https://artofproblemsolving.com/wiki/index.php?title=2016_USAMO_Problems/Problem_4&feed=atom&action=history
2016 USAMO Problems/Problem 4 - Revision history
2024-03-28T19:33:51Z
Revision history for this page on the wiki
MediaWiki 1.31.1
https://artofproblemsolving.com/wiki/index.php?title=2016_USAMO_Problems/Problem_4&diff=118526&oldid=prev
Ccx09: /* Solution 2 */
2020-02-28T04:13:18Z
<p><span dir="auto"><span class="autocomment">Solution 2</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 04:13, 28 February 2020</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;"><div>Step 3: <math>y=-x \implies 2f(4x)(f(x)-x^{2})=0</math>. Thus, <math>\forall x</math>, we have <math>f(4x)=0</math> or <math>f(x)=x^{2}</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>Step 3: <math>y=-x \implies 2f(4x)(f(x)-x^{2})=0</math>. Thus, <math>\forall x</math>, we have <math>f(4x)=0</math> or <math>f(x)=x^{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="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>Step 4: We now assume <math>f(x) \not = 0</math>, <math>x\not = 0</math>. We have <math>f(\frac{x}{4})=\frac{x^{2}}{16}</math>. Now, setting <math>x=y=\frac{x}{4}</math>, we have <math>f(\frac{x}{2}=\frac{x^{2}}{4}</math> or <math>f(\frac{x}{2})=0</math>. The former implies that <math>f(x)=0</math> or <math>x^{2}</math>. The latter implies that <math>f(x)=0</math> or <math>f(x)=\frac{x^{2}}{2}</math>. Assume the latter. <math>y=-2x \implies -\frac{3y^{2}}{2}f(7y)-{2y^{2}}f(5y)=\frac{x^{2}}{2}</math>. Clearly, this implies that <math>f(x)</math> is negative for some <math>m</math>. Now, we have <math>f(\frac{m}{4})=\frac{m^{2}}{16} \implies f(\frac{m}{2})=0,\frac{m^{2}}{4} \implies f(m) \geq 0</math>, which is a contradiction. Thus, <math>\forall x</math><math>f(x)=0</math> or <math>f(x)=x^{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>Step 4: We now assume <math>f(x) \not = 0</math>, <math>x\not = 0</math>. We have <math>f(\frac{x}{4})=\frac{x^{2}}{16}</math>. Now, setting <math>x=y=\frac{x}{4}</math>, we have <math>f(\frac{x}{2}<ins class="diffchange diffchange-inline">)</ins>=\frac{x^{2}}{4}</math> or <math>f(\frac{x}{2})=0</math>. The former implies that <math>f(x)=0</math> or <math>x^{2}</math>. The latter implies that <math>f(x)=0</math> or <math>f(x)=\frac{x^{2}}{2}</math>. Assume the latter. <math>y=-2x \implies -\frac{3y^{2}}{2}f(7y)-{2y^{2}}f(5y)=\frac{x^{2}}{2}</math>. Clearly, this implies that <math>f(x)</math> is negative for some <math>m</math>. Now, we have <math>f(\frac{m}{4})=\frac{m^{2}}{16} \implies f(\frac{m}{2})=0,\frac{m^{2}}{4} \implies f(m) \geq 0</math>, which is a contradiction. Thus, <math>\forall x</math><math>f(x)=0</math> or <math>f(x)=x^{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>Step 5: We now assume <math>f(x)=0</math>, <math>f(y)=y^{2}</math> for some <math>x,y \not = 0</math>. Let <math>m</math> be sufficiently large integer, let <math>z=|4^{m}x|</math> and take the absolute value of <math>y</math>(since the function is even). Choose <math>c</math> such that <math>3z-c=y</math>. Note that we have <math>\frac{c}{z}</math>~<math>3</math> and <math>\frac{y}{z}</math>~<math>0</math>. Note that <math>f(z)=0</math>. Now, <math>x=z, y=c \implies</math> LHS is positive, as the second term is positive and the first term is nonnegative and thus the right term is equal to <math>(z+c)^{4}</math>~<math>256z^{4}</math>. Now if <math>f(z-3c)=0</math>, the second term of the LHS/RHS clearly ~0 as <math>m \to \infty</math>. if <math>f(z-3c)=0</math>, then we have LHS/RHS ~ <math>0</math>, otherwise, we have LHS/RHS~<math>\frac{8^{2}\cdot 3z^{4}}{256z^{4}}</math>~<math>\frac{3}{4}</math>, a contradiction, as we're clearly not dividing by <math>0</math>, and we should have LHS/RHS=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>Step 5: We now assume <math>f(x)=0</math>, <math>f(y)=y^{2}</math> for some <math>x,y \not = 0</math>. Let <math>m</math> be sufficiently large integer, let <math>z=|4^{m}x|</math> and take the absolute value of <math>y</math>(since the function is even). Choose <math>c</math> such that <math>3z-c=y</math>. Note that we have <math>\frac{c}{z}</math>~<math>3</math> and <math>\frac{y}{z}</math>~<math>0</math>. Note that <math>f(z)=0</math>. Now, <math>x=z, y=c \implies</math> LHS is positive, as the second term is positive and the first term is nonnegative and thus the right term is equal to <math>(z+c)^{4}</math>~<math>256z^{4}</math>. Now if <math>f(z-3c)=0</math>, the second term of the LHS/RHS clearly ~0 as <math>m \to \infty</math>. if <math>f(z-3c)=0</math>, then we have LHS/RHS ~ <math>0</math>, otherwise, we have LHS/RHS~<math>\frac{8^{2}\cdot 3z^{4}}{256z^{4}}</math>~<math>\frac{3}{4}</math>, a contradiction, as we're clearly not dividing by <math>0</math>, and we should have LHS/RHS=1.</div></td></tr>
</table>
Ccx09
https://artofproblemsolving.com/wiki/index.php?title=2016_USAMO_Problems/Problem_4&diff=100070&oldid=prev
Blacksheep2003: /* Alternative Solution 1 */
2019-01-04T16:16:13Z
<p><span dir="auto"><span class="autocomment">Alternative Solution 1</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 16:16, 4 January 2019</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l46" >Line 46:</td>
