https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Mathpro1441&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T11:50:34ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=Bezout%27s_Lemma&diff=107001Bezout's Lemma2019-06-26T16:07:06Z<p>Mathpro1441: /* Proof */</p>
<hr />
<div>'''Bezout's Lemma''' states that if <math>x</math> and <math>y</math> are nonzero [[Integer|integers]] and <math>g = \gcd(x,y)</math>, then there exist integers <math>\alpha</math> and <math>\beta</math> such that <math>x\alpha+y\beta=g</math>. In other words, there exists a linear combination of <math>x</math> and <math>y</math> equal to <math>g</math>.<br />
<br />
Furthermore, <math>g</math> is the smallest positive integer that can be expressed in this form, i.e. <math>g = \min\{x\alpha+y\beta|\alpha,\beta\in\mathbb Z, x\alpha+y\beta > 0\}</math>.<br />
<br />
In particular, if <math>x</math> and <math>y</math> are [[relatively prime]] then there are integers <math>\alpha</math> and <math>\beta</math> for which <math>x\alpha+y\beta=1</math>.<br />
<br />
==Proof==<br />
Let <math>x = gx_1</math>, <math>y = gy_1</math>, and notice that <math>\gcd(x_1,y_1) = 1</math>.<br />
<br />
Since <math>\gcd(x_1,y_1)=1</math>, <math>\text{lcm}(x_1,y_1)=x_1y_1</math>. So <math>\alpha=y_1</math> is smallest positive <math>\alpha</math> for which <math>x_1\alpha\equiv 0\pmod{y}</math>. Now if for all integers <math>0\le a,b<y_1</math>, we have that <math>x_1a\not\equiv x_1b\pmod{y_1}</math>, then one of those <math>y_1</math> integers must be 1 from the [[Pigeonhole Principle]]. Assume for contradiction that <math>x_1a\equiv x_1b\pmod{y_1}</math>, and WLOG let <math>b>a</math>. Then, <math>x_1(b-a)\equiv 0\pmod {y_1}</math>, and so as we saw above this means <math>b-a\ge y_1</math> but this is impossible since <math>0\le a,b<y_1</math>. Thus there exists an <math>\alpha</math> such that <math>x_1\alpha\equiv 1\pmod{y_1}</math>.<br />
<br />
Therefore <math>y_1|(x_1\alpha-1)</math>, and so there exists an integer <math>\beta</math> such that <math>x_1\alpha - 1 = y_1\beta</math>, and so <math>x_1\alpha + y_1\beta = 1</math>. Now multiplying through by <math>g</math> gives, <math>gx_1\alpha + gy_1\beta = g</math>, or <math>x\alpha+y\beta = g</math>.<br />
<br />
Thus there does exist integers <math>\alpha</math> and <math>\beta</math> such that <math>x\alpha+y\beta=g</math>.<br />
<br />
Now to prove <math>g</math> is minimum, consider any positive integer <math>g' = x\alpha'+y\beta'</math>. As <math>g|x,y</math> we get <math>g|x\alpha'+y\beta' = g'</math>, and as <math>g</math> and <math>g'</math> are both positive integers this gives <math>g\le g'</math>. So <math>g</math> is indeed the minimum.<br />
<br />
==Generalization to Principal Ideal Domains==<br />
Bezout's Lemma can be generalized to [[Principal ideal domain|principal ideal domains]].<br />
<br />
Let <math>R</math> be a principal ideal domain, and consider any <math>x,y\in R</math>. Let <math>g = \gcd(x,y)</math>. Then there exist elements <math>r_1,r_2\in R</math> for which <math>xr_1+yr_2 = g</math>. Furthermore, <math>g</math> is the minimal such element (under divisibility), i.e. if <math>g' = xr_1'+yr_2'</math> then <math>g|g'</math>.<br />
<br />
Note that this statement is indeed a generalization of the previous statement, as the [[ring]] of integers, <math>\mathbb Z</math> is a principal ideal domain.<br />
<br />
==Proof==<br />
Consider the [[ideal]] <math>I = (x,y) = \{xr_1+yr_2|r_1,r_2\in R\}</math>. As <math>R</math> is a principal ideal domain, <math>I</math> must be principle, that is it must be generated by a single element, say <math>I = (g)</math>. Now from the definition of <math>I</math>, there must exist <math>r_1,r_2\in R</math> such that <math>g = xr_1+yr_2</math>. We now claim that <math>g = \gcd(x,y)</math>.<br />
<br />
First we prove the following simple fact: if <math>z\in I</math>, then <math>g|z</math>. To see this, note that if <math>z\in I = (g)</math>, then there must be some <math>r\in R</math> such that <math>z = rg</math>. But now by definition we have <math>g|z</math>.<br />
<br />
Now from this, as <math>x,y\in I</math>, we get that <math>g|x,y</math>. Furthermore, consider any <math>s\in R</math> with <math>s|x,y</math>. We clearly have that <math>s|xr_1+yr_2 = g</math>. So indeed <math>g = \gcd(x,y)</math>.<br />
<br />
Now we shall prove minimality. Let <math>g' = xr_1'+yr_2'</math>. Then as <math>g|x,y</math>, we have <math>g|xr_1'+yr_2' = g'</math>, as desired.<br />
<br />
==See also==<br />
[[Category:Number theory]]<br />
[[Category:Abstract algebra]]<br />
[[Category:Ring theory]]<br />
{{stub}}<br />
[[Category:Theorems]]</div>Mathpro1441https://artofproblemsolving.com/wiki/index.php?title=Bezout%27s_Lemma&diff=107000Bezout's Lemma2019-06-26T16:06:22Z<p>Mathpro1441: /* Proof */</p>
<hr />
