https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Jj+ca888&feedformat=atom AoPS Wiki - User contributions [en] 2022-01-27T21:51:48Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2019_USAJMO_Problems/Problem_3&diff=106988 2019 USAJMO Problems/Problem 3 2019-06-25T23:58:31Z <p>Jj ca888: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> &lt;math&gt;(*)&lt;/math&gt; Let &lt;math&gt;ABCD&lt;/math&gt; be a cyclic quadrilateral satisfying &lt;math&gt;AD^2+BC^2=AB^2&lt;/math&gt;. The diagonals of &lt;math&gt;ABCD&lt;/math&gt; intersect at &lt;math&gt;E&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be a point on side &lt;math&gt;\overline{AB}&lt;/math&gt; satisfying &lt;math&gt;\angle APD=\angle BPC&lt;/math&gt;. Show that line &lt;math&gt;PE&lt;/math&gt; bisects &lt;math&gt;\overline{CD}&lt;/math&gt;.<br /> <br /> <br /> Let &lt;math&gt;PE \cap DC = M&lt;/math&gt;. Also, let &lt;math&gt;N&lt;/math&gt; be the midpoint of &lt;math&gt;AB&lt;/math&gt;.<br /> <br /> Note that only one point &lt;math&gt;P&lt;/math&gt; satisfies the given angle condition. With this in mind, construct &lt;math&gt;P'&lt;/math&gt; with the following properties:<br /> <br /> &lt;math&gt;AP' \cdot AB = AD^2 \quad \text{and} \quad BP' \cdot AB = CD^2&lt;/math&gt;<br /> <br /> <br /> <br /> Claim:&lt;math&gt;P = P'&lt;/math&gt;<br /> <br /> Proof:<br /> <br /> The conditions imply the similarities &lt;math&gt;ADP \sim ABD&lt;/math&gt; and &lt;math&gt;BCP \sim BAC&lt;/math&gt; whence &lt;math&gt;\measuredangle APD = \measuredangle BDA = \measuredangle BCA = \measuredangle CPB&lt;/math&gt; as desired. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> Claim: &lt;math&gt;PE&lt;/math&gt; is a symmedian in &lt;math&gt;AEB&lt;/math&gt;<br /> <br /> Proof:<br /> <br /> We have <br /> <br /> &lt;cmath&gt;AP \cdot AB = AD^2 \iff AB^2 \cdot AP = AD^2 \cdot AB&lt;/cmath&gt;<br /> &lt;cmath&gt;\iff \left( \frac{AB}{AD} \right)^2 = \frac{AB}{AP}&lt;/cmath&gt;<br /> &lt;cmath&gt;\iff \left( \frac{AB}{AD} \right)^2 - 1 = \frac{AB}{AP} - 1&lt;/cmath&gt;<br /> &lt;cmath&gt;\iff \frac{AB^2 - AD^2}{AD^2} = \frac{BP}{AP}&lt;/cmath&gt;<br /> &lt;cmath&gt;\iff \left(\frac{BC}{AD} \right)^2 = \left(\frac{BE}{AE} \right)^2 = \frac{BP}{AP} &lt;/cmath&gt;<br /> <br /> as desired. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> Since &lt;math&gt;P&lt;/math&gt; is the isogonal conjugate of &lt;math&gt;N&lt;/math&gt;, &lt;math&gt;\measuredangle PEA = \measuredangle MEC = \measuredangle BEN&lt;/math&gt;. However &lt;math&gt;\measuredangle MEC = \measuredangle BEN&lt;/math&gt; implies that &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;CD&lt;/math&gt; from similar triangles, so we are done. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> ~sriraamster<br /> <br /> ==Solution 2==<br /> <br /> By monoticity, we can see that the point &lt;math&gt;P&lt;/math&gt; is unique. Therefore, if we find another point &lt;math&gt;P'&lt;/math&gt; with all the same properties as &lt;math&gt;P&lt;/math&gt;, then &lt;math&gt;P = P'&lt;/math&gt;<br /> <br /> Part 1) Let &lt;math&gt;N&lt;/math&gt; be a point on &lt;math&gt;\overline{AB}&lt;/math&gt; such that &lt;math&gt;AN\cdot AB = AD^2&lt;/math&gt;, and &lt;math&gt;BN \cdot AB = BC^2&lt;/math&gt;. Obviously &lt;math&gt;N&lt;/math&gt; exists because adding the two equations gives &lt;math&gt;AN\cdot AB + BN \cdot AB = AD^2 + BC^2 = AB^2&lt;/math&gt;, which is the problem statement. Notice that converse PoP gives&lt;cmath&gt;AN\cdot AB = AD^2 \implies \bigtriangleup ADN \sim \bigtriangleup ABD&lt;/cmath&gt;&lt;cmath&gt;BN\cdot AB = AD^2 \implies \bigtriangleup CBN \sim \bigtriangleup ABC&lt;/cmath&gt;Therefore, &lt;math&gt;\angle AND = \angle ADB = \angle ACB = \angle CNB&lt;/math&gt;, so &lt;math&gt;N&lt;/math&gt; does indeed satisfy all the conditions &lt;math&gt;P&lt;/math&gt; does, so &lt;math&gt;N = P&lt;/math&gt;. Hence, &lt;math&gt;\bigtriangleup ADP \sim \bigtriangleup ABD&lt;/math&gt; and &lt;math&gt;\bigtriangleup CBP \sim\bigtriangleup ABC&lt;/math&gt;.