https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_24&feed=atom&action=history
2013 AMC 12B Problems/Problem 24 - Revision history
2024-03-28T11:32:48Z
Revision history for this page on the wiki
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https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_24&diff=199544&oldid=prev
R00tsofunity at 04:08, 12 October 2023
2023-10-12T04:08:36Z
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 04:08, 12 October 2023</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>~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]</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:Isabelchen isabelchen]</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;">==Video Solution by MOP 2024==</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://youtu.be/y0s6OTQ7KfI</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;">~r00tsOfUnity</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>
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R00tsofunity
https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_24&diff=185145&oldid=prev
Isabelchen at 16:44, 29 December 2022
2022-12-29T16:44:03Z
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 16:44, 29 December 2022</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l95" >Line 95:</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>~ASAB</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>~ASAB</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 6==</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;">[[File:2013AMC12BProblem24Solution6.png|center|500px]]</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;"><math>\angle BXC = \angle ANC</math>, <math>\angle BCX = \angle ACN</math>, <math>\triangle BCX \sim \triangle ACN</math>, <math>\frac{CX}{CN} = \frac{BC}{AC}</math></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;"><math>\angle MBC = \angle BAC</math>, <math>\angle BCM = \angle ACB</math>, <math>\triangle BCM \sim \triangle ACB</math>, <math>\frac{BC}{AC} = \frac{CM}{BC}</math>, <math>BC = \sqrt{2}</math></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;">Let <math>BX = x</math>, <math>CN = CX + x</math>, <math>\frac{CX}{CX + x} = \frac{\sqrt{2}}{2}</math>, <math>2CX = CX \sqrt{2} + x \sqrt{2}</math>, <math>CX = x(\sqrt{2} + 1)</math></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;"><math>BC^2 = BX^2 + CX^2 - 2 \cdot BX \cdot CX \cdot \cos 120^\circ</math></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;"><math>2 = x^2(\sqrt{2} + 1)^2 + x^2 + x^2(\sqrt{2} + 1) = 5x^2 + 3x^2 \sqrt{2}</math></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;"><math>x^2 = \frac{2}{5 + 3 \sqrt{2} } = \boxed{\textbf{(A) }\frac{10-6\sqrt{2}}{7}}</math></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;">~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]</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 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>
<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>{{AMC12 box|year=2013|ab=B|num-b=23|num-a=25}}</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>{{AMC12 box|year=2013|ab=B|num-b=23|num-a=25}}</div></td></tr>
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Isabelchen
https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_24&diff=173616&oldid=prev
Asweatyasianboie at 02:24, 27 April 2022
2022-04-27T02:24:05Z
