https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Champion999&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T17:54:36ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_6&diff=1019692019 AMC 10A Problems/Problem 62019-02-13T02:18:33Z<p>Champion999: </p>
<hr />
<div>==Problem==<br />
<br />
For how many of the following types of quadrilaterals does there exist a point in the plane of the quadrilateral that is equidistant from all four vertices of the quadrilateral?<br />
*a square<br />
*a rectangle that is not a square<br />
*a rhombus that is not a square<br />
*a parallelogram that is not a rectangle or a rhombus<br />
*an isosceles trapezoid that is not a parallelogram<br />
<br />
<math>\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 5</math><br />
<br />
==Solution==<br />
This question is simply asking how many of the listed quadrilaterals are cyclic (since the point equidistant from all four vertices would be the center of the circumscribed circle). A square, a rectangle, and an isosceles trapezoid (that isn't a parallelogram) are all cyclic, and the other two are not. Thus, the answer is <math>3 \implies \boxed{\textbf{(C)}}.</math><br />
<br />
==Solution 2==<br />
We can use the process of elimination. Going down, we can see a square obviously applies. A rectangle that is not a square works as well. Both rhombi and parallelograms don't have a point where the lines are equidistant. But, isosceles trapezoids DO have a point, so the answer is <math>\boxed{3(C)}</math><br />
<br />
==Solution 3==<br />
The perpendicular bisector of a line segment is the locus of all points that are equidistant from the endpoints. The question then boils down to finding the shapes where the perpendicular bisectors of the sides all intersect at a point. This is true for a square, rectangle, and isosceles trapezoid, so the answer is <math>\boxed{\textbf{(C)}}</math>.<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2019|ab=A|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Champion999https://artofproblemsolving.com/wiki/index.php?title=User:Champion999&diff=95408User:Champion9992018-06-20T03:18:28Z<p>Champion999: Blanked the page</p>
<hr />
<div></div>Champion999https://artofproblemsolving.com/wiki/index.php?title=2016_USAJMO_Problems/Problem_5&diff=846852016 USAJMO Problems/Problem 52017-03-13T02:10:41Z<p>Champion999: </p>
<hr />
<div>== Problem ==<br />
<br />
Let <math>\triangle ABC</math> be an acute triangle, with <math>O</math> as its circumcenter. Point <math>H</math> is the foot of the perpendicular from <math>A</math> to line <math>\overleftrightarrow{BC}</math>, and points <math>P</math> and <math>Q</math> are the feet of the perpendiculars from <math>H</math> to the lines <math>\overleftrightarrow{AB}</math> and <math>\overleftrightarrow{AC}</math>, respectively.<br />
<br />
Given that <cmath>AH^2=2\cdot AO^2,</cmath>prove that the points <math>O,P,</math> and <math>Q</math> are collinear.<br />
<br />
== Solution 1==<br />
It is well-known that <math>AH\cdot 2AO=AB\cdot AC</math> (just use similar triangles or standard area formulas). Then by Power of a Point,<br />
<cmath>AP\cdot AB=AH^2=AQ\cdot AC</cmath> Consider the transformation <math>X\mapsto \Psi(X)</math> which dilates <math>X</math> from <math>A</math> by a factor of <math>\dfrac{AB}{AQ}=\dfrac{AC}{AP}</math> and reflects about the <math>A</math>-angle bisector. Then <math>\Psi(O)</math> clearly lies on <math>AH</math>, and its distance from <math>A</math> is <cmath>AO\cdot\frac{AB}{AQ}=AO\cdot\frac{AB}{\frac{AH^2}{AC}}=AO\cdot\frac{AB\cdot AC}{AH^2}=\frac{AO\cdot AH\cdot 2AO}{AH^2}=\frac{2AO^2}{AH}=AH</cmath> so <math>\Psi(O)=H</math>, hence we conclude that <math>O,P,Q</math> are collinear, as desired.<br />
<br />
== Solution 2==<br />
