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2019 AMC 10B Problems/Problem 16 - Revision history
2024-03-28T12:19:31Z
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Magnetoninja: /* Solution 5 (Short with Trig) */
2023-06-06T21:26:08Z
<p><span dir="auto"><span class="autocomment">Solution 5 (Short with Trig)</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 21:26, 6 June 2023</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l52" >Line 52:</td>
<td colspan="2" class="diff-lineno">Line 52:</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 5 (Short with Trig)==</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 5 (Short with Trig)==</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>\angle B=\theta_1</math>, then <math>\angle A=90-\theta_1</math>. Since <math>AC=AD</math>, <math>\angle ADC=90-\theta_1</math>. Similarly, <math>\angle BDE=\theta_1</math>. Then, <math>\angle EDC=180-\theta_1-(90-\theta_1)=90</math>. Therefore <math>\bigtriangleup CDE</math> is right. Let <math>AC=CD=4</math> and <math>DE=EB=3</math>, then <math>EC=5</math>. Let <math>\angle DEC=\angle ACD=\theta_2</math>. We know that <math>\cos \theta_2=\frac{3}{5}</math> so we can apply the Law of Cosines on <math>\bigtriangleup ACD</math> to find <math>AD=\sqrt{32-32\cdot{\frac{3}{5}}}=\sqrt{\frac{2}{5}\cdot{32}} \Longrightarrow \frac{8}{\sqrt{5}}</math>. Doing Pythagorean for <math>BA</math>, we get <math>4\sqrt{5}</math>. Then, <math>BD=4\sqrt{5}-\frac{8}{\sqrt{5}} \Longrightarrow \frac{12}{\sqrt{5}}</math><del class="diffchange diffchange-inline">. Then, </del>the requested ratio is <math>8:12=\boxed{\textbf{(A) } 2:3}</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>\angle B=\theta_1</math>, then <math>\angle A=90-\theta_1</math>. Since <math>AC=AD</math>, <math>\angle ADC=90-\theta_1</math>. Similarly, <math>\angle BDE=\theta_1</math>. Then, <math>\angle EDC=180-\theta_1-(90-\theta_1)=90</math>. Therefore <math>\bigtriangleup CDE</math> is right. Let <math>AC=CD=4</math> and <math>DE=EB=3</math>, then <math>EC=5</math>. Let <math>\angle DEC=\angle ACD=\theta_2</math>. We know that <math>\cos \theta_2=\frac{3}{5}</math> so we can apply the Law of Cosines on <math>\bigtriangleup ACD</math> to find <math>AD=\sqrt{32-32\cdot{\frac{3}{5}}}=\sqrt{\frac{2}{5}\cdot{32}} \Longrightarrow \frac{8}{\sqrt{5}}</math>. Doing Pythagorean for <math>BA</math>, we get <math>4\sqrt{5}</math>. Then, <math>BD=4\sqrt{5}-\frac{8}{\sqrt{5}} \Longrightarrow \frac{12}{\sqrt{5}}</math> <ins class="diffchange diffchange-inline">so </ins>the requested ratio is <math>8:12=\boxed{\textbf{(A) } 2:3}</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>~[https://artofproblemsolving.com/wiki/index.php/User:Magnetoninja Magnetoninja]</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>~[https://artofproblemsolving.com/wiki/index.php/User:Magnetoninja Magnetoninja]</div></td></tr>
</table>
Magnetoninja
https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_16&diff=194004&oldid=prev
Magnetoninja: /* Solution 5 (Short with Trig) */
2023-06-06T21:25:26Z
<p><span dir="auto"><span class="autocomment">Solution 5 (Short with Trig)</span></span></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 21:25, 6 June 2023</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l52" >Line 52:</td>
<td colspan="2" class="diff-lineno">Line 52:</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 5 (Short with Trig)==</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 5 (Short with Trig)==</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>\angle B=\<del class="diffchange diffchange-inline">theta</del></math>, then <math>\angle A=90-\<del class="diffchange diffchange-inline">theta</del></math>. Since <math>AC=AD</math>, <math>\angle ADC=90-\<del class="diffchange diffchange-inline">theta</del></math>. Similarly, <math>\angle BDE=\<del class="diffchange diffchange-inline">theta</del></math>. Then, <math>\angle