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 02:24, 27 April 2022</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l68" >Line 68:</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;"></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>==<del class="diffchange diffchange-inline">solution </del>4 ==</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">Solution </ins>4 ==</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>Since <math>\triangle BXN</math> is equilateral, let's assume the sides of them are all <math>a</math>, and denote the length of <math>XM</math> is <math>m</math>. Since <math>CN</math> bisects <math>\angle BCA</math>, applying the angle bisector theorem and we can get <math>BC=\frac{a}{m}</math>;<math>AN=2m</math>. Now applying LOC, we can get <math>(a+2m)^2+(a+m)^2-2(a+2m)(a+m)\cos\frac{\pi}{3}=1</math>. We get <math>a^2+3m^2+3am=1</math>. Now applying the Stewart theorem in <math>\triangle BAC</math>, we can find that <math>{\frac{a^2}{m^2}+(a+2m)^2=2(1+(a+m)^2)}</math>, after simplifying, we get <math>{\frac{a^2}{m^2}-a^2+2m^2=2}</math>. After observation, the main key for this problem is <math>a^2</math>, so we can solve <math>a</math> in term of <math>m</math>. Let's see the equation <math>{\frac{a^2}{m^2}-a^2+2m^2=2}</math>, we can find that <math>a=\sqrt{2}m</math> so <math>a^2=2m^2</math>. Now back solving the first equation we can get that <math>a=\frac{-3m+\sqrt{4-3m^2}}{m}</math>cuz the negative one can't work. After solving, we can get that <math>m^2=\frac{1}{5+3\sqrt{2}}</math> so <math>a^2=2m^2</math> and we get <math>a^2 = \boxed{\textbf{(A) } \frac{10-6\sqrt{2}}{7}}</math>~ bluesoul</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>Since <math>\triangle BXN</math> is equilateral, let's assume the sides of them are all <math>a</math>, and denote the length of <math>XM</math> is <math>m</math>. Since <math>CN</math> bisects <math>\angle BCA</math>, applying the angle bisector theorem and we can get <math>BC=\frac{a}{m}</math>;<math>AN=2m</math>. Now applying LOC, we can get <math>(a+2m)^2+(a+m)^2-2(a+2m)(a+m)\cos\frac{\pi}{3}=1</math>. We get <math>a^2+3m^2+3am=1</math>. Now applying the Stewart theorem in <math>\triangle BAC</math>, we can find that <math>{\frac{a^2}{m^2}+(a+2m)^2=2(1+(a+m)^2)}</math>, after simplifying, we get <math>{\frac{a^2}{m^2}-a^2+2m^2=2}</math>. After observation, the main key for this problem is <math>a^2</math>, so we can solve <math>a</math> in term of <math>m</math>. Let's see the equation <math>{\frac{a^2}{m^2}-a^2+2m^2=2}</math>, we can find that <math>a=\sqrt{2}m</math> so <math>a^2=2m^2</math>. Now back solving the first equation we can get that <math>a=\frac{-3m+\sqrt{4-3m^2}}{m}</math>cuz the negative one can't work. After solving, we can get that <math>m^2=\frac{1}{5+3\sqrt{2}}</math> so <math>a^2=2m^2</math> and we get <math>a^2 = \boxed{\textbf{(A) } \frac{10-6\sqrt{2}}{7}}</math>~ bluesoul</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 (Similar Triangles)==</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;">Denote the length of <math>AN</math> as <math>a</math> and the length of <math>NB</math> as <math>b.</math></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;">Let <math>M'</math> be the midpoint of <math>\overline{BC} .</math> Denote the intersection of <math>\overline{MM'}</math> and <math>\overline{CN}</math> as <math>X' .</math> Note that <math>MX' = \frac12 AN = \frac{a}{2}</math> and <math>CN = 2\cdot NX' .</math> As <math>\overline{MM'} || \overline{AB},</math> we have that <math>\triangle MXX' \sim \triangle BXN</math> or <math>\triangle MXX'</math> is equilateral and <math>XX' =MX' = \frac{a}{2}.</math> Thus, <math>CN = 2b+a</math> and <math>CX=a+b.</math></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;">Observe that </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;"><cmath>\triangle BXC\sim \triangle ANC \implies \frac{BX}{XC} =\frac{AN}{NC} \implies \frac{b}{b+a} = \frac{a}{2b+a} \implies a = \sqrt 2 b.</cmath></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;">By the angle bisector theorem, we have that <math>BC=\frac{2b}{a} = \sqrt 2.