<div>'''Bezout's Lemma''' states that if <math>x</math> and <math>y</math> are nonzero [[Integer|integers]] and <math>g = \gcd(x,y)</math>, then there exist integers <math>\alpha</math> and <math>\beta</math> such that <math>x\alpha+y\beta=g</math>. In other words, there exists a linear combination of <math>x</math> and <math>y</math> equal to <math>g</math>.<br />
<br />
Furthermore, <math>g</math> is the smallest positive integer that can be expressed in this form, i.e. <math>g = \min\{x\alpha+y\beta|\alpha,\beta\in\mathbb Z, x\alpha+y\beta > 0\}</math>.<br />
<br />
In particular, if <math>x</math> and <math>y</math> are [[relatively prime]] then there are integers <math>\alpha</math> and <math>\beta</math> for which <math>x\alpha+y\beta=1</math>.<br />
<br />
==Proof==<br />
Let <math>x = gx_1</math>, <math>y = gy_1</math>, and notice that <math>\gcd(x_1,y_1) = 1</math>.<br />
<br />
Since <math>\gcd(x_1,y_1)=1</math>, <math>\text{lcm}(x_1,y_1)=x_1y_1</math>. So <math>\alpha=y_1</math> is smallest positive <math>\alpha</math> for which <math>x\alpha\equiv 0\pmod{y}</math>. Now if for all integers <math>0\le a,b<y_1</math>, we have that <math>x_1a\not\equiv x_1b\pmod{y_1}</math>, then one of those <math>y_1</math> integers must be 1 from the [[Pigeonhole Principle]]. Assume for contradiction that <math>x_1a\equiv x_1b\pmod{y_1}</math>, and WLOG let <math>b>a</math>. Then, <math>x_1(b-a)\equiv 0\pmod {y_1}</math>, and so as we saw above this means <math>b-a\ge y_1</math> but this is impossible since <math>0\le a,b<y_1</math>. Thus there exists an <math>\alpha</math> such that <math>x_1\alpha\equiv 1\pmod{y_1}</math>.<br />
<br />
Therefore <math>y_1|(x_1\alpha-1)</math>, and so there exists an integer <math>\beta</math> such that <math>x_1\alpha - 1 = y_1\beta</math>, and so <math>x_1\alpha + y_1\beta = 1</math>. Now multiplying through by <math>g</math> gives, <math>gx_1\alpha + gy_1\beta = g</math>, or <math>x\alpha+y\beta = g</math>.<br />
<br />
Thus there does exist integers <math>\alpha</math> and <math>\beta</math> such that <math>x\alpha+y\beta=g</math>.<br />
<br />
Now to prove <math>g</math> is minimum, consider any positive integer <math>g' = x\alpha'+y\beta'</math>. As <math>g|x,y</math> we get <math>g|x\alpha'+y\beta' = g'</math>, and as <math>g</math> and <math>g'</math> are both positive integers this gives <math>g\le g'</math>. So <math>g</math> is indeed the minimum.<br />
<br />
==Generalization to Principal Ideal Domains==<br />
Bezout's Lemma can be generalized to [[Principal ideal domain|principal ideal domains]].<br />
<br />
Let <math>R</math> be a principal ideal domain, and consider any <math>x,y\in R</math>. Let <math>g = \gcd(x,y)</math>. Then there exist elements <math>r_1,r_2\in R</math> for which <math>xr_1+yr_2 = g</math>. Furthermore, <math>g</math> is the minimal such element (under divisibility), i.e. if <math>g' = xr_1'+yr_2'</math> then <math>g|g'</math>.<br />
<br />
Note that this statement is indeed a generalization of the previous statement, as the [[ring]] of integers, <math>\mathbb Z</math> is a principal ideal domain.<br />
<br />
==Proof==<br />
Consider the [[ideal]] <math>I = (x,y) = \{xr_1+yr_2|r_1,r_2\in R\}</math>. As <math>R</math> is a principal ideal domain, <math>I</math> must be principle, that is it must be generated by a single element, say <math>I = (g)</math>. Now from the definition of <math>I</math>, there must exist <math>r_1,r_2\in R</math> such that <math>g = xr_1+yr_2</math>. We now claim that <math>g = \gcd(x,y)</math>.<br />
<br />
First we prove the following simple fact: if <math>z\in I</math>, then <math>g|z</math>. To see this, note that if <math>z\in I = (g)</math>, then there must be some <math>r\in R</math> such that <math>z = rg</math>. But now by definition we have <math>g|z</math>.<br />
<br />
Now from this, as <math>x,y\in I</math>, we get that <math>g|x,y</math>. Furthermore, consider any <math>s\in R</math> with <math>s|x,y</math>. We clearly have that <math>s|xr_1+yr_2 = g</math>. So indeed <math>g = \gcd(x,y)</math>.<br />
<br />
Now we shall prove minimality. Let <math>g' = xr_1'+yr_2'</math>. Then as <math>g|x,y</math>, we have <math>g|xr_1'+yr_2' = g'</math>, as desired.<br />
<br />
==See also==<br />
[[Category:Number theory]]<br />
[[Category:Abstract algebra]]<br />
[[Category:Ring theory]]<br />
{{stub}}<br />
[[Category:Theorems]]</div>Mathpro1441https://artofproblemsolving.com/wiki/index.php?title=1967_AHSME_Problems/Problem_33&diff=1046251967 AHSME Problems/Problem 332019-03-17T20:39:49Z<p>Mathpro1441: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
<asy><br />
fill(circle((4,0),4),grey);<br />
fill((0,0)--(8,0)--(8,-4)--(0,-4)--cycle,white);<br />
fill(circle((7,0),1),white);<br />
fill(circle((3,0),3),white);<br />
draw((0,0)--(8,0),black+linewidth(1));<br />
draw((6,0)--(6,sqrt(12)),black+linewidth(1));<br />