<br />
We will use barycentric coordinates with respect to <math>\triangle ABC.</math> The given condition is equivalent to <math>(\sin B\sin C)^2=\frac{1}{2}.</math> Note that <cmath>O=(\sin(2A):\sin(2B):\sin(2C)), P=(\cos^2B,\sin^2B,0), Q=(\cos^2C,0,\sin^2C).</cmath> Therefore, we must show that <cmath>\begin{vmatrix}<br />
\sin(2A) & \sin(2B) & \sin(2C) \\ <br />
\cos^2B & \sin^2B & 0 \\ <br />
\cos^2C & 0 & \sin^2C \\ <br />
\end{vmatrix}=0.</cmath> Expanding, we must prove <cmath>\sin(2A)\sin^2B\sin^2C=\cos^2C\sin^2B\sin(2C)+\sin^2C\cos^2B\sin(2B)</cmath> <cmath>\frac{\sin(2A)}{2}=\sin^2B(1-\sin^2C)\sin(2C)+\sin^2C(1-\sin^2B)\sin(2B)</cmath> <br />
<cmath>\begin{align*}<br />
\frac{\sin(2A)+\sin(2B)+\sin(2C)}{2}&=\sin^2B\sin(2C)+\sin^2C\sin(2B)\\<br />
&=2\sin B\sin C(\sin B\cos C+\cos B\sin C) \\<br />
&=2\sin B\sin C\sin A.\end{align*}</cmath><br />
<br />
Let <math>x=e^{iA}, y=e^{iB}, z=e^{iC},</math> such that <math>xyz=-1.</math> The left side is equal to <cmath>\frac{x^2+y^2+z^2-\frac{1}{x^2}-\frac{1}{y^2}-\frac{1}{z^2}}{4i}.</cmath> The right side is equal to <br />
<cmath>\begin{align*}<br />
2\cdot \frac{x-\frac{1}{x}}{2i}\cdot \frac{y-\frac{1}{y}}{2i}\cdot \frac{z-\frac{1}{z}}{2i}&=\frac{xyz-\frac{1}{xyz}-\frac{xy}{z}-\frac{yz}{x}-\frac{xz}{y}+\frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy}}{-4i}\\<br />
&=\frac{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-x^2-y^2-z^2}{-4i},\end{align*}</cmath> which is equivalent to the left hand side. Therefore, the determinant is <math>0,</math> and <math>O,P,Q</math> are collinear. <math>\blacksquare</math><br />
<br />
<br />
== Solution 3 ==<br />
For convenience, let <math>a, b, c</math> denote the lengths of segments <math>BC, CA, AB,</math> respectively, and let <math>\alpha, \beta, \gamma</math> denote the measures of <math>\angle CAB, \angle ABC, \angle BCA,</math> respectively. Let <math>R</math> denote the circumradius of <math>\triangle ABC.</math><br />
<br />
Since the central angle <math>\angle AOB</math> subtends the same arc as the inscribed angle <math>\angle ACB</math> on the circumcircle of <math>\triangle ABC,</math> we have <math>\angle AOB = 2\gamma.</math> Note that <math>OA = OB,</math> so <math>\angle OAB = \angle OBA.</math> Thus, <math>\angle OAB = \frac{\pi}{2} - \gamma.</math> Similarly, one can show that <math>\angle OAC = \frac{\pi}{2} - \beta.</math> (One could probably cite this as well-known, but I have proved it here just in case.)<br />
<br />
Clearly, <math>AO = R.</math> Since <math>AH^2 = 2\cdot AO^2,</math> we have <math>AH = \sqrt{2}R.</math> Thus, <math>AH\cdot AO = \sqrt{2}R^2.</math><br />
<br />
Note that <math>AH = b\sin\gamma = c\sin\beta.</math> The Extended Law of Sines states that:<br />
<cmath>\frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{c}{\sin\gamma} = 2R.</cmath><br />
Therefore, <math>AH = \frac{bc}{2R} = \sqrt{2}R.</math> Thus, <math>bc = \sqrt{2}R^2.</math><br />
<br />
Since <math>\angle PHA = \beta</math> and <math>\angle QHA = \gamma,</math> we have:<br />
<cmath>AP = AH\sin\beta = c\sin^2\beta = \frac{b^2 c}{4R^2}</cmath><br />
<cmath>AQ = AH\sin\gamma = b\sin^2\gamma = \frac{bc^2}{4R^2}</cmath><br />
It follows that:<br />
<cmath>AP\cdot AQ = \frac{b^3 c^3}{16R^4} = \frac{16\sqrt{2}R^6}{16R^4} = \sqrt{2}R^2.</cmath><br />
We see that <math>AP\cdot AQ = AH\cdot AO.</math><br />
<br />
Rearranging <math>AP\cdot AQ = AH\cdot AO,</math> we get <math>\frac{AP}{AH} = \frac{AO}{AQ}.</math> We also have <math>\angle PAH = \angle OAQ = \frac{\pi}{2} - \beta,</math> so <math>\triangle PAH\sim\triangle OAQ</math> by SAS similarity. Thus, <math>\angle AOQ = \angle APH,</math> so <math>\angle AOQ</math> is a right angle.<br />
<br />