<td colspan="2" class="diff-lineno">Line 46:</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>==Alternative 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>==Alternative Solution 1==</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>From steps 1 and 2 of Solution 1 we have that <math>f(0)=0</math>, and <math>f(-y) \cdot f(y)=(f(y))^2</math>. Therefore, if <math>f(y) \ne 0</math>, then <math>f(y)=f(-y)</math>. Furthermore, setting <math>y=-x</math> gives us <math>(f(x)-x^2) \cdot f(4x) + (f(-x)-x^2) \cdot f(4x) = f(0) = 0</math>. The LHS can be factored as <math>(f(x)+f(-x)-2x^2) \cdot f(4x) = 0</math>. In particular, if <math>f(4x) \ne 0</math>, then we have <math>f(x)+f(-x)=2x^2</math>. However, since we have from step 2 that <math>f(x)=f(-x)</math>, assuming <math>f(x) \ne 0</math>, the equation becomes <math>2f(x)=2x^2</math>, so for every <math>x</math>, <math>f(x)</math> is equivalent to either <math>0</math> or <math>x^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>From steps 1 and 2 of Solution 1 we have that <math>f(0)=0</math>, and <math>f(-y) \cdot f(y)=(f(y))^2</math>. Therefore, if <math>f(y) \ne 0</math>, then <math>f(y)=f(-y)</math>. Furthermore, setting <math>y=-x</math> gives us <math>(f(x)-x^2) \cdot f(4x) + (f(-x)-x^2) \cdot f(4x) = f(0) = 0</math>. The LHS can be factored as <math>(f(x)+f(-x)-2x^2) \cdot f(4x) = 0</math>. In particular, if <math>f(4x) \ne 0</math>, then we have <math>f(x)+f(-x)=2x^2</math>. However, since we have from step 2 that <math>f(x)=f(-x)</math>, assuming <math>f(x) \ne 0</math>, the equation becomes <math>2f(x)=2x^2</math>, so for every <math>x</math>, <math>f(x)</math> is equivalent to either <math>0</math> or <math>x^2</math><ins class="diffchange diffchange-inline">. From step 6 of Solution 1, we can prove that <math>f(x)=0</math>, and <math>f(x)=x^2</math> are the only possible solutions</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>
Blacksheep2003
https://artofproblemsolving.com/wiki/index.php?title=2016_USAMO_Problems/Problem_4&diff=100067&oldid=prev
Blacksheep2003: /* Alternative Solution 1 */
2019-01-04T16:06:51Z
<p><span dir="auto"><span class="autocomment">Alternative Solution 1</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 16:06, 4 January 2019</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l46" >Line 46:</td>
<td colspan="2" class="diff-lineno">Line 46:</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>==Alternative 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>==Alternative Solution 1==</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>From steps 1 and 2 of Solution 1 we have that <math>f(0)=0</math>, and <math>f(-y) \cdot f(y)=(f(y))^2</math>. Therefore, if <math>f(y) \ne 0</math>, then <math>f(y)=f(-y)</math>. Furthermore, setting <math>y=-x</math> gives us <math>(f(x)-x^2) \cdot f(4x) + (f(-x)-x^2) \cdot f(4x) = f(0) = 0</math>. The LHS can be factored as <math>(f(x)+f(-x)-2x^2) \cdot f(4x) = 0</math>. In particular, if <math>f(4x) \ne 0</math>, then we have <math>f(x)+f(-x)=2x^2</math>. However, since we have from step 2 that <math>f(x)=f(-x)</math>, assuming <math>f(x) \ne 0</math>, the equation becomes <math>2f(x)=2x^2</math>, so <math>f(x)<del class="diffchange diffchange-inline">=</del>0<del class="diffchange diffchange-inline">, </del>x^2</math> <del class="diffchange diffchange-inline">are the only possible solutions</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>From steps 1 and 2 of Solution 1 we have that <math>f(0)=0</math>, and <math>f(-y) \cdot f(y)=(f(y))^2</math>. Therefore, if <math>f(y) \ne 0</math>, then <math>f(y)=f(-y)</math>. Furthermore, setting <math>y=-x</math> gives us <math>(f(x)-x^2) \cdot f(4x) + (f(-x)-x^2) \cdot f(4x) = f(0) = 0</math>. The LHS can be factored as <math>(f(x)+f(-x)-2x^2) \cdot f(4x) = 0</math>. In particular, if <math>f(4x) \ne 0</math>, then we have <math>f(x)+f(-x)=2x^2</math>. However, since we have from step 2 that <math>f(x)=f(-x)</math>, assuming <math>f(x) \ne 0</math>, the equation becomes <math>2f(x)=2x^2</math>, so <ins class="diffchange diffchange-inline">for every <math>x</math>, </ins><math>f(x)<ins class="diffchange diffchange-inline"></math> is equivalent to either <math></ins>0<ins class="diffchange diffchange-inline"></math> or <math></ins>x^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>==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>
Blacksheep2003
https://artofproblemsolving.com/wiki/index.php?title=2016_USAMO_Problems/Problem_4&diff=100046&oldid=prev
Blacksheep2003: /* Alternative Solution 1 */
2019-01-04T02:26:04Z
<p><span dir="auto"><span class="autocomment">Alternative Solution 1</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 02:26, 4 January 2019</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l46" >Line 46:</td>
<td colspan="2" class="diff-lineno">Line 46:</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>==Alternative 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>==Alternative Solution 1==</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>From steps 1 and 2 of Solution 1 we have that <math>f(0)=0</math>, and <math>f(-y) \cdot f(y)=(f(y))^2</math>. Therefore, if <math>f(y) \ne 0</math>, then <math>f(y)=f(-y)</math>. Furthermore, setting <math>y=-x</math> gives us <math>(f(x)-x^2) \cdot f(4x) + (f(-x)-x^2) \cdot f(4x) = f(0) = 0</math>. The LHS can be factored as <math>(f(x)+f(-x)-2x^2) \cdot f(4x) = 0</math>. In particular, if <math>f(4x) \ne 0</math>, then we have <math>f(x)+f(-x)=2x^2</math>. However, since we have from step 2 that <math>f(x)=f(-x)</math>, assuming <math>f(x) \ne 0</math>, the equation becomes <math>2f(x)=2x^2</math>, so <math>f(x)=0, x^2</math> are the only solutions.</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>From steps 1 and 2 of Solution 1 we have that <math>f(0)=0</math>, and <math>f(-y) \cdot f(y)=(f(y))^2</math>. Therefore, if <math>f(y) \ne 0</math>, then <math>f(y)=f(-y)</math>. Furthermore, setting <math>y=-x</math> gives us <math>(f(x)-x^2) \cdot f(4x) + (f(-x)-x^2) \cdot f(4x) = f(0) = 0</math>. The LHS can be factored as <math>(f(x)+f(-x)-2x^2) \cdot f(4x) = 0</math>. In particular, if <math>f(4x) \ne 0</math>, then we have <math>f(x)+f(-x)=2x^2</math>. However, since we have from step 2 that <math>f(x)=f(-x)</math>, assuming <math>f(x) \ne 0</math>, the equation becomes <math>2f(x)=2x^2</math>, so <math>f(x)=0, x^2</math> are the only <ins class="diffchange diffchange-inline">possible </ins>solutions.</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>
Blacksheep2003
https://artofproblemsolving.com/wiki/index.php?title=2016_USAMO_Problems/Problem_4&diff=100045&oldid=prev
Blacksheep2003: /* Alternative Solution 1 */
2019-01-04T02:24:37Z