MP("A", (0,0), W); MP("B", (8,0), E); MP("C", (6,0), S); MP("D",(6,sqrt(12)), N);<br />
</asy><br />
<br />
In this diagram semi-circles are constructed on diameters <math>\overline{AB}</math>, <math>\overline{AC}</math>, and <math>\overline{CB}</math>, so that they are mutually tangent. If <math>\overline{CD} \bot \overline{AB}</math>, then the ratio of the shaded area to the area of a circle with <math>\overline{CD}</math> as radius is:<br />
<br />
<math>\textbf{(A)}\ 1:2\qquad<br />
\textbf{(B)}\ 1:3\qquad<br />
\textbf{(C)}\ \sqrt{3}:7\qquad<br />
\textbf{(D)}\ 1:4\qquad<br />
\textbf{(E)}\ \sqrt{2}:6</math><br />
<br />
== Solution ==<br />
<br />
To make the problem much simpler while staying in the constraints of the problem, position point <math>C</math> halfway between <math>A</math> and <math>B</math>. Then, call <math>\overline{AC} = \overline{BC}=r</math> . The area of the shaded region is then <cmath>\frac{ \pi r^2 - \pi (r/2)^2 - \pi (r/2)^2}{2}=\frac{\pi r^2}{4}</cmath><br />
Because <math>\overline{CD}=r</math> the area of the circle with <math>\overline{CD}</math> as radius is <math>\pi r^2</math>.<br />
Our ratio is then <cmath>\frac{\pi r^2}{4} : \pi r^2 = 1:4</cmath><br />
<br />
Which corresponds with answer <math>\fbox{D}</math><br />
<br />
== See also ==<br />
{{AHSME box|year=1967|num-b=32|num-a=34}} <br />
<br />
[[Category: Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Mathpro1441https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_15&diff=1041452009 AMC 12A Problems/Problem 152019-03-08T11:53:43Z<p>Mathpro1441: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
For what value of <math>n</math> is <math>i + 2i^2 + 3i^3 + \cdots + ni^n = 48 + 49i</math>?<br />
<br />
Note: here <math>i = \sqrt { - 1}</math>.<br />
<br />
<math>\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 49 \qquad \textbf{(D)}\ 97 \qquad \textbf{(E)}\ 98</math><br />
<br />
== Solution 1 == <br />
We know that <math>i^x</math> cycles every <math>4</math> powers so we group the sum in <math>4</math>s. <br />
<cmath>i+2i^2+3i^3+4i^4=2-2i</cmath><br />
<cmath>5i^5+6i^6+7i^7+8i^8=2-2i</cmath><br />
<br />
We can postulate that every group of <math>4</math> is equal to <math>2-2i</math>.<br />
For 24 groups we thus, get <math>48-48i</math> as our sum. <br />
We know the solution must lie near<br />
The next term is the <math>24*4+1=97</math>th term. This term is equal to <math>97i</math> (first in a group of <math>4</math> so <math>i^{97}=i</math>) and our sum is now <math>48+49i</math> so <math>n=97</math> is our answer<br />
<br />
== Solution 2==<br />
<br />
Obviously, even powers of <math>i</math> are real and odd powers of <math>i</math> are imaginary. <br />
Hence the real part of the sum is <math>2i^2 + 4i^4 + 6i^6 + \ldots</math>, and <br />
the imaginary part is <math>i + 3i^3 + 5i^5 + \cdots</math>.<br />
<br />
Let's take a look at the real part first. We have <math>i^2=-1</math>, hence the real part simplifies to <math>-2+4-6+8-10+\cdots</math>.<br />
If there were an odd number of terms, we could pair them as follows: <math>-2 + (4-6) + (8-10) + \cdots</math>, hence the result would be negative. As we need the real part to be <math>48</math>, we must have an even number of terms. If we have an even number of terms, we can pair them as <math>(-2+4) + (-6+8) + \cdots</math>. Each parenthesis is equal to <math>2</math>, thus there are <math>24</math> of them, and the last value used is <math>96</math>. This happens for <math>n=96</math> and <math>n=97</math>. As <math>n=96</math> is not present as an option, we may conclude that the answer is <math>\boxed{97}</math>.<br />
<br />
In a complete solution, we should now verify which of <math>n=96</math> and <math>n=97</math> will give us the correct imaginary part.<br />
<br />
We can rewrite the imaginary part as follows: <math>i + 3i^3 + 5i^5 + \cdots = i(1 + 3i^2 + 5i^4 + \cdots) = i(1 - 3 + 5 - \cdots)</math>. We need to obtain <math>(1 - 3 + 5 - \cdots) = 49</math>. Once again we can repeat the same reasoning: If the number of terms were even, the left hand side would be negative, thus the number of terms is odd. The left hand side can then be rewritten as <math>1 + (-3+5) + (-7+9) + \cdots</math>. We need <math>24</math> parentheses, therefore the last value used is <math>97</math>. This happens when <math>n=97</math> or <math>n=98</math>, and we are done.<br />
<br />
== See Also ==<br />
<br />
{{AMC12 box|year=2009|ab=A|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Mathpro1441https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_15&diff=1041332009 AMC 12A Problems/Problem 152019-03-08T01:25:54Z<p>Mathpro1441: /* Solution */</p>
<hr />
<div>== Problem ==<br />
For what value of <math>n</math> is <math>i + 2i^2 + 3i^3 + \cdots + ni^n = 48 + 49i</math>?<br />
<br />
