https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Serpent+121&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T10:18:46ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_II_Problems/Problem_5&diff=1502472021 AIME II Problems/Problem 52021-03-25T11:36:51Z<p>Serpent 121: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
For positive real numbers <math>s</math>, let <math>\tau(s)</math> denote the set of all obtuse triangles that have area <math>s</math> and two sides with lengths <math>4</math> and <math>10</math>. The set of all <math>s</math> for which <math>\tau(s)</math> is nonempty, but all triangles in <math>\tau(s)</math> are congruent, is an interval <math>[a,b)</math>. Find <math>a^2+b^2</math>.<br />
<br />
==Solution 1==<br />
We start by defining a triangle. The two small sides MUST add to a larger sum than the long side. We are given 4 and 10 as the sides, so we know that the 3rd side is between 6 and 14, exclusive. We also have to consider the word OBTUSE triangles. That means that the two small sides squared is less than the 3rd side. So the triangles sides are between 6 and <math>\sqrt{84}</math> exclusive, and the larger bound is between <math>\sqrt{116}</math> and 14, exclusive. The area of these triangles are from 0 (straight line) to <math>2\sqrt{84}</math> on the first "small bound" and the larger bound is between 0 and 20. <br />
<math>0 < s < 2\sqrt{84}</math> is our first equation, and <math>0 < s < 20</math> is our 2nd equation. Therefore, the area is between <math>\sqrt{336}</math> and <math>\sqrt{400}</math>, so our final answer is <math>\boxed{736}</math>.<br />
<br />
~ARCTICTURN<br />
<br />
==Solution 2 (Casework: Detailed Explanation of Solution 1)==<br />
Every obtuse triangle must satisfy <i><b>both</b></i> of the following:<br />
<br />
* <b>Triangle Inequality Theorem:</b> If <math>a,b,</math> and <math>c</math> are the side-lengths of a triangle with <math>a\leq b\leq c,</math> then <math>a+b>c.</math><br />
<br />
* <b>Pythagorean Inequality Theorem:</b> If <math>a,b,</math> and <math>c</math> are the side-lengths of an obtuse triangle with <math>a\leq b<c,</math> then <math>a^2+b^2<c^2.</math><br />
<br />
For one such obtuse triangle, let <math>4,10,</math> and <math>x</math> be the side-lengths and <math>k</math> be the area. We will use casework on the longest side:<br />
<br />
<b>Case (1): The longest side has length <math>\boldsymbol{10.}</math></b><br />
<br />
By the Triangle Inequality Theorem, we have <math>4+x>10,</math> from which <math>x>6.</math><br />
<br />
By the Pythagorean Inequality Theorem, we get <math>4^2+x^2<10^2,</math> so that <math>x<\sqrt{84}.</math><br />
<br />
Taking the intersection produces <math>6<x<\sqrt{84}</math> for this case.<br />
<br />
At <math>x=6,</math> the obtuse triangle degenerates into a straight line with area <math>k=0;</math> at <math>x=\sqrt{84},</math> the obtuse triangle degenerates into a right triangle with area <math>k=\frac12\cdot4\cdot\sqrt{84}=2\sqrt{84}.</math> Together, we obtain <math>0<k<2\sqrt{84},</math> or <math>k\in\left(0,2\sqrt{84}\right).</math><br />
<br />
<b>Case (2): The longest side has length <math>\boldsymbol{x.}</math></b><br />
<br />
By the Triangle Inequality Theorem, we have <math>4+10>x,</math> from which <math>x<14.</math><br />