EDC=180-\<del class="diffchange diffchange-inline">theta</del>-(90-\<del class="diffchange diffchange-inline">theta</del>)=90</math>. Therefore <math>\bigtriangleup CDE</math> is right. Let <math>AC=CD=4</math> and <math>DE=EB=3</math>, then <math>EC=5</math>. Let <math>\angle DEC=\angle ACD=\<del class="diffchange diffchange-inline">alpha</del></math>. We know that <math>\cos \<del class="diffchange diffchange-inline">alpha</del>=\frac{3}{5}</math> so we can apply the Law of Cosines on <math>\bigtriangleup ACD</math> to find <math>AD=\sqrt{32-32\cdot{\frac{3}{5}}}=\sqrt{\frac{2}{5}\cdot{32}} \Longrightarrow \frac{8}{\sqrt{5}}</math>. Doing Pythagorean for <math>BA</math>, we get <math>4\sqrt{5}</math>. Then, <math>BD=4\sqrt{5}-\frac{8}{\sqrt{5}} \Longrightarrow \frac{12}{\sqrt{5}}</math>. Then, the requested ratio is <math>8:12=\boxed{\textbf{(A) } 2:3}</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>\angle B=\<ins class="diffchange diffchange-inline">theta_1</ins></math>, then <math>\angle A=90-\<ins class="diffchange diffchange-inline">theta_1</ins></math>. Since <math>AC=AD</math>, <math>\angle ADC=90-\<ins class="diffchange diffchange-inline">theta_1</ins></math>. Similarly, <math>\angle BDE=\<ins class="diffchange diffchange-inline">theta_1</ins></math>. Then, <math>\angle EDC=180-\<ins class="diffchange diffchange-inline">theta_1</ins>-(90-\<ins class="diffchange diffchange-inline">theta_1</ins>)=90</math>. Therefore <math>\bigtriangleup CDE</math> is right. Let <math>AC=CD=4</math> and <math>DE=EB=3</math>, then <math>EC=5</math>. Let <math>\angle DEC=\angle ACD=\<ins class="diffchange diffchange-inline">theta_2</ins></math>. We know that <math>\cos \<ins class="diffchange diffchange-inline">theta_2</ins>=\frac{3}{5}</math> so we can apply the Law of Cosines on <math>\bigtriangleup ACD</math> to find <math>AD=\sqrt{32-32\cdot{\frac{3}{5}}}=\sqrt{\frac{2}{5}\cdot{32}} \Longrightarrow \frac{8}{\sqrt{5}}</math>. Doing Pythagorean for <math>BA</math>, we get <math>4\sqrt{5}</math>. Then, <math>BD=4\sqrt{5}-\frac{8}{\sqrt{5}} \Longrightarrow \frac{12}{\sqrt{5}}</math>. Then, the requested ratio is <math>8:12=\boxed{\textbf{(A) } 2:3}</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>~[https://artofproblemsolving.com/wiki/index.php/User:Magnetoninja Magnetoninja]</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>~[https://artofproblemsolving.com/wiki/index.php/User:Magnetoninja Magnetoninja]</div></td></tr>
</table>
Magnetoninja
https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_16&diff=193483&oldid=prev
Magnetoninja: /* Solution 1 */
2023-05-25T04:20:20Z
<p><span dir="auto"><span class="autocomment">Solution 1</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 04:20, 25 May 2023</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l33" >Line 33:</td>
<td colspan="2" class="diff-lineno">Line 33:</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>Alternatively, once finding the length of <math>AD</math> one could use the Pythagorean Theorem to find <math>AB</math> and consequently <math>DB</math>, and then compute the ratio.</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>Alternatively, once finding the length of <math>AD</math> one could use the Pythagorean Theorem to find <math>AB</math> and consequently <math>DB</math>, and then compute the ratio.</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><del style="font-weight: bold; text-decoration: none;"></del></div></td><td colspan="2"> </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><del style="font-weight: bold; text-decoration: none;">~[https://artofproblemsolving.com/wiki/index.php/User:Magnetoninja Magnetoninja]</del></div></td><td colspan="2"> </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 2==</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 2==</div></td></tr>
</table>
Magnetoninja