Rearranging <math>AP\cdot AQ = AH\cdot AO,</math> we get <math>\frac{AP}{AO} = \frac{AO}{AH}.</math> We also have <math>\angle PAO = \angle HAQ = \frac{\pi}{2} - \gamma,</math> so <math>\triangle PAO\sim\triangle HAQ</math> by SAS similarity. Thus, <math>\angle AOP = \angle AQH,</math> so <math>\angle AOP</math> is a right angle.<br />
<br />
Since <math>\angle AOP</math> and <math>\angle AOQ</math> are both right angles, we get <math>\angle POQ = \pi,</math> so we conclude that <math>P, O, Q</math> are collinear, and we are done. (We also obtain the extra interesting fact that <math>AO\perp PQ.</math>)<br />
<br />
== Solution 4==<br />
Draw the altitude from <math>O</math> to <math>AB</math>, and let the foot of this altitude be <math>D</math>. <br />
<br />
Then, by the Right Triangle Altitude Theorem on triangle <math>AHB</math>, we have: <math>AB\cdot AP=AH^{2}</math>.<br />
<br />
Since <math>OD</math> is the perpendicular bisector of <math>AB</math>, <math>2\cdot AD = AB</math>.<br />
<br />
Substituting this into our previous equation gives <math>2\cdot AD \cdot AP = AH^{2}</math>, which equals <math>2\cdot AO^{2}</math> by the problem condition.<br />
<br />
Thus, <math>2\cdot AD\cdot AP = 2\cdot AO^{2} \implies AD\cdot AP = AO^{2}</math>.<br />
<br />
Again, by the Right Triangle Altitude Theorem, angle <math>AOP</math> is right.<br />
<br />
By dropping an altitude from <math>O</math> to <math>AC</math> and using the same method, we can find that angle <math>AOQ</math> is right. Since <math>\angle AOP=\angle AOQ=90</math>, <math>P</math>, <math>O</math>, <math>Q</math> are collinear and we are done.<br />
<br />
~champion999<br />
<br />
{{MAA Notice}}<br />
<br />
==See also==<br />
{{USAJMO newbox|year=2016|num-b=4|num-a=6}}</div>Champion999https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12B_Problems/Problem_19&diff=715092006 AMC 12B Problems/Problem 192015-08-10T13:47:00Z<p>Champion999: /* Alternate Solution */</p>
<hr />
<div>== Problem ==<br />
Mr. Jones has eight children of different ages. On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. "Look, daddy!" she exclaims. "That number is evenly divisible by the age of each of us kids!" "That's right," replies Mr. Jones, "and the last two digits just happen to be my age." Which of the following is not the age of one of Mr. Jones's children?<br />
<br />
<math><br />
\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8<br />
</math><br />
<br />
== Solution ==<br />
First, The number of the plate is divisible by <math>9</math> and in the form of<br />
<math>aabb</math>, <math>abba</math> or <math>abab</math>. <br />
<br />
We can conclude straight away that <math>a+b= 9</math> using the <math>9</math> divisibility rule. <br />
<br />
If <math>b=1</math>, the number is not divisible by <math>2</math> (unless it's <math>1818</math>, which is not divisible by <math>4</math>), which means there are no <math>2</math>, <math>4</math>, <math>6</math>, or <math>8</math> year olds on the car, but that can't be true, as that would mean there are less than <math>8</math> kids on the car. <br />
<br />
If <math>b=2</math>, then the only possible number is <math>7272</math>. <math>7272</math> is divisible by <math>4</math>, <math>6</math>, and <math>8</math>, but not by <math>5</math> and <math>7</math>, so that doesn't work. <br />
<br />
If <math>b=3</math>, then the only number is <math>6336</math>, also not divisible by <math>5</math> or <math>7</math>. <br />
<br />
If <math>b=4</math>, the only number is <math>5544</math>. It is divisible by <math>4</math>, <math>6</math>, <math>7</math>, and <math>8</math>. <br />
<br />
Therefore, we conclude that the answer is <math> \mathrm{(B)}\ 5 </math><br />
<br />