<br />
By the Pythagorean Inequality Theorem, we get <math>4^2+10^2<x^2,</math> so that <math>x>\sqrt{116}.</math><br />
<br />
Taking the intersection produces <math>\sqrt{116}<x<14</math> for this case.<br />
<br />
At <math>x=14,</math> the obtuse triangle degenerates into a straight line with area <math>k=0;</math> at <math>x=\sqrt{116},</math> the obtuse triangle degenerates into a right triangle with area <math>k=\frac12\cdot4\cdot10=20.</math> Together, we obtain <math>0<k<20,</math> or <math>k\in\left(0,20\right).</math><br />
<br />
<b>Answer</b><br />
<br />
It is possible for non-congruent obtuse triangles to have the same area. Therefore, all such positive real numbers <math>s</math> are in exactly one of <math>\left(0,2\sqrt{84}\right)</math> or <math>\left(0,20\right).</math> Taking the exclusive disjunction, the set of all such <math>s</math> is <cmath>[a,b)=\left(0,2\sqrt{84}\right)\oplus\left(0,20\right)=\left[2\sqrt{84},20\right),</cmath> from which <math>a^2+b^2=\boxed{736}.</math><br />
<br />
~MRENTHUSIASM<br />
<br />
==Solution 3==<br />
<br />
We have the diagram below.<br />
<br />
<asy><br />
<br />
draw((0,0)--(1,2*sqrt(3)));<br />
draw((1,2*sqrt(3))--(10,0));<br />
draw((10,0)--(0,0));<br />
label("A",(0,0),SW);<br />
label("B",(1,2*sqrt(3)),N);<br />
label("C",(10,0),SE);<br />
label("$\theta$",(0,0),NE);<br />
label("$\alpha$",(1,2*sqrt(3)),SSE);<br />
label("$4$",(0,0)--(1,2*sqrt(3)),WNW);<br />
label("$10$",(0,0)--(10,0),S);<br />
<br />
</asy><br />
<br />
We proceed by taking cases on the angles that can be obtuse, and finding the ranges for <math>s</math> that they yield .<br />
<br />
If angle <math>\theta</math> is obtuse, then we have that <math>s \in (0,20)</math>. This is because <math>s=20</math> is attained at <math>\theta = 90^{\circ}</math>, and the area of the triangle is strictly decreasing as <math>\theta</math> increases beyond <math>90^{\circ}</math>. This can be observed from<br />
<cmath>s=\frac{1}{2}(4)(10)\sin\theta</cmath>by noting that <math>\sin\theta</math> is decreasing in <math>\theta \in (90^{\circ},180^{\circ})</math>.<br />
<br />
Then, we note that if <math>\alpha</math> is obtuse, we have <math>s \in (0,4\sqrt{21})</math>. This is because we get <math>x=\sqrt{10^2-4^2}=\sqrt{84}=2\sqrt{21}</math> when <math>\alpha=90^{\circ}</math>, yileding <math>s=4\sqrt{21}</math>. Then, <math>s</math> is decreasing as <math>\alpha</math> increases by the same argument as before.<br />
<br />
<math>\angle{ACB}</math> cannot be obtuse since <math>AC>AB</math>.<br />
<br />
Now we have the intervals <math>s \in (0,20)</math> and <math>s \in (0,4\sqrt{21})</math> for the cases where <math>\theta</math> and <math>\alpha</math> are obtuse, respectively. We are looking for the <math>s</math> that are in exactly one of these intervals, and because <math>4\sqrt{21}<20</math>, the desired range is<br />
<cmath>s\in [4\sqrt{21},20)</cmath>giving <cmath>a^2+b^2=\boxed{736}\Box</cmath><br />
<br />
==See also==<br />
{{AIME box|year=2021|n=II|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Serpent 121https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_II_Problems/Problem_5&diff=1502462021 AIME II Problems/Problem 52021-03-25T11:35:18Z<p>Serpent 121: /* Solution 2 (Casework: Detailed Explanation of Solution 1) */</p>