https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_16&diff=193482&oldid=prev
Magnetoninja: /* Solution 5 (Short with Trig) */
2023-05-25T04:20:12Z
<p><span dir="auto"><span class="autocomment">Solution 5 (Short with Trig)</span></span></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 04:20, 25 May 2023</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l55" >Line 55:</td>
<td colspan="2" class="diff-lineno">Line 55:</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>Let <math>\angle B=\theta</math>, then <math>\angle A=90-\theta</math>. Since <math>AC=AD</math>, <math>\angle ADC=90-\theta</math>. Similarly, <math>\angle BDE=\theta</math>. Then, <math>\angle EDC=180-\theta-(90-\theta)=90</math>. Therefore <math>\bigtriangleup CDE</math> is right. Let <math>AC=CD=4</math> and <math>DE=EB=3</math>, then <math>EC=5</math>. Let <math>\angle DEC=\angle ACD=\alpha</math>. We know that <math>\cos \alpha=\frac{3}{5}</math> so we can apply the Law of Cosines on <math>\bigtriangleup ACD</math> to find <math>AD=\sqrt{32-32\cdot{\frac{3}{5}}}=\sqrt{\frac{2}{5}\cdot{32}} \Longrightarrow \frac{8}{\sqrt{5}}</math>. Doing Pythagorean for <math>BA</math>, we get <math>4\sqrt{5}</math>. Then, <math>BD=4\sqrt{5}-\frac{8}{\sqrt{5}} \Longrightarrow \frac{12}{\sqrt{5}}</math>. Then, the requested ratio is <math>8:12=\boxed{\textbf{(A) } 2:3}</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>Let <math>\angle B=\theta</math>, then <math>\angle A=90-\theta</math>. Since <math>AC=AD</math>, <math>\angle ADC=90-\theta</math>. Similarly, <math>\angle BDE=\theta</math>. Then, <math>\angle EDC=180-\theta-(90-\theta)=90</math>. Therefore <math>\bigtriangleup CDE</math> is right. Let <math>AC=CD=4</math> and <math>DE=EB=3</math>, then <math>EC=5</math>. Let <math>\angle DEC=\angle ACD=\alpha</math>. We know that <math>\cos \alpha=\frac{3}{5}</math> so we can apply the Law of Cosines on <math>\bigtriangleup ACD</math> to find <math>AD=\sqrt{32-32\cdot{\frac{3}{5}}}=\sqrt{\frac{2}{5}\cdot{32}} \Longrightarrow \frac{8}{\sqrt{5}}</math>. Doing Pythagorean for <math>BA</math>, we get <math>4\sqrt{5}</math>. Then, <math>BD=4\sqrt{5}-\frac{8}{\sqrt{5}} \Longrightarrow \frac{12}{\sqrt{5}}</math>. Then, the requested ratio is <math>8:12=\boxed{\textbf{(A) } 2:3}</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><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;">~[https://artofproblemsolving.com/wiki/index.php/User:Magnetoninja Magnetoninja]</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>==Video Solution 1==</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>==Video Solution 1==</div></td></tr>
</table>
Magnetoninja
https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_16&diff=193481&oldid=prev
Magnetoninja: /* Solution 1 */
2023-05-25T04:19:52Z
<p><span dir="auto"><span class="autocomment">Solution 1</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 04:19, 25 May 2023</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l33" >Line 33:</td>
<td colspan="2" class="diff-lineno">Line 33:</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>Alternatively, once finding the length of <math>AD</math> one could use the Pythagorean Theorem to find <math>AB</math> and consequently <math>DB</math>, and then compute the ratio.</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>Alternatively, once finding the length of <math>AD</math> one could use the Pythagorean Theorem to find <math>AB</math> and consequently <math>DB</math>, and then compute the ratio.</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;">~[https://artofproblemsolving.com/wiki/index.php/User:Magnetoninja Magnetoninja]</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>==Solution 2==</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 2==</div></td></tr>
</table>
Magnetoninja
https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_16&diff=193480&oldid=prev
Magnetoninja: /* Solution 5 */
