
2024-03-29T05:10:08Z
Revision history for this page on the wiki
MediaWiki 1.31.1
https://artofproblemsolving.com/wiki/index.php?title=1980_AHSME_Problems/Problem_5&diff=56288&oldid=prev
Nathan wailes at 16:47, 5 July 2013
2013-07-05T16:47:26Z
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 16:47, 5 July 2013</td>
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<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>{{AHSME box|year=1980|num-b=4|num-a=6}}</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>{{AHSME box|year=1980|num-b=4|num-a=6}}</div></td></tr>
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Nathan wailes
https://artofproblemsolving.com/wiki/index.php?title=1980_AHSME_Problems/Problem_5&diff=52018&oldid=prev
Claudiafeng at 23:58, 31 March 2013
2013-03-31T23:58:39Z
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<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)} \ \frac{\sqrt{3}}{2} \qquad \text{(B)} \ \frac{\sqrt{3}}{3} \qquad \text{(C)} \ \frac{\sqrt{2}}{2} \qquad \text{(D)} \ \frac12 \qquad \text{(E)} \ \frac23 </math></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)} \ \frac{\sqrt{3}}{2} \qquad \text{(B)} \ \frac{\sqrt{3}}{3} \qquad \text{(C)} \ \frac{\sqrt{2}}{2} \qquad \text{(D)} \ \frac12 \qquad \text{(E)} \ \frac23 </math></div></td></tr>
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<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 style="font-weight: bold; text-decoration: none;">== Solution ==</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 style="font-weight: bold; text-decoration: none;"></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 style="font-weight: bold; text-decoration: none;">We find that <math> m\angle PCQ=30^\circ </math>. Because it is a <math> 30^\circ-60^\circ-90^\circ </math> right triangle, we can let <math> PQ=x </math>, so <math> CQ=AQ=x\sqrt{3} </math>. Thus, <math> \frac{PQ}{AQ}=\frac{x}{x\sqrt{3}}=\frac{\sqrt{3}}{3}\Rightarrow\boxed{(B)} </math>.</ins></div></td></tr>
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Claudiafeng
https://artofproblemsolving.com/wiki/index.php?title=1980_AHSME_Problems/Problem_5&diff=47647&oldid=prev
Mrdavid445: Created page with "==Problem== If <math>AB</math> and <math>CD</math> are perpendicular diameters of circle <math>Q</math>, <math>P</math> in <math>\overline{AQ}</math>, and <math>\measuredangle Q..."
2012-07-16T17:09:03Z
<p>Created page with "==Problem== If <math>AB</math> and <math>CD</math> are perpendicular diameters of circle <math>Q</math>, <math>P</math> in <math>\overline{AQ}</math>, and <math>\measuredangle Q..."</p>
<p><b>New page</b></p><div>==Problem==<br />
<br />
If <math>AB</math> and <math>CD</math> are perpendicular diameters of circle <math>Q</math>, <math>P</math> in <math>\overline{AQ}</math>, and <math>\measuredangle QPC = 60^\circ</math>, then the length of <math>PQ</math> divided by the length of <math>AQ</math> is <br />
<br />
<asy><br />
defaultpen(linewidth(0.7)+fontsize(10));<br />
pair A=(-1,0), B=(1,0), C=(0,1), D=(0,-1), Q=origin, P=(-0.5,0);<br />
draw(P--C--D^^A--B^^Circle(Q,1));<br />
label("$A$", A, W);<br />
label("$B$", B, E);<br />
label("$C$", C, N);<br />
label("$D$", D, S);<br />
label("$P$", P, S);<br />
label("$Q$", Q, SE);<br />
label("$60^\circ$", P+0.0.5*dir(30), dir(30));</asy><br />
<br />
<math> \text{(A)} \ \frac{\sqrt{3}}{2} \qquad \text{(B)} \ \frac{\sqrt{3}}{3} \qquad \text{(C)} \ \frac{\sqrt{2}}{2} \qquad \text{(D)} \ \frac12 \qquad \text{(E)} \ \frac23 </math></div>
Mrdavid445