<hr />
<div>==Problem==<br />
For positive real numbers <math>s</math>, let <math>\tau(s)</math> denote the set of all obtuse triangles that have area <math>s</math> and two sides with lengths <math>4</math> and <math>10</math>. The set of all <math>s</math> for which <math>\tau(s)</math> is nonempty, but all triangles in <math>\tau(s)</math> are congruent, is an interval <math>[a,b)</math>. Find <math>a^2+b^2</math>.<br />
<br />
==Solution 1==<br />
We start by defining a triangle. The two small sides MUST add to a larger sum than the long side. We are given 4 and 10 as the sides, so we know that the 3rd side is between 6 and 14, exclusive. We also have to consider the word OBTUSE triangles. That means that the two small sides squared is less than the 3rd side. So the triangles sides are between 6 and <math>\sqrt{84}</math> exclusive, and the larger bound is between <math>\sqrt{116}</math> and 14, exclusive. The area of these triangles are from 0 (straight line) to <math>2\sqrt{84}</math> on the first "small bound" and the larger bound is between 0 and 20. <br />
<math>0 < s < 2\sqrt{84}</math> is our first equation, and <math>0 < s < 20</math> is our 2nd equation. Therefore, the area is between <math>\sqrt{336}</math> and <math>\sqrt{400}</math>, so our final answer is <math>\boxed{736}</math>.<br />
<br />
~ARCTICTURN<br />
<br />
==Solution 2 (Casework: Detailed Explanation of Solution 1)==<br />
Every obtuse triangle must satisfy <i><b>both</b></i> of the following:<br />
<br />
* <b>Triangle Inequality Theorem:</b> If <math>a,b,</math> and <math>c</math> are the side-lengths of a triangle with <math>a\leq b\leq c,</math> then <math>a+b>c.</math><br />
<br />
* <b>Pythagorean Inequality Theorem:</b> If <math>a,b,</math> and <math>c</math> are the side-lengths of an obtuse triangle with <math>a\leq b<c,</math> then <math>a^2+b^2<c^2.</math><br />
<br />
For one such obtuse triangle, let <math>4,10,</math> and <math>x</math> be the side-lengths and <math>k</math> be the area. We will use casework on the longest side:<br />
<br />
<b>Case (1): The longest side has length <math>\boldsymbol{10.}</math></b><br />
<br />
By the Triangle Inequality Theorem, we have <math>4+x>10,</math> from which <math>x>6.</math><br />
<br />
By the Pythagorean Inequality Theorem, we get <math>4^2+x^2<10^2,</math> so that <math>x<\sqrt{84}.</math><br />
<br />
Taking the intersection produces <math>6<x<\sqrt{84}</math> for this case.<br />
<br />
At <math>x=6,</math> the obtuse triangle degenerates into a straight line with area <math>k=0;</math> at <math>x=\sqrt{84},</math> the obtuse triangle degenerates into a right triangle with area <math>k=\frac12\cdot4\cdot\sqrt{84}=2\sqrt{84}.</math> Together, we obtain <math>0<k<2\sqrt{84},</math> or <math>k\in\left(0,2\sqrt{84}\right).</math><br />
<br />
<b>Case (2): The longest side has length <math>\boldsymbol{x.}</math></b><br />
<br />
By the Triangle Inequality Theorem, we have <math>4+10>x,</math> from which <math>x<14.</math><br />
<br />
By the Pythagorean Inequality Theorem, we get <math>4^2+10^2<x^2,</math> so that <math>x>\sqrt{116}.</math><br />
<br />
Taking the intersection produces <math>\sqrt{116}<x<14</math> for this case.<br />
<br />