<br /> <br /> Part 2) Define &lt;math&gt;G&lt;/math&gt; as the midpoint of &lt;math&gt;CD&lt;/math&gt;. Furthermore, create a point &lt;math&gt;X&lt;/math&gt; such that &lt;math&gt;DX || DC&lt;/math&gt; and &lt;math&gt;CX || ED&lt;/math&gt;. Obviously &lt;math&gt;XCED&lt;/math&gt; must be a parallelogram. Now we set up for Jacobi's. The problem already gives us that &lt;math&gt;\angle APD = \angle CPB&lt;/math&gt;, which is good for starters. Furthermore, &lt;math&gt;\bigtriangleup ADP \sim \bigtriangleup ABD&lt;/math&gt; tells us that&lt;cmath&gt;\angle ADP = \angle ABD = \angle ACD = \angle XDC&lt;/cmath&gt;This gives us our second needed angle equivalence. Lastly, &lt;math&gt;\bigtriangleup CBP \sim\bigtriangleup ABC&lt;/math&gt; will give&lt;cmath&gt;\angle BCP = \angle BAC = \angle BDC = \angle XCD&lt;/cmath&gt;which is our last necessary angle equivalence to apply Jacobi's. Finally, applying Jacobi's tells us that &lt;math&gt;AC&lt;/math&gt;, &lt;math&gt;BD&lt;/math&gt;, and &lt;math&gt;XP&lt;/math&gt; are concurrent &lt;math&gt;\implies&lt;/math&gt; &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;P&lt;/math&gt; collinear. Additionally, since parallelogram diagonals bisect each other, &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;G&lt;/math&gt;, and &lt;math&gt;E&lt;/math&gt; are collinear, so finally we obtain that &lt;math&gt;P&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;G&lt;/math&gt; are collinear, as desired.<br /> <br /> -jj_ca888<br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2019|num-b=2|num-a=4}}</div> Jj ca888 https://artofproblemsolving.com/wiki/index.php?title=2019_USAJMO_Problems/Problem_3&diff=106987 2019 USAJMO Problems/Problem 3 2019-06-25T23:56:44Z <p>Jj ca888: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> &lt;math&gt;(*)&lt;/math&gt; Let &lt;math&gt;ABCD&lt;/math&gt; be a cyclic quadrilateral satisfying &lt;math&gt;AD^2+BC^2=AB^2&lt;/math&gt;. The diagonals of &lt;math&gt;ABCD&lt;/math&gt; intersect at &lt;math&gt;E&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be a point on side &lt;math&gt;\overline{AB}&lt;/math&gt; satisfying &lt;math&gt;\angle APD=\angle BPC&lt;/math&gt;. Show that line &lt;math&gt;PE&lt;/math&gt; bisects &lt;math&gt;\overline{CD}&lt;/math&gt;.<br /> <br /> <br /> Let &lt;math&gt;PE \cap DC = M&lt;/math&gt;. Also, let &lt;math&gt;N&lt;/math&gt; be the midpoint of &lt;math&gt;AB&lt;/math&gt;.<br /> <br /> Note that only one point &lt;math&gt;P&lt;/math&gt; satisfies the given angle condition. With this in mind, construct &lt;math&gt;P'&lt;/math&gt; with the following properties:<br /> <br /> &lt;math&gt;AP' \cdot AB = AD^2 \quad \text{and} \quad BP' \cdot AB = CD^2&lt;/math&gt;<br /> <br /> <br /> <br /> Claim:&lt;math&gt;P = P'&lt;/math&gt;<br /> <br /> Proof:<br /> <br /> The conditions imply the similarities &lt;math&gt;ADP \sim ABD&lt;/math&gt; and &lt;math&gt;BCP \sim BAC&lt;/math&gt; whence &lt;math&gt;\measuredangle APD = \measuredangle BDA = \measuredangle BCA = \measuredangle CPB&lt;/math&gt; as desired. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> Claim: &lt;math&gt;PE&lt;/math&gt; is a symmedian in &lt;math&gt;AEB&lt;/math&gt;<br /> <br /> Proof:<br /> <br /> We have <br /> <br /> &lt;cmath&gt;AP \cdot AB = AD^2 \iff AB^2 \cdot AP = AD^2 \cdot AB&lt;/cmath&gt;<br /> &lt;cmath&gt;\iff \left( \frac{AB}{AD} \right)^2 = \frac{AB}{AP}&lt;/cmath&gt;<br /> &lt;cmath&gt;\iff \left( \frac{AB}{AD} \right)^2 - 1 = \frac{AB}{AP} - 1&lt;/cmath&gt;<br /> &lt;cmath&gt;\iff \frac{AB^2 - AD^2}{AD^2} = \frac{BP}{AP}&lt;/cmath&gt;<br /> &lt;cmath&gt;\iff \left(\frac{BC}{AD} \right)^2 = \left(\frac{BE}{AE} \right)^2 = \frac{BP}{AP} &lt;/cmath&gt;<br /> <br /> as desired. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> Since &lt;math&gt;P&lt;/math&gt; is the isogonal conjugate of &lt;math&gt;N&lt;/math&gt;, &lt;math&gt;\measuredangle PEA = \measuredangle MEC = \measuredangle BEN&lt;/math&gt;. However &lt;math&gt;\measuredangle MEC = \measuredangle BEN&lt;/math&gt; implies that &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;CD&lt;/math&gt; from similar triangles, so we are done. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> ~sriraamster<br /> <br /> ==Solution 2==<br /> <br /> By monoticity, we can see that the point &lt;math&gt;P&lt;/math&gt; is unique. Therefore, if we find another point &lt;math&gt;P'&lt;/math&gt; with all the same properties as &lt;math&gt;P&lt;/math&gt;, then &lt;math&gt;P = P'&lt;/math&gt;<br /> <br /> Part 1) Let &lt;math&gt;N&lt;/math&gt; be a point on &lt;math&gt;\overline{AB}&lt;/math&gt; such that &lt;math&gt;AN\cdot AB = AD^2&lt;/math&gt;, and &lt;math&gt;BN \cdot AB = BC^2&lt;/math&gt;. Obviously &lt;math&gt;N&lt;/math&gt; exists because adding the two equations gives &lt;math&gt;AN\cdot AB + BN \cdot AB = AD^2 + BC^2 = AB^2&lt;/math&gt;, which is the problem statement. Notice that converse PoP gives&lt;cmath&gt;AN\cdot AB = AD^2 \implies \bigtriangleup ADN \sim \bigtriangleup ABD&lt;/cmath&gt;&lt;cmath&gt;BN\cdot AB = AD^2 \implies \bigtriangleup CBN \sim \bigtriangleup ABC&lt;/cmath&gt;Therefore, &lt;math&gt;\angle AND = \angle ADB = \angle ACB = \angle CNB&lt;/math&gt;, so &lt;math&gt;N&lt;/math&gt; does indeed satisfy all the conditions &lt;math&gt;P&lt;/math&gt; does, so &lt;math&gt;N = P&lt;/math&gt;. Hence, &lt;math&gt;\bigtriangleup ADP \sim \bigtriangleup ABD&lt;/math&gt; and &lt;math&gt;\bigtriangleup CBP \sim\bigtriangleup ABC&lt;/math&gt;.