Note: here <math>i = \sqrt { - 1}</math>.<br />
<br />
<math>\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 49 \qquad \textbf{(D)}\ 97 \qquad \textbf{(E)}\ 98</math><br />
<br />
== Solution 1 == <br />
We know that <math>i^x</math> cycles every <math>4</math> numbers so we group the sum in <math>4</math>s. <br />
<cmath>i+2i^2+3i^3+4i^4=2-2i</cmath><br />
<cmath>5i^5+6i^6+7i^7+8i^8=2-2i</cmath><br />
<br />
We can postulate that every group of <math>4</math> is equal to <math>2-2i</math>.<br />
For 24 groups we thus, get <math>48-48i</math> as our sum. <br />
We know the solution must lie near<br />
The next term is the <math>24*4+1=97</math>th term. This term is equal to <math>97i</math> (first in a group of <math>4</math>) and our sum is now <math>48+49i</math> so <math>n=97</math> is our answer<br />
<br />
<br />
== Solution 2==<br />
<br />
Obviously, even powers of <math>i</math> are real and odd powers of <math>i</math> are imaginary. <br />
Hence the real part of the sum is <math>2i^2 + 4i^4 + 6i^6 + \ldots</math>, and <br />
the imaginary part is <math>i + 3i^3 + 5i^5 + \cdots</math>.<br />
<br />
Let's take a look at the real part first. We have <math>i^2=-1</math>, hence the real part simplifies to <math>-2+4-6+8-10+\cdots</math>.<br />
If there were an odd number of terms, we could pair them as follows: <math>-2 + (4-6) + (8-10) + \cdots</math>, hence the result would be negative. As we need the real part to be <math>48</math>, we must have an even number of terms. If we have an even number of terms, we can pair them as <math>(-2+4) + (-6+8) + \cdots</math>. Each parenthesis is equal to <math>2</math>, thus there are <math>24</math> of them, and the last value used is <math>96</math>. This happens for <math>n=96</math> and <math>n=97</math>. As <math>n=96</math> is not present as an option, we may conclude that the answer is <math>\boxed{97}</math>.<br />
<br />
In a complete solution, we should now verify which of <math>n=96</math> and <math>n=97</math> will give us the correct imaginary part.<br />
<br />
We can rewrite the imaginary part as follows: <math>i + 3i^3 + 5i^5 + \cdots = i(1 + 3i^2 + 5i^4 + \cdots) = i(1 - 3 + 5 - \cdots)</math>. We need to obtain <math>(1 - 3 + 5 - \cdots) = 49</math>. Once again we can repeat the same reasoning: If the number of terms were even, the left hand side would be negative, thus the number of terms is odd. The left hand side can then be rewritten as <math>1 + (-3+5) + (-7+9) + \cdots</math>. We need <math>24</math> parentheses, therefore the last value used is <math>97</math>. This happens when <math>n=97</math> or <math>n=98</math>, and we are done.<br />
<br />
== See Also ==<br />
<br />
{{AMC12 box|year=2009|ab=A|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Mathpro1441https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_10&diff=1010682013 AMC 12A Problems/Problem 102019-01-29T12:20:38Z<p>Mathpro1441: /* Solution 2 */</p>
<hr />
<div>== Problem==<br />
<br />
Let <math>S</math> be the set of positive integers <math>n</math> for which <math>\tfrac{1}{n}</math> has the repeating decimal representation <math>0.\overline{ab} = 0.ababab\cdots,</math> with <math>a</math> and <math>b</math> different digits. What is the sum of the elements of <math>S</math>?<br />
<br />
<math> \textbf{(A)}\ 11\qquad\textbf{(B)}\ 44\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 143\qquad\textbf{(E)}\ 155\qquad </math><br />
<br />
==Solution 1==<br />
Note that <math>\frac{1}{11} = 0.\overline{09}</math>.<br />
<br />
Dividing by 3 gives <math>\frac{1}{33} = 0.\overline{03}</math>, and dividing by 9 gives <math>\frac{1}{99} = 0.\overline{01}</math>.<br />
<br />
<math>S = \{11, 33, 99\}</math><br />
<br />
<math>11 + 33 + 99 = 143</math><br />
<br />
The answer must be at least <math>143</math>, but cannot be <math>155</math> since no <math>n \le 12</math> other than <math>11</math> satisfies the conditions, so the answer is <math>143</math>.<br />
<br />
==Solution 2==<br />
Let us begin by working with the condition <math>0.\overline{ab} = 0.ababab\cdots,</math>. Let <math>x = 0.ababab\cdots</math>. So, <math>100x-x = ab \Rightarrow x = \frac{ab}{99}</math>. In order for this fraction <math>x</math> to be in the form <math>\frac{1}{n}</math>, <math>99</math> must be a multiple of <math>ab</math>. Hence the possibilities of <math>ab</math> are <math>1,3,9,11,33,99</math>. Checking each of these, <math>\frac{1}{99} = 0.\overline{01}, \frac{3}{99}=\frac{1}{33} = 0.\overline{03}, \frac{9}{99}=\frac{1}{11} = 0.\overline{09}, \frac{11}{99}=\frac{1}{9} = 0.\overline{1}, \frac{33}{99} =\frac{1}{3}= 0.\overline{3},</math> and <math>\frac{99}{99} = 1</math>. So the only values of <math>n</math> that have distinct <math>a</math> and <math>b</math> are <math>11,33,</math> and <math>99</math>. So, <math>11+33+99= \boxed{\textbf{(D)} 143}</math><br />