'''NOTE''': Automatically, since there are 8 children and all of their ages are less than or equal to 9 and are different, the answer choices can be narrowed down to <math>5</math> or <math>8</math>.<br />
<br />
== Alternate Solution ==<br />
<br />
We know that the number of the plate is divisible by <math>9</math> and in the form <math>aabb</math>, <math>abba</math>, or <math>abab</math> for distinct digits <math>a,b</math>.<br />
<br />
Using the divisibility rule for <math>9</math>, we can conclude that <math>a+b\equiv 0\pmod{9}</math>.<br />
<br />
We also know that the number of the plate is even, because you can only discard one number from the integers <math>1</math> through <math>9</math>, inclusive (<math>8</math> children, oldest is <math>9</math>), and there's always going to be an even number left.<br />
<br />
If one of the children was <math>5</math> years old, then the plate number would be divisible by both <math>5</math> and <math>2</math>. Thus, the units digit must be <math>0</math>. <br />
Then, the possible form of the plate number would be <math>aa00</math>, <math>0bb0</math>, or <math>a0a0</math>. The first case is not possible because <math>00</math> is not a possible age for the father.<br />
<br />
We have the remaining forms <math>0bb0</math> and <math>a0a0</math>.<br />
<br />
<math>\textbf{Case 1: 0bb0:}</math><br />
We know that <math>2b\equiv 0\pmod{9}</math>, so <math>b</math> has to be either <math>0</math> or <math>9</math>. <math>b</math> can't be <math>0</math> because <math>a, b</math> are distinct. So, <math>b=9</math>. However, the number <math>0990</math> is not divisible by both <math>4</math> and <math>7</math>, so whichever number you discard, the other one will still be there. This creates a contradiction, so Case <math>1</math> cannot be true.<br />
<br />
<math>\textbf{Case 2: a0a0:}</math><br />
Similarly to Case <math>1</math>, we can determine that <math>a</math> has to be <math>9</math>. However, the number <math>9090</math> is not divisible by both <math>4</math> and <math>7</math>. Using the same logic in case <math>1</math>, we conclude that Case <math>2</math> cannot be true.<br />
<br />
Since we have disproven all of our cases, we know that it is impossible for one of the children to be <math> \mathrm{(B)}\ 5 </math> years old.<br />
<br />
'''NOTE''': This might look tedious, but it only took me around 30 seconds to do on paper.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2006|ab=B|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Champion999https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12B_Problems/Problem_19&diff=715082006 AMC 12B Problems/Problem 192015-08-10T13:45:53Z<p>Champion999: /* Alternate Solution */</p>
<hr />
<div>== Problem ==<br />
Mr. Jones has eight children of different ages. On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. "Look, daddy!" she exclaims. "That number is evenly divisible by the age of each of us kids!" "That's right," replies Mr. Jones, "and the last two digits just happen to be my age." Which of the following is not the age of one of Mr. Jones's children?<br />
<br />
<math><br />
\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8<br />
</math><br />
<br />
== Solution ==<br />
First, The number of the plate is divisible by <math>9</math> and in the form of<br />
<math>aabb</math>, <math>abba</math> or <math>abab</math>. <br />
<br />
We can conclude straight away that <math>a+b= 9</math> using the <math>9</math> divisibility rule. <br />
<br />
If <math>b=1</math>, the number is not divisible by <math>2</math> (unless it's <math>1818</math>, which is not divisible by <math>4</math>), which means there are no <math>2</math>, <math>4</math>, <math>6</math>, or <math>8</math> year olds on the car, but that can't be true, as that would mean there are less than <math>8</math> kids on the car. <br />
