https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Sachi2019&feedformat=atom AoPS Wiki - User contributions [en] 2021-09-17T01:06:47Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=User:Sachi2019&diff=124984 User:Sachi2019 2020-06-11T17:15:37Z <p>Sachi2019: Created page with &quot;[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]&quot;</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> Sachi2019 https://artofproblemsolving.com/wiki/index.php?title=Angle_Bisector_Theorem&diff=121621 Angle Bisector Theorem 2020-04-24T17:51:02Z <p>Sachi2019: /* Proof */</p> <hr /> <div>{{WotWAnnounce|week=June 6-12}}<br /> <br /> == Introduction ==<br /> The '''Angle Bisector Theorem''' states that given [[triangle]] &lt;math&gt;\triangle ABC&lt;/math&gt; and [[angle bisector]] AD, where D is on side BC, then &lt;math&gt; \frac cm = \frac bn &lt;/math&gt;. It follows that &lt;math&gt; \frac cb = \frac mn &lt;/math&gt;. Likewise, the [[converse]] of this theorem holds as well.<br /> <br /> <br /> Further by combining with [[Stewart's Theorem]] it can be shown that &lt;math&gt;AD^2 = b\cdot c - m \cdot n&lt;/math&gt;<br /> <br /> &lt;asy&gt; 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(&quot;$A$&quot;,A,(1,1));label(&quot;$B$&quot;,B,(-1,-1));label(&quot;$C$&quot;,C,(1,-1));label(&quot;$D$&quot;,D,(0,-1)); dot(A^^B^^C^^D,blue);label(&quot;$b$&quot;,(A+C)/2,(1,0));label(&quot;$c$&quot;,(A+B)/2,(0,1));label(&quot;$m$&quot;,(B+D)/2,(0,-1));label(&quot;$n$&quot;,(D+C)/2,(0,-1)); &lt;/asy&gt;<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 &lt;math&gt; BD = 2, CD = 5,&lt;/math&gt; and &lt;math&gt; AB + AC = 10 &lt;/math&gt;, find AB and AC.&lt;br&gt; '''''Solution:''''' By the angle bisector theorem, &lt;math&gt; \frac{AB}2 = \frac{AC}5&lt;/math&gt; or &lt;math&gt; AB = \frac 25 AC &lt;/math&gt;. Plugging this into &lt;math&gt; AB + AC = 10 &lt;/math&gt; and solving for AC gives &lt;math&gt; AC = \frac{50}7&lt;/math&gt;. We can plug this back in to find &lt;math&gt; AB = \frac{20}7 &lt;/math&gt;.<br /> # In triangle ABC, let P be a point on BC and let &lt;math&gt; AB = 20, AC = 10, BP = \frac{20\sqrt{3}}3, CP = \frac{10\sqrt{3}}3 &lt;/math&gt;. Find the value of &lt;math&gt; m\angle BAP - m\angle CAP &lt;/math&gt;. &lt;br&gt; '''''Solution:''''' First, we notice that &lt;math&gt; \frac{AB}{BP}=\frac{AC}{CP} &lt;/math&gt;. 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> Sachi2019 https://artofproblemsolving.com/wiki/index.php?title=Angle_Bisector_Theorem&diff=121620 Angle Bisector Theorem 2020-04-24T17:50:46Z <p>Sachi2019: /* Proof */</p> <hr /> <div>{{WotWAnnounce|week=June 6-12}}<br /> <br /> == Introduction ==<br /> The '''Angle Bisector Theorem''' states that given [[triangle]] &lt;math&gt;\triangle ABC&lt;/math&gt; and [[angle bisector]] AD, where D is on side BC, then &lt;math&gt; \frac cm = \frac bn &lt;/math&gt;. It follows that &lt;math&gt; \frac cb = \frac mn &lt;/math&gt;. Likewise, the [[converse]] of this theorem holds as well.<br /> <br /> <br /> Further by combining with [[Stewart's Theorem]] it can be shown that &lt;math&gt;AD^2 = b\cdot c - m \cdot n&lt;/math&gt;<br /> <br /> &lt;asy&gt; 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(&quot;$A$&quot;,A,(1,1));label(&quot;$B$&quot;,B,(-1,-1));label(&quot;$C$&quot;,C,(1,-1));label(&quot;$D$&quot;,D,(0,-1)); dot(A^^B^^C^^D,blue);label(&quot;$b$&quot;,(A+C)/2,(1,0));label(&quot;$c$&quot;,(A+B)/2,(0,1));label(&quot;$m$&quot;,(B+D)/2,(0,-1));label(&quot;$n$&quot;,(D+C)/2,(0,-1)); &lt;/asy&gt;<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 &lt;math&gt; BD = 2, CD = 5,&lt;/math&gt; and &lt;math&gt; AB + AC = 10 &lt;/math&gt;, find AB and AC.