<br /> <br /> Part 2) Define &lt;math&gt;G&lt;/math&gt; as the midpoint of &lt;math&gt;CD&lt;/math&gt;. Furthermore, create a point &lt;math&gt;X&lt;/math&gt; such that &lt;math&gt;DX || DC&lt;/math&gt; and &lt;math&gt;CX || ED&lt;/math&gt;. Obviously &lt;math&gt;XCED&lt;/math&gt; must be a parallelogram. Now we set up for Jacobi's. The problem already gives us that &lt;math&gt;\angle APD = \angle CPB&lt;/math&gt;, which is good for starters. Furthermore, &lt;math&gt;\bigtriangleup ADP \sim \bigtriangleup ABD&lt;/math&gt; tells us that&lt;cmath&gt;\angle ADP = \angle ABD = \angle ACD = \angle XDC&lt;/cmath&gt;This gives us our second needed angle equivalence. Lastly, &lt;math&gt;\bigtriangleup CBP \sim\bigtriangleup ABC&lt;/math&gt; will give&lt;cmath&gt;\angle BCP = \angle BAC = \angle BDC = \angle XCD&lt;/cmath&gt;which is our last necessary angle equivalence to apply Jacobi's. Finally, applying Jacobi's tells us that &lt;math&gt;AC&lt;/math&gt;, &lt;math&gt;BD&lt;/math&gt;, and &lt;math&gt;XP&lt;/math&gt; are concurrent &lt;math&gt;\implies&lt;/math&gt; &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;P&lt;/math&gt; collinear. Additionally, since parallelogram diagonals bisect each other, &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;G&lt;/math&gt;, and &lt;math&gt;E&lt;/math&gt; are collinear, so finally we obtain that &lt;math&gt;P&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;G&lt;/math&gt; are collinear, as desired.<br /> <br /> <br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2019|num-b=2|num-a=4}}</div> Jj ca888 https://artofproblemsolving.com/wiki/index.php?title=2019_USAJMO_Problems/Problem_3&diff=106986 2019 USAJMO Problems/Problem 3 2019-06-25T23:56:27Z <p>Jj ca888: /* Solution */</p> <hr /> <div>==Problem==<br /> &lt;math&gt;(*)&lt;/math&gt; Let &lt;math&gt;ABCD&lt;/math&gt; be a cyclic quadrilateral satisfying &lt;math&gt;AD^2+BC^2=AB^2&lt;/math&gt;. The diagonals of &lt;math&gt;ABCD&lt;/math&gt; intersect at &lt;math&gt;E&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be a point on side &lt;math&gt;\overline{AB}&lt;/math&gt; satisfying &lt;math&gt;\angle APD=\angle BPC&lt;/math&gt;. Show that line &lt;math&gt;PE&lt;/math&gt; bisects &lt;math&gt;\overline{CD}&lt;/math&gt;.<br /> <br /> <br /> Let &lt;math&gt;PE \cap DC = M&lt;/math&gt;. Also, let &lt;math&gt;N&lt;/math&gt; be the midpoint of &lt;math&gt;AB&lt;/math&gt;.<br /> <br /> Note that only one point &lt;math&gt;P&lt;/math&gt; satisfies the given angle condition. With this in mind, construct &lt;math&gt;P'&lt;/math&gt; with the following properties:<br /> <br /> &lt;math&gt;AP' \cdot AB = AD^2 \quad \text{and} \quad BP' \cdot AB = CD^2&lt;/math&gt;<br /> <br /> <br /> <br /> Claim:&lt;math&gt;P = P'&lt;/math&gt;<br /> <br /> Proof:<br /> <br /> The conditions imply the similarities &lt;math&gt;ADP \sim ABD&lt;/math&gt; and &lt;math&gt;BCP \sim BAC&lt;/math&gt; whence &lt;math&gt;\measuredangle APD = \measuredangle BDA = \measuredangle BCA = \measuredangle CPB&lt;/math&gt; as desired. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> Claim: &lt;math&gt;PE&lt;/math&gt; is a symmedian in &lt;math&gt;AEB&lt;/math&gt;<br /> <br /> Proof:<br /> <br /> We have <br /> <br /> &lt;cmath&gt;AP \cdot AB = AD^2 \iff AB^2 \cdot AP = AD^2 \cdot AB&lt;/cmath&gt;<br /> &lt;cmath&gt;\iff \left( \frac{AB}{AD} \right)^2 = \frac{AB}{AP}&lt;/cmath&gt;<br /> &lt;cmath&gt;\iff \left( \frac{AB}{AD} \right)^2 - 1 = \frac{AB}{AP} - 1&lt;/cmath&gt;<br /> &lt;cmath&gt;\iff \frac{AB^2 - AD^2}{AD^2} = \frac{BP}{AP}&lt;/cmath&gt;<br /> &lt;cmath&gt;\iff \left(\frac{BC}{AD} \right)^2 = \left(\frac{BE}{AE} \right)^2 = \frac{BP}{AP} &lt;/cmath&gt;<br /> <br /> as desired. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> Since &lt;math&gt;P&lt;/math&gt; is the isogonal conjugate of &lt;math&gt;N&lt;/math&gt;, &lt;math&gt;\measuredangle PEA = \measuredangle MEC = \measuredangle BEN&lt;/math&gt;. However &lt;math&gt;\measuredangle MEC = \measuredangle BEN&lt;/math&gt; implies that &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;CD&lt;/math&gt; from similar triangles, so we are done. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> ~sriraamster<br /> <br /> ==Solution 2==<br /> <br /> By monoticity, we can see that the point &lt;math&gt;P&lt;/math&gt; is unique. Therefore, if we find another point &lt;math&gt;P'&lt;/math&gt; with all the same properties as &lt;math&gt;P&lt;/math&gt;, then &lt;math&gt;P = P'&lt;/math&gt;<br /> <br /> Part 1) Let &lt;math&gt;N&lt;/math&gt; be a point on &lt;math&gt;\overline{AB}&lt;/math&gt; such that &lt;math&gt;AN\cdot AB = AD^2&lt;/math&gt;, and &lt;math&gt;BN \cdot AB = BC^2&lt;/math&gt;. Obviously &lt;math&gt;N&lt;/math&gt; exists because adding the two equations gives &lt;math&gt;AN\cdot AB + BN \cdot AB = AD^2 + BC^2 = AB^2&lt;/math&gt;, which is the problem statement. Notice that converse PoP gives&lt;cmath&gt;AN\cdot AB = AD^2 \implies \bigtriangleup ADN \sim \bigtriangleup ABD&lt;/cmath&gt;&lt;cmath&gt;BN\cdot AB = AD^2 \implies \bigtriangleup CBN \sim \bigtriangleup ABC&lt;/cmath&gt;Therefore, &lt;math&gt;\angle AND = \angle ADB = \angle ACB = \angle CNB&lt;/math&gt;, so &lt;math&gt;N&lt;/math&gt; does indeed satisfy all the conditions &lt;math&gt;P&lt;/math&gt; does, so &lt;math&gt;N = P&lt;/math&gt;. Hence, &lt;math&gt;\bigtriangleup ADP \sim \bigtriangleup ABD&lt;/math&gt; and &lt;math&gt;\bigtriangleup CBP \sim\bigtriangleup ABC&lt;/math&gt;.<br /> <br /> Part 2) Define &lt;math&gt;G&lt;/math&gt; as the midpoint of &lt;math&gt;CD&lt;/math&gt;. Furthermore, create a point &lt;math&gt;X&lt;/math&gt; such that &lt;math&gt;DX || DC&lt;/math&gt; and &lt;math&gt;CX || ED&lt;/math&gt;. Obviously &lt;math&gt;XCED&lt;/math&gt; must be a parallelogram. Now we set up for Jacobi's. The problem already gives us that &lt;math&gt;\angle APD = \angle CPB&lt;/math&gt;, which is good for starters. Furthermore, &lt;math&gt;\bigtriangleup ADP \sim \bigtriangleup ABD&lt;/math&gt; tells us that&lt;cmath&gt;\angle ADP = \angle ABD = \angle ACD = \angle XDC&lt;/cmath&gt;This gives us our second needed angle equivalence. Lastly, &lt;math&gt;\bigtriangleup CBP \sim\bigtriangleup ABC&lt;/math&gt; will give&lt;cmath&gt;\angle BCP = \angle BAC = \angle BDC = \angle XCD&lt;/cmath&gt;which is our last necessary angle equivalence to apply Jacobi's. Finally, applying Jacobi's tells us that &lt;math&gt;AC&lt;/math&gt;, &lt;math&gt;BD&lt;/math&gt;, and &lt;math&gt;XP&lt;/math&gt; are concurrent &lt;math&gt;\implies&lt;/math&gt; &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;P&lt;/math&gt; collinear. Additionally, since parallelogram diagonals bisect each other, &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;G&lt;/math&gt;, and &lt;math&gt;E&lt;/math&gt; are collinear, so finally we obtain that &lt;math&gt;P&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;G&lt;/math&gt; are collinear, as desired.<br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2019|num-b=2|num-a=4}}</div> Jj ca888 https://artofproblemsolving.com/wiki/index.php?title=User:Jj_ca888&diff=103846 User:Jj ca888 2019-02-26T05:10:37Z <p>Jj ca888: Created page with &quot;Andrew wen&quot;</p> <hr /> <div>Andrew wen</div> Jj ca888 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_15&diff=103832 2012 AIME II Problems/Problem 15 2019-02-26T01:22:50Z <p>Jj ca888: /* Solution Using Only Elementary Geometry Methods */</p> <hr /> <div>== Problem 15 ==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is inscribed in circle &lt;math&gt;\omega&lt;/math&gt; with &lt;math&gt;AB=5&lt;/math&gt;, &lt;math&gt;BC=7&lt;/math&gt;, and &lt;math&gt;AC=3&lt;/math&gt;. The bisector of angle &lt;math&gt;A&lt;/math&gt; meets side &lt;math&gt;\overline{BC}&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and circle &lt;math&gt;\omega&lt;/math&gt; at a second point &lt;math&gt;E&lt;/math&gt;. Let &lt;math&gt;\gamma&lt;/math&gt; be the circle with diameter &lt;math&gt;\overline{DE}&lt;/math&gt;. Circles &lt;math&gt;\omega&lt;/math&gt; and &lt;math&gt;\gamma&lt;/math&gt; meet at &lt;math&gt;E&lt;/math&gt; and a second point &lt;math&gt;F&lt;/math&gt;. Then &lt;math&gt;AF^2 = \frac mn&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 Using Only Elementary Geometry Methods ==<br /> Let &lt;math&gt;P&lt;/math&gt; be the unique point on circle &lt;math&gt;\omega&lt;/math&gt; such that &lt;math&gt;\bigtriangleup BCP&lt;/math&gt; is equilateral. Point &lt;math&gt;E&lt;/math&gt; must be the point on &lt;math&gt;\omega&lt;/math&gt; such that &lt;math&gt;\angle PED&lt;/math&gt; is right. We extend line &lt;math&gt;ED&lt;/math&gt; to hit circle &lt;math&gt;\omega&lt;/math&gt; again at &lt;math&gt;X&lt;/math&gt;.