</math></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 apply the Law of Cosines on <math>\triangle BXC</math> as follows:</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;"><cmath></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;">BC^2=BX^2+XC^2-2BX\cdot XC\cdot \cos120^\circ</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;"></cmath></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;"><cmath>\begin{align*}</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;">2&=b^2+(\sqrt 2+1)^2b^2 +(\sqrt2 + 1)b^2 \\</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;">&=b^2(5+3\sqrt 2)</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;">\end{align*}</cmath></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;">or</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;"><cmath>\boxed{b^2=\textbf{(A) } \frac{10-6\sqrt{2}}{7}}</cmath></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;">~ASAB</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>== 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>
<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>{{AMC12 box|year=2013|ab=B|num-b=23|num-a=25}}</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>{{AMC12 box|year=2013|ab=B|num-b=23|num-a=25}}</div></td></tr>
</table>
Asweatyasianboie
https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_24&diff=162324&oldid=prev
Bluesoul at 15:09, 16 September 2021
2021-09-16T15:09:35Z
<p></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 15:09, 16 September 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l69" >Line 69:</td>
<td colspan="2" class="diff-lineno">Line 69:</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 4 ==</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 ==</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>Since <math>\triangle BXN</math> is equilateral, let's assume the sides of them are all <math>a</math>, and denote the length of <math>XM</math> is <math>m</math>. Since <math>CN</math> bisects <math>\angle BCA</math>, applying the angle bisector theorem and we can get <math>BC=\frac{a}{m}</math>;<math>AN=2m</math>. Now applying LOC, we can get <math>(a+2m)^2+(a+m)^2-2(a+2m)(a+m)\cos\frac{\pi}{3}=1</math>. We get <math>a^2+3m^2+3am=1</math>. Now applying the Stewart theorem in <math>\triangle BAC</math>, we can find that <math>{\<del class="diffchange diffchange-inline">(</del>frac{a^2}{m^2}+(a+2m)^2=2(1+(a+m)^2)}</math>, after simplifying, we get <math>{\frac{a^2}{m^2}-a^2+2m^2=2}</math>. After observation, the main key for this problem is <math>a^2</math>, so we can solve <math>a</math> in term of <math>m</math>. Let's see the equation <math>{\frac{a^2}{m^2}-a^2+2m^2=2}</math>, we can find that <math>a=\sqrt{2}m</math> so <math>a^2=2m^2</math>. Now back solving the first equation we can get that <math>a=\frac{-3m+\sqrt{4-3m^2}}{m}</math>cuz the negative one can't work. After solving, we can get that <math>m^2=\frac{1}{5+3\sqrt{2}}</math> so <math>a^2=2m^2</math> and we get <math>a^2 = \boxed{\textbf{(A) } \frac{10-6\sqrt{2}}{7}}</math>~ bluesoul</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>Since <math>\triangle BXN</math> is equilateral, let's assume the sides of them are all <math>a</math>, and denote the length of <math>XM</math> is <math>m</math>. Since <math>CN</math> bisects <math>\angle BCA</math>, applying the angle bisector theorem and we can get <math>BC=\frac{a}{m}</math>;<math>AN=2m</math>. Now applying LOC, we can get <math>(a+2m)^2+(a+m)^2-2(a+2m)(a+m)\cos\frac{\pi}{3}=1</math>. We get <math>a^2+3m^2+3am=1</math>. Now applying the Stewart theorem in <math>\triangle BAC</math>, we can find that <math>{\frac{a^2}{m^2}+(a+2m)^2=2(1+(a+m)^2)}</math>, after simplifying, we get <math>{\frac{a^2}{m^2}-a^2+2m^2=2}</math>. After observation, the main key for this problem is <math>a^2</math>, so we can solve <math>a</math> in term of <math>m</math>. Let's see the equation <math>{\frac{a^2}{m^2}-a^2+2m^2=2}</math>, we can find that <math>a=\sqrt{2}m</math> so <math>a^2=2m^2</math>. Now back solving the first equation we can get that <math>a=\frac{-3m+\sqrt{4-3m^2}}{m}</math>cuz the negative one can't work. After solving, we can get that <math>m^2=\frac{1}{5+3\sqrt{2}}</math> so <math>a^2=2m^2</math> and we get <math>a^2 = \boxed{\textbf{(A) } \frac{10-6\sqrt{2}}{7}}</math>~ bluesoul</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>== 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>