&lt;br&gt; '''''Solution:''''' By the angle bisector theorem, &lt;math&gt; \frac{AB}2 = \frac{AC}5&lt;/math&gt; or &lt;math&gt; AB = \frac 25 AC &lt;/math&gt;. Plugging this into &lt;math&gt; AB + AC = 10 &lt;/math&gt; and solving for AC gives &lt;math&gt; AC = \frac{50}7&lt;/math&gt;. We can plug this back in to find &lt;math&gt; AB = \frac{20}7 &lt;/math&gt;.<br /> # In triangle ABC, let P be a point on BC and let &lt;math&gt; AB = 20, AC = 10, BP = \frac{20\sqrt{3}}3, CP = \frac{10\sqrt{3}}3 &lt;/math&gt;. Find the value of &lt;math&gt; m\angle BAP - m\angle CAP &lt;/math&gt;. &lt;br&gt; '''''Solution:''''' First, we notice that &lt;math&gt; \frac{AB}{BP}=\frac{AC}{CP} &lt;/math&gt;. 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> Sachi2019 https://artofproblemsolving.com/wiki/index.php?title=Angle_Bisector_Theorem&diff=121619 Angle Bisector Theorem 2020-04-24T17:49:42Z <p>Sachi2019: /* Proof */</p> <hr /> <div>{{WotWAnnounce|week=June 6-12}}<br /> <br /> == Introduction ==<br /> The '''Angle Bisector Theorem''' states that given [[triangle]] &lt;math&gt;\triangle ABC&lt;/math&gt; and [[angle bisector]] AD, where D is on side BC, then &lt;math&gt; \frac cm = \frac bn &lt;/math&gt;. It follows that &lt;math&gt; \frac cb = \frac mn &lt;/math&gt;. Likewise, the [[converse]] of this theorem holds as well.<br /> <br /> <br /> Further by combining with [[Stewart's Theorem]] it can be shown that &lt;math&gt;AD^2 = b\cdot c - m \cdot n&lt;/math&gt;<br /> <br /> &lt;asy&gt; 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(&quot;$A$&quot;,A,(1,1));label(&quot;$B$&quot;,B,(-1,-1));label(&quot;$C$&quot;,C,(1,-1));label(&quot;$D$&quot;,D,(0,-1)); dot(A^^B^^C^^D,blue);label(&quot;$b$&quot;,(A+C)/2,(1,0));label(&quot;$c$&quot;,(A+B)/2,(0,1));label(&quot;$m$&quot;,(B+D)/2,(0,-1));label(&quot;$n$&quot;,(D+C)/2,(0,-1)); &lt;/asy&gt;<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 &lt;math&gt; BD = 2, CD = 5,&lt;/math&gt; and &lt;math&gt; AB + AC = 10 &lt;/math&gt;, find AB and AC.&lt;br&gt; '''''Solution:''''' By the angle bisector theorem, &lt;math&gt; \frac{AB}2 = \frac{AC}5&lt;/math&gt; or &lt;math&gt; AB = \frac 25 AC &lt;/math&gt;. Plugging this into &lt;math&gt; AB + AC = 10 &lt;/math&gt; and solving for AC gives &lt;math&gt; AC = \frac{50}7&lt;/math&gt;. We can plug this back in to find &lt;math&gt; AB = \frac{20}7 &lt;/math&gt;.<br /> # In triangle ABC, let P be a point on BC and let &lt;math&gt; AB = 20, AC = 10, BP = \frac{20\sqrt{3}}3, CP = \frac{10\sqrt{3}}3 &lt;/math&gt;. Find the value of &lt;math&gt; m\angle BAP - m\angle CAP &lt;/math&gt;. &lt;br&gt; '''''Solution:''''' First, we notice that &lt;math&gt; \frac{AB}{BP}=\frac{AC}{CP} &lt;/math&gt;. 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> Sachi2019 https://artofproblemsolving.com/wiki/index.php?title=Angle_Bisector_Theorem&diff=121618 Angle Bisector Theorem 2020-04-24T17:49:22Z <p>Sachi2019: /* Proof */</p> <hr /> <div>{{WotWAnnounce|week=June 6-12}}<br /> <br /> == Introduction ==<br /> The '''Angle Bisector Theorem''' states that given [[triangle]] &lt;math&gt;\triangle ABC&lt;/math&gt; and [[angle bisector]] AD, where D is on side BC, then &lt;math&gt; \frac cm = \frac bn &lt;/math&gt;. It follows that &lt;math&gt; \frac cb = \frac mn &lt;/math&gt;. Likewise, the [[converse]] of this theorem holds as well.<br /> <br /> <br /> Further by combining with [[Stewart's Theorem]] it can be shown that &lt;math&gt;AD^2 = b\cdot c - m \cdot n&lt;/math&gt;<br /> <br /> &lt;asy&gt; 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(&quot;$A$&quot;,A,(1,1));label(&quot;$B$&quot;,B,(-1,-1));label(&quot;$C$&quot;,C,(1,-1));label(&quot;$D$&quot;,D,(0,-1)); dot(A^^B^^C^^D,blue);label(&quot;$b$&quot;,(A+C)/2,(1,0));label(&quot;$c$&quot;,(A+B)/2,(0,1));label(&quot;$m$&quot;,(B+D)/2,(0,-1));label(&quot;$n$&quot;,(D+C)/2,(0,-1)); &lt;/asy&gt;<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 &lt;math&gt; BD = 2, CD = 5,&lt;/math&gt; and &lt;math&gt; AB + AC = 10 &lt;/math&gt;, find AB and AC.