2023-05-25T04:18:33Z
<p><span dir="auto"><span class="autocomment">Solution 5</span></span></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 04:18, 25 May 2023</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l50" >Line 50:</td>
<td colspan="2" class="diff-lineno">Line 50:</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>WLOG, <math>AC = CD = 4</math> and <math>DE = EB = 3</math>. Notice that in <math>\triangle ACB</math>, we have <math>m\angle BAC + m\angle ABC = 90^{\circ}</math>. Since <math>AC = CD</math> and <math>DE = EB</math>, we find that <math>m\angle DAC = m\angle ADC</math> and <math>m\angle DBE = m\angle BDE</math>, so <math>m\angle ADC + m\angle BDE = 90^{\circ}</math> and <math>\angle EDC</math> is right. Therefore, <math>CE = 5</math> by 3-4-5 triangle, <math>CB = 8</math> and <math>AB = 4\sqrt{5}</math>. Define point F such that <math>CF</math> is an altitude; we know the area of the whole triangle is <math>16</math> and we know the hypotenuse is <math>4\sqrt{5}</math>, so <math>CF = \frac{16}{4\sqrt{5}}\cdot2=\frac{8}{\sqrt{5}}</math>. By the geometric mean theorem, <math>x\left(4\sqrt{5}-x\right)=4\sqrt{5}x-x^{2}=\left(\frac{8}{\sqrt{5}}\right)^{2}=\frac{64}{5}</math>. Solving the quadratic we get <math>x=\frac{10\sqrt{5}\pm6\sqrt{5}}{5}</math>, so <math>x=\frac{4\sqrt{5}}{5} or \frac{16\sqrt{5}}{5}</math>. For now, assume <math>x=\frac{4\sqrt{5}}{5}</math>. Then <math>FB=4\sqrt{5}-\frac{4\sqrt{5}}{5}=\frac{16\sqrt{5}}{5}</math>. <math>CF</math> splits <math>AD</math> into two parts (quick congruence by Leg-Angle) so <math>FD = AF = \frac{4\sqrt{5}}{5}</math> and <math>DB = FB - FD = \frac{16\sqrt{5}}{5}-\frac{4\sqrt{5}}{5}=\frac{12\sqrt{5}}{5}</math>. <math>AD = 2\cdot\frac{4\sqrt{5}}{5}=\frac{8\sqrt{5}}{5}</math>. Now we know <math>AD</math> and <math>DB</math>, we can find <math>\frac{AD}{DB}=\frac{\frac{8\sqrt{5}}{5}}{\frac{12\sqrt{5}}{5}}=\frac{8\sqrt{5}}{5}\cdot\frac{5}{12\sqrt{5}}=\frac{8}{12}=\frac{2}{3}</math> or <math>\boxed{\textbf{(A) } 2:3}</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>WLOG, <math>AC = CD = 4</math> and <math>DE = EB = 3</math>. Notice that in <math>\triangle ACB</math>, we have <math>m\angle BAC + m\angle ABC = 90^{\circ}</math>. Since <math>AC = CD</math> and <math>DE = EB</math>, we find that <math>m\angle DAC = m\angle ADC</math> and <math>m\angle DBE = m\angle BDE</math>, so <math>m\angle ADC + m\angle BDE = 90^{\circ}</math> and <math>\angle EDC</math> is right. Therefore, <math>CE = 5</math> by 3-4-5 triangle, <math>CB = 8</math> and <math>AB = 4\sqrt{5}</math>. Define point F such that <math>CF</math> is an altitude; we know the area of the whole triangle is <math>16</math> and we know the hypotenuse is <math>4\sqrt{5}</math>, so <math>CF = \frac{16}{4\sqrt{5}}\cdot2=\frac{8}{\sqrt{5}}</math>. By the geometric mean theorem, <math>x\left(4\sqrt{5}-x\right)=4\sqrt{5}x-x^{2}=\left(\frac{8}{\sqrt{5}}\right)^{2}=\frac{64}{5}</math>. Solving the quadratic we get <math>x=\frac{10\sqrt{5}\pm6\sqrt{5}}{5}</math>, so <math>x=\frac{4\sqrt{5}}{5} or \frac{16\sqrt{5}}{5}</math>. For now, assume <math>x=\frac{4\sqrt{5}}{5}</math>. Then <math>FB=4\sqrt{5}-\frac{4\sqrt{5}}{5}=\frac{16\sqrt{5}}{5}</math>. <math>CF</math> splits <math>AD</math> into two parts (quick congruence by Leg-Angle) so <math>FD = AF = \frac{4\sqrt{5}}{5}</math> and <math>DB = FB - FD = \frac{16\sqrt{5}}{5}-\frac{4\sqrt{5}}{5}=\frac{12\sqrt{5}}{5}</math>. <math>AD = 2\cdot\frac{4\sqrt{5}}{5}=\frac{8\sqrt{5}}{5}</math>. Now we know <math>AD</math> and <math>DB</math>, we can find <math>\frac{AD}{DB}=\frac{\frac{8\sqrt{5}}{5}}{\frac{12\sqrt{5}}{5}}=\frac{8\sqrt{5}}{5}\cdot\frac{5}{12\sqrt{5}}=\frac{8}{12}=\frac{2}{3}</math> or <math>\boxed{\textbf{(A) } 2:3}</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="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>==Solution 5==</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>==Solution 