https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Sachi2019&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T04:51:25ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=User:Sachi2019&diff=124984User:Sachi20192020-06-11T17:15:37Z<p>Sachi2019: Created page with "[b]Buncha Mock Test Scores[/b] [list] [*] TMC 12A - 139.5 [*] TMC 12B - 115.5 [*] AoPS Mathcounts Nationals - 29 [*] CMC 12A Year 3 - 117 [*] CIME Year 3 - 10[/list]"</p>
<hr />
<div>[b]Buncha Mock Test Scores[/b]<br />
[list]<br />
[*] TMC 12A - 139.5<br />
[*] TMC 12B - 115.5<br />
[*] AoPS Mathcounts Nationals - 29 <br />
[*] CMC 12A Year 3 - 117<br />
[*] CIME Year 3 - 10[/list]</div>Sachi2019https://artofproblemsolving.com/wiki/index.php?title=Angle_Bisector_Theorem&diff=121621Angle Bisector Theorem2020-04-24T17:51:02Z<p>Sachi2019: /* Proof */</p>
<hr />
<div>{{WotWAnnounce|week=June 6-12}}<br />
<br />
== Introduction ==<br />
The '''Angle Bisector Theorem''' states that given [[triangle]] <math>\triangle ABC</math> and [[angle bisector]] AD, where D is on side BC, then <math> \frac cm = \frac bn </math>. It follows that <math> \frac cb = \frac mn </math>. Likewise, the [[converse]] of this theorem holds as well.<br />
<br />
<br />
Further by combining with [[Stewart's Theorem]] it can be shown that <math>AD^2 = b\cdot c - m \cdot n</math><br />
<br />
<asy> size(200); defaultpen(fontsize(12)); real a,b,c,d; pair A=(1,9), B=(-11,0), C=(4,0), D; b = abs(C-A); c = abs(B-A); D = (b*B+c*C)/(b+c); draw(A--B--C--A--D,black); MA(B,A,D,2,green); MA(D,A,C,2,green); label("$A$",A,(1,1));label("$B$",B,(-1,-1));label("$C$",C,(1,-1));label("$D$",D,(0,-1)); dot(A^^B^^C^^D,blue);label("$b$",(A+C)/2,(1,0));label("$c$",(A+B)/2,(0,1));label("$m$",(B+D)/2,(0,-1));label("$n$",(D+C)/2,(0,-1)); </asy><br />
<br />
== Proof ==<br />
<br />
By LoS on ACD and ABD, <br />
<br />
AB/BD=sin(BDA)/sin(BAD) ... (1) and<br />
AC/AD=sin(ADC)/sin(DAC) ... (2)<br />
Well, we also know that BDA and ADC add to 180 degrees. I think that means that we can use sin(180-x)=sin(x) here. Doing so, we see that<br />
sin(BDA)=sin(ADC)<br />
I noticed that these are the numerators of (1) and (2) respectively. Since BAD and DAC are equal, then you get the equation for the bisector angle theorem. ~ sachi2019<br />
<br />
== Examples ==<br />
<br />
# Let ABC be a triangle with angle bisector AD with D on line segment BC. If <math> BD = 2, CD = 5,</math> and <math> AB + AC = 10 </math>, find AB and AC.<br> '''''Solution:''''' By the angle bisector theorem, <math> \frac{AB}2 = \frac{AC}5</math> or <math> AB = \frac 25 AC </math>. Plugging this into <math> AB + AC = 10 </math> and solving for AC gives <math> AC = \frac{50}7</math>. We can plug this back in to find <math> AB = \frac{20}7 </math>.<br />
# In triangle ABC, let P be a point on BC and let <math> AB = 20, AC = 10, BP = \frac{20\sqrt{3}}3, CP = \frac{10\sqrt{3}}3 </math>. Find the value of <math> m\angle BAP - m\angle CAP </math>. <br> '''''Solution:''''' First, we notice that <math> \frac{AB}{BP}=\frac{AC}{CP} </math>. Thus, AP is the angle bisector of angle A, making our answer 0.<br />
# Part '''(b)''', [[1959 IMO Problems/Problem 5]].<br />
<br />
== See also ==<br />
* [[Angle bisector]]<br />
* [[Geometry]]<br />
* [[Stewart's Theorem]]<br />
<br />
[[Category:Geometry]]<br />
<br />
[[Category:Theorems]]</div>Sachi2019https://artofproblemsolving.com/wiki/index.php?title=Angle_Bisector_Theorem&diff=121620Angle Bisector Theorem2020-04-24T17:50:46Z<p>Sachi2019: /* Proof */</p>
<hr />
<div>{{WotWAnnounce|week=June 6-12}}<br />
<br />
== Introduction ==<br />
The '''Angle Bisector Theorem''' states that given [[triangle]] <math>\triangle ABC</math> and [[angle bisector]] AD, where D is on side BC, then <math> \frac cm = \frac bn </math>. It follows that <math> \frac cb = \frac mn </math>. Likewise, the [[converse]] of this theorem holds as well.<br />
<br />
<br />
Further by combining with [[Stewart's Theorem]] it can be shown that <math>AD^2 = b\cdot c - m \cdot n</math><br />