At <math>x=14,</math> the obtuse triangle degenerates into a straight line with area <math>k=0;</math> at <math>x=\sqrt{116},</math> the obtuse triangle degenerates into a right triangle with area <math>k=\frac12\cdot4\cdot10=20.</math> Together, we obtain <math>0<k<20,</math> or <math>k\in\left(0,20\right).</math><br />
<br />
<b>Answer</b><br />
<br />
It is possible for non-congruent obtuse triangles to have the same area. Therefore, all such positive real numbers <math>s</math> are in exactly one of <math>\left(0,2\sqrt{84}\right)</math> or <math>\left(0,20\right).</math> Taking the exclusive disjunction, the set of all such <math>s</math> is <cmath>[a,b)=\left(0,2\sqrt{84}\right)\oplus\left(0,20\right)=\left[2\sqrt{84},20\right),</cmath> from which <math>a^2+b^2=\boxed{736}.</math><br />
<br />
~MRENTHUSIASM<br />
<br />
==Solution 3==<br />
<br />
We have the diagram below.<br />
<br />
[asy]<br />
<br />
draw((0,0)--(1,2*sqrt(3)));<br />
draw((1,2*sqrt(3))--(10,0));<br />
draw((10,0)--(0,0));<br />
label("A",(0,0),SW);<br />
label("B",(1,2*sqrt(3)),N);<br />
label("C",(10,0),SE);<br />
label("<math>\theta</math>",(0,0),NE);<br />
label("<math>\alpha</math>",(1,2*sqrt(3)),SSE);<br />
label("<math>4</math>",(0,0)--(1,2*sqrt(3)),WNW);<br />
label("<math>10</math>",(0,0)--(10,0),S);<br />
<br />
[/asy]<br />
<br />
We proceed by taking cases on the angles that can be obtuse, and finding the ranges for <math>s</math> that they yield .<br />
<br />
If angle <math>\theta</math> is obtuse, then we have that <math>s \in (0,20)</math>. This is because <math>s=20</math> is attained at <math>\theta = 90^{\circ}</math>, and the area of the triangle is strictly decreasing as <math>\theta</math> increases beyond <math>90^{\circ}</math>. This can be observed from<br />
<cmath>s=\frac{1}{2}(4)(10)\sin\theta</cmath>by noting that <math>\sin\theta</math> is decreasing in <math>\theta \in (90^{\circ},180^{\circ})</math>.<br />
<br />
Then, we note that if <math>\alpha</math> is obtuse, we have <math>s \in (0,4\sqrt{21})</math>. This is because we get <math>x=\sqrt{10^2-4^2}=\sqrt{84}=2\sqrt{21}</math> when <math>\alpha=90^{\circ}</math>, yileding <math>s=4\sqrt{21}</math>. Then, <math>s</math> is decreasing as <math>\alpha</math> increases by the same argument as before.<br />
<br />
<math>\angle{ACB}</math> cannot be obtuse since <math>AC>AB</math>.<br />
<br />
Now we have the intervals <math>s \in (0,20)</math> and <math>s \in (0,4\sqrt{21})</math> for the cases where <math>\theta</math> and <math>\alpha</math> are obtuse, respectively. We are looking for the <math>s</math> that are in exactly one of these intervals, and because <math>4\sqrt{21}<20</math>, the desired range is<br />
<cmath>s\in [4\sqrt{21},20)</cmath>giving <cmath>a^2+b^2=\boxed{736}\Box</cmath><br />
<br />
==See also==<br />
{{AIME box|year=2021|n=II|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Serpent 121https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_3&diff=1465962021 AMC 10B Problems/Problem 32021-02-14T10:42:15Z<p>Serpent 121: /* Solution 2 (Fast and not rigorous) */</p>
<hr />
<div>==Problem==<br />
<br />
In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the <math>28</math> students in the program, <math>25\%</math> of the juniors and <math>10\%</math> of the seniors are on the debate team. How many juniors are in the program?<br />