<br /> <br /> Claim 1: &lt;math&gt;X&lt;/math&gt; is the midpoint of minor arc &lt;math&gt;\overarc{BC}&lt;/math&gt;. <br /> Proof: Notice that &lt;math&gt;\angle PEX&lt;/math&gt; and &lt;math&gt;\angle PAX&lt;/math&gt; are subtended by the same arc, thus they are both equal to 90 degrees.<br /> <br /> Now it is obvious that &lt;math&gt;\bigtriangleup PED \sim \bigtriangleup ADX&lt;/math&gt;. After this, we length chase with similar triangles and Pythag, which yields &lt;math&gt;AP = 8&lt;/math&gt;, &lt;math&gt;EX = \frac{56}{\sqrt{57}}&lt;/math&gt;, &lt;math&gt;AX = \frac{2}{\sqrt{3}}&lt;/math&gt;, &lt;math&gt;PE = \frac{14}{\sqrt{19}}&lt;/math&gt;, and &lt;math&gt;PX = \frac{14}{\sqrt{3}}&lt;/math&gt;. We can tie this all together with Ptolemy's theorem for cyclic quadrilaterals, and we find that &lt;math&gt;AE = \frac{30}{\sqrt{19}}&lt;/math&gt;. Thus, &lt;math&gt;AE^2 = \frac{900}{19}&lt;/math&gt;, and the answer is &lt;math&gt;\boxed{919}&lt;/math&gt;<br /> -jj_ca888<br /> <br /> == Solution 1==<br /> Use the angle bisector theorem to find &lt;math&gt;CD=\frac{21}{8}&lt;/math&gt;, &lt;math&gt;BD=\frac{35}{8}&lt;/math&gt;, and use the Stewart's Theorem to find &lt;math&gt;AD=\frac{15}{8}&lt;/math&gt;. Use Power of the Point to find &lt;math&gt;DE=\frac{49}{8}&lt;/math&gt;, and so &lt;math&gt;AE=8&lt;/math&gt;. Use law of cosines to find &lt;math&gt;\angle CAD = \frac{\pi} {3}&lt;/math&gt;, hence &lt;math&gt;\angle BAD = \frac{\pi}{3}&lt;/math&gt; as well, and &lt;math&gt;\triangle BCE&lt;/math&gt; is equilateral, so &lt;math&gt;BC=CE=BE=7&lt;/math&gt;.<br /> <br /> I'm sure there is a more elegant solution from here, but instead we'll do some hairy law of cosines:<br /> <br /> &lt;math&gt;AE^2 = AF^2 + EF^2 - 2 \cdot AF \cdot EF \cdot \cos \angle AFE.&lt;/math&gt; (1)<br /> <br /> &lt;math&gt;AF^2 = AE^2 + EF^2 - 2 \cdot AE \cdot EF \cdot \cos \angle AEF.&lt;/math&gt; Adding these two and simplifying we get:<br /> <br /> &lt;math&gt;EF = AF \cdot \cos \angle AFE + AE \cdot \cos \angle AEF&lt;/math&gt; (2). Ah, but &lt;math&gt;\angle AFE = \angle ACE&lt;/math&gt; (since &lt;math&gt;F&lt;/math&gt; lies on &lt;math&gt;\omega&lt;/math&gt;), and we can find &lt;math&gt;cos \angle ACE&lt;/math&gt; using the law of cosines:<br /> <br /> &lt;math&gt;AE^2 = AC^2 + CE^2 - 2 \cdot AC \cdot CE \cdot \cos \angle ACE&lt;/math&gt;, and plugging in &lt;math&gt;AE = 8, AC = 3, BE = BC = 7,&lt;/math&gt; we get &lt;math&gt;\cos \angle ACE = -1/7 = \cos \angle AFE&lt;/math&gt;.<br /> <br /> Also, &lt;math&gt;\angle AEF = \angle DEF&lt;/math&gt;, and &lt;math&gt;\angle DFE = \pi/2&lt;/math&gt; (since &lt;math&gt;F&lt;/math&gt; is on the circle &lt;math&gt;\gamma&lt;/math&gt; with diameter &lt;math&gt;DE&lt;/math&gt;), so &lt;math&gt;\cos \angle AEF = EF/DE = 8 \cdot EF/49&lt;/math&gt;. <br /> <br /> Plugging in all our values into equation (2), we get:<br /> <br /> &lt;math&gt;EF = -\frac{AF}{7} + 8 \cdot \frac{8EF}{49}&lt;/math&gt;, or &lt;math&gt;EF = \frac{7}{15} \cdot AF&lt;/math&gt;.<br /> <br /> Finally, we plug this into equation (1), yielding:<br /> <br /> &lt;math&gt;8^2 = AF^2 + \frac{49}{225} \cdot AF^2 - 2 \cdot AF \cdot \frac{7AF}{15} \cdot \frac{-1}{7}&lt;/math&gt;. Thus,<br /> <br /> &lt;math&gt;64 = \frac{AF^2}{225} \cdot (225+49+30),&lt;/math&gt; or &lt;math&gt;AF^2 = \frac{900}{19}.&lt;/math&gt; The answer is &lt;math&gt;\boxed{919}&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> <br /> Let &lt;math&gt;a = BC&lt;/math&gt;, &lt;math&gt;b = CA&lt;/math&gt;, &lt;math&gt;c = AB&lt;/math&gt; for convenience. We claim that &lt;math&gt;AF&lt;/math&gt; is a symmedian. Indeed, let &lt;math&gt;M&lt;/math&gt; be the midpoint of segment &lt;math&gt;BC&lt;/math&gt;. Since &lt;math&gt;\angle EAB=\angle EAC&lt;/math&gt;, it follows that &lt;math&gt;EB = EC&lt;/math&gt; and consequently &lt;math&gt;EM\perp BC&lt;/math&gt;. Therefore, &lt;math&gt;M\in \gamma&lt;/math&gt;. Now let &lt;math&gt;G = FD\cap \omega&lt;/math&gt;. Since &lt;math&gt;EG&lt;/math&gt; is a diameter, &lt;math&gt;G&lt;/math&gt; lies on the perpendicular bisector of &lt;math&gt;BC&lt;/math&gt;; hence &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;G&lt;/math&gt; are collinear. From &lt;math&gt;\angle DAG = \angle DMG = 90&lt;/math&gt;, it immediately follows that quadrilateral &lt;math&gt;ADMG&lt;/math&gt; is cyclic. Therefore, &lt;math&gt;\angle MAD = \angle MGD=\angle EAF&lt;/math&gt;, implying that &lt;math&gt;AF&lt;/math&gt; is a symmedian, as claimed.