<br />
<asy> size(200); defaultpen(fontsize(12)); real a,b,c,d; pair A=(1,9), B=(-11,0), C=(4,0), D; b = abs(C-A); c = abs(B-A); D = (b*B+c*C)/(b+c); draw(A--B--C--A--D,black); MA(B,A,D,2,green); MA(D,A,C,2,green); label("$A$",A,(1,1));label("$B$",B,(-1,-1));label("$C$",C,(1,-1));label("$D$",D,(0,-1)); dot(A^^B^^C^^D,blue);label("$b$",(A+C)/2,(1,0));label("$c$",(A+B)/2,(0,1));label("$m$",(B+D)/2,(0,-1));label("$n$",(D+C)/2,(0,-1)); </asy><br />
<br />
== Proof ==<br />
<br />
By LoS on ACD and ABD, \\*<br />
\newline<br />
AB/BD=sin(BDA)/sin(BAD) ... (1)<br />
\newline<br />
AC/AD=sin(ADC)/sin(DAC) ... (2)<br />
Well, we also know that BDA and ADC add to 180 degrees. I think that means that we can use sin(180-x)=sin(x) here. Doing so, we see that<br />
sin(BDA)=sin(ADC)<br />
I noticed that these are the numerators of (1) and (2) respectively. Since BAD and DAC are equal, then you get the equation for the bisector angle theorem. ~ sachi2019<br />
<br />
== Examples ==<br />
<br />
# Let ABC be a triangle with angle bisector AD with D on line segment BC. If <math> BD = 2, CD = 5,</math> and <math> AB + AC = 10 </math>, find AB and AC.<br> '''''Solution:''''' By the angle bisector theorem, <math> \frac{AB}2 = \frac{AC}5</math> or <math> AB = \frac 25 AC </math>. Plugging this into <math> AB + AC = 10 </math> and solving for AC gives <math> AC = \frac{50}7</math>. We can plug this back in to find <math> AB = \frac{20}7 </math>.<br />
# In triangle ABC, let P be a point on BC and let <math> AB = 20, AC = 10, BP = \frac{20\sqrt{3}}3, CP = \frac{10\sqrt{3}}3 </math>. Find the value of <math> m\angle BAP - m\angle CAP </math>. <br> '''''Solution:''''' First, we notice that <math> \frac{AB}{BP}=\frac{AC}{CP} </math>. Thus, AP is the angle bisector of angle A, making our answer 0.<br />
# Part '''(b)''', [[1959 IMO Problems/Problem 5]].<br />
<br />
== See also ==<br />
* [[Angle bisector]]<br />
* [[Geometry]]<br />
* [[Stewart's Theorem]]<br />
<br />
[[Category:Geometry]]<br />
<br />
[[Category:Theorems]]</div>Sachi2019https://artofproblemsolving.com/wiki/index.php?title=Angle_Bisector_Theorem&diff=121619Angle Bisector Theorem2020-04-24T17:49:42Z<p>Sachi2019: /* Proof */</p>
<hr />
<div>{{WotWAnnounce|week=June 6-12}}<br />
<br />
== Introduction ==<br />
The '''Angle Bisector Theorem''' states that given [[triangle]] <math>\triangle ABC</math> and [[angle bisector]] AD, where D is on side BC, then <math> \frac cm = \frac bn </math>. It follows that <math> \frac cb = \frac mn </math>. Likewise, the [[converse]] of this theorem holds as well.<br />
<br />
<br />
Further by combining with [[Stewart's Theorem]] it can be shown that <math>AD^2 = b\cdot c - m \cdot n</math><br />
<br />
<asy> size(200); defaultpen(fontsize(12)); real a,b,c,d; pair A=(1,9), B=(-11,0), C=(4,0), D; b = abs(C-A); c = abs(B-A); D = (b*B+c*C)/(b+c); draw(A--B--C--A--D,black); MA(B,A,D,2,green); MA(D,A,C,2,green); label("$A$",A,(1,1));label("$B$",B,(-1,-1));label("$C$",C,(1,-1));label("$D$",D,(0,-1)); dot(A^^B^^C^^D,blue);label("$b$",(A+C)/2,(1,0));label("$c$",(A+B)/2,(0,1));label("$m$",(B+D)/2,(0,-1));label("$n$",(D+C)/2,(0,-1)); </asy><br />
<br />
== Proof ==<br />
<br />
By LoS on ACD and ABD,<br />
AB/BD=sin(BDA)/sin(BAD) ... (1)<br />
AC/AD=sin(ADC)/sin(DAC) ... (2)<br />
Well, we also know that BDA and ADC add to 180 degrees. I think that means that we can use sin(180-x)=sin(x) here. Doing so, we see that<br />
sin(BDA)=sin(ADC)<br />
I noticed that these are the numerators of (1) and (2) respectively. Since BAD and DAC are equal, then you get the equation for the bisector angle theorem. ~ sachi2019<br />
<br />
== Examples ==<br />
<br />