<p><span dir="auto"><span class="autocomment">Alternative Solution 1</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 02:24, 4 January 2019</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l46" >Line 46:</td>
<td colspan="2" class="diff-lineno">Line 46:</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>==Alternative 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>==Alternative Solution 1==</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>From steps 1 and 2 of Solution 1 we have that <math>f(0)=0</math>, and <math>f(-y) \cdot f(y)=(f(y))^2</math><del class="diffchange diffchange-inline">, therefore</del>, if <math>f(y) \ne 0</math>, then <math>f(y)=f(-y)</math>. Furthermore, setting <math>y=-x</math> gives us <math>(f(x)-x^2) \cdot f(4x) + (f(-x)-x^2) \cdot f(4x) = f(0) = 0</math>. The LHS can be factored as <math>(f(x)+f(-x)-2x^2) \cdot f(4x) = 0</math>. <del class="diffchange diffchange-inline">If </del><math>f(4x) \ne 0</math>, then we have <math>f(x)+f(-x)=2x^2</math>. However, since <math>f(x)=f(-x)</math>, assuming <math>f(x) \ne 0</math>, the equation becomes <math>2f(x)=2x^2</math>, so <math>f(x)=0, x^2</math> are the only solutions.</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>From steps 1 and 2 of Solution 1 we have that <math>f(0)=0</math>, and <math>f(-y) \cdot f(y)=(f(y))^2</math><ins class="diffchange diffchange-inline">. Therefore</ins>, if <math>f(y) \ne 0</math>, then <math>f(y)=f(-y)</math>. Furthermore, setting <math>y=-x</math> gives us <math>(f(x)-x^2) \cdot f(4x) + (f(-x)-x^2) \cdot f(4x) = f(0) = 0</math>. The LHS can be factored as <math>(f(x)+f(-x)-2x^2) \cdot f(4x) = 0</math>. <ins class="diffchange diffchange-inline">In particular, if </ins><math>f(4x) \ne 0</math>, then we have <math>f(x)+f(-x)=2x^2</math>. However, since <ins class="diffchange diffchange-inline">we have from step 2 that </ins><math>f(x)=f(-x)</math>, assuming <math>f(x) \ne 0</math>, the equation becomes <math>2f(x)=2x^2</math>, so <math>f(x)=0, x^2</math> are the only solutions.</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>
Blacksheep2003
https://artofproblemsolving.com/wiki/index.php?title=2016_USAMO_Problems/Problem_4&diff=100044&oldid=prev
Blacksheep2003: /* Alternative Solution 1 */
2019-01-04T02:22:32Z
<p><span dir="auto"><span class="autocomment">Alternative Solution 1</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 02:22, 4 January 2019</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l46" >Line 46:</td>
<td colspan="2" class="diff-lineno">Line 46:</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>==Alternative 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>==Alternative Solution 1==</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>From steps 1 and 2 of Solution 1 we have that <math>f(0)=0</math>, and <math>f(-y)f(y)=(f(y))^2</math>, <del class="diffchange diffchange-inline">so </del>if <math>f(y) \ne 0</math>, then <math>f(y)=f(-y)</math>. Furthermore, setting <math>y=-x</math> gives us <math>(f(x)-x^2) \cdot f(4x) + (f(-x)-x^2) \cdot f(4x) = f(0) = 0</math>. The LHS can be factored as <math>(f(x)+f(-x)-2x^2) \cdot f(4x) = 0</math>. <del class="diffchange diffchange-inline">Therefore, if </del><math>f(4x) \ne 0</math>, then we have <del class="diffchange diffchange-inline"><math>f(x)+f(-x)-2x^2=0</math>, or </del><math>f(x)+f(-x)=2x^2</math>. However, since <del class="diffchange diffchange-inline">from step 2 we have that </del><math>f(x)=f(-x)</math> assuming <math>f(x) \ne 0</math>, the equation becomes <math>2f(x)=2x^2</math>, so <math>f(x)=0, x^2</math> are the only solutions.</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>From steps 1 and 2 of Solution 1 we have that <math>f(0)=0</math>, and <math>f(-y) <ins class="diffchange diffchange-inline">\cdot </ins>f(y)=(f(y))^2</math>, <ins class="diffchange diffchange-inline">therefore, </ins>if <math>f(y) \ne 0</math>, then <math>f(y)=f(-y)</math>. Furthermore, setting <math>y=-x</math> gives us <math>(f(x)-x^2) \cdot f(4x) + (f(-x)-x^2) \cdot f(4x) = f(0) = 0</math>. The LHS can be factored as <math>(f(x)+f(-x)-2x^2) \cdot f(4x) = 0</math>. <ins class="diffchange diffchange-inline">If </ins><math>f(4x) \ne 0</math>, then we have <math>f(x)+f(-x)=2x^2</math>. However, since <math>f(x)=f(-x)</math><ins class="diffchange diffchange-inline">, </ins>assuming <math>f(x) \ne 0</math>, the equation becomes <math>2f(x)=2x^2</math>, so <math>f(x)=0, x^2</math> are the only solutions.</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>
Blacksheep2003
https://artofproblemsolving.com/wiki/index.php?title=2016_USAMO_Problems/Problem_4&diff=100043&oldid=prev
Blacksheep2003: /* Alternative Solution 1 */
2019-01-04T02:19:53Z
<p><span dir="auto"><span class="autocomment">Alternative Solution 1</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 02:19, 4 January 2019</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l46" >Line 46:</td>
<td colspan="2" class="diff-lineno">Line 46:</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>==Alternative 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>==Alternative Solution 1==</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>From steps 1 and 2 we have that<del class="diffchange diffchange-inline">, setting <math>x=y=0</math>, </del><math>f(0)=0</math>, and<del class="diffchange diffchange-inline">, setting <math>x=0</math>, </del><math>f(-y)f(y)=(f(y))^2</math>, so if <math>f(y) \ne 0</math>, then <math>f(y)=f(-y)</math>. Furthermore, setting <math>y=-x</math> gives us <math>(f(x)-x^2) \cdot f(4x) + (f(-x)-x^2) \cdot f(4x) = f(0) = 0</math>. The LHS can be factored as <math>(f(x)+f(-x)-2x^2) \cdot f(4x) = 0</math>. Therefore, if <math>f(4x) \ne 0</math>, then we have <math>f(x)+f(-x)-2x^2=0</math>, or <math>f(x)+f(-x)=2x^2</math>. However, since from step 2 we have that <math>f(x)=f(-x)</math> assuming <math>f(x) \ne 0</math>, the equation becomes <math>2f(x)=2x^2</math>, so <math>f(x)=0, x^2</math> are the only solutions.  </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>From steps 1 and 2 <ins class="diffchange diffchange-inline">of Solution 1 </ins>we have that <math>f(0)=0</math>, and <math>f(-y)f(y)=(f(y))^2</math>, so if <math>f(y) \ne 0</math>, then <math>f(y)=f(-y)</math>. Furthermore, setting <math>y=-x</math> gives us <math>(f(x)-x^2) \cdot f(4x) + (f(-x)-x^2) \cdot f(4x) = f(0) = 0</math>. The LHS can be factored as <math>(f(x)+f(-x)-2x^2) \cdot f(4x) = 0</math>. Therefore, if <math>f(4x) \ne 0</math>, then we have <math>f(x)+f(-x)-2x^2=0</math>, or <math>f(x)+f(-x)=2x^2</math>. However, since from step 2 we have that <math>f(x)=f(-x)</math> assuming <math>f(x) \ne 0</math>, the equation becomes <math>2f(x)=2x^2</math>, so <math>f(x)=0, x^2</math> are the only solutions.</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td 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>