<br />
If <math>b=2</math>, then the only possible number is <math>7272</math>. <math>7272</math> is divisible by <math>4</math>, <math>6</math>, and <math>8</math>, but not by <math>5</math> and <math>7</math>, so that doesn't work. <br />
<br />
If <math>b=3</math>, then the only number is <math>6336</math>, also not divisible by <math>5</math> or <math>7</math>. <br />
<br />
If <math>b=4</math>, the only number is <math>5544</math>. It is divisible by <math>4</math>, <math>6</math>, <math>7</math>, and <math>8</math>. <br />
<br />
Therefore, we conclude that the answer is <math> \mathrm{(B)}\ 5 </math><br />
<br />
'''NOTE''': Automatically, since there are 8 children and all of their ages are less than or equal to 9 and are different, the answer choices can be narrowed down to <math>5</math> or <math>8</math>.<br />
<br />
== Alternate Solution ==<br />
<br />
We know that the number of the plate is divisible by <math>9</math> and in the form <math>aabb</math>, <math>abba</math>, or <math>abab</math> for distinct digits <math>a,b</math>.<br />
<br />
Using the divisibility rule for <math>9</math>, we can conclude that <math>a+b\equiv 0\pmod{9}</math>.<br />
<br />
We also know that the number of the plate is even, because you can only discard one number from the integers <math>1</math> through <math>9</math>, inclusive (<math>8</math> children, oldest is <math>9</math>), and there's always going to be an even number left.<br />
<br />
If one of the children was <math>5</math> years old, then the plate number would be divisible by both <math>5</math> and <math>2</math>. Thus, the units digit must be <math>0</math>. <br />
Then, the possible form of the plate number would be <math>aa00</math>, <math>0bb0</math>, or <math>a0a0</math>. The first case is not possible because <math>00</math> is not a possible age for the father.<br />
<br />
We have the remaining forms <math>0bb0</math> and <math>a0a0</math>.<br />
<math>\textbf{Case 1: 0bb0}</math><br />
We know that <math>2b\equiv 0\pmod{9}</math>, so <math>b</math> has to be either <math>0</math> or <math>9</math>. <math>b</math> can't be <math>0</math> because <math>a, b</math> are distinct. So, <math>b=9</math>. However, the number <math>0990</math> is not divisible by both <math>4</math> and <math>7</math>, so whichever number you discard, the other one will still be there. This creates a contradiction, so Case <math>1</math> cannot be true.<br />
<br />
<math>\textbf{Case 2: a0a0}</math><br />
Similarly to Case <math>1</math>, we can determine that <math>a</math> has to be <math>9</math>. However, the number <math>9090</math> is not divisible by both <math>4</math> and <math>7</math>. Using the same logic in case <math>1</math>, we conclude that Case <math>2</math> cannot be true.<br />
<br />
Since we have disproven all of our cases, we know that it is impossible for one of the children to be <math> \mathrm{(B)}\ 5 </math> years old.<br />
<br />
'''NOTE''': This might look tedious, but it only took me around 30 seconds to do on paper.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2006|ab=B|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Champion999https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12B_Problems/Problem_19&diff=715072006 AMC 12B Problems/Problem 192015-08-10T13:45:23Z<p>Champion999: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Mr. Jones has eight children of different ages. On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. "Look, daddy!" she exclaims. "That number is evenly divisible by the age of each of us kids!" "That's right," replies Mr. Jones, "and the last two digits just happen to be my age." Which of the following is not the age of one of Mr. Jones's children?<br />
<br />
<math><br />