&lt;br&gt; '''''Solution:''''' By the angle bisector theorem, &lt;math&gt; \frac{AB}2 = \frac{AC}5&lt;/math&gt; or &lt;math&gt; AB = \frac 25 AC &lt;/math&gt;. Plugging this into &lt;math&gt; AB + AC = 10 &lt;/math&gt; and solving for AC gives &lt;math&gt; AC = \frac{50}7&lt;/math&gt;. We can plug this back in to find &lt;math&gt; AB = \frac{20}7 &lt;/math&gt;.<br /> # In triangle ABC, let P be a point on BC and let &lt;math&gt; AB = 20, AC = 10, BP = \frac{20\sqrt{3}}3, CP = \frac{10\sqrt{3}}3 &lt;/math&gt;. Find the value of &lt;math&gt; m\angle BAP - m\angle CAP &lt;/math&gt;. &lt;br&gt; '''''Solution:''''' First, we notice that &lt;math&gt; \frac{AB}{BP}=\frac{AC}{CP} &lt;/math&gt;. 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> Sachi2019 https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_2&diff=120661 2020 AIME I Problems/Problem 2 2020-04-06T22:57:45Z <p>Sachi2019: /* Solution */</p> <hr /> <div><br /> == Problem ==<br /> There is a unique positive real number &lt;math&gt;x&lt;/math&gt; such that the three numbers &lt;math&gt;\log_8{2x}&lt;/math&gt;, &lt;math&gt;\log_4{x}&lt;/math&gt;, and &lt;math&gt;\log_2{x}&lt;/math&gt;, in that order, form a geometric progression with positive common ratio. The number &lt;math&gt;x&lt;/math&gt; can be written as &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> == Solution ==<br /> Since these form a geometric series, &lt;math&gt;\frac{\log_2{x}}{\log_4{x}}&lt;/math&gt; is the common ratio. Rewriting this, we get &lt;math&gt;\frac{\log_x{4}}{\log_x{2}} = \log_2{4} = 2&lt;/math&gt; by base change formula. Therefore, the common ratio is 2. Now &lt;math&gt;\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}&lt;/math&gt;<br /> <br /> &lt;math&gt;\implies -\frac{1}{6}\log_2{x} = \frac{2}{3} \implies \log_2{x} = -4 \implies x = \frac{1}{16}&lt;/math&gt;. Therefore, &lt;math&gt;1 + 16 = \boxed{017}&lt;/math&gt;.<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 &lt;math&gt;x=2^y&lt;/math&gt;, we can obtain three terms of a geometric sequence through logarithm properties. The three terms are &lt;cmath&gt;\frac{y+1}{3}, \frac{y}{2}, y.&lt;/cmath&gt; 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: &lt;cmath&gt;\frac{y^2+y}{3} = \frac{y^2}{4},&lt;/cmath&gt; which can be solved to reveal &lt;math&gt;y = -4&lt;/math&gt;. Therefore, &lt;math&gt;x = 2^{-4} = \frac{1}{16}&lt;/math&gt;, so our answer is &lt;math&gt;\boxed{017}&lt;/math&gt;.<br /> <br /> -molocyxu<br /> <br /> ==Solution 3==<br /> Let &lt;math&gt;r&lt;/math&gt; be the common ratio. We have &lt;cmath&gt; r = \frac{\log_4{(x)}}{\log_8{(2x)}} = \frac{\log_2{(x)}}{\log_4{(x)}}&lt;/cmath&gt;<br /> Hence we obtain &lt;cmath&gt; (\log_4{(x)})(\log_4{(x)}) = (\log_8{(2x)})(\log_2{(x)})&lt;/cmath&gt;<br /> Ideally we change everything to base &lt;math&gt;64&lt;/math&gt; and we can get: &lt;cmath&gt; (\log_{64}{(x^3)})(\log_{64}{(x^3)}) = (\log_{64}{(x^6)})(\log_{64}{(4x^2)})&lt;/cmath&gt;<br /> Now divide to get: &lt;cmath&gt;\frac{\log_{64}{(x^3)}}{\log_{64}{(4x^2)}} = \frac{\log_{64}{(x^6)}}{\log_{64}{(x^3)}}&lt;/cmath&gt;<br /> By change-of-base we obtain: &lt;cmath&gt;\log_{(4x^2)}{(x^3)} = \log_{(x^3)}{(x^6)} = 2&lt;/cmath&gt;<br /> Hence &lt;math&gt;(4x^2)^2 = x^3 \rightarrow 16x^4 = x^3 \rightarrow x = \frac{1}{16}&lt;/math&gt; and we have &lt;math&gt;1+16 = \boxed{017}&lt;/math&gt; as desired.