<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>Step 1: <math>x=y=0 \implies f(0)=0</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>Step 1: <math>x=y=0 \implies f(0)=0</math></div></td></tr>
</table>
Blacksheep2003
https://artofproblemsolving.com/wiki/index.php?title=2016_USAMO_Problems/Problem_4&diff=100042&oldid=prev
Blacksheep2003 at 02:18, 4 January 2019
2019-01-04T02:18:47Z
<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 02:18, 4 January 2019</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l45" >Line 45:</td>
<td colspan="2" class="diff-lineno">Line 45:</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>'''Step 7:''' By Step 6 and the second observation from Step 4, the only possible solutions are <math>f \equiv 0</math> and <math>f(x) = x^2</math> for all <math>x \in \mathbb{R}.</math> It's easy to check that both of these work, so we're done.</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>'''Step 7:''' By Step 6 and the second observation from Step 4, the only possible solutions are <math>f \equiv 0</math> and <math>f(x) = x^2</math> for all <math>x \in \mathbb{R}.</math> It's easy to check that both of these work, so we're done.</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>==<del class="diffchange diffchange-inline">Alternate </del>Solution 1==</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">Alternative </ins>Solution 1==</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>From steps 1 and 2 we have that, setting <math>x=y=0</math>, <math>f(0)=0</math>, and, setting <math>x=0</math>, <math>f(-y)f(y)=(f(y))^2</math>, so if <math>f(y) \ne 0</math>, then <math>f(y)=f(-y)</math>. Furthermore, setting <math>y=-x</math> gives us <math>(f(x)-x^2) \cdot f(4x) + (f(-x)-x^2) \cdot f(4x) = f(0) = 0</math>. The LHS can be factored as <math>(f(x)+f(-x)-2x^2) \cdot f(4x) = 0</math>. Therefore, if <math>f(4x) \ne 0</math>, then we have <math>f(x)+f(-x)-2x^2=0</math>, or <math>f(x)+f(-x)=2x^2</math>. However, since from step 2 we have that <math>f(x)=f(-x)</math> assuming f(x) \ne 0<math>, the equation becomes <<del class="diffchange diffchange-inline">/</del>math>2f(x)=2x^2<math>, so <<del class="diffchange diffchange-inline">/</del>math>f(x)=0, x^2<math> are the only solutions.  </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>From steps 1 and 2 we have that, setting <math>x=y=0</math>, <math>f(0)=0</math>, and, setting <math>x=0</math>, <math>f(-y)f(y)=(f(y))^2</math>, so if <math>f(y) \ne 0</math>, then <math>f(y)=f(-y)</math>. Furthermore, setting <math>y=-x</math> gives us <math>(f(x)-x^2) \cdot f(4x) + (f(-x)-x^2) \cdot f(4x) = f(0) = 0</math>. The LHS can be factored as <math>(f(x)+f(-x)-2x^2) \cdot f(4x) = 0</math>. Therefore, if <math>f(4x) \ne 0</math>, then we have <math>f(x)+f(-x)-2x^2=0</math>, or <math>f(x)+f(-x)=2x^2</math>. However, since from step 2 we have that <math>f(x)=f(-x)</math> assuming <ins class="diffchange diffchange-inline"><math></ins>f(x) \ne 0<<ins class="diffchange diffchange-inline">/</ins>math>, the equation becomes <math>2f(x)=2x^2<<ins class="diffchange diffchange-inline">/</ins>math>, so <math>f(x)=0, x^2<<ins class="diffchange diffchange-inline">/</ins>math> are the only solutions.  </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 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>
<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>Step 1: <<del class="diffchange diffchange-inline">/</del>math>x=y=0 \implies f(0)=0<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>Step 1: <math>x=y=0 \implies f(0)=0<<ins class="diffchange diffchange-inline">/</ins>math></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="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>Step 2: <<del class="diffchange diffchange-inline">/</del>math>x=0 \implies f(y)f(-y)=f(y)^{2}<math>. Now, assume <<del class="diffchange diffchange-inline">/</del>math>y \not = 0<math>. Then, if <<del class="diffchange diffchange-inline">/</del>math>f(y)=0<math>, we substitute in <<del class="diffchange diffchange-inline">/</del>math>-y<math> to get <<del class="diffchange diffchange-inline">/</del>math>f(y)f(-y)=f(-y)^{2}<math>, or <<del class="diffchange diffchange-inline">/</del>math>f(y)=f(-y)=0<math>. Otherwise, we divide both sides by <<del class="diffchange diffchange-inline">/</del>math>f(y)<math> to get <<del class="diffchange diffchange-inline">/</del>math>f(y)=f(-y)<math>. If <<del class="diffchange diffchange-inline">/</del>math>y=0<math>, we obviously have <<del class="diffchange diffchange-inline">/</del>math>f(0)=f(0)<math>. Thus, the function is even.</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>Step 2: <math>x=0 \implies f(y)f(-y)=f(y)^{2}<<ins class="diffchange diffchange-inline">/</ins>math>. Now, assume <math>y \not = 0<<ins class="diffchange diffchange-inline">/</ins>math>. Then, if <math>f(y)=0<<ins class="diffchange diffchange-inline">/</ins>math>, we substitute in <math>-y<<ins class="diffchange diffchange-inline">/</ins>math> to get <math>f(y)f(-y)=f(-y)^{2}<<ins class="diffchange diffchange-inline">/</ins>math>, or <math>f(y)=f(-y)=0<<ins class="diffchange diffchange-inline">/</ins>math>. Otherwise, we divide both sides by <math>f(y)<<ins class="diffchange diffchange-inline">/</ins>math> to get <math>f(y)=f(-y)<<ins class="diffchange diffchange-inline">/</ins>math>. If <math>y=0<<ins class="diffchange diffchange-inline">/</ins>math>, we obviously have <math>f(0)=f(0)<<ins class="diffchange diffchange-inline">/</ins>math>. Thus, the function is even.