<br />
<br />
==Solution 3==<br />
<br />
Notice that we have <math>\frac{100}{n}= ab.\overline{ab}</math> <br />
<br />
We can subtract <math>\frac{1}{n}=00.\overline{ab}</math> to get <cmath>\frac{99}{n}=ab</cmath><br />
<br />
From this we determine <math>n</math> must be a positive factor of <math>99</math><br />
<br />
<br />
The factors of <math>99</math> are <math>1,3,9,11,33,</math> and <math>99</math>.<br />
<br />
For <math>n=1,3,</math> and <math>9</math> however, they yield <math>ab=99,33</math> and <math>11</math> which doesn't satisfy <math>a</math> and <math>b</math> being distinct. <br />
<br />
For <math>n=11,33</math> and <math>99</math> we have <math>ab=09,03</math> and <math>01</math>. (Notice that <math>a</math> or <math>b</math> can be zero)<br />
<br />
The sum of these <math>n</math> are <math>11+33+99=143</math><br />
<br />
<math>\boxed{\textbf{(D)} 143}</math><br />
<br />
== See also ==<br />
{{AMC12 box|year=2013|ab=A|num-b=9|num-a=11}}<br />
{{MAA Notice}}</div>Mathpro1441https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_4&diff=974242016 AMC 12B Problems/Problem 42018-08-27T21:07:20Z<p>Mathpro1441: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
The ratio of the measures of two acute angles is <math>5:4</math>, and the complement of one of these two angles is twice as large as the complement of the other. What is the sum of the degree measures of the two angles?<br />
<br />
<math>\textbf{(A)}\ 75\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 135\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 270</math><br />
<br />
==Solution==<br />
By: dragonfly<br />
<br />
We set up equations to find each angle. The larger angle will be represented as <math>x</math> and the smaller angle will be represented as <math>y</math>, in degrees. This implies that <br />
<br />
<math>4x=5y</math><br />
<br />
and<br />
<br />
<math>2\times(90-x)=90-y</math><br />
<br />
since the larger the original angle, the smaller the complement.<br />
<br />
We then find that <math>x=75</math> and <math>y=60</math>, and their sum is <math>\boxed{\textbf{(C)}\ 135}</math><br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=B|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Mathpro1441https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_12&diff=974182016 AMC 12A Problems/Problem 122018-08-27T01:21:04Z<p>Mathpro1441: /* Solution 2 */</p>
<hr />
<div>==Problem 12==<br />
<br />
In <math>\triangle ABC</math>, <math>AB = 6</math>, <math>BC = 7</math>, and <math>CA = 8</math>. Point <math>D</math> lies on <math>\overline{BC}</math>, and <math>\overline{AD}</math> bisects <math>\angle BAC</math>. Point <math>E</math> lies on <math>\overline{AC}</math>, and <math>\overline{BE}</math> bisects <math>\angle ABC</math>. The bisectors intersect at <math>F</math>. What is the ratio <math>AF</math> : <math>FD</math>?<br />
<br />
<asy> pair A = (0,0), B=(6,0), C=intersectionpoints(Circle(A,8),Circle(B,7))[0], F=incenter(A,B,C), D=extension(A,F,B,C),E=extension(B,F,A,C); draw(A--B--C--A--D^^B--E); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,NE); label("$E$",E,NW); label("$F$",F,1.5*N); </asy><br />
<br />
<math>\textbf{(A)}\ 3:2\qquad\textbf{(B)}\ 5:3\qquad\textbf{(C)}\ 2:1\qquad\textbf{(D)}\ 7:3\qquad\textbf{(E)}\ 5:2</math><br />
<br />
== Solution 1==<br />
<br />
Applying the angle bisector theorem to <math>\triangle ABC</math> with <math>\angle CAB</math> being bisected by <math>AD</math>, we have<br />
<br />
<cmath>\frac{CD}{AC}=\frac{BD}{AB}.</cmath><br />
<br />
Thus, we have <br />
<br />
<cmath>\frac{CD}{8}=\frac{BD}{6},</cmath><br />
<br />
and cross multiplying and dividing by <math>2</math> gives us<br />
<br />
<cmath>3\cdot CD=4\cdot BD.</cmath><br />
<br />
<br />
Since <math>CD+BD=BC=7</math>, we can substitute <math>CD=7-BD</math> into the former equation. Therefore, we get <math>3(7-BD)=4BD</math>, so <math>BD=3</math>.<br />
<br />
<br />
Apply the angle bisector theorem again to <math>\triangle ABD</math> with <math>\angle ABC</math> being bisected. This gives us<br />
<br />
<cmath>\frac{AB}{AF}=\frac{BD}{FD},</cmath><br />
<br />
and since <math>AB=6</math> and <math>BD=3</math>, we have<br />
<br />
<cmath>\frac{6}{AF}=\frac{3}{FD}.</cmath><br />
<br />
Cross multiplying and dividing by <math>3</math> gives us <br />
<br />
<cmath>AF=2\cdot FD,</cmath><br />
<br />
and dividing by <math>FD</math> gives us<br />
<br />
<cmath>\frac{AF}{FD}=\frac{2}{1}.</cmath><br />
<br />
Therefore,<br />
<br />
<cmath>AF:FD=\frac{AF}{FD}=\frac{2}{1}=\boxed{\textbf{(C)}\; 2 : 1}.</cmath><br />
<br />
== Solution 2==<br />
<br />
By the angle bisector theorem, <math>\frac{AB}{AE} = \frac{CB}{CE}</math><br />
<br />
<math>\frac{6}{AE} = \frac{7}{8 - AE}</math> so <math>AE = \frac{48}{13}</math><br />
<br />
Similarly, <math>CD = 4</math>.<br />