5 <ins class="diffchange diffchange-inline">(Short with Trig)</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>Let <math>\angle B=\theta</math>, then <math>\angle A=90-\theta</math>. Since <math>AC=AD</math>, <math>\angle ADC=90-\theta</math>. Similarly, <math>\angle BDE=\theta</math>. Then, <math>\angle EDC=180-\theta-(90-\theta)=90</math>. Therefore <math>\bigtriangleup CDE</math> is right. Let <math>AC=CD=4</math> and <math>DE=EB=3</math>, then <math>EC=5</math>. Let <math>\angle DEC=\angle ACD=\alpha</math>. We know that <math>\cos \alpha=\frac{3}{5}</math> so we can apply the Law of Cosines on <math>\bigtriangleup ACD</math> to find <math>AD=\sqrt{32-32\cdot{\frac{3}{5}}}=\sqrt{\frac{2}{5}\cdot{32}} \Longrightarrow \frac{8}{\sqrt{5}}</math>. Doing Pythagorean for <math>BA</math>, we get <math>4\sqrt{5}</math>. Then, <math>BD=4\sqrt{5}-\frac{8}{\sqrt{5}} \Longrightarrow \frac{12}{\sqrt{5}}</math>. Then, the requested ratio is <math>8:12=\boxed{\textbf{(A) } 2:3}</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>Let <math>\angle B=\theta</math>, then <math>\angle A=90-\theta</math>. Since <math>AC=AD</math>, <math>\angle ADC=90-\theta</math>. Similarly, <math>\angle BDE=\theta</math>. Then, <math>\angle EDC=180-\theta-(90-\theta)=90</math>. Therefore <math>\bigtriangleup CDE</math> is right. Let <math>AC=CD=4</math> and <math>DE=EB=3</math>, then <math>EC=5</math>. Let <math>\angle DEC=\angle ACD=\alpha</math>. We know that <math>\cos \alpha=\frac{3}{5}</math> so we can apply the Law of Cosines on <math>\bigtriangleup ACD</math> to find <math>AD=\sqrt{32-32\cdot{\frac{3}{5}}}=\sqrt{\frac{2}{5}\cdot{32}} \Longrightarrow \frac{8}{\sqrt{5}}</math>. Doing Pythagorean for <math>BA</math>, we get <math>4\sqrt{5}</math>. Then, <math>BD=4\sqrt{5}-\frac{8}{\sqrt{5}} \Longrightarrow \frac{12}{\sqrt{5}}</math>. Then, the requested ratio is <math>8:12=\boxed{\textbf{(A) } 2:3}</math>.</div></td></tr>
</table>
Magnetoninja
https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_16&diff=193479&oldid=prev
Magnetoninja: /* Solution 5 */
2023-05-25T04:18:16Z
<p><span dir="auto"><span class="autocomment">Solution 5</span></span></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 04:18, 25 May 2023</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l52" >Line 52:</td>
<td colspan="2" class="diff-lineno">Line 52:</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 5==</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 5==</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>We know that</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">Let <math>\angle B=\theta</math>, then <math>\angle A=90-\theta</math>. Since <math>AC=AD</math>, <math>\angle ADC=90-\theta</math>. Similarly, <math>\angle BDE=\theta</math>. Then, <math>\angle EDC=180-\theta-(90-\theta)=90</math>. Therefore <math>\bigtriangleup CDE</math> is right. Let <math>AC=CD=4</math> and <math>DE=EB=3</math>, then <math>EC=5</math>. Let <math>\angle DEC=\angle ACD=\alpha</math>. </ins>We know that <ins class="diffchange diffchange-inline"><math>\cos \alpha=\frac{3}{5}</math> so we can apply the Law of Cosines on <math>\bigtriangleup ACD</math> to find <math>AD=\sqrt{32-32\cdot{\frac{3}{5}}}=\sqrt{\frac{2}{5}\cdot{32}} \Longrightarrow \frac{8}{\sqrt{5}}</math>. Doing Pythagorean for <math>BA</math>, we get <math>4\sqrt{5}</math>. Then, <math>BD=4\sqrt{5}-\frac{8}{\sqrt{5}} \Longrightarrow \frac{12}{\sqrt{5}}</math>. Then, the requested ratio is <math>8:12=\boxed{\textbf{(A) } 2:3}</math>.</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>==Video Solution 1==</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>==Video Solution 1==</div></td></tr>
</table>
Magnetoninja
https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_16&diff=193477&oldid=prev
Magnetoninja: /* Solution 4 (a bit long) */
2023-05-25T03:54:52Z