# Let ABC be a triangle with angle bisector AD with D on line segment BC. If <math> BD = 2, CD = 5,</math> and <math> AB + AC = 10 </math>, find AB and AC.<br> '''''Solution:''''' By the angle bisector theorem, <math> \frac{AB}2 = \frac{AC}5</math> or <math> AB = \frac 25 AC </math>. Plugging this into <math> AB + AC = 10 </math> and solving for AC gives <math> AC = \frac{50}7</math>. We can plug this back in to find <math> AB = \frac{20}7 </math>.<br />
# In triangle ABC, let P be a point on BC and let <math> AB = 20, AC = 10, BP = \frac{20\sqrt{3}}3, CP = \frac{10\sqrt{3}}3 </math>. Find the value of <math> m\angle BAP - m\angle CAP </math>. <br> '''''Solution:''''' First, we notice that <math> \frac{AB}{BP}=\frac{AC}{CP} </math>. Thus, AP is the angle bisector of angle A, making our answer 0.<br />
# Part '''(b)''', [[1959 IMO Problems/Problem 5]].<br />
<br />
== See also ==<br />
* [[Angle bisector]]<br />
* [[Geometry]]<br />
* [[Stewart's Theorem]]<br />
<br />
[[Category:Geometry]]<br />
<br />
[[Category:Theorems]]</div>Sachi2019https://artofproblemsolving.com/wiki/index.php?title=Angle_Bisector_Theorem&diff=121618Angle Bisector Theorem2020-04-24T17:49:22Z<p>Sachi2019: /* Proof */</p>
<hr />
<div>{{WotWAnnounce|week=June 6-12}}<br />
<br />
== Introduction ==<br />
The '''Angle Bisector Theorem''' states that given [[triangle]] <math>\triangle ABC</math> and [[angle bisector]] AD, where D is on side BC, then <math> \frac cm = \frac bn </math>. It follows that <math> \frac cb = \frac mn </math>. Likewise, the [[converse]] of this theorem holds as well.<br />
<br />
<br />
Further by combining with [[Stewart's Theorem]] it can be shown that <math>AD^2 = b\cdot c - m \cdot n</math><br />
<br />
<asy> size(200); defaultpen(fontsize(12)); real a,b,c,d; pair A=(1,9), B=(-11,0), C=(4,0), D; b = abs(C-A); c = abs(B-A); D = (b*B+c*C)/(b+c); draw(A--B--C--A--D,black); MA(B,A,D,2,green); MA(D,A,C,2,green); label("$A$",A,(1,1));label("$B$",B,(-1,-1));label("$C$",C,(1,-1));label("$D$",D,(0,-1)); dot(A^^B^^C^^D,blue);label("$b$",(A+C)/2,(1,0));label("$c$",(A+B)/2,(0,1));label("$m$",(B+D)/2,(0,-1));label("$n$",(D+C)/2,(0,-1)); </asy><br />
<br />
== Proof ==<br />
<br />
By LoS on ACD and ABD,<br />
AB/BD=sin(BDA)/sin(BAD) ... (1)<br />
AC/AD=sin(ADC)/sin(DAC) ... (2)<br />
Well, we also know that BDA and ADC add to 180 degrees. I think that means that we can use sin(180-x)=sin(x) here. Doing so, we see that<br />
sin(BDA)=sin(ADC)<br />
I noticed that these are the numerators of (1) and (2) respectively. Since BAD and DAC are equal, then you get the equation for the bisector angle theorem.<br />
<br />
== Examples ==<br />
<br />
# Let ABC be a triangle with angle bisector AD with D on line segment BC. If <math> BD = 2, CD = 5,</math> and <math> AB + AC = 10 </math>, find AB and AC.<br> '''''Solution:''''' By the angle bisector theorem, <math> \frac{AB}2 = \frac{AC}5</math> or <math> AB = \frac 25 AC </math>. Plugging this into <math> AB + AC = 10 </math> and solving for AC gives <math> AC = \frac{50}7</math>. We can plug this back in to find <math> AB = \frac{20}7 </math>.<br />
# In triangle ABC, let P be a point on BC and let <math> AB = 20, AC = 10, BP = \frac{20\sqrt{3}}3, CP = \frac{10\sqrt{3}}3 </math>. Find the value of <math> m\angle BAP - m\angle CAP </math>. <br> '''''Solution:''''' First, we notice that <math> \frac{AB}{BP}=\frac{AC}{CP} </math>. Thus, AP is the angle bisector of angle A, making our answer 0.<br />
# Part '''(b)''', [[1959 IMO Problems/Problem 5]].<br />
<br />
== See also ==<br />
* [[Angle bisector]]<br />
* [[Geometry]]<br />
* [[Stewart's Theorem]]<br />
<br />
[[Category:Geometry]]<br />
<br />
[[Category:Theorems]]</div>Sachi2019https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_2&diff=1206612020 AIME I Problems/Problem 22020-04-06T22:57:45Z<p>Sachi2019: /* Solution */</p>
<hr />
<div><br />
== Problem ==<br />
There is a unique positive real number <math>x</math> such that the three numbers <math>\log_8{2x}</math>, <math>\log_4{x}</math>, and <math>\log_2{x}</math>, in that order, form a geometric progression with positive common ratio. The number <math>x</math> can be written as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.<br />
<br />
== Solution ==<br />