<br /> <br /> The rest is standard; here's a quick way to finish. From above, quadrilateral &lt;math&gt;ABFC&lt;/math&gt; is harmonic, so &lt;math&gt;\dfrac{FB}{FC}=\dfrac{AB}{BC}=\dfrac{c}{a}&lt;/math&gt;. In conjunction with &lt;math&gt;\triangle ABF\sim\triangle AMC&lt;/math&gt;, it follows that &lt;math&gt;AF^2=\dfrac{b^2c^2}{AM^2}=\dfrac{4b^2c^2}{2b^2+2c^2-a^2}&lt;/math&gt;. (Notice that this holds for all triangles &lt;math&gt;ABC&lt;/math&gt;.) To finish, substitute &lt;math&gt;a = 7&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt;, &lt;math&gt;c=5&lt;/math&gt; to obtain &lt;math&gt;AF^2=\dfrac{900}{19}\implies 900+19=\boxed{919}&lt;/math&gt; as before.<br /> <br /> '''-Solution by thecmd999'''<br /> <br /> ==Solution 3==<br /> &lt;asy&gt;<br /> size(6cm);<br /> pair E,X,B,C,A,D,M,F,R,I;<br /> real z=sqrt(3)*14/3;<br /> real y=2*sqrt(3)/21;<br /> real x=224*sqrt(3)/57;<br /> E=(z,0);<br /> X=(0,0);<br /> D=(sqrt(3)*7/6,-7/8);<br /> M=(sqrt(3)*7/6,0);<br /> B=z/2*dir(60);<br /> C=z/2*dir(300);<br /> A=(y,-8/7);<br /> F=(x,-sqrt(3)*x/4);<br /> R=circumcenter(A,B,C);<br /> I=circumcenter(M,E,F);<br /> draw(E--X);<br /> draw(A--E);<br /> draw(A--B);<br /> draw(A--C);<br /> draw(B--C);<br /> draw(A--F);<br /> draw(X--F);<br /> draw(E--F);<br /> draw(circumcircle(A,B,C));<br /> draw(circumcircle(M,F,E));<br /> dot(D);<br /> dot(F);<br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(E);<br /> dot(X);<br /> dot(R);<br /> dot(I);<br /> label(&quot;$A$&quot;,A,dir(220));<br /> label(&quot;$B$&quot;,B,dir(110));<br /> label(&quot;$C$&quot;,C,dir(250));<br /> label(&quot;$D$&quot;,D,dir(60));<br /> label(&quot;$E$&quot;,E,dir(0));<br /> label(&quot;$F$&quot;,F,dir(315));<br /> label(&quot;$X$&quot;,X,dir(180));<br /> &lt;/asy&gt;<br /> First of all, use the [[Angle Bisector Theorem]] to find that &lt;math&gt;BD=35/8&lt;/math&gt; and &lt;math&gt;CD=21/8&lt;/math&gt;, and use [[Stewart's Theorem]] to find that &lt;math&gt;AD=15/8&lt;/math&gt;. Then use [[Power of a Point Theorem|Power of a Point]] to find that &lt;math&gt;DE=49/8&lt;/math&gt;. Then use the [[Circumradius|circumradius of a triangle]] formula to find that the length of the circumradius of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;\frac{7\sqrt{3}}{3}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;DE&lt;/math&gt; is the diameter of circle &lt;math&gt;\gamma&lt;/math&gt;, &lt;math&gt;\angle DFE&lt;/math&gt; is &lt;math&gt;90^\circ&lt;/math&gt;. Extending &lt;math&gt;DF&lt;/math&gt; to intersect circle &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;X&lt;/math&gt;, we find that &lt;math&gt;XE&lt;/math&gt; is the diameter of the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt; (since &lt;math&gt;\angle DFE&lt;/math&gt; is &lt;math&gt;90^\circ&lt;/math&gt;). Therefore, &lt;math&gt;XE=\frac{14\sqrt{3}}{3}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;EF=x&lt;/math&gt;, &lt;math&gt;XD=a&lt;/math&gt;, and &lt;math&gt;DF=b&lt;/math&gt;. Then, by the [[Pythagorean Theorem]],<br /> <br /> &lt;cmath&gt;x^2+b^2=\left(\frac{49}{8}\right)^2=\frac{2401}{64}&lt;/cmath&gt;<br /> <br /> and<br /> <br /> &lt;cmath&gt;x^2+(a+b)^2=\left(\frac{14\sqrt{3}}{3}\right)^2=\frac{196}{3}.&lt;/cmath&gt;<br /> <br /> Subtracting the first equation from the second, the &lt;math&gt;x^2&lt;/math&gt; term cancels out and we obtain:<br /> <br /> &lt;cmath&gt;(a+b)^2-b^2=\frac{196}{3}-\frac{2401}{64}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;a^2+2ab = \frac{5341}{192}.&lt;/cmath&gt;<br /> <br /> By Power of a Point, &lt;math&gt;ab=BD \cdot DC=735/64=2205/192&lt;/math&gt;, so<br /> <br /> &lt;cmath&gt;a^2+2 \cdot \frac{2205}{192}=\frac{5341}{192}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;a^2=\frac{931}{192}.&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;a=XD&lt;/math&gt;, &lt;math&gt;XD=\frac{7\sqrt{19}}{8\sqrt{3}}&lt;/math&gt;.