https://artofproblemsolving.com/wiki/index.php?title=1983_AHSME_Problems/Problem_19&feed=atom&action=history
1983 AHSME Problems/Problem 19 - Revision history
2024-03-19T09:54:29Z
Revision history for this page on the wiki
MediaWiki 1.31.1
https://artofproblemsolving.com/wiki/index.php?title=1983_AHSME_Problems/Problem_19&diff=103577&oldid=prev
Sevenoptimus: Fixed formatting
2019-02-20T04:57:51Z
<p>Fixed formatting</p>
<table class="diff diff-contentalign-left" data-mw="interface">
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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 04:57, 20 February 2019</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l13" >Line 13:</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>==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="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>Let <math>AD = y</math>. Since <math>AD</math> bisects <math>\angle{BAC}</math>, the Angle Bisector Theorem gives <math>\frac{DB}{CD} = \frac{AB}{AC} = 2</math>, so let <math>CD = x</math> and <math>DB = 2x</math>. Applying the Law of Cosines to <math>\triangle CAD</math> gives <math>x^2 = 3^2 + y^2 - 3y</math>, and to <math>\triangle DAB</math> gives <math>(2x)^2 = 6^2 + y^2 - 6y</math>. Subtracting <math>4</math> times the first equation from the second equation therefore yields <math>0 = 6y - 3y^2 \Rightarrow y(y-2) = 0</math>, so <math>y</math> is <math>0</math> or <math>2</math>. But since <math>y \neq 0</math> (<math>y</math> is the length of a side of a triangle), <math>y</math> must be <math>2</math>, so the answer is <math>\boxed{\textbf{A}}</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>Let <math>AD = y</math>. Since <math>AD</math> bisects <math>\angle{BAC}</math>, the Angle Bisector Theorem gives <math>\frac{DB}{CD} = \frac{AB}{AC} = 2</math>, so let <math>CD = x</math> and <math>DB = 2x</math>. Applying the Law of Cosines to <math>\triangle CAD</math> gives <math>x^2 = 3^2 + y^2 - 3y</math>, and to <math>\triangle DAB</math> gives <math>(2x)^2 = 6^2 + y^2 - 6y</math>. Subtracting <math>4</math> times the first equation from the second equation therefore yields <math>0 = 6y - 3y^2 \Rightarrow y(y-2) = 0</math>, so <math>y</math> is <math>0</math> or <math>2</math>. But since <math>y \neq 0</math> (<math>y</math> is the length of a side of a triangle), <math>y</math> must be <math>2</math>, so the answer is <math>\boxed{\textbf{<ins class="diffchange diffchange-inline">(</ins>A<ins class="diffchange diffchange-inline">)</ins>} <ins class="diffchange diffchange-inline">\ 2</ins>}</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>
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Sevenoptimus
https://artofproblemsolving.com/wiki/index.php?title=1983_AHSME_Problems/Problem_19&diff=103576&oldid=prev
Sevenoptimus: Improved clarity and added box at the bottom
2019-02-20T04:57:22Z
<p>Improved clarity and added box at the bottom</p>
<table class="diff diff-contentalign-left" data-mw="interface">
<col class="diff-marker" />
<col class="diff-content" />
<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<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 04:57, 20 February 2019</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l13" >Line 13:</td>
<td colspan="2" class="diff-lineno">Line 13:</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="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>Let <math>AD = y</math>. Since <math>AD</math> bisects <math>\angle{BAC}</math>, the Angle Bisector Theorem gives <math>\frac{DB}{CD} = \frac{AB}{AC} = 2</math>, so <del class="diffchange diffchange-inline">write </del><math>CD = x</math> and <math>DB = 2x</math>. Applying the Law of Cosines to <math>\triangle CAD</math> gives <math>x^2 = 3^2 + y^2 - 3y</math>, and to <math>\triangle DAB</math> gives <math>(2x)^2 = 6^2 + y^2 - 6y</math>. Subtracting <math>4</math> times the first equation from the second equation therefore yields <math>0 = 6y - 3y^2 \Rightarrow y(y-2) = 0</math>, so <math>y</math> is <math>0</math> or <math>2</math>. But since <math>y \neq 0</math> (<math>y</math> is the length of a side of a triangle), <math>y</math> must be <math>2</math>, so the answer is <math>\boxed{\textbf{A}}</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>Let <math>AD = y</math>. Since <math>AD</math> bisects <math>\angle{BAC}</math>, the Angle Bisector Theorem gives <math>\frac{DB}{CD} = \frac{AB}{AC} = 2</math>, so <ins class="diffchange diffchange-inline">let </ins><math>CD = x</math> and <math>DB = 2x</math>. Applying the Law of Cosines to <math>\triangle CAD</math> gives <math>x^2 = 3^2 + y^2 - 3y</math>, and to <math>\triangle DAB</math> gives <math>(2x)^2 = 6^2 + y^2 - 6y</math>. Subtracting <math>4</math> times the first equation from the second equation therefore yields <math>0 = 6y - 3y^2 \Rightarrow y(y-2) = 0</math>, so <math>y</math> is <math>0</math> or <math>2</math>. But since <math>y \neq 0</math> (<math>y</math> is the length of a side of a triangle), <math>y</math> must be <math>2</math>, so the answer is <math>\boxed{\textbf{A}}</math>.</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> </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">==See Also==</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">{{AHSME box|year=1983|num-b=18|num-a=20}}</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> </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">{{MAA Notice}}</ins></div></td></tr>
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Sevenoptimus
https://artofproblemsolving.com/wiki/index.php?title=1983_AHSME_Problems/Problem_19&diff=100962&oldid=prev
Sevenoptimus: Added a solution
2019-01-27T21:00:42Z
<p>Added a solution</p>
<p><b>New page</b></p><div>==Problem==<br />
<br />
Point <math>D</math> is on side <math>CB</math> of triangle <math>ABC</math>. If <br />
<math>\angle{CAD} = \angle{DAB} = 60^\circ\mbox{, }AC = 3\mbox{ and }AB = 6</math>, <br />
then the length of <math>AD</math> is <br />
<br />
<math>\textbf{(A)} \ 2 \qquad <br />
\textbf{(B)} \ 2.5 \qquad <br />
\textbf{(C)} \ 3 \qquad <br />
\textbf{(D)} \ 3.5 \qquad <br />
\textbf{(E)} \ 4 </math> <br />
<br />
==Solution==<br />
<br />
Let <math>AD = y</math>. Since <math>AD</math> bisects <math>\angle{BAC}</math>, the Angle Bisector Theorem gives <math>\frac{DB}{CD} = \frac{AB}{AC} = 2</math>, so write <math>CD = x</math> and <math>DB = 2x</math>. Applying the Law of Cosines to <math>\triangle CAD</math> gives <math>x^2 = 3^2 + y^2 - 3y</math>, and to <math>\triangle DAB</math> gives <math>(2x)^2 = 6^2 + y^2 - 6y</math>. Subtracting <math>4</math> times the first equation from the second equation therefore yields <math>0 = 6y - 3y^2 \Rightarrow y(y-2) = 0</math>, so <math>y</math> is <math>0</math> or <math>2</math>. But since <math>y \neq 0</math> (<math>y</math> is the length of a side of a triangle), <math>y</math> must be <math>2</math>, so the answer is <math>\boxed{\textbf{A}}</math>.</div>
Sevenoptimus