Since these form a geometric series, <math>\frac{\log_2{x}}{\log_4{x}}</math> is the common ratio. Rewriting this, we get <math>\frac{\log_x{4}}{\log_x{2}} = \log_2{4} = 2</math> by base change formula. Therefore, the common ratio is 2. Now <math>\frac{\log_4{x}}{\log_8{2x}} = 2 \implies \log_4{x} = 2\log_8{2} + 2\log_8{x} \implies \frac{1}{2}\log_2{x} = \frac{2}{3} + \frac{2}{3}\log_2{x}</math><br />
<br />
<math>\implies -\frac{1}{6}\log_2{x} = \frac{2}{3} \implies \log_2{x} = -4 \implies x = \frac{1}{16}</math>. Therefore, <math>1 + 16 = \boxed{017}</math>.<br />
<br />
~ JHawk0224<br />
<br />
See here for a video solution:<br />
<br />
https://youtu.be/nPL7nUXnRbo<br />
<br />
==Solution 2==<br />
If we set <math>x=2^y</math>, we can obtain three terms of a geometric sequence through logarithm properties. The three terms are <cmath>\frac{y+1}{3}, \frac{y}{2}, y.</cmath> In a three-term geometric sequence, the middle term squared is equal to the product of the other two terms, so we obtain the following: <cmath>\frac{y^2+y}{3} = \frac{y^2}{4},</cmath> which can be solved to reveal <math>y = -4</math>. Therefore, <math>x = 2^{-4} = \frac{1}{16}</math>, so our answer is <math>\boxed{017}</math>.<br />
<br />
-molocyxu<br />
<br />
==Solution 3==<br />
Let <math>r</math> be the common ratio. We have <cmath> r = \frac{\log_4{(x)}}{\log_8{(2x)}} = \frac{\log_2{(x)}}{\log_4{(x)}}</cmath><br />
Hence we obtain <cmath> (\log_4{(x)})(\log_4{(x)}) = (\log_8{(2x)})(\log_2{(x)})</cmath><br />
Ideally we change everything to base <math>64</math> and we can get: <cmath> (\log_{64}{(x^3)})(\log_{64}{(x^3)}) = (\log_{64}{(x^6)})(\log_{64}{(4x^2)})</cmath><br />
Now divide to get: <cmath>\frac{\log_{64}{(x^3)}}{\log_{64}{(4x^2)}} = \frac{\log_{64}{(x^6)}}{\log_{64}{(x^3)}}</cmath><br />
By change-of-base we obtain: <cmath>\log_{(4x^2)}{(x^3)} = \log_{(x^3)}{(x^6)} = 2</cmath><br />
Hence <math>(4x^2)^2 = x^3 \rightarrow 16x^4 = x^3 \rightarrow x = \frac{1}{16}</math> and we have <math>1+16 = \boxed{017}</math> as desired.<br />
<br />
~skyscraper<br />
<br />
==Solution 4 (Exponents > Logarithms)==<br />
Let <math>r</math> be the common ratio, and let <math>a</math> be the starting term (<math>a=\log_{8}{(2x)}</math>). We then have: <cmath>\log_{8}{(2x)}=a, \log_{4}{(x)}=ar, \log_{2}{(x)}=ar^2</cmath> Rearranging these equations gives: <cmath>8^a=2x, 4^{ar}=x, 2^{ar^2}=x</cmath><br />
Deal with the last two equations first: Setting them equal gives: <cmath>4^{ar}=2^{ar^2} \Rightarrow 2^{2ar}=2^{ar^2}</cmath> Using LTE results in: <cmath>2ar=ar^2 \Rightarrow r=2</cmath> Using this value of <math>r</math>, substitute into the first and second equations (or the first and third, it doesn't really matter) to get: <cmath>8^a=2x, 4^{2a}=x</cmath> Changing these to a common base gives: <cmath>2^{3a}=2x, 2^{4a}=x</cmath> Dividing the first equation by 2 on both sides yields: <cmath>2^{3a-1}=x</cmath> Setting these equations equal to each other and applying LTE again gives: <cmath>3a-1=4a \Rightarrow a=-1</cmath> Substituting this back into the first equation gives: <cmath>8^{-1}=2x \Rightarrow 2x=\frac{1}{8} \Rightarrow x=\frac{1}{16}</cmath> Therefore, <math>m+n=1+16=\boxed{017}</math><br />
<br />
~IAmTheHazard<br />
<br />
==Solution 5==<br />
<br />
We can relate the logarithms as follows:<br />
<br />
<cmath>\frac{\log_4{x}}{\log_8{(2x)}}=\frac{\log_2{x}}{\log_4{x}}</cmath><br />
<cmath>\log_8{(2x)}\log_2{x}=\log_4{x}\log_4{x}</cmath><br />
<br />
Now we can convert all logarithm bases to <math>2</math> using the identity <math>\log_a{b}=\log_{a^c}{b^c}</math>:<br />
<br />
<cmath>\log_2{\sqrt[3]{2x}}\log_2{x}=\log_2{\sqrt{x}}\log_2{\sqrt{x}}</cmath><br />
<br />
We can solve for <math>x</math> as follows:<br />
<br />
<cmath>\frac{1}{3}\log_2{(2x)}\log_2{x}=\frac{1}{4}\log_2{x}\log_2{x}</cmath><br />
<cmath>\frac{1}{3}\log_2{(2x)}=\frac{1}{4}\log_2{x}</cmath><br />
<cmath>\frac{1}{3}\log_2{2}+\frac{1}{3}\log_2{x}=\frac{1}{4}\log_2{x}</cmath><br />