<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>{{AMC12 box|year=2013|ab=B|num-b=23|num-a=25}}</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>{{AMC12 box|year=2013|ab=B|num-b=23|num-a=25}}</div></td></tr>
</table>
Bluesoul
https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_24&diff=162323&oldid=prev
Bluesoul at 15:08, 16 September 2021
2021-09-16T15:08:40Z
<p></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 15:08, 16 September 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l69" >Line 69:</td>
<td colspan="2" class="diff-lineno">Line 69:</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 4 ==</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 ==</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>Since <math>\triangle BXN</math> is equilateral, let's assume the sides of them are all <math>a</math>, and denote the length of <math>XM</math> is <math>m</math>. Since <math>CN</math> bisects <math>\angle BCA</math>, applying the angle bisector theorem and we can get <math>BC=\frac{a}{m}</math>;<math>AN=2m</math>. Now applying LOC, we can get <math>(a+2m)^2+(a+m)^2-2(a+2m)(a+m)\cos\frac{\pi}{3}=1</math>. We get <math>a^2+3m^2+3am=1</math>. Now applying the Stewart theorem in <math>\triangle BAC</math>, we can find that <math>{\(frac{a^2}{m^2}+(a+2m)^2=2<del class="diffchange diffchange-inline">[</del>1+(a+m)^2<del class="diffchange diffchange-inline">]</del>}</math>, after simplifying, we get <math>{\frac{a^2}{m^2}-a^2+2m^2=2}</math>. After observation, the main key for this problem is <math>a^2</math>, so we can solve <math>a</math> in term of <math>m</math>. Let's see the equation <math>{\frac{a^2}{m^2}-a^2+2m^2=2}</math>, we can find that <math>a=\sqrt{2}m</math> so <math>a^2=2m^2</math>. Now back solving the first equation we can get that <math>a=\frac{-3m+\sqrt{4-3m^2}}{m}</math>cuz the negative one can't work. After solving, we can get that <math>m^2=\frac{1}{5+3\sqrt{2}}</math> so <math>a^2=2m^2</math> and we get <math>a^2 = \boxed{\textbf{(A) } \frac{10-6\sqrt{2}}{7}}</math>~ bluesoul</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>Since <math>\triangle BXN</math> is equilateral, let's assume the sides of them are all <math>a</math>, and denote the length of <math>XM</math> is <math>m</math>. Since <math>CN</math> bisects <math>\angle BCA</math>, applying the angle bisector theorem and we can get <math>BC=\frac{a}{m}</math>;<math>AN=2m</math>. Now applying LOC, we can get <math>(a+2m)^2+(a+m)^2-2(a+2m)(a+m)\cos\frac{\pi}{3}=1</math>. We get <math>a^2+3m^2+3am=1</math>. Now applying the Stewart theorem in <math>\triangle BAC</math>, we can find that <math>{\(frac{a^2}{m^2}+(a+2m)^2=2<ins class="diffchange diffchange-inline">(</ins>1+(a+m)^2<ins class="diffchange diffchange-inline">)</ins>}</math>, after simplifying, we get <math>{\frac{a^2}{m^2}-a^2+2m^2=2}</math>. After observation, the main key for this problem is <math>a^2</math>, so we can solve <math>a</math> in term of <math>m</math>. Let's see the equation <math>{\frac{a^2}{m^2}-a^2+2m^2=2}</math>, we can find that <math>a=\sqrt{2}m</math> so <math>a^2=2m^2</math>. Now back solving the first equation we can get that <math>a=\frac{-3m+\sqrt{4-3m^2}}{m}</math>cuz the negative one can't work. After solving, we can get that <math>m^2=\frac{1}{5+3\sqrt{2}}</math> so <math>a^2=2m^2</math> and we get <math>a^2 = \boxed{\textbf{(A) } \frac{10-6\sqrt{2}}{7}}</math>~ bluesoul</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>== 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>