<p><span dir="auto"><span class="autocomment">Solution 4 (a bit long)</span></span></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 03:54, 25 May 2023</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l49" >Line 49:</td>
<td colspan="2" class="diff-lineno">Line 49:</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 4 (a bit long)==</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 4 (a bit long)==</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>WLOG, <math>AC = CD = 4</math> and <math>DE = EB = 3</math>. Notice that in <math>\triangle ACB</math>, we have <math>m\angle BAC + m\angle ABC = 90^{\circ}</math>. Since <math>AC = CD</math> and <math>DE = EB</math>, we find that <math>m\angle DAC = m\angle ADC</math> and <math>m\angle DBE = m\angle BDE</math>, so <math>m\angle ADC + m\angle BDE = 90^{\circ}</math> and <math>\angle EDC</math> is right. Therefore, <math>CE = 5</math> by 3-4-5 triangle, <math>CB = 8</math> and <math>AB = 4\sqrt{5}</math>. Define point F such that <math>CF</math> is an altitude; we know the area of the whole triangle is <math>16</math> and we know the hypotenuse is <math>4\sqrt{5}</math>, so <math>CF = \frac{16}{4\sqrt{5}}\cdot2=\frac{8}{\sqrt{5}}</math>. By the geometric mean theorem, <math>x\left(4\sqrt{5}-x\right)=4\sqrt{5}x-x^{2}=\left(\frac{8}{\sqrt{5}}\right)^{2}=\frac{64}{5}</math>. Solving the quadratic we get <math>x=\frac{10\sqrt{5}\pm6\sqrt{5}}{5}</math>, so <math>x=\frac{4\sqrt{5}}{5} or \frac{16\sqrt{5}}{5}</math>. For now, assume <math>x=\frac{4\sqrt{5}}{5}</math>. Then <math>FB=4\sqrt{5}-\frac{4\sqrt{5}}{5}=\frac{16\sqrt{5}}{5}</math>. <math>CF</math> splits <math>AD</math> into two parts (quick congruence by Leg-Angle) so <math>FD = AF = \frac{4\sqrt{5}}{5}</math> and <math>DB = FB - FD = \frac{16\sqrt{5}}{5}-\frac{4\sqrt{5}}{5}=\frac{12\sqrt{5}}{5}</math>. <math>AD = 2\cdot\frac{4\sqrt{5}}{5}=\frac{8\sqrt{5}}{5}</math>. Now we know <math>AD</math> and <math>DB</math>, we can find <math>\frac{AD}{DB}=\frac{\frac{8\sqrt{5}}{5}}{\frac{12\sqrt{5}}{5}}=\frac{8\sqrt{5}}{5}\cdot\frac{5}{12\sqrt{5}}=\frac{8}{12}=\frac{2}{3}</math> or <math>\boxed{\textbf{(A) } 2:3}</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>WLOG, <math>AC = CD = 4</math> and <math>DE = EB = 3</math>. Notice that in <math>\triangle ACB</math>, we have <math>m\angle BAC + m\angle ABC = 90^{\circ}</math>. Since <math>AC = CD</math> and <math>DE = EB</math>, we find that <math>m\angle DAC = m\angle ADC</math> and <math>m\angle DBE = m\angle BDE</math>, so <math>m\angle ADC + m\angle BDE = 90^{\circ}</math> and <math>\angle EDC</math> is right. Therefore, <math>CE = 5</math> by 3-4-5 triangle, <math>CB = 8</math> and <math>AB = 4\sqrt{5}</math>. Define point F such that <math>CF</math> is an altitude; we know the area of the whole triangle is <math>16</math> and we know the hypotenuse is <math>4\sqrt{5}</math>, so <math>CF = \frac{16}{4\sqrt{5}}\cdot2=\frac{8}{\sqrt{5}}</math>. By the geometric mean theorem, <math>x\left(4\sqrt{5}-x\right)=4\sqrt{5}x-x^{2}=\left(\frac{8}{\sqrt{5}}\right)^{2}=\frac{64}{5}</math>. Solving the quadratic we get <math>x=\frac{10\sqrt{5}\pm6\sqrt{5}}{5}</math>, so <math>x=\frac{4\sqrt{5}}{5} or \frac{16\sqrt{5}}{5}</math>. For now, assume <math>x=\frac{4\sqrt{5}}{5}</math>. Then <math>FB=4\sqrt{5}-\frac{4\sqrt{5}}{5}=\frac{16\sqrt{5}}{5}</math>. <math>CF</math> splits <math>AD</math> into two parts (quick congruence by Leg-Angle) so <math>FD = AF = \frac{4\sqrt{5}}{5}</math> and <math>DB = FB - FD = \frac{16\sqrt{5}}{5}-\frac{4\sqrt{5}}{5}=\frac{12\sqrt{5}}{5}</math>. <math>AD = 2\cdot\frac{4\sqrt{5}}{5}=\frac{8\sqrt{5}}{5}</math>. Now we know <math>AD</math> and <math>DB</math>, we can find <math>\frac{AD}{DB}=\frac{\frac{8\sqrt{5}}{5}}{\frac{12\sqrt{5}}{5}}=\frac{8\sqrt{5}}{5}\cdot\frac{5}{12\sqrt{5}}=\frac{8}{12}=\frac{2}{3}</math> or <math>\boxed{\textbf{(A) } 2:3}</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><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;">==Solution 5==</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 know that</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>==Video Solution 1==</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>==Video Solution 1==</div></td></tr>