\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8<br />
</math><br />
<br />
== Solution ==<br />
First, The number of the plate is divisible by <math>9</math> and in the form of<br />
<math>aabb</math>, <math>abba</math> or <math>abab</math>. <br />
<br />
We can conclude straight away that <math>a+b= 9</math> using the <math>9</math> divisibility rule. <br />
<br />
If <math>b=1</math>, the number is not divisible by <math>2</math> (unless it's <math>1818</math>, which is not divisible by <math>4</math>), which means there are no <math>2</math>, <math>4</math>, <math>6</math>, or <math>8</math> year olds on the car, but that can't be true, as that would mean there are less than <math>8</math> kids on the car. <br />
<br />
If <math>b=2</math>, then the only possible number is <math>7272</math>. <math>7272</math> is divisible by <math>4</math>, <math>6</math>, and <math>8</math>, but not by <math>5</math> and <math>7</math>, so that doesn't work. <br />
<br />
If <math>b=3</math>, then the only number is <math>6336</math>, also not divisible by <math>5</math> or <math>7</math>. <br />
<br />
If <math>b=4</math>, the only number is <math>5544</math>. It is divisible by <math>4</math>, <math>6</math>, <math>7</math>, and <math>8</math>. <br />
<br />
Therefore, we conclude that the answer is <math> \mathrm{(B)}\ 5 </math><br />
<br />
'''NOTE''': Automatically, since there are 8 children and all of their ages are less than or equal to 9 and are different, the answer choices can be narrowed down to <math>5</math> or <math>8</math>.<br />
<br />
== Alternate Solution ==<br />
<br />
We know that the number of the plate is divisible by <math>9</math> and in the form <math>aabb</math>, <math>abba</math>, or <math>abab</math> for distinct digits <math>a,b</math>.<br />
<br />
Using the divisibility rule for <math>9</math>, we can conclude that <math>a+b\equiv 0\pmod{9}</math>.<br />
<br />
We also know that the number of the plate is even, because you can only discard one number from the integers <math>1</math> through <math>9</math>, inclusive (<math>8</math> children, oldest is <math>9</math>), and there's always going to be an even number left.<br />
<br />
If one of the children was <math>5</math> years old, then the plate number would be divisible by both <math>5</math> and <math>2</math>. Thus, the units digit must be <math>0</math>. <br />
Then, the possible form of the plate number would be <math>aa00</math>, <math>0bb0</math>, or <math>a0a0</math>. The first case is not possible because <math>00</math> is not a possible age for the father.<br />
<br />
We have the remaining forms <math>0bb0</math> and <math>a0a0</math>.<br />
<math>\textbf{Case 1: </math>0bb0<math>}</math><br />
We know that <math>2b\equiv 0\pmod{9}</math>, so <math>b</math> has to be either <math>0</math> or <math>9</math>. <math>b</math> can't be <math>0</math> because <math>a, b</math> are distinct. So, <math>b=9</math>. However, the number <math>0990</math> is not divisible by both <math>4</math> and <math>7</math>, so whichever number you discard, the other one will still be there. This creates a contradiction, so Case <math>1</math> cannot be true.<br />
<br />
<math>\textbf{Case 2: </math>a0a0<math>}</math><br />
Similarly to Case <math>1</math>, we can determine that <math>a</math> has to be <math>9</math>. However, the number <math>9090</math> is not divisible by both <math>4</math> and <math>7</math>. Using the same logic in case <math>1</math>, we conclude that Case <math>2</math> cannot be true.<br />
<br />
Since we have disproven all of our cases, we know that it is impossible for one of the children to be <math> \mathrm{(B)}\ 5 </math> years old.<br />
<br />