<br />
Now, we use [[mass points]]. Assign point <math>C</math> a mass of <math>1</math>.<br />
<br />
<math>mC \cdot CD = mB \cdot DB</math> , so <math>mB = \frac{4}{3}</math><br />
<br />
Similarly, <math>A</math> will have a mass of <math>\frac{7}{6}</math><br />
<br />
<math>mD = mC + mB = 1 + \frac{4}{3} = \frac{7}{3}</math><br />
<br />
So <math>\frac{AF}{AD} = \frac{mD}{mA} = \boxed{\textbf{(C)}\; 2 : 1}</math><br />
<br />
== Solution 3==<br />
<br />
Denote <math>[\triangle{ABC}]</math> as the area of triangle ABC and let <math>r</math> be the inradius. Also, as above, use the angle bisector theorem to find that <math>BD = 3</math>. There are two ways to continue from here:<br />
<br />
<math>1.</math> Note that <math>F</math> is the incenter. Then, <math>\frac{AF}{FD} = \frac{[\triangle{AFB}]}{[\triangle{BFD}]} = \frac{AB * \frac{r}{2}}{BD * \frac{r}{2}} = \frac{AB}{BD} = \boxed{\textbf{(C)}\; 2 : 1}</math><br />
<br />
<math>2.</math> Apply the angle bisector theorem on <math>\triangle{ABD}</math> to get <math>\frac{AF}{FD} = \frac{AB}{BD} = \frac{6}{3} = \boxed{\textbf{(C)}\; 2 : 1}</math><br />
<br />
==Solution 4==<br />
Draw the third angle bisector, and denote the point where this bisector intersects AB as P. Using angle bisector theorem, we see <math>AE=48/13 , EC=56/13, AP=16/5, PB=14/5</math>. Applying Van Aubel's theorem, <math>AF/FD=(48/13)/(56/13) + (16/5)/(14/5)=(6/7)+(8/7)=14/7=2/1</math>, and so the answer is <math>\boxed{\textbf{(C)}\; 2 : 1}</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=A|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Mathpro1441https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_7&diff=973902017 AMC 10B Problems/Problem 72018-08-25T14:25:07Z<p>Mathpro1441: /* Solution 2 (Quicker, Make Approximations) */</p>
<hr />
<div>==Problem==<br />
Samia set off on her bicycle to visit her friend, traveling at an average speed of <math>17</math> kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at <math>5</math> kilometers per hour. In all it took her <math>44</math> minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?<br />
<br />
<math>\textbf{(A)}\ 2.0\qquad\textbf{(B)}\ 2.2\qquad\textbf{(C)}\ 2.8\qquad\textbf{(D)}\ 3.4\qquad\textbf{(E)}\ 4.4</math><br />
<br />
==Solution 1==<br />
Let's call the distance that Samia had to travel in total as <math>2x</math>, so that we can avoid fractions. We know that the length of the bike ride and how far she walked are equal, so they are both <math>\frac{2x}{2}</math>, or <math>x</math>.<br />
<cmath></cmath><br />
She bikes at a rate of <math>17</math> kph, so she travels the distance she bikes in <math>\frac{x}{17}</math> hours. She walks at a rate of <math>5</math> kph, so she travels the distance she walks in <math>\frac{x}{5}</math> hours.<br />
<cmath></cmath><br />
The total time is <math>\frac{x}{17}+\frac{x}{5} = \frac{22x}{85}</math>. This is equal to <math>\frac{44}{60} = \frac{11}{15}</math> of an hour. Solving for <math>x</math>, we have:<br />
<cmath></cmath><br />
<cmath>\frac{22x}{85} = \frac{11}{15}</cmath><br />
<cmath>\frac{2x}{85} = \frac{1}{15}</cmath><br />
<cmath>30x = 85</cmath><br />
<cmath>6x = 17</cmath><br />
<cmath>x = \frac{17}{6}</cmath><br />
<cmath></cmath><br />
Since <math>x</math> is the distance of how far Samia traveled by both walking and biking, and we want to know how far Samia walked to the nearest tenth, we have that Samia walked about <math>\boxed{\bold{(C)} 2.8}</math>.<br />
<br />
==Solution 2 (Quick, Make Approximations)==<br />
<br />
Notice that Samia walks <math>\frac {17}{5}</math> times slower than she bikes, and that she walks and bikes the same distance. Thus, the fraction of the total time that Samia will be walking is<br />
<cmath>\frac{\frac{17}{5}}{\frac{17}{5}+\frac{5}{5}} = \frac{17}{22}</cmath> <br />
Then, multiply this by the time<br />
<cmath>\frac{17}{22} \cdot 44 \text{minutes} = 34 \text{minutes} </cmath><br />
34 minutes is a little greater than <math>\frac {1}{2}</math> of an hour so Samia traveled <br />
<cmath>\sim \frac {1}{2} \cdot 5 = 2.5 \text{kilometers}</cmath> The answer choice a little greater than 2.5 is <math>\boxed{\bold{(C)} 2.8}</math>. (Note that we could've multiplied <math>\frac {34}{60}=\frac {17}{30}</math> by <math>5</math> and gotten the exact answer as well)<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2017|ab=B|num-b=6|num-a=8}}<br />
{{AMC12 box|year=2017|ab=B|before=First Problem|num-a=5}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Algebra Problems]]</div>Mathpro1441https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_7&diff=973892017 AMC 10B Problems/Problem 72018-08-25T14:22:14Z<p>Mathpro1441: /* Solution 2 (Quicker, Make Approximations) */</p>
<hr />
<div>==Problem==<br />