</table>
Magnetoninja
https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_16&diff=180217&oldid=prev
Pi is 3.14: /* Video Solution 2 */
2022-11-05T18:14:37Z
<p><span dir="auto"><span class="autocomment">Video Solution 2</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 18:14, 5 November 2022</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l55" >Line 55:</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>~IceMatrix</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>~IceMatrix</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>== Video Solution <del class="diffchange diffchange-inline">2</del>==</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>== Video Solution <ins class="diffchange diffchange-inline">by OmegaLearn </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;"><div>https://youtu.be/4_x1sgcQCp4?t=4245</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>https://youtu.be/4_x1sgcQCp4?t=4245</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>
</table>
Pi is 3.14
https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_16&diff=175171&oldid=prev
Lvluo at 19:46, 21 June 2022
2022-06-21T19:46:04Z
<p></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 19:46, 21 June 2022</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l16" >Line 16:</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>draw((0,5)--(2.5,3),black);</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>draw((0,5)--(2.5,3),black);</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>dot((0,5));</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>dot((0,5));</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;">label ("$B$", (0,8),N);</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;">label ("$C$", (0,0),SW);</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;">label ("$A$", (4,0),SE);</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;">label ("$E$", (0,5),W);</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;">label ("$D$", (2.5,3),NE);</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;"><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;"><div>~ By Little Mouse</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>~ By Little Mouse</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> </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"><!-- EDITED BY LVLUO --></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;"><div>==Solution 1==</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 1==</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>Without loss of generality, let <math>AC = CD = 4</math> and <math>DE = EB = 3</math>. Let <math>\angle A = \alpha</math> and <math>\angle B = \beta = 90^{\circ} - \alpha</math>. As <math>\triangle ACD</math> and <math>\triangle DEB</math> are isosceles, <math>\angle ADC = \alpha</math> and <math>\angle BDE = \beta</math>. Then <math>\angle CDE = 180^{\circ} - \alpha - \beta = 90^{\circ}</math>, so <math>\triangle CDE</math> is a <math>3-4-5</math> triangle with <math>CE = 5</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>Without loss of generality, let <math>AC = CD = 4</math> and <math>DE = EB = 3</math>. Let <math>\angle A = \alpha</math> and <math>\angle B = \beta = 90^{\circ} - \alpha</math>. As <math>\triangle ACD</math> and <math>\triangle DEB</math> are isosceles, <math>\angle ADC = \alpha</math> and <math>\angle BDE = \beta</math>. Then <math>\angle CDE = 180^{\circ} - \alpha - \beta = 90^{\circ}</math>, so <math>\triangle CDE</math> is a <math>3-4-5</math> triangle with <math>CE = 5</math>.</div></td></tr>
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Lvluo