<br /> <br /> Because &lt;math&gt;\angle EXF&lt;/math&gt; and &lt;math&gt;\angle EAF&lt;/math&gt; intercept the same arc in circle &lt;math&gt;\omega&lt;/math&gt; and the same goes for &lt;math&gt;\angle XFA&lt;/math&gt; and &lt;math&gt;\angle XEA&lt;/math&gt;, &lt;math&gt;\angle EXF\cong\angle EAF&lt;/math&gt; and &lt;math&gt;\angle XFA\cong\angle XEA&lt;/math&gt;. Therefore, &lt;math&gt;\triangle XDE\sim\triangle ADF&lt;/math&gt; by [[Similarity (geometry)|AA Similarity]]. Since side lengths in similar triangles are proportional,<br /> <br /> &lt;cmath&gt;\frac{AF}{\frac{15}{8}}=\frac{\frac{14\sqrt{3}}{3}}{\frac{7\sqrt{19}}{8\sqrt{3}}}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\frac{AF}{\frac{15}{8}}=\frac{16}{\sqrt{19}}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;AF \cdot \sqrt{19} = 30&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;AF = \frac{30}{\sqrt{19}}.&lt;/cmath&gt;<br /> <br /> However, the problem asks for &lt;math&gt;AF^2&lt;/math&gt;, so &lt;math&gt;AF^2 = \frac{900}{19}\implies 900 + 19 = \boxed{919}&lt;/math&gt;.<br /> <br /> '''-Solution by TheBoomBox77'''<br /> <br /> ==Solution 4==<br /> It can be verified with law of cosines that &lt;math&gt;\angle BAC=120^\circ.&lt;/math&gt; Also, &lt;math&gt;E&lt;/math&gt; is the midpoint of major arc &lt;math&gt;BC&lt;/math&gt; so &lt;math&gt;BE=CE,&lt;/math&gt; and &lt;math&gt;\angle BEC=60.&lt;/math&gt; Thus &lt;math&gt;CBE&lt;/math&gt; is equilateral. Notice now that &lt;math&gt;\angle BFC=\angle BFE= 60.&lt;/math&gt; But &lt;math&gt;\angle DFE=90&lt;/math&gt; so &lt;math&gt;FD&lt;/math&gt; bisects &lt;math&gt;\angle BFC.&lt;/math&gt; Thus, &lt;math&gt;\frac{BF}{CF}=\frac{BD}{CD}=\frac{BA}{CA}=\frac{5}{3}.&lt;/math&gt; <br /> <br /> Let &lt;math&gt;BF=5a, CF=3a.&lt;/math&gt; By law of cosines on &lt;math&gt;BFC&lt;/math&gt; we find &lt;math&gt;a\sqrt{5^2+3^2-5*3}=a\sqrt{19}=7.&lt;/math&gt; But by ptolemy on &lt;math&gt;BFCA&lt;/math&gt;, &lt;math&gt;15a+15a=7*AF,&lt;/math&gt; so &lt;math&gt;AF= \frac{30}{\sqrt{19}},&lt;/math&gt; so &lt;math&gt;AF^2=\frac{900}{19}&lt;/math&gt; and the answer is &lt;math&gt;900+19=\boxed{919}&lt;/math&gt;<br /> <br /> ~abacadaea<br /> <br /> == See Also ==<br /> {{AIME box|year=2012|n=II|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Jj ca888 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_15&diff=103831 2012 AIME II Problems/Problem 15 2019-02-26T01:09:03Z <p>Jj ca888: /* Solution Using Only Elementary Geometry Methods */</p> <hr /> <div>== Problem 15 ==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is inscribed in circle &lt;math&gt;\omega&lt;/math&gt; with &lt;math&gt;AB=5&lt;/math&gt;, &lt;math&gt;BC=7&lt;/math&gt;, and &lt;math&gt;AC=3&lt;/math&gt;. The bisector of angle &lt;math&gt;A&lt;/math&gt; meets side &lt;math&gt;\overline{BC}&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and circle &lt;math&gt;\omega&lt;/math&gt; at a second point &lt;math&gt;E&lt;/math&gt;. Let &lt;math&gt;\gamma&lt;/math&gt; be the circle with diameter &lt;math&gt;\overline{DE}&lt;/math&gt;. Circles &lt;math&gt;\omega&lt;/math&gt; and &lt;math&gt;\gamma&lt;/math&gt; meet at &lt;math&gt;E&lt;/math&gt; and a second point &lt;math&gt;F&lt;/math&gt;. Then &lt;math&gt;AF^2 = \frac mn&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 Using Only Elementary Geometry Methods ==<br /> Let &lt;math&gt;P&lt;/math&gt; be the unique point on circle &lt;math&gt;\omega&lt;/math&gt; such that &lt;math&gt;\bigtriangleup BCP&lt;/math&gt; is equilateral. Point &lt;math&gt;E&lt;/math&gt; must be the point on &lt;math&gt;\omega&lt;/math&gt; such that &lt;math&gt;\angle PED&lt;/math&gt; is right. We extend line &lt;math&gt;ED&lt;/math&gt; to hit circle &lt;math&gt;\omega&lt;/math&gt; again at &lt;math&gt;X&lt;/math&gt;.<br /> <br /> Claim 1: &lt;math&gt;X&lt;/math&gt; is the midpoint of minor arc &lt;math&gt;\overarc{BC}&lt;/math&gt;. <br /> Proof: Notice that &lt;math&gt;\angle PEX&lt;/math&gt; and &lt;math&gt;\angle PAX&lt;/math&gt; are subtended by the same arc, thus they are both equal to 90 degrees.