<br />
<math>\textbf{(A)} ~5 \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~8 \qquad\textbf{(D)} ~11 \qquad\textbf{(E)} ~20</math><br />
<br />
==Solution 1==<br />
Say there are <math>j</math> juniors and <math>s</math> seniors in the program. Converting percentages to fractions, <math>\frac{j}{4}</math> and <math>\frac{s}{10}</math> are on the debate team, and since an equal number of juniors and seniors are on the debate team, <math>\frac{j}{4} = \frac{s}{10}.</math><br />
<br />
Cross-multiplying and simplifying we get <math>5j=2s.</math> Additionally, since there are <math>28</math> students in the program, <math>j+s = 28.</math> It is now a matter of solving the system of equations <cmath>5j=2s</cmath><cmath>j+s=28,</cmath> and the solution is <math>j = 8, s = 20.</math> Since we want the number of juniors, the answer is <cmath>\boxed{(C) \text{ } 8}.</cmath><br />
<br />
-PureSwag<br />
<br />
==Solution 2 (Fast and not rigorous)==<br />
We immediately see that <math>E</math> is the only possible amount of seniors, as <math>10\%</math> can only correspond with an answer choice ending with <math>0</math>. Thus the number of seniors is <math>20</math> and the number of juniors is <math>28-20=8\rightarrow \boxed{C}</math>. ~samrocksnature<br />
<br />
==Solution 3==<br />
Since there are an equal number of juniors and seniors on the debate team, suppose there are <math>x</math> juniors and <math>x</math> seniors. This number represents <math>25\% =\frac{1}{4}</math> of the juniors and <math>10\%= \frac{1}{10}</math> of the seniors, which tells us that there are <math>4x</math> juniors and <math>10x</math> seniors. There are <math>28</math> juniors and seniors in the program altogether, so we get <br />
<cmath>10x+4x=28,</cmath><br />
<cmath>14x=28,</cmath><br />
<cmath>x=2. </cmath><br />
Which means there are <math>4x=8</math> juniors on the debate team, <math>\boxed{\text{(C)} \, 8}</math>.<br />
<br />
== Video Solution by OmegaLearn (System of Equations) ==<br />
https://youtu.be/BtEF-hJBGV8<br />
<br />
<br />
{{AMC10 box|year=2021|ab=B|num-b=2|num-a=4}}</div>Serpent 121https://artofproblemsolving.com/wiki/index.php?title=1974_USAMO_Problems/Problem_2&diff=1233111974 USAMO Problems/Problem 22020-05-30T11:21:46Z<p>Serpent 121: </p>
<hr />
<div>==Problem==<br />
Prove that if <math>a</math>, <math>b</math>, and <math>c</math> are positive real numbers, then<br />
<center><math>a^ab^bc^c\ge (abc)^{(a+b+c)/3}</math></center><br />
<br />
==Solution 1==<br />
Consider the function <math>f(x)=x\ln{x}</math>. <math>f''(x)=\frac{1}{x}>0</math> for <math>x>0</math>; therefore, it is a convex function and we can apply [[Jensen's Inequality]]:<br />
<center><math>\frac{a\ln{a}+b\ln{b}+c\ln{c}}{3}\ge \left(\frac{a+b+c}{3}\right)\ln\left(\frac{a+b+c}{3}\right)</math></center><br />
Apply [[AM-GM]] to get<br />
<center><math>\frac{a+b+c}{3}\ge \sqrt[3]{abc}</math></center><br />
which implies<br />
<center><math>\frac{a\ln{a}+b\ln{b}+c\ln{c}}{3}\ge \left(\frac{a+b+c}{3}\right)\ln\left(\sqrt[3]{abc}\right)</math></center><br />
Rearranging,<br />
<center><math>a\ln{a}+b\ln{b}+c\ln{c}\ge\left(\frac{a+b+c}{3}\right)\ln\left(abc\right)</math></center><br />
Because <math>f(x) = e^x</math> is an increasing function, we can conclude that:<br />