Samia set off on her bicycle to visit her friend, traveling at an average speed of <math>17</math> kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at <math>5</math> kilometers per hour. In all it took her <math>44</math> minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?<br />
<br />
<math>\textbf{(A)}\ 2.0\qquad\textbf{(B)}\ 2.2\qquad\textbf{(C)}\ 2.8\qquad\textbf{(D)}\ 3.4\qquad\textbf{(E)}\ 4.4</math><br />
<br />
==Solution 1==<br />
Let's call the distance that Samia had to travel in total as <math>2x</math>, so that we can avoid fractions. We know that the length of the bike ride and how far she walked are equal, so they are both <math>\frac{2x}{2}</math>, or <math>x</math>.<br />
<cmath></cmath><br />
She bikes at a rate of <math>17</math> kph, so she travels the distance she bikes in <math>\frac{x}{17}</math> hours. She walks at a rate of <math>5</math> kph, so she travels the distance she walks in <math>\frac{x}{5}</math> hours.<br />
<cmath></cmath><br />
The total time is <math>\frac{x}{17}+\frac{x}{5} = \frac{22x}{85}</math>. This is equal to <math>\frac{44}{60} = \frac{11}{15}</math> of an hour. Solving for <math>x</math>, we have:<br />
<cmath></cmath><br />
<cmath>\frac{22x}{85} = \frac{11}{15}</cmath><br />
<cmath>\frac{2x}{85} = \frac{1}{15}</cmath><br />
<cmath>30x = 85</cmath><br />
<cmath>6x = 17</cmath><br />
<cmath>x = \frac{17}{6}</cmath><br />
<cmath></cmath><br />
Since <math>x</math> is the distance of how far Samia traveled by both walking and biking, and we want to know how far Samia walked to the nearest tenth, we have that Samia walked about <math>\boxed{\bold{(C)} 2.8}</math>.<br />
<br />
==Solution 2 (Quicker, Make Approximations)==<br />
<br />
Notice that Samia walks <math>\frac {17}{5}</math> times slower than she bikes, and that she walks and bikes the same distance. Thus, the fraction of the total time that Samia will be walking is<br />
<cmath>\frac{\frac{17}{5}}{\frac{17}{5}+\frac{5}{5}} = \frac{17}{22}</cmath> <br />
Then, multiply this by the time<br />
<cmath>\frac{17}{22} \cdot 44 \text{minutes} = 34 \text{minutes} </cmath><br />
34 minutes is a little greater than <math>\frac {1}{2}</math> of an hour so Samia traveled <br />
<cmath>\sim \frac {1}{2} \cdot 5 = 2.5 \text{kilometers}</cmath> The answer choice a little greater than 2.5 is <math>\boxed{\bold{(C)} 2.8}</math>. (Note that we could've multiplied <math>\frac {34}{60}=\frac {17}{30}</math> by <math>5</math> and gotten the exact answer as well)<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2017|ab=B|num-b=6|num-a=8}}<br />
{{AMC12 box|year=2017|ab=B|before=First Problem|num-a=5}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Algebra Problems]]</div>Mathpro1441https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_7&diff=973882017 AMC 10B Problems/Problem 72018-08-25T14:15:58Z<p>Mathpro1441: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
Samia set off on her bicycle to visit her friend, traveling at an average speed of <math>17</math> kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at <math>5</math> kilometers per hour. In all it took her <math>44</math> minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?<br />
<br />
<math>\textbf{(A)}\ 2.0\qquad\textbf{(B)}\ 2.2\qquad\textbf{(C)}\ 2.8\qquad\textbf{(D)}\ 3.4\qquad\textbf{(E)}\ 4.4</math><br />
<br />
==Solution 1==<br />
Let's call the distance that Samia had to travel in total as <math>2x</math>, so that we can avoid fractions. We know that the length of the bike ride and how far she walked are equal, so they are both <math>\frac{2x}{2}</math>, or <math>x</math>.<br />
<cmath></cmath><br />
She bikes at a rate of <math>17</math> kph, so she travels the distance she bikes in <math>\frac{x}{17}</math> hours. She walks at a rate of <math>5</math> kph, so she travels the distance she walks in <math>\frac{x}{5}</math> hours.<br />
<cmath></cmath><br />
The total time is <math>\frac{x}{17}+\frac{x}{5} = \frac{22x}{85}</math>. This is equal to <math>\frac{44}{60} = \frac{11}{15}</math> of an hour. Solving for <math>x</math>, we have:<br />
<cmath></cmath><br />
<cmath>\frac{22x}{85} = \frac{11}{15}</cmath><br />
<cmath>\frac{2x}{85} = \frac{1}{15}</cmath><br />
<cmath>30x = 85</cmath><br />
<cmath>6x = 17</cmath><br />
<cmath>x = \frac{17}{6}</cmath><br />
<cmath></cmath><br />
Since <math>x</math> is the distance of how far Samia traveled by both walking and biking, and we want to know how far Samia walked to the nearest tenth, we have that Samia walked about <math>\boxed{\bold{(C)} 2.8}</math>.<br />
<br />
==Solution 2 (Quicker, Make Approximations)==<br />
<br />
Notice that Samia walks <math>\frac {17}{5}</math> times slower than she bikes, and that she walks and bikes the same distance. Thus, the fraction of the total time that Samia will be walking is <br />
<cmath>\frac{\frac{17}{5}}{\frac{17}{5}+\frac{5}{5}} = \frac{17}{22}</cmath><br />
Then, multiply this by the time:<br />