We get <math>x=\frac{1}{16}</math>. Verifying that the common ratio is positive, we find the answer of <math>\boxed{017}</math>.<br />
<br />
~QIDb602<br />
<br />
<br />
==Solution 6==<br />
<br />
If the numbers are in a geometric sequence, the middle term must be the geometric mean of the surrounding terms. We can rewrite the first two logarithmic expressions as <math>\frac{1+\log_2{x}}{3}</math> and <math>\frac{1}{2}\log_2{x}</math>, respectively. Therefore:<br />
<cmath>\frac{1}{2}\log_2{x}=\sqrt{\left(\frac{1+\log_2{x}}{3}\right)\left(\log_2{x}\right)}</cmath><br />
Let <math>n=\log_2{x}</math>. We can rewrite the expression as:<br />
<cmath>\frac{n}{2}=\sqrt{\frac{n(n+1)}{3}}</cmath><br />
<cmath>\frac{n^2}{4}=\frac{n(n+1)}{3}</cmath><br />
<cmath>4n(n+1)=3n^2</cmath><br />
<cmath>4n^2+4n=3n^2</cmath><br />
<cmath>n^2+4n=0</cmath><br />
<cmath>n(n+4)=0</cmath><br />
<cmath>n=0 \text{ and } -4</cmath><br />
Zero does not work in this case, so we consider <math>n=-4</math>: <math>\log_2{x}=-4 \rightarrow x=\frac{1}{16}</math>. Therefore, <math>1+16=\boxed{017}</math>.<br />
<br />
~Bowser498<br />
<br />
See here for a video solution:<br />
<br />
https://youtu.be/nPL7nUXnRbo<br />
<br />
==See Also==<br />
<br />
{{AIME box|year=2020|n=I|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Sachi2019https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_2&diff=1206602020 AIME I Problems/Problem 22020-04-06T22:56:38Z<p>Sachi2019: /* Solution 6 */</p>
<hr />
<div><br />
== Problem ==<br />
There is a unique positive real number <math>x</math> such that the three numbers <math>\log_8{2x}</math>, <math>\log_4{x}</math>, and <math>\log_2{x}</math>, in that order, form a geometric progression with positive common ratio. The number <math>x</math> can be written as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.<br />
<br />
== Solution ==<br />
Since these form a geometric series, <math>\frac{\log_2{x}}{\log_4{x}}</math> is the common ratio. Rewriting this, we get <math>\frac{\log_x{4}}{\log_x{2}} = \log_2{4} = 2</math> by base change formula. Therefore, the common ratio is 2. Now <math>\frac{\log_4{x}}{\log_8{2x}} = 2 \implies \log_4{x} = 2\log_8{2} + 2\log_8{x} \implies \frac{1}{2}\log_2{x} = \frac{2}{3} + \frac{2}{3}\log_2{x}</math><br />
<br />
<math>\implies -\frac{1}{6}\log_2{x} = \frac{2}{3} \implies \log_2{x} = -4 \implies x = \frac{1}{16}</math>. Therefore, <math>1 + 16 = \boxed{017}</math>.<br />
<br />
~ JHawk0224<br />
<br />
==Solution 2==<br />
If we set <math>x=2^y</math>, we can obtain three terms of a geometric sequence through logarithm properties. The three terms are <cmath>\frac{y+1}{3}, \frac{y}{2}, y.</cmath> In a three-term geometric sequence, the middle term squared is equal to the product of the other two terms, so we obtain the following: <cmath>\frac{y^2+y}{3} = \frac{y^2}{4},</cmath> which can be solved to reveal <math>y = -4</math>. Therefore, <math>x = 2^{-4} = \frac{1}{16}</math>, so our answer is <math>\boxed{017}</math>.<br />
<br />
-molocyxu<br />
<br />
==Solution 3==<br />
Let <math>r</math> be the common ratio. We have <cmath> r = \frac{\log_4{(x)}}{\log_8{(2x)}} = \frac{\log_2{(x)}}{\log_4{(x)}}</cmath><br />
Hence we obtain <cmath> (\log_4{(x)})(\log_4{(x)}) = (\log_8{(2x)})(\log_2{(x)})</cmath><br />
Ideally we change everything to base <math>64</math> and we can get: <cmath> (\log_{64}{(x^3)})(\log_{64}{(x^3)}) = (\log_{64}{(x^6)})(\log_{64}{(4x^2)})</cmath><br />
Now divide to get: <cmath>\frac{\log_{64}{(x^3)}}{\log_{64}{(4x^2)}} = \frac{\log_{64}{(x^6)}}{\log_{64}{(x^3)}}</cmath><br />
By change-of-base we obtain: <cmath>\log_{(4x^2)}{(x^3)} = \log_{(x^3)}{(x^6)} = 2</cmath><br />
Hence <math>(4x^2)^2 = x^3 \rightarrow 16x^4 = x^3 \rightarrow x = \frac{1}{16}</math> and we have <math>1+16 = \boxed{017}</math> as desired.<br />
<br />
~skyscraper<br />
<br />
==Solution 4 (Exponents > Logarithms)==<br />