</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>.</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>.</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>Step 3: <<del class="diffchange diffchange-inline">/</del>math>y=-x \implies 2f(4x)(f(x)-x^{2})=0<math>. Thus, <<del class="diffchange diffchange-inline">/</del>math>\forall x<math>, we have <<del class="diffchange diffchange-inline">/</del>math>f(4x)=0<math> or <<del class="diffchange diffchange-inline">/</del>math>f(x)=x^{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>Step 3: <math>y=-x \implies 2f(4x)(f(x)-x^{2})=0<<ins class="diffchange diffchange-inline">/</ins>math>. Thus, <math>\forall x<<ins class="diffchange diffchange-inline">/</ins>math>, we have <math>f(4x)=0<<ins class="diffchange diffchange-inline">/</ins>math> or <math>f(x)=x^{2}<<ins class="diffchange diffchange-inline">/</ins>math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="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>Step 4: We now assume <<del class="diffchange diffchange-inline">/</del>math>f(x) \not = 0<math>, <<del class="diffchange diffchange-inline">/</del>math>x\not = 0<math>. We have <<del class="diffchange diffchange-inline">/</del>math>f(\frac{x}{4})=\frac{x^{2}}{16}<math>. Now, setting <<del class="diffchange diffchange-inline">/</del>math>x=y=\frac{x}{4}<math>, we have <<del class="diffchange diffchange-inline">/</del>math>f(\frac{x}{2}=\frac{x^{2}}{4}<math> or <<del class="diffchange diffchange-inline">/</del>math>f(\frac{x}{2})=0<math>. The former implies that <<del class="diffchange diffchange-inline">/</del>math>f(x)=0<math> or <<del class="diffchange diffchange-inline">/</del>math>x^{2}<math>. The latter implies that <<del class="diffchange diffchange-inline">/</del>math>f(x)=0<math> or <<del class="diffchange diffchange-inline">/</del>math>f(x)=\frac{x^{2}}{2}<math>. Assume the latter. <<del class="diffchange diffchange-inline">/</del>math>y=-2x \implies -\frac{3y^{2}}{2}f(7y)-{2y^{2}}f(5y)=\frac{x^{2}}{2}<math>. Clearly, this implies that <<del class="diffchange diffchange-inline">/</del>math>f(x)<math> is negative for some <<del class="diffchange diffchange-inline">/</del>math>m<math>. Now, we have <<del class="diffchange diffchange-inline">/</del>math>f(\frac{m}{4})=\frac{m^{2}}{16} \implies f(\frac{m}{2})=0,\frac{m^{2}}{4} \implies f(m) \geq 0<math>, which is a contradiction. Thus, <<del class="diffchange diffchange-inline">/</del>math>\forall x<math><<del class="diffchange diffchange-inline">/</del>math>f(x)=0<math> or <<del class="diffchange diffchange-inline">/</del>math>f(x)=x^{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>Step 4: We now assume <math>f(x) \not = 0<<ins class="diffchange diffchange-inline">/</ins>math>, <math>x\not = 0<<ins class="diffchange diffchange-inline">/</ins>math>. We have <math>f(\frac{x}{4})=\frac{x^{2}}{16}<<ins class="diffchange diffchange-inline">/</ins>math>. Now, setting <math>x=y=\frac{x}{4}<<ins class="diffchange diffchange-inline">/</ins>math>, we have <math>f(\frac{x}{2}=\frac{x^{2}}{4}<<ins class="diffchange diffchange-inline">/</ins>math> or <math>f(\frac{x}{2})=0<<ins class="diffchange diffchange-inline">/</ins>math>. The former implies that <math>f(x)=0<<ins class="diffchange diffchange-inline">/</ins>math> or <math>x^{2}<<ins class="diffchange diffchange-inline">/</ins>math>. The latter implies that <math>f(x)=0<<ins class="diffchange diffchange-inline">/</ins>math> or <math>f(x)=\frac{x^{2}}{2}<<ins class="diffchange diffchange-inline">/</ins>math>. Assume the latter. <math>y=-2x \implies -\frac{3y^{2}}{2}f(7y)-{2y^{2}}f(5y)=\frac{x^{2}}{2}<<ins class="diffchange diffchange-inline">/</ins>math>. Clearly, this implies that <math>f(x)<<ins class="diffchange diffchange-inline">/</ins>math> is negative for some <math>m<<ins class="diffchange diffchange-inline">/</ins>math>. Now, we have <math>f(\frac{m}{4})=\frac{m^{2}}{16} \implies f(\frac{m}{2})=0,\frac{m^{2}}{4} \implies f(m) \geq 0<<ins class="diffchange diffchange-inline">/</ins>math>, which is a contradiction. Thus, <math>\forall x<<ins class="diffchange diffchange-inline">/</ins>math><math>f(x)=0<<ins class="diffchange diffchange-inline">/</ins>math> or <math>f(x)=x^{2}<<ins class="diffchange diffchange-inline">/</ins>math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="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>Step 5: We now assume <<del class="diffchange diffchange-inline">/</del>math>f(x)=0<math>, <<del class="diffchange diffchange-inline">/</del>math>f(y)=y^{2}<math> for some <<del class="diffchange diffchange-inline">/</del>math>x,y \not = 0<math>. Let <<del class="diffchange diffchange-inline">/</del>math>m<math> be sufficiently large integer, let <<del class="diffchange diffchange-inline">/</del>math>z=|4^{m}x|<math> and take the absolute value of <<del class="diffchange diffchange-inline">/</del>math>y<math>(since the function is even). Choose <<del class="diffchange diffchange-inline">/</del>math>c<math> such that <<del class="diffchange diffchange-inline">/</del>math>3z-c=y<math>. Note that we have <<del class="diffchange diffchange-inline">/</del>math>\frac{c}{z}<math>~<<del class="diffchange diffchange-inline">/</del>math>3<math> and <<del class="diffchange diffchange-inline">/</del>math>\frac{y}{z}<math>~<<del class="diffchange diffchange-inline">/</del>math>0<math>. Note that <<del class="diffchange diffchange-inline">/</del>math>f(z)=0<math>. Now, <<del class="diffchange diffchange-inline">/</del>math>x=z, y=c \implies<math> LHS is positive, as the second term is positive and the first term is nonnegative and thus the right term is equal to <<del class="diffchange diffchange-inline">/</del>math>(z+c)^{4}<math>~<<del class="diffchange diffchange-inline">/</del>math>256z^{4}<math>. Now if <<del class="diffchange diffchange-inline">/</del>math>f(z-3c)=0<math>, the second term of the LHS/RHS clearly ~0 as <<del class="diffchange diffchange-inline">/</del>math>m \to \infty<math>. if <<del class="diffchange diffchange-inline">/</del>math>f(z-3c)=0<math>, then we have LHS/RHS ~ <<del class="diffchange diffchange-inline">/</del>math>0<math>, otherwise, we have LHS/RHS~<<del class="diffchange diffchange-inline">/</del>math>\frac{8^{2}\cdot 3z^{4}}{256z^{4}}<math>~<<del class="diffchange diffchange-inline">/</del>math>\frac{3}{4}<math>, a contradiction, as we're clearly not dividing by </math><del class="diffchange diffchange-inline">0$</del>, and we should have LHS/RHS=1.