<cmath>\frac{17}{22} \cdot 44 \text{minutes} = 34 \text{minutes} </cmath><br />
34 minutes is a little greater than <math>\frac {1}{2}</math> of an hour so Samia traveled <br />
<cmath>\sim \frac {1}{2} \cdot 5 = 2.5 \text{kilometers}</cmath> The answer choice a little greater than 2.5 is <math>\boxed{\bold{(C)} 2.8}</math>. (Note that we could've multiplied <math>\frac {34}{60}=\frac {17}{30}</math> by <math>5</math> and gotten the exact answer as well)<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2017|ab=B|num-b=6|num-a=8}}<br />
{{AMC12 box|year=2017|ab=B|before=First Problem|num-a=5}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Algebra Problems]]</div>Mathpro1441https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_3&diff=973322018 AMC 12B Problems/Problem 32018-08-21T21:07:59Z<p>Mathpro1441: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
<br />
A line with slope 2 intersects a line with slope 6 at the point <math>(40,30)</math>. What is the distance between the <math>x</math>-intercepts of these two lines? <br />
<br />
<math>(\text{A}) 5 \qquad (\text{B}) 10 \qquad (\text{C}) 20 \qquad (\text{D}) 25 \qquad (\text{E}) 50</math><br />
<br />
<br />
==Solution 1==<br />
<br />
<br />
Using the slope-intercept form, we get the equations <math>y-30 = 6(x-40)</math> and <math>y-30 = 2(x-40)</math>. Simplifying, we get <math>6x-y=210</math> and <math>2x-y=50</math>. Letting <math>y=0</math> in both equations and solving for <math>x</math> gives the <math>x</math>-intercepts: <math>x=35</math> and <math>x=25</math>, respectively. Thus the distance between them is <math>35-25 = 10 \Rightarrow \boxed{(\text{B}) 10}<br />
\indent</math><br />
<br />
==Solution 2==<br />
<br />
In order for the line with slope <math>2</math> to travel "up" <math>30</math> units (from <math>y=0</math>), it must have traveled <math>30/2=15</math> units to the right. Thus, the <math>x</math>-intercept is at <math>x=40-15=25</math>. As for the line with slope <math>6</math>, in order for it to travel "up" <math>30</math> units it must have traveled <math>30/6=5</math> units to the right. Thus its <math>x</math>-intercept is at <math>x=40-5=35</math>. Then the distance between them is <math>35-25=10 \Rightarrow \boxed{(\text{B}) 10}<br />
\indent</math><br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2018|ab=B|num-b=2|num-a=4}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Algebra Problems]]</div>Mathpro1441https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_13&diff=972182018 AMC 10A Problems/Problem 132018-08-13T17:19:30Z<p>Mathpro1441: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
<br />
A paper triangle with sides of lengths 3,4, and 5 inches, as shown, is folded so that point <math>A</math> falls on point <math>B</math>. What is the length in inches of the crease?<br />
<asy><br />
draw((0,0)--(4,0)--(4,3)--(0,0));<br />
label("$A$", (0,0), SW);<br />
label("$B$", (4,3), NE);<br />
label("$C$", (4,0), SE);<br />
label("$4$", (2,0), S);<br />
label("$3$", (4,1.5), E);<br />
label("$5$", (2,1.5), NW);<br />
fill(origin--(0,0)--(4,3)--(4,0)--cycle, gray);<br />
</asy><br />
<math>\textbf{(A) } 1+\frac12 \sqrt2 \qquad \textbf{(B) } \sqrt3 \qquad \textbf{(C) } \frac74 \qquad \textbf{(D) } \frac{15}{8} \qquad \textbf{(E) } 2 </math><br />
<br />
==Solution 1==<br />
<br />
First, we need to realize that the crease line is just the perpendicular bisector of side <math>AB</math>, the hypotenuse of right triangle <math>\triangle ABC</math>. Call the midpoint of <math>AB</math> point <math>D</math>. Draw this line and call the intersection point with <math>AC</math> as <math>E</math>. Now, <math>\triangle ACB</math> is similar to <math>\triangle ADE</math> by <math>AA</math> similarity. Setting up the ratios, we find that<br />
<cmath>\frac{BC}{AC}=\frac{DE}{AD} \Rightarrow \frac{3}{4}=\frac{DE}{\frac{5}{2}} \Rightarrow DE=\frac{15}{8}.</cmath><br />
Thus, our answer is <math>\boxed{\textbf{D) } \frac{15}{8}}</math>.<br />
<br />
~Nivek<br />
<br />
==Solution 2==<br />
<br />
Use the ruler and graph paper you brought to quickly draw a 3-4-5 triangle of any scale (don't trust the diagram in the booklet). Very carefully fold the acute vertices together and make a crease. Measure the crease with the ruler. If you were reasonably careful, you should see that it measures somewhat more than <math>\frac{7}{4}</math> units and somewhat less than <math>2</math> units. The only answer choice in range is <math>\boxed{\textbf{D) } \frac{15}{8}}</math>.<br />
<br />
This is pretty much a cop-out, but it's allowed in the rules technically.<br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2018|ab=A|num-b=12|num-a=14}}<br />
{{AMC12 box|year=2018|ab=A|num-b=10|num-a=12}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Geometry Problems]]</div>Mathpro1441