Let <math>r</math> be the common ratio, and let <math>a</math> be the starting term (<math>a=\log_{8}{(2x)}</math>). We then have: <cmath>\log_{8}{(2x)}=a, \log_{4}{(x)}=ar, \log_{2}{(x)}=ar^2</cmath> Rearranging these equations gives: <cmath>8^a=2x, 4^{ar}=x, 2^{ar^2}=x</cmath><br />
Deal with the last two equations first: Setting them equal gives: <cmath>4^{ar}=2^{ar^2} \Rightarrow 2^{2ar}=2^{ar^2}</cmath> Using LTE results in: <cmath>2ar=ar^2 \Rightarrow r=2</cmath> Using this value of <math>r</math>, substitute into the first and second equations (or the first and third, it doesn't really matter) to get: <cmath>8^a=2x, 4^{2a}=x</cmath> Changing these to a common base gives: <cmath>2^{3a}=2x, 2^{4a}=x</cmath> Dividing the first equation by 2 on both sides yields: <cmath>2^{3a-1}=x</cmath> Setting these equations equal to each other and applying LTE again gives: <cmath>3a-1=4a \Rightarrow a=-1</cmath> Substituting this back into the first equation gives: <cmath>8^{-1}=2x \Rightarrow 2x=\frac{1}{8} \Rightarrow x=\frac{1}{16}</cmath> Therefore, <math>m+n=1+16=\boxed{017}</math><br />
<br />
~IAmTheHazard<br />
<br />
==Solution 5==<br />
<br />
We can relate the logarithms as follows:<br />
<br />
<cmath>\frac{\log_4{x}}{\log_8{(2x)}}=\frac{\log_2{x}}{\log_4{x}}</cmath><br />
<cmath>\log_8{(2x)}\log_2{x}=\log_4{x}\log_4{x}</cmath><br />
<br />
Now we can convert all logarithm bases to <math>2</math> using the identity <math>\log_a{b}=\log_{a^c}{b^c}</math>:<br />
<br />
<cmath>\log_2{\sqrt[3]{2x}}\log_2{x}=\log_2{\sqrt{x}}\log_2{\sqrt{x}}</cmath><br />
<br />
We can solve for <math>x</math> as follows:<br />
<br />
<cmath>\frac{1}{3}\log_2{(2x)}\log_2{x}=\frac{1}{4}\log_2{x}\log_2{x}</cmath><br />
<cmath>\frac{1}{3}\log_2{(2x)}=\frac{1}{4}\log_2{x}</cmath><br />
<cmath>\frac{1}{3}\log_2{2}+\frac{1}{3}\log_2{x}=\frac{1}{4}\log_2{x}</cmath><br />
We get <math>x=\frac{1}{16}</math>. Verifying that the common ratio is positive, we find the answer of <math>\boxed{017}</math>.<br />
<br />
~QIDb602<br />
<br />
<br />
==Solution 6==<br />
<br />
If the numbers are in a geometric sequence, the middle term must be the geometric mean of the surrounding terms. We can rewrite the first two logarithmic expressions as <math>\frac{1+\log_2{x}}{3}</math> and <math>\frac{1}{2}\log_2{x}</math>, respectively. Therefore:<br />
<cmath>\frac{1}{2}\log_2{x}=\sqrt{\left(\frac{1+\log_2{x}}{3}\right)\left(\log_2{x}\right)}</cmath><br />
Let <math>n=\log_2{x}</math>. We can rewrite the expression as:<br />
<cmath>\frac{n}{2}=\sqrt{\frac{n(n+1)}{3}}</cmath><br />
<cmath>\frac{n^2}{4}=\frac{n(n+1)}{3}</cmath><br />
<cmath>4n(n+1)=3n^2</cmath><br />
<cmath>4n^2+4n=3n^2</cmath><br />
<cmath>n^2+4n=0</cmath><br />
<cmath>n(n+4)=0</cmath><br />
<cmath>n=0 \text{ and } -4</cmath><br />
Zero does not work in this case, so we consider <math>n=-4</math>: <math>\log_2{x}=-4 \rightarrow x=\frac{1}{16}</math>. Therefore, <math>1+16=\boxed{017}</math>.<br />
<br />
~Bowser498<br />
<br />
See here for a video solution:<br />
<br />
https://youtu.be/nPL7nUXnRbo<br />
<br />
==See Also==<br />
<br />
{{AIME box|year=2020|n=I|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Sachi2019https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_1&diff=1206592020 AIME I Problems/Problem 12020-04-06T22:55:28Z<p>Sachi2019: /* Solution 2 */</p>
<hr />
<div><br />
== Problem ==<br />
In <math>\triangle ABC</math> with <math>AB=AC,</math> point <math>D</math> lies strictly between <math>A</math> and <math>C</math> on side <math>\overline{AC},</math> and point <math>E</math> lies strictly between <math>A</math> and <math>B</math> on side <math>\overline{AB}</math> such that <math>AE=ED=DB=BC.</math> The degree measure of <math>\angle ABC</math> is <math>\tfrac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n.</math><br />
<br />
== Solution 1==<br />
<asy><br />
size(10cm);<br />
pair A, B, C, D, F;<br />
A = (0, tan(3 * pi / 7));<br />
B = (1, 0);<br />
C = (-1, 0);<br />
F = rotate(90/7, A) * (A - (0, 2));<br />
D = rotate(900/7, F) * A;<br />
<br />
draw(A -- B -- C -- cycle);<br />
draw(F -- D);<br />