</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>Step 5: We now assume <math>f(x)=0<<ins class="diffchange diffchange-inline">/</ins>math>, <math>f(y)=y^{2}<<ins class="diffchange diffchange-inline">/</ins>math> for some <math>x,y \not = 0<<ins class="diffchange diffchange-inline">/</ins>math>. Let <math>m<<ins class="diffchange diffchange-inline">/</ins>math> be sufficiently large integer, let <math>z=|4^{m}x|<<ins class="diffchange diffchange-inline">/</ins>math> and take the absolute value of <math>y<<ins class="diffchange diffchange-inline">/</ins>math>(since the function is even). Choose <math>c<<ins class="diffchange diffchange-inline">/</ins>math> such that <math>3z-c=y<<ins class="diffchange diffchange-inline">/</ins>math>. Note that we have <math>\frac{c}{z}<<ins class="diffchange diffchange-inline">/</ins>math>~<math>3<<ins class="diffchange diffchange-inline">/</ins>math> and <math>\frac{y}{z}<<ins class="diffchange diffchange-inline">/</ins>math>~<math>0<<ins class="diffchange diffchange-inline">/</ins>math>. Note that <math>f(z)=0<<ins class="diffchange diffchange-inline">/</ins>math>. Now, <math>x=z, y=c \implies<<ins class="diffchange diffchange-inline">/</ins>math> LHS is positive, as the second term is positive and the first term is nonnegative and thus the right term is equal to <math>(z+c)^{4}<<ins class="diffchange diffchange-inline">/</ins>math>~<math>256z^{4}<<ins class="diffchange diffchange-inline">/</ins>math>. Now if <math>f(z-3c)=0<<ins class="diffchange diffchange-inline">/</ins>math>, the second term of the LHS/RHS clearly ~0 as <math>m \to \infty<<ins class="diffchange diffchange-inline">/</ins>math>. if <math>f(z-3c)=0<<ins class="diffchange diffchange-inline">/</ins>math>, then we have LHS/RHS ~ <math>0<<ins class="diffchange diffchange-inline">/</ins>math>, otherwise, we have LHS/RHS~<math>\frac{8^{2}\cdot 3z^{4}}{256z^{4}}<<ins class="diffchange diffchange-inline">/</ins>math>~<math>\frac{3}{4}<<ins class="diffchange diffchange-inline">/</ins>math>, a contradiction, as we're clearly not dividing by <ins class="diffchange diffchange-inline"><math>0</ins></math>, and we should have LHS/RHS=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;"></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>
</table>
Blacksheep2003
https://artofproblemsolving.com/wiki/index.php?title=2016_USAMO_Problems/Problem_4&diff=100041&oldid=prev
Blacksheep2003 at 02:17, 4 January 2019
2019-01-04T02:17:29Z
<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 02:17, 4 January 2019</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l45" >Line 45:</td>
<td colspan="2" class="diff-lineno">Line 45:</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>'''Step 7:''' By Step 6 and the second observation from Step 4, the only possible solutions are <math>f \equiv 0</math> and <math>f(x) = x^2</math> for all <math>x \in \mathbb{R}.</math> It's easy to check that both of these work, so we're done.</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>'''Step 7:''' By Step 6 and the second observation from Step 4, the only possible solutions are <math>f \equiv 0</math> and <math>f(x) = x^2</math> for all <math>x \in \mathbb{R}.</math> It's easy to check that both of these work, so we're done.</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;">==Alternate Solution 1==</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;">From steps 1 and 2 we have that, setting <math>x=y=0</math>, <math>f(0)=0</math>, and, setting <math>x=0</math>, <math>f(-y)f(y)=(f(y))^2</math>, so if <math>f(y) \ne 0</math>, then <math>f(y)=f(-y)</math>. Furthermore, setting <math>y=-x</math> gives us <math>(f(x)-x^2) \cdot f(4x) + (f(-x)-x^2) \cdot f(4x) = f(0) = 0</math>. The LHS can be factored as <math>(f(x)+f(-x)-2x^2) \cdot f(4x) = 0</math>. Therefore, if <math>f(4x) \ne 0</math>, then we have <math>f(x)+f(-x)-2x^2=0</math>, or <math>f(x)+f(-x)=2x^2</math>. However, since from step 2 we have that <math>f(x)=f(-x)</math> assuming f(x) \ne 0<math>, the equation becomes </math>2f(x)=2x^2<math>, so </math>f(x)=0, x^2<math> are the only solutions. </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 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>
<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>Step 1: <math>x=y=0 \implies f(0)=0<<del class="diffchange diffchange-inline">/</del>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>Step 1: <<ins class="diffchange diffchange-inline">/</ins>math>x=y=0 \implies f(0)=0<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>Step 2: <math>x=0 \implies f(y)f(-y)=f(y)^{2}<<del class="diffchange diffchange-inline">/</del>math>. Now, assume <math>y \not = 0<<del class="diffchange diffchange-inline">/</del>math>. Then, if <math>f(y)=0<<del class="diffchange diffchange-inline">/</del>math>, we substitute in <math>-y<<del class="diffchange diffchange-inline">/</del>math> to get <math>f(y)f(-y)=f(-y)^{2}<<del class="diffchange diffchange-inline">/</del>math>, or <math>f(y)=f(-y)=0<<del class="diffchange diffchange-inline">/</del>math>. Otherwise, we divide both sides by <math>f(y)<<del class="diffchange diffchange-inline">/</del>math> to get <math>f(y)=f(-y)<<del class="diffchange diffchange-inline">/</del>math>. If <math>y=0<<del class="diffchange diffchange-inline">/</del>math>, we obviously have <math>f(0)=f(0)<<del class="diffchange diffchange-inline">/</del>math>. Thus, the function is even.