<br /> <br /> ~skyscraper<br /> <br /> ==Solution 4 (Exponents &gt; Logarithms)==<br /> Let &lt;math&gt;r&lt;/math&gt; be the common ratio, and let &lt;math&gt;a&lt;/math&gt; be the starting term (&lt;math&gt;a=\log_{8}{(2x)}&lt;/math&gt;). We then have: &lt;cmath&gt;\log_{8}{(2x)}=a, \log_{4}{(x)}=ar, \log_{2}{(x)}=ar^2&lt;/cmath&gt; Rearranging these equations gives: &lt;cmath&gt;8^a=2x, 4^{ar}=x, 2^{ar^2}=x&lt;/cmath&gt;<br /> Deal with the last two equations first: Setting them equal gives: &lt;cmath&gt;4^{ar}=2^{ar^2} \Rightarrow 2^{2ar}=2^{ar^2}&lt;/cmath&gt; Using LTE results in: &lt;cmath&gt;2ar=ar^2 \Rightarrow r=2&lt;/cmath&gt; Using this value of &lt;math&gt;r&lt;/math&gt;, substitute into the first and second equations (or the first and third, it doesn't really matter) to get: &lt;cmath&gt;8^a=2x, 4^{2a}=x&lt;/cmath&gt; Changing these to a common base gives: &lt;cmath&gt;2^{3a}=2x, 2^{4a}=x&lt;/cmath&gt; Dividing the first equation by 2 on both sides yields: &lt;cmath&gt;2^{3a-1}=x&lt;/cmath&gt; Setting these equations equal to each other and applying LTE again gives: &lt;cmath&gt;3a-1=4a \Rightarrow a=-1&lt;/cmath&gt; Substituting this back into the first equation gives: &lt;cmath&gt;8^{-1}=2x \Rightarrow 2x=\frac{1}{8} \Rightarrow x=\frac{1}{16}&lt;/cmath&gt; Therefore, &lt;math&gt;m+n=1+16=\boxed{017}&lt;/math&gt;<br /> <br /> ~IAmTheHazard<br /> <br /> ==Solution 5==<br /> <br /> We can relate the logarithms as follows:<br /> <br /> &lt;cmath&gt;\frac{\log_4{x}}{\log_8{(2x)}}=\frac{\log_2{x}}{\log_4{x}}&lt;/cmath&gt;<br /> &lt;cmath&gt;\log_8{(2x)}\log_2{x}=\log_4{x}\log_4{x}&lt;/cmath&gt;<br /> <br /> Now we can convert all logarithm bases to &lt;math&gt;2&lt;/math&gt; using the identity &lt;math&gt;\log_a{b}=\log_{a^c}{b^c}&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;\log_2{\sqrt{2x}}\log_2{x}=\log_2{\sqrt{x}}\log_2{\sqrt{x}}&lt;/cmath&gt;<br /> <br /> We can solve for &lt;math&gt;x&lt;/math&gt; as follows:<br /> <br /> &lt;cmath&gt;\frac{1}{3}\log_2{(2x)}\log_2{x}=\frac{1}{4}\log_2{x}\log_2{x}&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{1}{3}\log_2{(2x)}=\frac{1}{4}\log_2{x}&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{1}{3}\log_2{2}+\frac{1}{3}\log_2{x}=\frac{1}{4}\log_2{x}&lt;/cmath&gt;<br /> We get &lt;math&gt;x=\frac{1}{16}&lt;/math&gt;. Verifying that the common ratio is positive, we find the answer of &lt;math&gt;\boxed{017}&lt;/math&gt;.<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 &lt;math&gt;\frac{1+\log_2{x}}{3}&lt;/math&gt; and &lt;math&gt;\frac{1}{2}\log_2{x}&lt;/math&gt;, respectively. Therefore:<br /> &lt;cmath&gt;\frac{1}{2}\log_2{x}=\sqrt{\left(\frac{1+\log_2{x}}{3}\right)\left(\log_2{x}\right)}&lt;/cmath&gt;<br /> Let &lt;math&gt;n=\log_2{x}&lt;/math&gt;. We can rewrite the expression as:<br /> &lt;cmath&gt;\frac{n}{2}=\sqrt{\frac{n(n+1)}{3}}&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{n^2}{4}=\frac{n(n+1)}{3}&lt;/cmath&gt;<br /> &lt;cmath&gt;4n(n+1)=3n^2&lt;/cmath&gt;<br /> &lt;cmath&gt;4n^2+4n=3n^2&lt;/cmath&gt;<br /> &lt;cmath&gt;n^2+4n=0&lt;/cmath&gt;<br /> &lt;cmath&gt;n(n+4)=0&lt;/cmath&gt;<br /> &lt;cmath&gt;n=0 \text{ and } -4&lt;/cmath&gt;<br /> Zero does not work in this case, so we consider &lt;math&gt;n=-4&lt;/math&gt;: &lt;math&gt;\log_2{x}=-4 \rightarrow x=\frac{1}{16}&lt;/math&gt;. Therefore, &lt;math&gt;1+16=\boxed{017}&lt;/math&gt;.<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> Sachi2019 https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_2&diff=120660 2020 AIME I Problems/Problem 2 2020-04-06T22:56:38Z <p>Sachi2019: /* Solution 6 */</p> <hr /> <div><br /> == Problem ==<br /> There is a unique positive real number &lt;math&gt;x&lt;/math&gt; such that the three numbers &lt;math&gt;\log_8{2x}&lt;/math&gt;, &lt;math&gt;\log_4{x}&lt;/math&gt;, and &lt;math&gt;\log_2{x}&lt;/math&gt;, in that order, form a geometric progression with positive common ratio. The number &lt;math&gt;x&lt;/math&gt; can be written as &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> == Solution ==<br /> Since these form a geometric series, &lt;math&gt;\frac{\log_2{x}}{\log_4{x}}&lt;/math&gt; is the common ratio. Rewriting this, we get &lt;math&gt;\frac{\log_x{4}}{\log_x{2}} = \log_2{4} = 2&lt;/math&gt; by base change formula. Therefore, the common ratio is 2. Now &lt;math&gt;\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}&lt;/math&gt;<br /> <br /> &lt;math&gt;\implies -\frac{1}{6}\log_2{x} = \frac{2}{3} \implies \log_2{x} = -4 \implies x = \frac{1}{16}&lt;/math&gt;. Therefore, &lt;math&gt;1 + 16 = \boxed{017}&lt;/math&gt;.