<center><math>e^{a\ln{a}+b\ln{b}+c\ln{c}}\ge{e}^{\ln\left(abc\right)(a+b+c)/3}</math></center><br />
which simplifies to the desired inequality.<br />
<br />
==Solution 2==<br />
Note that <math>(a^ab^bc^c)(a^bb^cc^a)(a^cb^ac^b)=\left((abc)^{(a+b+c)/3}\right)^3</math>.<br />
<br />
So if we can prove that <math>a^ab^bc^c\ge a^bb^cc^a</math> and <math>a^ab^bc^c\ge a^cb^ac^b</math>, then we are done.<br />
<br />
WLOG let <math>a\ge b\ge c</math>.<br />
<br />
Note that <math>(a^ab^bc^c)\cdot \left(\dfrac{c}{a}\right)^{a-b}\cdot \left(\dfrac{c}{b}\right)^{b-c}=a^bb^cc^a</math>. Since <math>\dfrac{c}{a} \le 1</math>, <math>\dfrac{c}{b} \le 1</math>, <math>a-b \ge 0</math>, and <math>b-c \ge 0</math>, it follows that <math>a^ab^bc^c \ge a^bb^cc^a</math>.<br />
<br />
Note that <math>(a^ab^bc^c)\cdot \left(\dfrac{b}{a}\right)^{a-b}\cdot \left(\dfrac{c}{a}\right)^{b-c}=a^cb^ac^b</math>. Since <math>\dfrac{b}{a} \le 1</math>, <math>\dfrac{c}{a} \le 1</math>, <math>a-b \ge 0</math>, and <math>b-c \ge 0</math>, it follows that <math>a^ab^bc^c \ge a^cb^ac^b</math>.<br />
<br />
Thus, <math>(a^ab^bc^c)^3\ge (a^ab^bc^c)(a^bb^cc^a)(a^cb^ac^b)=\left((abc)^{(a+b+c)/3}\right)^3</math>, and cube-rooting both sides gives <math>a^ab^bc^c\ge (abc)^{(a+b+c)/3}</math> as desired.<br />
<br />
==Solution 3==<br />
<br />
WLOG let <math>a\ge b\ge c</math>. Let <math>b = ax</math> and <math>c = ay</math>, where <math>x \ge 1</math> and <math>y \ge 1</math>.<br />
<br />
We want to prove that <math>(a)^{a}(ax)^{ax}(ay)^{ay} \ge (a \cdot ax \cdot ay)^{\frac{a + ax + ay}{3}}</math>.<br />
<br />
Simplifying and combining terms on each side, we get <math>a^{a + ax + ay}x^{ax}y^{ay} \ge a^{a + ax + ay}(xy)^{\frac{a + ax + ay}{3}}</math>.<br />
<br />
Since <math>a > 0</math>, we can divide out <math>a^{a + ax + ay}</math> to get <math>x^{ax}y^{ay} \ge (xy)^{\frac{a + ax + ay}{3}}</math>.<br />
<br />
Take the <math>a</math>th root of each side and then cube both sides to get <math>x^{3x}y^{3y} \ge (xy)^{1 + x + y}</math>.<br />
<br />
This simplifies to <math>x^{2x-1}y^{2y-1} \ge x^{y}y^{x}</math>.<br />
<br />
Since <math>2x - 1 \ge x</math> and <math>2y - 1 \ge y</math>, we only need to prove <math>x^{x}y^{y} \ge x^{y}y^{x}</math> for our given <math>x, y</math>.<br />
<br />
WLOG, let <math>y \ge x</math> and <math> y =kx</math> for <math>k \ge 1</math>. Then our expression becomes <br />
<br />
<math>x^{x}(xk)^{xk} \ge x^{xk}(xk)^{x}</math><br />
<br />
<math>x^{x+xk}k^{xk} \ge x^{x+xk}k^{x}</math><br />
<br />
<math>k^{xk} \ge k^{x}</math><br />
<br />
<math>k^k \ge k</math><br />
<br />
This is clearly true for <math>k \ge 1</math>.<br />
<br />
==Solution 4==<br />
WLOG let <math>a\ge b\ge c</math>. Then sequence <math>(a,b,c)</math> majorizes <math>(\frac{a+b+c}{3},\frac{a+b+c}{3},\frac{a+b+c}{3})</math>. Thus by Muirhead's Inequality, we have <math>\sum_{sym} a^ab^bc^c \ge \sum_{sym} a^{\frac{a+b+c}{3}}b^{\frac{a+b+c}{3}}c^{\frac{a+b+c}{3}}</math>, so <math>a^ab^bc^c \ge (abc)^{\frac{a+b+c}{3}}</math>.<br />
<br />
==Solution 5==<br />
Let <math>x=\frac{a}{\sqrt[3]{abc}},</math> <math>y=\frac{b}{\sqrt[3]{abc}}</math> and <math>z=\frac{c}{\sqrt[3]{abc}}.</math> Then <math>xyz=1</math> and a straightforward calculation reduces the problem to <br />