<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>{{AMC12 box|year=2013|ab=B|num-b=23|num-a=25}}</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>{{AMC12 box|year=2013|ab=B|num-b=23|num-a=25}}</div></td></tr>
</table>
Bluesoul
https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_24&diff=162321&oldid=prev
Bluesoul at 15:06, 16 September 2021
2021-09-16T15:06:37Z
<p></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 15:06, 16 September 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l69" >Line 69:</td>
<td colspan="2" class="diff-lineno">Line 69:</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 4 ==</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 ==</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>Since <math>\triangle BXN</math> is equilateral, let's assume the sides of them are all <math>a</math>, and denote the length of <math>XM</math> is <math>m</math>. Since <math>CN</math> bisects <math>\angle BCA</math>, applying the angle bisector theorem and we can get <math>BC=\frac{a}{m}</math>;<math>AN=2m</math>. Now applying LOC, we can get <math>(a+2m)^2+(a+m)^2-2(a+2m)(a+m)\cos\frac{\pi}{3}=1</math>. We get <math>a^2+3m^2+3am=1</math>. Now applying the Stewart theorem in <math>\triangle BAC</math>, we can find that <math>{\(frac{a^2}{m^2}+(a+2m)^2=2[1+(a+m)^2]}</math>, after simplifying, we get <math>{\frac{a^2}{m^2}-a^2+2m^2=2}</math>. After observation, the main key for this problem is <math>a^2</math>, so we can solve <math>a</math> in term of <math>m</math>. Let's see the equation <math>{\frac{a^2}{m^2}-a^2+2m^2=2}</math>, we can find that <math>a=\sqrt{2}m</math> so <math>a^2=2m^2</math>. Now back solving the first equation we can get that <math>a=\frac{-3m+\sqrt{4-3m^2}}{m}</math>cuz the negative one can't work. After solving, we can get that <math>m^2=\frac{1}{5+3\sqrt{2}}</math> so <math>a^2=2m^2=\frac{10-6\sqrt{2}}{7}<del class="diffchange diffchange-inline">.</del>}</math> ~ bluesoul</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>Since <math>\triangle BXN</math> is equilateral, let's assume the sides of them are all <math>a</math>, and denote the length of <math>XM</math> is <math>m</math>. Since <math>CN</math> bisects <math>\angle BCA</math>, applying the angle bisector theorem and we can get <math>BC=\frac{a}{m}</math>;<math>AN=2m</math>. Now applying LOC, we can get <math>(a+2m)^2+(a+m)^2-2(a+2m)(a+m)\cos\frac{\pi}{3}=1</math>. We get <math>a^2+3m^2+3am=1</math>. Now applying the Stewart theorem in <math>\triangle BAC</math>, we can find that <math>{\(frac{a^2}{m^2}+(a+2m)^2=2[1+(a+m)^2]}</math>, after simplifying, we get <math>{\frac{a^2}{m^2}-a^2+2m^2=2}</math>. After observation, the main key for this problem is <math>a^2</math>, so we can solve <math>a</math> in term of <math>m</math>. Let's see the equation <math>{\frac{a^2}{m^2}-a^2+2m^2=2}</math>, we can find that <math>a=\sqrt{2}m</math> so <math>a^2=2m^2</math>. Now back solving the first equation we can get that <math>a=\frac{-3m+\sqrt{4-3m^2}}{m}</math>cuz the negative one can't work. After solving, we can get that <math>m^2=\frac{1}{5+3\sqrt{2}}</math> so <math>a^2=2m<ins class="diffchange diffchange-inline">^2</math> and we get <math>a</ins>^2 = <ins class="diffchange diffchange-inline">\boxed{\textbf{(A) } </ins>\frac{10-6\sqrt{2}}{7}}</math>~ bluesoul</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>== 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>