'''NOTE''': This might look tedious, but it only took me around 30 seconds to do on paper.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2006|ab=B|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Champion999https://artofproblemsolving.com/wiki/index.php?title=User:Champion999&diff=71210User:Champion9992015-07-18T01:56:27Z<p>Champion999: </p>
<hr />
<div>Goals for 9th grade:<br />
<br />
AMC 10: 138<br />
AIME: 10<br />
JMO: 7<br />
<br />
Goals for 10th grade:<br />
<br />
AMC 10: 144<br />
AIME: 12<br />
JMO: HM MOP?<br />
<br />
Goals for 11th grade:<br />
<br />
AMC 12: 138<br />
AIME: 13<br />
AMO: 14 MOP?<br />
<br />
Goals for 12th grade:<br />
<br />
AMC 12: 144<br />
AIME: 14<br />
AMO: HM<br />
MOP?</div>Champion999https://artofproblemsolving.com/wiki/index.php?title=User:Champion999&diff=70909User:Champion9992015-06-28T17:34:53Z<p>Champion999: </p>
<hr />
<div>Goals for 9th grade:<br />
<br />
AMC 10: 138<br />
AIME: 10<br />
JMO: 7<br />
<br />
Goals for 10th grade:<br />
<br />
AMC 10: 144<br />
AIME: 12<br />
JMO: HM<br />
<br />
Goals for 11th grade:<br />
<br />
AMC 12: 138<br />
AIME: 13<br />
AMO: 14<br />
<br />
Goals for 12th grade:<br />
<br />
AMC 12: 144<br />
AIME: 14<br />
AMO: HM<br />
MOP?</div>Champion999https://artofproblemsolving.com/wiki/index.php?title=User:Champion999&diff=70908User:Champion9992015-06-28T17:32:36Z<p>Champion999: </p>
<hr />
<div>Goals: <br />
\begin{tabular}{c|c|c|c}Grade & AMC10/12 & AIME & USA(J)MO \\ \hline<br />
9th & 138 & 10 & 7 \\<br />
10th & 144 & 12 & HM \\<br />
11th & 138 & 13 & 14\\<br />
12th & 144 & 14 & HM \\<br />
\end{tabular}</div>Champion999https://artofproblemsolving.com/wiki/index.php?title=User:Champion999&diff=70907User:Champion9992015-06-28T17:24:29Z<p>Champion999: </p>
<hr />
<div>Goals:<br />
9th grade: AMC 10:<math>\mathbf{138}</math> AIME: <math>\mathbf{10}</math> JMO: <math>\textbf{\ge 7}</math> <br />
10th grade: AMC 10:<math>\mathbf{144}</math> AIME: <math>\mathbf{12}</math> JMO: <math>\textbf{HM?}</math> <br />
11th grade: AMC 12:<math>\mathbf{138}</math> AIME: <math>\mathbf{13}</math> AMO: <math>\textbf{\ge 14}</math> <br />
12th grade: AMC 12:<math>\mathbf{144}</math> AIME: <math>\mathbf{14}</math> AMO: <math>\textbf{HM?}</math><br />
<br />
\begin{center}<br />
\begin{tabular}{||c c c c||} <br />
\hline<br />
AMC 10 or 12 & AIME & USA(J)MO \\ [0.5ex] <br />
\hline\hline<br />
138 & 10 & \ge 7 \\ <br />
\hline<br />
144 & 12 & HM \\<br />
\hline<br />
138 & 13 & \ge 14 \\<br />
\hline<br />
144 & 14 & HM \\ [1ex] <br />
\hline<br />
\end{tabular}<br />
\end{center}</div>Champion999https://artofproblemsolving.com/wiki/index.php?title=User:Champion999&diff=70906User:Champion9992015-06-28T17:19:49Z<p>Champion999: </p>
<hr />
<div>Goals:<br />
9th grade: AMC 10:<math>\mathbf{138}</math> AIME: <math>\mathbf{10}</math> JMO: <math>\textbf{\ge 7}</math> <br />
10th grade: AMC 10:<math>\mathbf{144}</math> AIME: <math>\mathbf{12}</math> JMO: <math>\textbf{HM?}</math> <br />
11th grade: AMC 12:<math>\mathbf{138}</math> AIME: <math>\mathbf{13}</math> AMO: <math>\textbf{\ge 14}</math> <br />
12th grade: AMC 12:<math>\mathbf{144}</math> AIME: <math>\mathbf{14}</math> AMO: <math>\textbf{HM?}</math></div>Champion999https://artofproblemsolving.com/wiki/index.php?title=1950_AHSME_Problems/Problem_20&diff=701721950 AHSME Problems/Problem 202015-05-10T15:42:14Z<p>Champion999: /* Solution 2 */</p>
<hr />
<div>== Problem==<br />
<br />
When <math>x^{13}+1</math> is divided by <math>x-1</math>, the remainder is:<br />
<br />
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{None of these answers} </math><br />
<br />
==Solution==<br />
===Solution 1===<br />
<br />
Use synthetic division, and get that the remainder is <math>\boxed{\mathrm{(D)}\ 2.}</math><br />
<br />
===Solution 2===<br />
<br />