<cmath>x^xy^yz^z \ge 1.</cmath><br />
WLOG, assume <math>x\ge y\ge z.</math> Then <math>x\ge 1,</math> <math>z\le 1</math> and <math>xy=\frac{1}{z} \ge 1.</math> Therefore,<br />
<cmath> x^xy^yz^z=x^{x-y}(xy)^{y-z}(xyz)^z \ge 1.</cmath><br />
<br />
J.Z.<br />
<br />
==Solution 6==<br />
<br />
Cubing both sides of the given inequality gives <cmath>a^{3a}b^{3b}c^{3c}\ge(abc)^{a+b+c}</cmath><br />
If we take <math>a^{3a}b^{3b}c^{3c}</math> as the product of <math>3a</math> <math>a</math>'s, <math>3b</math> <math>b</math>'s, and <math>3c</math> <math>c's</math>, we get that <br />
<br />
<center><br />
<math>\sqrt[3(a+b+c)]{a^{3a}b^{3b}c^{3c}}\ge\frac{3(a+b+c)}{\frac{3a}{a}+\frac{3b}{b}+\frac{3c}{c}}=\frac{a+b+c}{3}\ge\sqrt[3]{abc}</math><br />
</center><br />
<br />
by GM-HM, as desired.<br />
<br />
==Solution 7==<br />
<br />
Replacing <math>a</math> with <math>ak</math>, <math>b</math> with <math>bk</math>, and <math>c</math> with <math>ck</math>, for some positive real <math>k</math> we get:<br />
<math>a^{ak}k^{ak}b^{bk}k^{bk}c^{ck}k^{ck} = a^{ak}b^{bk}c^{ck}k^{ak+bk+ck} \ge {k^3abc}^{{ak+bk+ck}/3} = k^{ak+bk+ck}{abc}^{{ak+bk+ck}/3}</math><br />
This means that this inequality is homogeneous since both sides have the same power of <math>k</math> as a factor. Since the inequality is homogeneous, we can scale <math>abc</math> so that their product is <math>1</math>, i.e. <math>abc = 1</math>. This makes the inequality turn into something much more nicer to deal with. Now we have to prove:<br />
<math>a^ab^bc^c \ge 1</math> given that <math>abc = 1</math>.<br />
Note that <math>a^a \ge a</math> even if <math>a \le 1</math>. Therefore <math>a^a \ge a</math>, <math>b^b \ge b</math>, and <math>c^c \ge c</math>. Multiplying these together we get:<br />
<math>a^ab^bc^c \ge abc = 1</math>. This proves the desired result. Equality holds when <math>a = b = c = 1</math>.<br />
<br />
==Solution 8 (Rearrangement)==<br />
Let <math>\log_2a=x</math>, <math>\log_2b=y</math> and <math>\log_2c=z</math>, and WLOG <math>a\ge b\ge c >0</math>. <br />
Then we have both <math>a\ge b\ge c</math> and <math>x\ge y \ge z</math>.<br />
By the rearrangement inequality, <br />
<math>ax+by+cz\ge ay + bz +cx</math> and <br />
<math>ax+by +cz\ge az+bx+cy</math>.<br />
Summing, <br />
<math>2(ax+by+cz)\ge a(y+z)+b(x+z)+c(x+y)</math><br />
Adding <math>ax+by+cz</math>,we get<br />
<math>ax+by+cz\ge \frac{(a+b+c)(x+y+z)}{3}</math>.<br />
Now we substitute back for <math>x,y,z</math> to get:<br />
<math>\log_2a^ab^bc^c\ge \log_2(abc)^{\frac{a+b+c}{3}}</math>.<br />
Raising <math>2</math> to the power of each side, we get<br />
<math>a^ab^bc^c\ge (abc)^{\frac{a+b+c}{3}}</math><br />
<br />
{{alternate solutions}}<br />
<br />
== See Also ==<br />
{{USAMO box|year=1974|num-b=1|num-a=3}}<br />
*[http://www.mathlinks.ro/viewtopic.php?t=102633 Simple Olympiad Inequality]<br />
*[http://www.mathlinks.ro/viewtopic.php?t=98846 Hard inequality]<br />
*[http://www.mathlinks.ro/viewtopic.php?t=85663 Inequality]<br />
*[http://www.mathlinks.ro/Forum/viewtopic.php?t=82706 Some q's on usamo write ups]<br />
*[http://www.mathlinks.ro/viewtopic.php?t=213258 ineq]<br />
*[http://www.mathlinks.ro/Forum/viewtopic.php?t=46247 exponents (generalization)]<br />
{{MAA Notice}}<br />
<br />