<br /> <br /> ~ JHawk0224<br /> <br /> ==Solution 2==<br /> If we set &lt;math&gt;x=2^y&lt;/math&gt;, we can obtain three terms of a geometric sequence through logarithm properties. The three terms are &lt;cmath&gt;\frac{y+1}{3}, \frac{y}{2}, y.&lt;/cmath&gt; 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: &lt;cmath&gt;\frac{y^2+y}{3} = \frac{y^2}{4},&lt;/cmath&gt; which can be solved to reveal &lt;math&gt;y = -4&lt;/math&gt;. Therefore, &lt;math&gt;x = 2^{-4} = \frac{1}{16}&lt;/math&gt;, so our answer is &lt;math&gt;\boxed{017}&lt;/math&gt;.<br /> <br /> -molocyxu<br /> <br /> ==Solution 3==<br /> Let &lt;math&gt;r&lt;/math&gt; be the common ratio. We have &lt;cmath&gt; r = \frac{\log_4{(x)}}{\log_8{(2x)}} = \frac{\log_2{(x)}}{\log_4{(x)}}&lt;/cmath&gt;<br /> Hence we obtain &lt;cmath&gt; (\log_4{(x)})(\log_4{(x)}) = (\log_8{(2x)})(\log_2{(x)})&lt;/cmath&gt;<br /> Ideally we change everything to base &lt;math&gt;64&lt;/math&gt; and we can get: &lt;cmath&gt; (\log_{64}{(x^3)})(\log_{64}{(x^3)}) = (\log_{64}{(x^6)})(\log_{64}{(4x^2)})&lt;/cmath&gt;<br /> Now divide to get: &lt;cmath&gt;\frac{\log_{64}{(x^3)}}{\log_{64}{(4x^2)}} = \frac{\log_{64}{(x^6)}}{\log_{64}{(x^3)}}&lt;/cmath&gt;<br /> By change-of-base we obtain: &lt;cmath&gt;\log_{(4x^2)}{(x^3)} = \log_{(x^3)}{(x^6)} = 2&lt;/cmath&gt;<br /> Hence &lt;math&gt;(4x^2)^2 = x^3 \rightarrow 16x^4 = x^3 \rightarrow x = \frac{1}{16}&lt;/math&gt; and we have &lt;math&gt;1+16 = \boxed{017}&lt;/math&gt; as desired.<br /> <br /> ~skyscraper<br /> <br /> ==Solution 4 (Exponents &gt; Logarithms)==<br /> Let &lt;math&gt;r&lt;/math&gt; be the common ratio, and let &lt;math&gt;a&lt;/math&gt; be the starting term (&lt;math&gt;a=\log_{8}{(2x)}&lt;/math&gt;). We then have: &lt;cmath&gt;\log_{8}{(2x)}=a, \log_{4}{(x)}=ar, \log_{2}{(x)}=ar^2&lt;/cmath&gt; Rearranging these equations gives: &lt;cmath&gt;8^a=2x, 4^{ar}=x, 2^{ar^2}=x&lt;/cmath&gt;<br /> Deal with the last two equations first: Setting them equal gives: &lt;cmath&gt;4^{ar}=2^{ar^2} \Rightarrow 2^{2ar}=2^{ar^2}&lt;/cmath&gt; Using LTE results in: &lt;cmath&gt;2ar=ar^2 \Rightarrow r=2&lt;/cmath&gt; Using this value of &lt;math&gt;r&lt;/math&gt;, substitute into the first and second equations (or the first and third, it doesn't really matter) to get: &lt;cmath&gt;8^a=2x, 4^{2a}=x&lt;/cmath&gt; Changing these to a common base gives: &lt;cmath&gt;2^{3a}=2x, 2^{4a}=x&lt;/cmath&gt; Dividing the first equation by 2 on both sides yields: &lt;cmath&gt;2^{3a-1}=x&lt;/cmath&gt; Setting these equations equal to each other and applying LTE again gives: &lt;cmath&gt;3a-1=4a \Rightarrow a=-1&lt;/cmath&gt; Substituting this back into the first equation gives: &lt;cmath&gt;8^{-1}=2x \Rightarrow 2x=\frac{1}{8} \Rightarrow x=\frac{1}{16}&lt;/cmath&gt; Therefore, &lt;math&gt;m+n=1+16=\boxed{017}&lt;/math&gt;<br /> <br /> ~IAmTheHazard<br /> <br /> ==Solution 5==<br /> <br /> We can relate the logarithms as follows:<br /> <br /> &lt;cmath&gt;\frac{\log_4{x}}{\log_8{(2x)}}=\frac{\log_2{x}}{\log_4{x}}&lt;/cmath&gt;<br /> &lt;cmath&gt;\log_8{(2x)}\log_2{x}=\log_4{x}\log_4{x}&lt;/cmath&gt;<br /> <br /> Now we can convert all logarithm bases to &lt;math&gt;2&lt;/math&gt; using the identity &lt;math&gt;\log_a{b}=\log_{a^c}{b^c}&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;\log_2{\sqrt{2x}}\log_2{x}=\log_2{\sqrt{x}}\log_2{\sqrt{x}}&lt;/cmath&gt;<br /> <br /> We can solve for &lt;math&gt;x&lt;/math&gt; as follows:<br /> <br /> &lt;cmath&gt;\frac{1}{3}\log_2{(2x)}\log_2{x}=\frac{1}{4}\log_2{x}\log_2{x}&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{1}{3}\log_2{(2x)}=\frac{1}{4}\log_2{x}&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{1}{3}\log_2{2}+\frac{1}{3}\log_2{x}=\frac{1}{4}\log_2{x}&lt;/cmath&gt;<br /> We get &lt;math&gt;x=\frac{1}{16}&lt;/math&gt;. Verifying that the common ratio is positive, we find the answer of &lt;math&gt;\boxed{017}&lt;/math&gt;.