<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>{{AMC12 box|year=2013|ab=B|num-b=23|num-a=25}}</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>{{AMC12 box|year=2013|ab=B|num-b=23|num-a=25}}</div></td></tr>
</table>
Bluesoul
https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_24&diff=162320&oldid=prev
Bluesoul at 15:05, 16 September 2021
2021-09-16T15:05:42Z
<p></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 15:05, 16 September 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l69" >Line 69:</td>
<td colspan="2" class="diff-lineno">Line 69:</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 4 ==</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 ==</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>Since <math>\triangle BXN</math> is equilateral, let's assume the sides of them are all <math>a</math>, and denote the length of <math>XM</math> is <math>m</math>. Since <math>CN</math> bisects <math>\angle BCA</math>, applying the angle bisector theorem and we can get <math>BC=\frac{a}{m}</math>;<math>AN=2m</math>. Now applying LOC, we can get <math>(a+2m)^2+(a+m)^2-2(a+2m)(a+m)\cos\frac{\pi}{3}=1</math>. We get <math>a^2+3m^2+3am=1</math>. Now applying the Stewart theorem in <math>\triangle BAC</math>, we can find that <math>{\(frac{a^2}{m^2}+(a+2m)^2=2[1+(a+m)^2]}</math>, after simplifying, we get <math>{\frac{a^2}{m^2}-a^2+2m^2=2}</math>. After observation, the main key for this problem is <math>a^2</math>, so we can solve <math>a</math> in term of <math>m</math>. Let's see the equation <math>{\frac{a^2}{m^2}-a^2+2m^2=2}</math>, we can find that <math>a=\sqrt{2}m</math> so <math>a^2=2m^2</math>. Now back solving the first equation we can get that <math>a=\frac{-3m+\sqrt{4-3m^2}}{m}</math>cuz the negative one can't work. After solving, we can get that <math>m^2=\frac{1}{5+3\sqrt{2}}</math> so <del class="diffchange diffchange-inline">$</del>a^2=2m^2=\frac{10-6\sqrt{2}}{7}.}~ bluesoul</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>Since <math>\triangle BXN</math> is equilateral, let's assume the sides of them are all <math>a</math>, and denote the length of <math>XM</math> is <math>m</math>. Since <math>CN</math> bisects <math>\angle BCA</math>, applying the angle bisector theorem and we can get <math>BC=\frac{a}{m}</math>;<math>AN=2m</math>. Now applying LOC, we can get <math>(a+2m)^2+(a+m)^2-2(a+2m)(a+m)\cos\frac{\pi}{3}=1</math>. We get <math>a^2+3m^2+3am=1</math>. Now applying the Stewart theorem in <math>\triangle BAC</math>, we can find that <math>{\(frac{a^2}{m^2}+(a+2m)^2=2[1+(a+m)^2]}</math>, after simplifying, we get <math>{\frac{a^2}{m^2}-a^2+2m^2=2}</math>. After observation, the main key for this problem is <math>a^2</math>, so we can solve <math>a</math> in term of <math>m</math>. Let's see the equation <math>{\frac{a^2}{m^2}-a^2+2m^2=2}</math>, we can find that <math>a=\sqrt{2}m</math> so <math>a^2=2m^2</math>. Now back solving the first equation we can get that <math>a=\frac{-3m+\sqrt{4-3m^2}}{m}</math>cuz the negative one can't work. After solving, we can get that <math>m^2=\frac{1}{5+3\sqrt{2}}</math> so <ins class="diffchange diffchange-inline"><math></ins>a^2=2m^2=\frac{10-6\sqrt{2}}{7}.}<ins class="diffchange diffchange-inline"></math> </ins>~ bluesoul</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>== 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>
<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>{{AMC12 box|year=2013|ab=B|num-b=23|num-a=25}}</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>{{AMC12 box|year=2013|ab=B|num-b=23|num-a=25}}</div></td></tr>
</table>
Bluesoul
https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_24&diff=162319&oldid=prev
Bluesoul at 15:05, 16 September 2021
2021-09-16T15:05:08Z
<p></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 15:05, 16 September 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l69" >Line 69:</td>