By the remainder theorem, the remainder is equal to the expression <math>x^{13}+1</math> when <math>x=1.</math> This gives the answer of <math> \boxed{(\mathrm{D})\ 2.} </math><br />
<br />
EDIT: This solution is invalid because x can't be 1.<br />
<br />
==See Also==<br />
<br />
{{AHSME 50p box|year=1950|num-b=19|num-a=21}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Champion999https://artofproblemsolving.com/wiki/index.php?title=1993_AJHSME_Problems/Problem_21&diff=701711993 AJHSME Problems/Problem 212015-05-10T15:32:38Z<p>Champion999: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
If the length of a rectangle is increased by <math>20\% </math> and its width is increased by <math>50\% </math>, then the area is increased by <br />
<br />
<math>\text{(A)}\ 10\% \qquad \text{(B)}\ 30\% \qquad \text{(C)}\ 70\% \qquad \text{(D)}\ 80\% \qquad \text{(E)}\ 100\% </math><br />
<br />
==Solution==<br />
If a rectangle had dimensions <math>10 \times 10</math> and area <math>100</math>, then its new dimensions would be <math>12 \times 15</math> and area <math>180</math>. The area is increased by <math>180-100=80</math> or <math>80/100 = \boxed{\text{(D)}\ 80\%}</math>.<br />
<br />
==Alternate Solution==<br />
Let the dimensions of the rectangle be <math>x \times y</math>. This rectangle has area <math>xy</math>. The new dimensions would be <math>1.2x \times 1.5y</math>, so the area is <math>=1.8xy</math>, which is <math>\boxed{80\%}</math> more than the original area.<br />
<br />
==See Also==<br />
{{AJHSME box|year=1993|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Champion999https://artofproblemsolving.com/wiki/index.php?title=User:Champion999&diff=63243User:Champion9992014-08-29T01:09:22Z<p>Champion999: </p>
<hr />
<div>He's a champion. NUFF SAID<br />
<br />
FTW Rating: Currently 2235, hopes to get to 2325 someday.<br />
<br />
Very bad typist, unlike (minimario/zetaintegrator, Iamteehee, qwerty137, mikechen, ML01, and many others)</div>Champion999https://artofproblemsolving.com/wiki/index.php?title=User:Champion999&diff=62425User:Champion9992014-06-30T19:26:55Z<p>Champion999: </p>
<hr />
<div>He's a champion. NUFF SAID<br />
<br />
FTW Rating: Currently 2108, hopes to get to 2325 someday.<br />
<br />
Very bad typist, unlike (minimario/zetaintegrator, Iamteehee, qwerty137, mikechen, ML01, and many others)</div>Champion999https://artofproblemsolving.com/wiki/index.php?title=User:Champion999&diff=62424User:Champion9992014-06-30T19:26:34Z<p>Champion999: </p>
<hr />
<div>He's a champion. NUFF SAID<br />
<br />
FTW Rating: Currently 2108, hopes to get to 2325 someday.<br />
<br />
Very bad typist, unlike (minimario/zetaintegrator, Iamteehee, qwerty137, mikechen, ML01, and many others)<br />
<br />
:)</div>Champion999https://artofproblemsolving.com/wiki/index.php?title=User:Champion999&diff=62423User:Champion9992014-06-30T19:26:27Z<p>Champion999: </p>
<hr />
<div>He's a champion. NUFF SAID<br />
<br />
FTW Rating: Currently 2108, hopes to get to 2325 someday.<br />
<br />
Very bad typist, unlike (minimario/zetaintegrator, Iamteehee, qwerty137, mikechen, ML01, and many others)<br />
<br />
:wallbash_red:</div>Champion999https://artofproblemsolving.com/wiki/index.php?title=User:Champion999&diff=62422User:Champion9992014-06-30T19:26:16Z<p>Champion999: </p>
<hr />
<div>He's a champion. NUFF SAID<br />
<br />
FTW Rating: Currently 2108, hopes to get to 2325 someday.<br />
<br />
Very bad typist, unlike (minimario/zetaintegrator, Iamteehee, qwerty137, mikechen, ML01, and many others)<br />
<br />
:wallbash:</div>Champion999https://artofproblemsolving.com/wiki/index.php?title=User:Champion999&diff=62421User:Champion9992014-06-30T19:23:48Z<p>Champion999: Created page with "I'm a champion!"</p>
<hr />
<div>I'm a champion!</div>Champion999