<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 &lt;math&gt;\frac{1+\log_2{x}}{3}&lt;/math&gt; and &lt;math&gt;\frac{1}{2}\log_2{x}&lt;/math&gt;, respectively. Therefore:<br /> &lt;cmath&gt;\frac{1}{2}\log_2{x}=\sqrt{\left(\frac{1+\log_2{x}}{3}\right)\left(\log_2{x}\right)}&lt;/cmath&gt;<br /> Let &lt;math&gt;n=\log_2{x}&lt;/math&gt;. We can rewrite the expression as:<br /> &lt;cmath&gt;\frac{n}{2}=\sqrt{\frac{n(n+1)}{3}}&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{n^2}{4}=\frac{n(n+1)}{3}&lt;/cmath&gt;<br /> &lt;cmath&gt;4n(n+1)=3n^2&lt;/cmath&gt;<br /> &lt;cmath&gt;4n^2+4n=3n^2&lt;/cmath&gt;<br /> &lt;cmath&gt;n^2+4n=0&lt;/cmath&gt;<br /> &lt;cmath&gt;n(n+4)=0&lt;/cmath&gt;<br /> &lt;cmath&gt;n=0 \text{ and } -4&lt;/cmath&gt;<br /> Zero does not work in this case, so we consider &lt;math&gt;n=-4&lt;/math&gt;: &lt;math&gt;\log_2{x}=-4 \rightarrow x=\frac{1}{16}&lt;/math&gt;. Therefore, &lt;math&gt;1+16=\boxed{017}&lt;/math&gt;.<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> Sachi2019 https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_1&diff=120659 2020 AIME I Problems/Problem 1 2020-04-06T22:55:28Z <p>Sachi2019: /* Solution 2 */</p> <hr /> <div><br /> == Problem ==<br /> In &lt;math&gt;\triangle ABC&lt;/math&gt; with &lt;math&gt;AB=AC,&lt;/math&gt; point &lt;math&gt;D&lt;/math&gt; lies strictly between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; on side &lt;math&gt;\overline{AC},&lt;/math&gt; and point &lt;math&gt;E&lt;/math&gt; lies strictly between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; on side &lt;math&gt;\overline{AB}&lt;/math&gt; such that &lt;math&gt;AE=ED=DB=BC.&lt;/math&gt; The degree measure of &lt;math&gt;\angle ABC&lt;/math&gt; is &lt;math&gt;\tfrac{m}{n},&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n.&lt;/math&gt;<br /> <br /> == Solution 1==<br /> &lt;asy&gt;<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(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, E);<br /> label(&quot;$C$&quot;, C, W);<br /> label(&quot;$D$&quot;, D, W);<br /> label(&quot;$E$&quot;, F, E);<br /> &lt;/asy&gt;<br /> <br /> If we set &lt;math&gt;\angle{BAC}&lt;/math&gt; to &lt;math&gt;x&lt;/math&gt;, we can find all other angles through these two properties:<br /> 1. Angles in a triangle sum to &lt;math&gt;180^{\circ}&lt;/math&gt;.<br /> 2. The base angles of an isoceles triangle are congruent.<br /> <br /> Now we angle chase. &lt;math&gt;\angle{ADE}=\angle{EAD}=x&lt;/math&gt;, &lt;math&gt;\angle{AED} = 180-2x&lt;/math&gt;, &lt;math&gt;\angle{BED}=\angle{EBD}=2x&lt;/math&gt;, &lt;math&gt;\angle{EDB} = 180-4x&lt;/math&gt;, &lt;math&gt;\angle{BDC} = \angle{BCD} = 3x&lt;/math&gt;, &lt;math&gt;\angle{CBD} = 180-6x&lt;/math&gt;. Since &lt;math&gt;AB = AC&lt;/math&gt; as given by the problem, &lt;math&gt;\angle{ABC} = \angle{ACB}&lt;/math&gt;, so &lt;math&gt;180-4x=3x&lt;/math&gt;. Therefore, &lt;math&gt;x = 180/7^{\circ}&lt;/math&gt;, and our desired angle is &lt;cmath&gt;180-4\left(\frac{180}{7}\right) = \frac{540}{7}&lt;/cmath&gt; for an answer of &lt;math&gt;\boxed{547}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;\angle{BAC}&lt;/math&gt; be &lt;math&gt;x&lt;/math&gt; in degrees. &lt;math&gt;\angle{ADE}=x&lt;/math&gt;.