draw(D -- B);<br />
<br />
label("$A$", A, N);<br />
label("$B$", B, E);<br />
label("$C$", C, W);<br />
label("$D$", D, W);<br />
label("$E$", F, E);<br />
</asy><br />
<br />
If we set <math>\angle{BAC}</math> to <math>x</math>, we can find all other angles through these two properties:<br />
1. Angles in a triangle sum to <math>180^{\circ}</math>.<br />
2. The base angles of an isoceles triangle are congruent.<br />
<br />
Now we angle chase. <math>\angle{ADE}=\angle{EAD}=x</math>, <math>\angle{AED} = 180-2x</math>, <math>\angle{BED}=\angle{EBD}=2x</math>, <math>\angle{EDB} = 180-4x</math>, <math>\angle{BDC} = \angle{BCD} = 3x</math>, <math>\angle{CBD} = 180-6x</math>. Since <math>AB = AC</math> as given by the problem, <math>\angle{ABC} = \angle{ACB}</math>, so <math>180-4x=3x</math>. Therefore, <math>x = 180/7^{\circ}</math>, and our desired angle is <cmath>180-4\left(\frac{180}{7}\right) = \frac{540}{7}</cmath> for an answer of <math>\boxed{547}</math>.<br />
<br />
==Solution 2==<br />
Let <math>\angle{BAC}</math> be <math>x</math> in degrees. <math>\angle{ADE}=x</math>.<br />
By Exterior Angle Theorem on triangle <math>AED</math>, <math>\angle{BED}=2x</math>.<br />
By Exterior Angle Theorem on triangle <math>ADB</math>, <math>\angle{BDC}=3x</math>.<br />
This tells us <math>\angle{BCA}=\angle{ABC}=3x</math> and <math>3x+3x+x=180</math>.<br />
Thus <math>x=\frac{180}{7}</math> and we want <math>\angle{ABC}=3x=\frac{540}{7}</math> to get an answer of <math>\boxed{547}</math>.<br />
<br />
<br />
https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_25<br />
(Almost Mirrored)<br />
<br />
See here for a video solution:<br />
<br />
https://youtu.be/4XkA0DwuqYk<br />
<br />
==See Also==<br />
<br />
{{AIME box|year=2020|n=I|before=First Problem|num-a=2}}<br />
{{MAA Notice}}</div>Sachi2019https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_23&diff=1145072019 AMC 8 Problems/Problem 232020-01-08T21:48:35Z<p>Sachi2019: /* Solution 1 */</p>
<hr />
<div>==Problem 23==<br />
After Euclid High School's last basketball game, it was determined that <math>\frac{1}{4}</math> of the team's points were scored by Alexa and <math>\frac{2}{7}</math> were scored by Brittany. Chelsea scored <math>15</math> points. None of the other <math>7</math> team members scored more than <math>2</math> points What was the total number of points scored by the other <math>7</math> team members?<br />
<br />
<math>\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14</math><br />
<br />
==Solution 1==<br />
Since <math>\frac{\text{total points}}{4}</math> and <math>\frac{2(\text{total points})}{7}</math> are integers, we have <math>28 | \text{total points}</math>. We see that the number of points scored by the other team members is less than or equal to <math>14</math> and greater than or equal to <math>0</math>. We let the total number of points be <math>t</math> and the total number of points scored by the other team members, which means that <math>\frac{t}{4} + \frac{2t}{7} + 15 + x = t \quad \implies \quad 0 \le \frac{13t}{28} - 15 = x \le 14</math>, which means <math>15 \le \frac{13t}{28} \le 29</math>. The only value of <math>t</math> that satisfies all conditions listed is <math>56</math>, so <math>x=\boxed{\textbf{(B)} 11}</math>. - juliankuang (lol im smart.)<br />
<br />
==Solution 2==<br />
Starting from the above equation <math>\frac{t}{4}+\frac{2t}{7} + 15 + x = t</math> where <math>t</math> is the total number of points scored and <math>x\le 14</math> is the number of points scored by the remaining 7 team members, we can simplify to obtain the Diophantine equation <math>x+15 = \frac{13}{28}t</math>, or <math>28x+28\cdot 15=13t</math>. Since <math>t</math> is necessarily divisible by 28, let <math>t=28u</math> where <math>u \ge 0</math> and divide by 28 to obtain <math>x + 15 = 13u</math>. Then it is easy to see <math>u=2</math> (<math>t=56</math>) is the only candidate, giving <math>x=\boxed{\textbf{(B)} 11}</math>. -scrabbler94<br />
<br />