[[Category:Olympiad Algebra Problems]]<br />
[[Category:Olympiad Inequality Problems]]</div>Serpent 121https://artofproblemsolving.com/wiki/index.php?title=1969_AHSME_Problems/Problem_20&diff=1230541969 AHSME Problems/Problem 202020-05-27T12:51:04Z<p>Serpent 121: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
<br />
Let <math>P</math> equal the product of 3,659,893,456,789,325,678 and 342,973,489,379,256. The number of digits in <math>P</math> is:<br />
<br />
<math>\text{(A) } 36\quad<br />
\text{(B) } 35\quad<br />
\text{(C) } 34\quad<br />
\text{(D) } 33\quad<br />
\text{(E) } 32</math><br />
<br />
== Solution 1 ==<br />
Through inspection, we see that the two digit number <math>33^{2}=1089=4</math> digits.<br />
Notice that any number that has the form <math>33abcdefg.......</math> multiplied by another <math>33qwertyu.........</math> will have its number of digits equal to the sum of the original numbers' digits.<br />
<br />
In this case, we see that the first number has <math>19</math> digits, and the second number has <math>15</math> digits.<br />
<br />
Note: this applies for numbers <math>33--->99</math><br />
<br />
Hence, the answer is <math>19+15=34</math> digits <math>\implies \fbox{C}</math><br />
<br />
== Solution 2 ==<br />
We can approximate the product with <math>10^{18} * 3.6 *10^{14} *3.4</math> Now observe that <math>3.6*3.4>10</math>, so we can further simplify the product with <math>10^{18}*10^{14}*10=10^{33}</math> which means the product has <math>\fbox{34 (C)}</math> digits.<br />
<br />
-serpent_121<br />
<br />
== See also ==<br />
{{AHSME 35p box|year=1969|num-b=19|num-a=21}} <br />
<br />
[[Category: Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Serpent 121https://artofproblemsolving.com/wiki/index.php?title=1969_AHSME_Problems/Problem_20&diff=1177771969 AHSME Problems/Problem 202020-02-14T08:39:37Z<p>Serpent 121: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
Let <math>P</math> equal the product of 3,659,893,456,789,325,678 and 342,973,489,379,256. The number of digits in <math>P</math> is:<br />
<br />
<math>\text{(A) } 36\quad<br />
\text{(B) } 35\quad<br />
\text{(C) } 34\quad<br />
\text{(D) } 33\quad<br />
\text{(E) } 32</math><br />
<br />
== Solution 1 ==<br />
Through inspection, we see that the two digit number <math>33^{2}=1089=4</math> digits.<br />
Notice that any number that has the form <math>33abcdefg.......</math> multiplied by another <math>33qwertyu.........</math> will have its number of digits equal to the sum of the original numbers' digits.<br />
<br />
In this case, we see that the first number has <math>19</math> digits, and the second number has <math>15</math> digits.<br />
<br />
Note: this applies for numbers <math>33--->99</math><br />
<br />
Hence, the answer is <math>19+15=34</math> digits <math>\implies \fbox{C}</math><br />
<br />
== Solution 2 ==<br />
We can approximate the product with <math>10^{18} * 3.6 *10^{14} *3.4</math> Now observe that <math>3.6*3.4>10</math>, so we can further simplify the product with <math>10^{18}*10^{14}*10=10^{33}</math> which means the product has <math>\fbox{34 (C)}</math> digits.<br />
<br />
== See also ==<br />
{{AHSME 35p box|year=1969|num-b=19|num-a=21}} <br />
<br />
[[Category: Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Serpent 121