<td colspan="2" class="diff-lineno">Line 69:</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 4 ==</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 ==</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>Since <math>\triangle BXN</math> is equilateral, let's assume the sides of them are all <math>a</math>, and denote the length of <math>XM</math> is <math>m</math>. Since <math>CN</math> bisects <math>\angle BCA</math>, applying the angle bisector theorem and we can get <math>BC=\frac{a}{m}</math>;<math>AN=2m</math>. Now applying LOC, we can get <math>(a+2m)^2+(a+m)^2-2(a+2m)(a+m)\cos\frac{\pi}{3}=1</math>. We get <math>a^2+3m^2+3am=1</math>. Now applying the Stewart theorem in <math>\triangle BAC</math>, we can find that <math>{\(frac{a^2}{m^2}+(a+2m)^2=2[1+(a+m)^2]}</math>, after simplifying, we get <math>{\frac{a^2}{m^2}-a^2+2m^2=2}</math>. After observation, the main key for this problem is <math>a^2</math>, so we can solve <math>a</math> in term of <math>m</math>. Let's see the equation <math>{\frac{a^2}{m^2}-a^2+2m^2=2}</math>, we can find that <math>a=\sqrt{2}m</math> so <math>a^2=2m^2</math>. Now back solving the first equation we can get that <math>a=\frac{-3m+\sqrt{4-3m^2}}{m}</math>cuz the negative one can't work. After solving, we can get that <math>m^2=\frac{1}{5+3\sqrt{2}}</math> so $a^2=2m^2=<del class="diffchange diffchange-inline">\boxed{\textbf{(A) }</del>\frac{10-6\sqrt{2}}{7}.}~ bluesoul</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>Since <math>\triangle BXN</math> is equilateral, let's assume the sides of them are all <math>a</math>, and denote the length of <math>XM</math> is <math>m</math>. Since <math>CN</math> bisects <math>\angle BCA</math>, applying the angle bisector theorem and we can get <math>BC=\frac{a}{m}</math>;<math>AN=2m</math>. Now applying LOC, we can get <math>(a+2m)^2+(a+m)^2-2(a+2m)(a+m)\cos\frac{\pi}{3}=1</math>. We get <math>a^2+3m^2+3am=1</math>. Now applying the Stewart theorem in <math>\triangle BAC</math>, we can find that <math>{\(frac{a^2}{m^2}+(a+2m)^2=2[1+(a+m)^2]}</math>, after simplifying, we get <math>{\frac{a^2}{m^2}-a^2+2m^2=2}</math>. After observation, the main key for this problem is <math>a^2</math>, so we can solve <math>a</math> in term of <math>m</math>. Let's see the equation <math>{\frac{a^2}{m^2}-a^2+2m^2=2}</math>, we can find that <math>a=\sqrt{2}m</math> so <math>a^2=2m^2</math>. Now back solving the first equation we can get that <math>a=\frac{-3m+\sqrt{4-3m^2}}{m}</math>cuz the negative one can't work. After solving, we can get that <math>m^2=\frac{1}{5+3\sqrt{2}}</math> so $a^2=2m^2=\frac{10-6\sqrt{2}}{7}.}~ bluesoul</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>== 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>
<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>{{AMC12 box|year=2013|ab=B|num-b=23|num-a=25}}</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>{{AMC12 box|year=2013|ab=B|num-b=23|num-a=25}}</div></td></tr>
</table>
Bluesoul
https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_24&diff=162318&oldid=prev
Bluesoul at 15:03, 16 September 2021
2021-09-16T15:03:54Z
<p></p>
<table class="diff diff-contentalign-left" data-mw="interface">
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<col class="diff-marker" />
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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 15:03, 16 September 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l67" >Line 67:</td>