<br /> By Exterior Angle Theorem on triangle &lt;math&gt;AED&lt;/math&gt;, &lt;math&gt;\angle{BED}=2x&lt;/math&gt;.<br /> By Exterior Angle Theorem on triangle &lt;math&gt;ADB&lt;/math&gt;, &lt;math&gt;\angle{BDC}=3x&lt;/math&gt;.<br /> This tells us &lt;math&gt;\angle{BCA}=\angle{ABC}=3x&lt;/math&gt; and &lt;math&gt;3x+3x+x=180&lt;/math&gt;.<br /> Thus &lt;math&gt;x=\frac{180}{7}&lt;/math&gt; and we want &lt;math&gt;\angle{ABC}=3x=\frac{540}{7}&lt;/math&gt; to get an answer of &lt;math&gt;\boxed{547}&lt;/math&gt;.<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> Sachi2019 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_23&diff=114507 2019 AMC 8 Problems/Problem 23 2020-01-08T21:48:35Z <p>Sachi2019: /* Solution 1 */</p> <hr /> <div>==Problem 23==<br /> After Euclid High School's last basketball game, it was determined that &lt;math&gt;\frac{1}{4}&lt;/math&gt; of the team's points were scored by Alexa and &lt;math&gt;\frac{2}{7}&lt;/math&gt; were scored by Brittany. Chelsea scored &lt;math&gt;15&lt;/math&gt; points. None of the other &lt;math&gt;7&lt;/math&gt; team members scored more than &lt;math&gt;2&lt;/math&gt; points What was the total number of points scored by the other &lt;math&gt;7&lt;/math&gt; team members?<br /> <br /> &lt;math&gt;\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Since &lt;math&gt;\frac{\text{total points}}{4}&lt;/math&gt; and &lt;math&gt;\frac{2(\text{total points})}{7}&lt;/math&gt; are integers, we have &lt;math&gt;28 | \text{total points}&lt;/math&gt;. We see that the number of points scored by the other team members is less than or equal to &lt;math&gt;14&lt;/math&gt; and greater than or equal to &lt;math&gt;0&lt;/math&gt;. We let the total number of points be &lt;math&gt;t&lt;/math&gt; and the total number of points scored by the other team members, which means that &lt;math&gt;\frac{t}{4} + \frac{2t}{7} + 15 + x = t \quad \implies \quad 0 \le \frac{13t}{28} - 15 = x \le 14&lt;/math&gt;, which means &lt;math&gt;15 \le \frac{13t}{28} \le 29&lt;/math&gt;. The only value of &lt;math&gt;t&lt;/math&gt; that satisfies all conditions listed is &lt;math&gt;56&lt;/math&gt;, so &lt;math&gt;x=\boxed{\textbf{(B)} 11}&lt;/math&gt;. - juliankuang (lol im smart.)<br /> <br /> ==Solution 2==<br /> Starting from the above equation &lt;math&gt;\frac{t}{4}+\frac{2t}{7} + 15 + x = t&lt;/math&gt; where &lt;math&gt;t&lt;/math&gt; is the total number of points scored and &lt;math&gt;x\le 14&lt;/math&gt; is the number of points scored by the remaining 7 team members, we can simplify to obtain the Diophantine equation &lt;math&gt;x+15 = \frac{13}{28}t&lt;/math&gt;, or &lt;math&gt;28x+28\cdot 15=13t&lt;/math&gt;. Since &lt;math&gt;t&lt;/math&gt; is necessarily divisible by 28, let &lt;math&gt;t=28u&lt;/math&gt; where &lt;math&gt;u \ge 0&lt;/math&gt; and divide by 28 to obtain &lt;math&gt;x + 15 = 13u&lt;/math&gt;. Then it is easy to see &lt;math&gt;u=2&lt;/math&gt; (&lt;math&gt;t=56&lt;/math&gt;) is the only candidate, giving &lt;math&gt;x=\boxed{\textbf{(B)} 11}&lt;/math&gt;. -scrabbler94<br /> <br /> ==Solution 3==<br /> Fakesolve: We first start by setting the total number of points as &lt;math&gt;28&lt;/math&gt;, since &lt;math&gt;\text{lcm}(4,7) = 28&lt;/math&gt;. However, we see that this does not work since we surpass the number of points just with the information given (&lt;math&gt;28\cdot\frac{1}{4}+28\cdot\frac{2}{7} + 15 &gt; 28&lt;/math&gt;). Next, we assume that the total number of points scored is &lt;math&gt;56&lt;/math&gt;. With this, we have that Alexa, Brittany, and Chelsea score: &lt;math&gt;56\cdot\frac{1}{4}+56\cdot\frac{2}{7} + 15 = 45&lt;/math&gt;, and thus, the other seven players would have scored a total of &lt;math&gt;56-45 = \boxed{\textbf{(B)} 11}&lt;/math&gt; (We see that this works since we could have &lt;math&gt;4&lt;/math&gt; of them score &lt;math&gt;2&lt;/math&gt; points, and the other &lt;math&gt;3&lt;/math&gt; of them score &lt;math&gt;1&lt;/math&gt; point) -aops5234<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=22|num-a=24}}<br /> <br /> {{MAA Notice}}</div> Sachi2019 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_23&diff=114506 2019 AMC 8 Problems/Problem 23 2020-01-08T21:47:32Z <p>Sachi2019: /* Solution 1 */</p> <hr /> <div>==Problem 23==<br /> After Euclid High School's last basketball game, it was determined that &lt;math&gt;\frac{1}{4}&lt;/math&gt; of the team's points were scored by Alexa and &lt;math&gt;\frac{2}{7}&lt;/math&gt; were scored by Brittany. Chelsea scored &lt;math&gt;15&lt;/math&gt; points. None of the other &lt;math&gt;7&lt;/math&gt; team members scored more than &lt;math&gt;2&lt;/math&gt; points What was the total number of points scored by the other &lt;math&gt;7&lt;/math&gt; team members?<br /> <br /> &lt;math&gt;\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Since &lt;math&gt;\frac{\text{total points}}{4}&lt;/math&gt; and &lt;math&gt;\frac{2(\text{total points})}{7}&lt;/math&gt; are integers, we have &lt;math&gt;28 | \text{total points}&lt;/math&gt;. We see that the number of points scored by the other team members is less than or equal to &lt;math&gt;14&lt;/math&gt; and greater than or equal to &lt;math&gt;0&lt;/math&gt;. We let the total number of points be &lt;math&gt;t&lt;/math&gt; and the total number of points scored by the other team members, which means that &lt;math&gt;\frac{t}{4} + \frac{2t}{7} + 15 + x = t \quad \implies \quad 0 \le \frac{13t}{28} - 15 = x \le 14&lt;/math&gt;, which means &lt;math&gt;15 \le \frac{13t}{28} \le 29&lt;/math&gt;. The only value of &lt;math&gt;t&lt;/math&gt; that satisfies all conditions listed is &lt;math&gt;56&lt;/math&gt;, so &lt;math&gt;x=\boxed{\textbf{(B)} 11}&lt;/math&gt;. - juliankuang (i mean wahta efa solutinaef)<br /> <br /> ==Solution 2==<br /> Starting from the above equation &lt;math&gt;\frac{t}{4}+\frac{2t}{7} + 15 + x = t&lt;/math&gt; where &lt;math&gt;t&lt;/math&gt; is the total number of points scored and &lt;math&gt;x\le 14&lt;/math&gt; is the number of points scored by the remaining 7 team members, we can simplify to obtain the Diophantine equation &lt;math&gt;x+15 = \frac{13}{28}t&lt;/math&gt;, or &lt;math&gt;28x+28\cdot 15=13t&lt;/math&gt;. Since &lt;math&gt;t&lt;/math&gt; is necessarily divisible by 28, let &lt;math&gt;t=28u&lt;/math&gt; where &lt;math&gt;u \ge 0&lt;/math&gt; and divide by 28 to obtain &lt;math&gt;x + 15 = 13u&lt;/math&gt;. Then it is easy to see &lt;math&gt;u=2&lt;/math&gt; (&lt;math&gt;t=56&lt;/math&gt;) is the only candidate, giving &lt;math&gt;x=\boxed{\textbf{(B)} 11}&lt;/math&gt;. -scrabbler94<br /> <br /> ==Solution 3==<br /> Fakesolve: We first start by setting the total number of points as &lt;math&gt;28&lt;/math&gt;, since &lt;math&gt;\text{lcm}(4,7) = 28&lt;/math&gt;. However, we see that this does not work since we surpass the number of points just with the information given (&lt;math&gt;28\cdot\frac{1}{4}+28\cdot\frac{2}{7} + 15 &gt; 28&lt;/math&gt;). Next, we assume that the total number of points scored is &lt;math&gt;56&lt;/math&gt;. With this, we have that Alexa, Brittany, and Chelsea score: &lt;math&gt;56\cdot\frac{1}{4}+56\cdot\frac{2}{7} + 15 = 45&lt;/math&gt;, and thus, the other seven players would have scored a total of &lt;math&gt;56-45 = \boxed{\textbf{(B)} 11}&lt;/math&gt; (We see that this works since we could have &lt;math&gt;4&lt;/math&gt; of them score &lt;math&gt;2&lt;/math&gt; points, and the other &lt;math&gt;3&lt;/math&gt; of them score &lt;math&gt;1&lt;/math&gt; point) -aops5234<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=22|num-a=24}}<br /> <br /> {{MAA Notice}}</div> Sachi2019