</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>Step 2: <<ins class="diffchange diffchange-inline">/</ins>math>x=0 \implies f(y)f(-y)=f(y)^{2}<math>. Now, assume <<ins class="diffchange diffchange-inline">/</ins>math>y \not = 0<math>. Then, if <<ins class="diffchange diffchange-inline">/</ins>math>f(y)=0<math>, we substitute in <<ins class="diffchange diffchange-inline">/</ins>math>-y<math> to get <<ins class="diffchange diffchange-inline">/</ins>math>f(y)f(-y)=f(-y)^{2}<math>, or <<ins class="diffchange diffchange-inline">/</ins>math>f(y)=f(-y)=0<math>. Otherwise, we divide both sides by <<ins class="diffchange diffchange-inline">/</ins>math>f(y)<math> to get <<ins class="diffchange diffchange-inline">/</ins>math>f(y)=f(-y)<math>. If <<ins class="diffchange diffchange-inline">/</ins>math>y=0<math>, we obviously have <<ins class="diffchange diffchange-inline">/</ins>math>f(0)=f(0)<math>. Thus, the function is even.</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>.</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>.</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>Step 3: <math>y=-x \implies 2f(4x)(f(x)-x^{2})=0<<del class="diffchange diffchange-inline">/</del>math>. Thus, <math>\forall x<<del class="diffchange diffchange-inline">/</del>math>, we have <math>f(4x)=0<<del class="diffchange diffchange-inline">/</del>math> or <math>f(x)=x^{2}<<del class="diffchange diffchange-inline">/</del>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>Step 3: <<ins class="diffchange diffchange-inline">/</ins>math>y=-x \implies 2f(4x)(f(x)-x^{2})=0<math>. Thus, <<ins class="diffchange diffchange-inline">/</ins>math>\forall x<math>, we have <<ins class="diffchange diffchange-inline">/</ins>math>f(4x)=0<math> or <<ins class="diffchange diffchange-inline">/</ins>math>f(x)=x^{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="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>Step 4: We now assume <math>f(x) \not = 0<<del class="diffchange diffchange-inline">/</del>math>, <math>x\not = 0<<del class="diffchange diffchange-inline">/</del>math>. We have <math>f(\frac{x}{4})=\frac{x^{2}}{16}<<del class="diffchange diffchange-inline">/</del>math>. Now, setting <math>x=y=\frac{x}{4}<<del class="diffchange diffchange-inline">/</del>math>, we have <math>f(\frac{x}{2}=\frac{x^{2}}{4}<<del class="diffchange diffchange-inline">/</del>math> or <math>f(\frac{x}{2})=0<<del class="diffchange diffchange-inline">/</del>math>. The former implies that <math>f(x)=0<<del class="diffchange diffchange-inline">/</del>math> or <math>x^{2}<<del class="diffchange diffchange-inline">/</del>math>. The latter implies that <math>f(x)=0<<del class="diffchange diffchange-inline">/</del>math> or <math>f(x)=\frac{x^{2}}{2}<<del class="diffchange diffchange-inline">/</del>math>. Assume the latter. <math>y=-2x \implies -\frac{3y^{2}}{2}f(7y)-{2y^{2}}f(5y)=\frac{x^{2}}{2}<<del class="diffchange diffchange-inline">/</del>math>. Clearly, this implies that <math>f(x)<<del class="diffchange diffchange-inline">/</del>math> is negative for some <math>m<<del class="diffchange diffchange-inline">/</del>math>. Now, we have <math>f(\frac{m}{4})=\frac{m^{2}}{16} \implies f(\frac{m}{2})=0,\frac{m^{2}}{4} \implies f(m) \geq 0<<del class="diffchange diffchange-inline">/</del>math>, which is a contradiction. Thus, <math>\forall x<<del class="diffchange diffchange-inline">/</del>math><math>f(x)=0<<del class="diffchange diffchange-inline">/</del>math> or <del class="diffchange diffchange-inline"><math>f(x)=x^{2}</math>.</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>Step 4: We now assume <<ins class="diffchange diffchange-inline">/</ins>math>f(x) \not = 0<math>, <<ins class="diffchange diffchange-inline">/</ins>math>x\not = 0<math>. We have <<ins class="diffchange diffchange-inline">/</ins>math>f(\frac{x}{4})=\frac{x^{2}}{16}<math>. Now, setting <<ins class="diffchange diffchange-inline">/</ins>math>x=y=\frac{x}{4}<math>, we have <<ins class="diffchange diffchange-inline">/</ins>math>f(\frac{x}{2}=\frac{x^{2}}{4}<math> or <<ins class="diffchange diffchange-inline">/</ins>math>f(\frac{x}{2})=0<math>. The former implies that <<ins class="diffchange diffchange-inline">/</ins>math>f(x)=0<math> or <<ins class="diffchange diffchange-inline">/</ins>math>x^{2}<math>. The latter implies that <<ins class="diffchange diffchange-inline">/</ins>math>f(x)=0<math> or <<ins class="diffchange diffchange-inline">/</ins>math>f(x)=\frac{x^{2}}{2}<math>. Assume the latter. <<ins class="diffchange diffchange-inline">/</ins>math>y=-2x \implies -\frac{3y^{2}}{2}f(7y)-{2y^{2}}f(5y)=\frac{x^{2}}{2}<math>. Clearly, this implies that <<ins class="diffchange diffchange-inline">/</ins>math>f(x)<math> is negative for some <<ins class="diffchange diffchange-inline">/</ins>math>m<math>. Now, we have <<ins class="diffchange diffchange-inline">/</ins>math>f(\frac{m}{4})=\frac{m^{2}}{16} \implies f(\frac{m}{2})=0,\frac{m^{2}}{4} \implies f(m) \geq 0<math>, which is a contradiction. Thus, <<ins class="diffchange diffchange-inline">/</ins>math>\forall x<math><<ins class="diffchange diffchange-inline">/</ins>math>f(x)=0<math> or </math>f(x)=x^{2}<math>.