<td colspan="2" class="diff-lineno">Line 67:</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><cmath>\left (b\sqrt{2} - \frac{b}{2} \right)^2 + \left(\frac{b\sqrt{3}}{2} \right)^2 = \left(2 - \sqrt{2} \right)^2 \implies b^2 = \boxed{\textbf{(A) } \frac{10-6\sqrt{2}}{7}}</cmath>.</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><cmath>\left (b\sqrt{2} - \frac{b}{2} \right)^2 + \left(\frac{b\sqrt{3}}{2} \right)^2 = \left(2 - \sqrt{2} \right)^2 \implies b^2 = \boxed{\textbf{(A) } \frac{10-6\sqrt{2}}{7}}</cmath>.</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 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 4 ==</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;">Since <math>\triangle BXN</math> is equilateral, let's assume the sides of them are all <math>a</math>, and denote the length of <math>XM</math> is <math>m</math>. Since <math>CN</math> bisects <math>\angle BCA</math>, applying the angle bisector theorem and we can get <math>BC=\frac{a}{m}</math>;<math>AN=2m</math>. Now applying LOC, we can get <math>(a+2m)^2+(a+m)^2-2(a+2m)(a+m)\cos\frac{\pi}{3}=1</math>. We get <math>a^2+3m^2+3am=1</math>. Now applying the Stewart theorem in <math>\triangle BAC</math>, we can find that <math>{\(frac{a^2}{m^2}+(a+2m)^2=2[1+(a+m)^2]}</math>, after simplifying, we get <math>{\frac{a^2}{m^2}-a^2+2m^2=2}</math>. After observation, the main key for this problem is <math>a^2</math>, so we can solve <math>a</math> in term of <math>m</math>. Let's see the equation <math>{\frac{a^2}{m^2}-a^2+2m^2=2}</math>, we can find that <math>a=\sqrt{2}m</math> so <math>a^2=2m^2</math>. Now back solving the first equation we can get that <math>a=\frac{-3m+\sqrt{4-3m^2}}{m}</math>cuz the negative one can't work. After solving, we can get that <math>m^2=\frac{1}{5+3\sqrt{2}}</math> so $a^2=2m^2=\boxed{\textbf{(A) }\frac{10-6\sqrt{2}}{7}.}~ bluesoul</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>== 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>
<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>{{AMC12 box|year=2013|ab=B|num-b=23|num-a=25}}</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>{{AMC12 box|year=2013|ab=B|num-b=23|num-a=25}}</div></td></tr>
</table>
Bluesoul
https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_24&diff=131922&oldid=prev
Jbala at 17:16, 16 August 2020
2020-08-16T17:16:33Z
<p></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 17:16, 16 August 2020</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="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>ABC</math> be a triangle where <math>M</math> is the midpoint of <math>\overline{AC}</math>, and <math>\overline{CN}</math> is the angle bisector of <math>\angle{ACB}</math> with <math>N</math> on <math>\overline{AB}</math>. Let <math>X</math> be the intersection of the median <math>\overline{BM}</math> and the bisector <math>\overline{CN}</math>. In addition <math>\triangle BXN</math> is equilateral with <math>AC=2</math>. What is <math><del class="diffchange diffchange-inline">BN</del>^2</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>ABC</math> be a triangle where <math>M</math> is the midpoint of <math>\overline{AC}</math>, and <math>\overline{CN}</math> is the angle bisector of <math>\angle{ACB}</math> with <math>N</math> on <math>\overline{AB}</math>. Let <math>X</math> be the intersection of the median <math>\overline{BM}</math> and the bisector <math>\overline{CN}</math>. In addition <math>\triangle BXN</math> is equilateral with <math>AC=2</math>. What is <math><ins class="diffchange diffchange-inline">BX</ins>^2</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><math>\textbf{(A)}\  \frac{10-6\sqrt{2}}{7} \qquad \textbf{(B)}\ \frac{2}{9} \qquad \textbf{(C)}\ \frac{5\sqrt{2}-3\sqrt{3}}{8} \qquad \textbf{(D)}\ \frac{\sqrt{2}}{6} \qquad \textbf{(E)}\ \frac{3\sqrt{3}-4}{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><math>\textbf{(A)}\  \frac{10-6\sqrt{2}}{7} \qquad \textbf{(B)}\ \frac{2}{9} \qquad \textbf{(C)}\ \frac{5\sqrt{2}-3\sqrt{3}}{8} \qquad \textbf{(D)}\ \frac{\sqrt{2}}{6} \qquad \textbf{(E)}\ \frac{3\sqrt{3}-4}{5}</math></div></td></tr>
</table>
Jbala