==Solution 3==<br />
Fakesolve: We first start by setting the total number of points as <math>28</math>, since <math>\text{lcm}(4,7) = 28</math>. However, we see that this does not work since we surpass the number of points just with the information given (<math>28\cdot\frac{1}{4}+28\cdot\frac{2}{7} + 15 > 28</math>). Next, we assume that the total number of points scored is <math>56</math>. With this, we have that Alexa, Brittany, and Chelsea score: <math>56\cdot\frac{1}{4}+56\cdot\frac{2}{7} + 15 = 45</math>, and thus, the other seven players would have scored a total of <math>56-45 = \boxed{\textbf{(B)} 11}</math> (We see that this works since we could have <math>4</math> of them score <math>2</math> points, and the other <math>3</math> of them score <math>1</math> point) -aops5234<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=22|num-a=24}}<br />
<br />
{{MAA Notice}}</div>Sachi2019https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_23&diff=1145062019 AMC 8 Problems/Problem 232020-01-08T21:47:32Z<p>Sachi2019: /* Solution 1 */</p>
<hr />
<div>==Problem 23==<br />
After Euclid High School's last basketball game, it was determined that <math>\frac{1}{4}</math> of the team's points were scored by Alexa and <math>\frac{2}{7}</math> were scored by Brittany. Chelsea scored <math>15</math> points. None of the other <math>7</math> team members scored more than <math>2</math> points What was the total number of points scored by the other <math>7</math> team members?<br />
<br />
<math>\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14</math><br />
<br />
==Solution 1==<br />
Since <math>\frac{\text{total points}}{4}</math> and <math>\frac{2(\text{total points})}{7}</math> are integers, we have <math>28 | \text{total points}</math>. We see that the number of points scored by the other team members is less than or equal to <math>14</math> and greater than or equal to <math>0</math>. We let the total number of points be <math>t</math> and the total number of points scored by the other team members, which means that <math>\frac{t}{4} + \frac{2t}{7} + 15 + x = t \quad \implies \quad 0 \le \frac{13t}{28} - 15 = x \le 14</math>, which means <math>15 \le \frac{13t}{28} \le 29</math>. The only value of <math>t</math> that satisfies all conditions listed is <math>56</math>, so <math>x=\boxed{\textbf{(B)} 11}</math>. - juliankuang (i mean wahta efa solutinaef)<br />
<br />
==Solution 2==<br />
Starting from the above equation <math>\frac{t}{4}+\frac{2t}{7} + 15 + x = t</math> where <math>t</math> is the total number of points scored and <math>x\le 14</math> is the number of points scored by the remaining 7 team members, we can simplify to obtain the Diophantine equation <math>x+15 = \frac{13}{28}t</math>, or <math>28x+28\cdot 15=13t</math>. Since <math>t</math> is necessarily divisible by 28, let <math>t=28u</math> where <math>u \ge 0</math> and divide by 28 to obtain <math>x + 15 = 13u</math>. Then it is easy to see <math>u=2</math> (<math>t=56</math>) is the only candidate, giving <math>x=\boxed{\textbf{(B)} 11}</math>. -scrabbler94<br />
<br />
==Solution 3==<br />
Fakesolve: We first start by setting the total number of points as <math>28</math>, since <math>\text{lcm}(4,7) = 28</math>. However, we see that this does not work since we surpass the number of points just with the information given (<math>28\cdot\frac{1}{4}+28\cdot\frac{2}{7} + 15 > 28</math>). Next, we assume that the total number of points scored is <math>56</math>. With this, we have that Alexa, Brittany, and Chelsea score: <math>56\cdot\frac{1}{4}+56\cdot\frac{2}{7} + 15 = 45</math>, and thus, the other seven players would have scored a total of <math>56-45 = \boxed{\textbf{(B)} 11}</math> (We see that this works since we could have <math>4</math> of them score <math>2</math> points, and the other <math>3</math> of them score <math>1</math> point) -aops5234<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=22|num-a=24}}<br />
<br />
{{MAA Notice}}</div>Sachi2019