</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 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 class="diffchange diffchange-inline">Step 5: We now assume <math>f(x)=0</del></<del class="diffchange diffchange-inline">math>, <</del>math>f(<del class="diffchange diffchange-inline">y)=y^{2}</math> for some <math></del>x<del class="diffchange diffchange-inline">,y \not = 0</math>. Let <math>m</math> be sufficiently large integer, let <math>z=|4^{m}x|</math> and take the absolute value of <math>y</math>(since the function is even). Choose <math>c</math> such that <math>3z-c=y</math>. Note that we have <math>\frac{c}{z}</math>~<math>3</math> and <math>\frac{y}{z}</math>~<math>0</math>. Note that <math>f(z</del>)=<del class="diffchange diffchange-inline">0</math>. Now, <math></del>x<del class="diffchange diffchange-inline">=z, y=c \implies</math> LHS is positive, as the second term is positive and the first term is nonnegative and thus the right term is equal to <math>(z+c)^{4}</math>~<math>256z^{4}</math>. Now if <math>f(z-3c)=0</math>, the second term of the LHS/RHS clearly ~0 as <math>m \to \infty</math>. if <math>f(z-3c)=0</math>, then we have LHS/RHS ~ <math>0</math>, otherwise, we have LHS/RHS~<math>\frac{8</del>^{2<del class="diffchange diffchange-inline">}\cdot 3z^{4}}{256z^{4}</del>}<<del class="diffchange diffchange-inline">/</del>math><del class="diffchange diffchange-inline">~<math>\frac{3}{4}</math>, a contradiction, as we're clearly not dividing by <math>0</math>, and we should have LHS/RHS=1</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 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;">Step 5: We now assume </math>f(x)=0<math>, </math>f(y)=y^{2}<math> for some </math>x,y \not = 0<math>. Let </math>m<math> be sufficiently large integer, let </math>z=|4^{m}x|<math> and take the absolute value of </math>y<math>(since the function is even). Choose </math>c<math> such that </math>3z-c=y<math>. Note that we have </math>\frac{c}{z}<math>~</math>3<math> and </math>\frac{y}{z}<math>~</math>0<math>. Note that </math>f(z)=0<math>. Now, </math>x=z, y=c \implies<math> LHS is positive, as the second term is positive and the first term is nonnegative and thus the right term is equal to </math>(z+c)^{4}<math>~</math>256z^{4}<math>. Now if </math>f(z-3c)=0<math>, the second term of the LHS/RHS clearly ~0 as </math>m \to \infty<math>. if </math>f(z-3c)=0<math>, then we have LHS/RHS ~ </math>0<math>, otherwise, we have LHS/RHS~</math>\frac{8^{2}\cdot 3z^{4}}{256z^{4}}<math>~</math>\frac{3}{4}<math>, a contradiction, as we're clearly not dividing by </math>0$, and we should have LHS/RHS=1.</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;"></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>
Blacksheep2003
https://artofproblemsolving.com/wiki/index.php?title=2016_USAMO_Problems/Problem_4&diff=97385&oldid=prev
V Enhance: Delete the obviously wrong sol3
2018-08-25T08:24:52Z
<p>Delete the obviously wrong sol3</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 08:24, 25 August 2018</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l56" >Line 56:</td>
<td colspan="2" class="diff-lineno">Line 56:</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>Step 5: We now assume <math>f(x)=0</math>, <math>f(y)=y^{2}</math> for some <math>x,y \not = 0</math>. Let <math>m</math> be sufficiently large integer, let <math>z=|4^{m}x|</math> and take the absolute value of <math>y</math>(since the function is even). Choose <math>c</math> such that <math>3z-c=y</math>. Note that we have <math>\frac{c}{z}</math>~<math>3</math> and <math>\frac{y}{z}</math>~<math>0</math>. Note that <math>f(z)=0</math>. Now, <math>x=z, y=c \implies</math> LHS is positive, as the second term is positive and the first term is nonnegative and thus the right term is equal to <math>(z+c)^{4}</math>~<math>256z^{4}</math>. Now if <math>f(z-3c)=0</math>, the second term of the LHS/RHS clearly ~0 as <math>m \to \infty</math>. if <math>f(z-3c)=0</math>, then we have LHS/RHS ~ <math>0</math>, otherwise, we have LHS/RHS~<math>\frac{8^{2}\cdot 3z^{4}}{256z^{4}}</math>~<math>\frac{3}{4}</math>, a contradiction, as we're clearly not dividing by <math>0</math>, and we should have LHS/RHS=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>Step 5: We now assume <math>f(x)=0</math>, <math>f(y)=y^{2}</math> for some <math>x,y \not = 0</math>. Let <math>m</math> be sufficiently large integer, let <math>z=|4^{m}x|</math> and take the absolute value of <math>y</math>(since the function is even). Choose <math>c</math> such that <math>3z-c=y</math>. Note that we have <math>\frac{c}{z}</math>~<math>3</math> and <math>\frac{y}{z}</math>~<math>0</math>. Note that <math>f(z)=0</math>. Now, <math>x=z, y=c \implies</math> LHS is positive, as the second term is positive and the first term is nonnegative and thus the right term is equal to <math>(z+c)^{4}</math>~<math>256z^{4}</math>. Now if <math>f(z-3c)=0</math>, the second term of the LHS/RHS clearly ~0 as <math>m \to \infty</math>. if <math>f(z-3c)=0</math>, then we have LHS/RHS ~ <math>0</math>, otherwise, we have LHS/RHS~<math>\frac{8^{2}\cdot 3z^{4}}{256z^{4}}</math>~<math>\frac{3}{4}</math>, a contradiction, as we're clearly not dividing by <math>0</math>, and we should have LHS/RHS=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;"></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 style="font-weight: bold; text-decoration: none;">==Solution Simple Substitution (3)==</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;">Each of these steps is rather intuitive. Obviously the function can only be solved by substitution.</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="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;">Substitute <math>x=0, y=0</math> to get <math>f(0)^2+f(0)^2=f(0)^2.</math> Clearly <math>f(0)^2=0, f(0)=0</math>.</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;"></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;">Substitute <math>x=0</math> only to get <math>f(y)f(-y)=f(y)^2</math>. Either <math>f(y)=0, f(-y)=f(y)</math>. In the first case a solution is <math>\boxed{f(x)=0}</math>. In the second, we know <math>f</math> is even.</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;"></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;">Finally, substitute <math>y=-x</math>. Then <math>(f(x)-x^2)f(4x)+(f(y)-x^2)f(4x)=0</math>, so <math>2(f(x)-x^2)f(4x)=0</math>. If <math>f(x)-x^2=0, \boxed{f(x)=x^2}</math>. If <math>f(4x)=0</math>, we've already found the solution. Substitution of <math>f</math> for both cases works, and because <math>f</math> has to apply for all real numbers, we are done.</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>{{MAA Notice}}</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>{{MAA Notice}}</div></td></tr>
</table>
V Enhance