<br /> <br /> Now it is obvious that &lt;math&gt;\bigtriangleup PED \sim \bigtriangleup ADX&lt;/math&gt;. After this, we length chase with similar triangles and Pythag, which yields &lt;math&gt;AP = 8&lt;/math&gt;, &lt;math&gt;EX = \frac{56}{\sqrt{57}}&lt;/math&gt;, &lt;math&gt;AX = \frac{2}{\sqrt{3}}&lt;/math&gt;, &lt;math&gt;PE = \frac{14}{\sqrt{19}}&lt;/math&gt;, and &lt;math&gt;PX = \frac{14}{\sqrt{3}}&lt;/math&gt;. We can tie this all together with Ptolemy's theorem for cyclic quadrilaterals, and we find that &lt;math&gt;AE = \frac{30}{\sqrt{19}}&lt;/math&gt;. Thus, &lt;math&gt;AE^2 = \frac{900}{19}&lt;/math&gt;, and the answer is &lt;math&gt;\boxed{919}&lt;/math&gt;<br /> <br /> == Solution 1==<br /> Use the angle bisector theorem to find &lt;math&gt;CD=\frac{21}{8}&lt;/math&gt;, &lt;math&gt;BD=\frac{35}{8}&lt;/math&gt;, and use the Stewart's Theorem to find &lt;math&gt;AD=\frac{15}{8}&lt;/math&gt;. Use Power of the Point to find &lt;math&gt;DE=\frac{49}{8}&lt;/math&gt;, and so &lt;math&gt;AE=8&lt;/math&gt;. Use law of cosines to find &lt;math&gt;\angle CAD = \frac{\pi} {3}&lt;/math&gt;, hence &lt;math&gt;\angle BAD = \frac{\pi}{3}&lt;/math&gt; as well, and &lt;math&gt;\triangle BCE&lt;/math&gt; is equilateral, so &lt;math&gt;BC=CE=BE=7&lt;/math&gt;.<br /> <br /> I'm sure there is a more elegant solution from here, but instead we'll do some hairy law of cosines:<br /> <br /> &lt;math&gt;AE^2 = AF^2 + EF^2 - 2 \cdot AF \cdot EF \cdot \cos \angle AFE.&lt;/math&gt; (1)<br /> <br /> &lt;math&gt;AF^2 = AE^2 + EF^2 - 2 \cdot AE \cdot EF \cdot \cos \angle AEF.&lt;/math&gt; Adding these two and simplifying we get:<br /> <br /> &lt;math&gt;EF = AF \cdot \cos \angle AFE + AE \cdot \cos \angle AEF&lt;/math&gt; (2). Ah, but &lt;math&gt;\angle AFE = \angle ACE&lt;/math&gt; (since &lt;math&gt;F&lt;/math&gt; lies on &lt;math&gt;\omega&lt;/math&gt;), and we can find &lt;math&gt;cos \angle ACE&lt;/math&gt; using the law of cosines:<br /> <br /> &lt;math&gt;AE^2 = AC^2 + CE^2 - 2 \cdot AC \cdot CE \cdot \cos \angle ACE&lt;/math&gt;, and plugging in &lt;math&gt;AE = 8, AC = 3, BE = BC = 7,&lt;/math&gt; we get &lt;math&gt;\cos \angle ACE = -1/7 = \cos \angle AFE&lt;/math&gt;.<br /> <br /> Also, &lt;math&gt;\angle AEF = \angle DEF&lt;/math&gt;, and &lt;math&gt;\angle DFE = \pi/2&lt;/math&gt; (since &lt;math&gt;F&lt;/math&gt; is on the circle &lt;math&gt;\gamma&lt;/math&gt; with diameter &lt;math&gt;DE&lt;/math&gt;), so &lt;math&gt;\cos \angle AEF = EF/DE = 8 \cdot EF/49&lt;/math&gt;. <br /> <br /> Plugging in all our values into equation (2), we get:<br /> <br /> &lt;math&gt;EF = -\frac{AF}{7} + 8 \cdot \frac{8EF}{49}&lt;/math&gt;, or &lt;math&gt;EF = \frac{7}{15} \cdot AF&lt;/math&gt;.<br /> <br /> Finally, we plug this into equation (1), yielding:<br /> <br /> &lt;math&gt;8^2 = AF^2 + \frac{49}{225} \cdot AF^2 - 2 \cdot AF \cdot \frac{7AF}{15} \cdot \frac{-1}{7}&lt;/math&gt;. Thus,<br /> <br /> &lt;math&gt;64 = \frac{AF^2}{225} \cdot (225+49+30),&lt;/math&gt; or &lt;math&gt;AF^2 = \frac{900}{19}.&lt;/math&gt; The answer is &lt;math&gt;\boxed{919}&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> <br /> Let &lt;math&gt;a = BC&lt;/math&gt;, &lt;math&gt;b = CA&lt;/math&gt;, &lt;math&gt;c = AB&lt;/math&gt; for convenience. We claim that &lt;math&gt;AF&lt;/math&gt; is a symmedian. Indeed, let &lt;math&gt;M&lt;/math&gt; be the midpoint of segment &lt;math&gt;BC&lt;/math&gt;. Since &lt;math&gt;\angle EAB=\angle EAC&lt;/math&gt;, it follows that &lt;math&gt;EB = EC&lt;/math&gt; and consequently &lt;math&gt;EM\perp BC&lt;/math&gt;. Therefore, &lt;math&gt;M\in \gamma&lt;/math&gt;. Now let &lt;math&gt;G = FD\cap \omega&lt;/math&gt;. Since &lt;math&gt;EG&lt;/math&gt; is a diameter, &lt;math&gt;G&lt;/math&gt; lies on the perpendicular bisector of &lt;math&gt;BC&lt;/math&gt;; hence &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;G&lt;/math&gt; are collinear. From &lt;math&gt;\angle DAG = \angle DMG = 90&lt;/math&gt;, it immediately follows that quadrilateral &lt;math&gt;ADMG&lt;/math&gt; is cyclic. Therefore, &lt;math&gt;\angle MAD = \angle MGD=\angle EAF&lt;/math&gt;, implying that &lt;math&gt;AF&lt;/math&gt; is a symmedian, as claimed.<br /> <br /> The rest is standard; here's a quick way to finish. From above, quadrilateral &lt;math&gt;ABFC&lt;/math&gt; is harmonic, so &lt;math&gt;\dfrac{FB}{FC}=\dfrac{AB}{BC}=\dfrac{c}{a}&lt;/math&gt;. In conjunction with &lt;math&gt;\triangle ABF\sim\triangle AMC&lt;/math&gt;, it follows that &lt;math&gt;AF^2=\dfrac{b^2c^2}{AM^2}=\dfrac{4b^2c^2}{2b^2+2c^2-a^2}&lt;/math&gt;. (Notice that this holds for all triangles &lt;math&gt;ABC&lt;/math&gt;.) To finish, substitute &lt;math&gt;a = 7&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt;, &lt;math&gt;c=5&lt;/math&gt; to obtain &lt;math&gt;AF^2=\dfrac{900}{19}\implies 900+19=\boxed{919}&lt;/math&gt; as before.<br /> <br /> '''-Solution by thecmd999'''<br /> <br /> ==Solution 3==<br /> &lt;asy&gt;<br /> size(6cm);<br /> pair E,X,B,C,A,D,M,F,R,I;<br /> real z=sqrt(3)*14/3;<br /> real y=2*sqrt(3)/21;<br /> real x=224*sqrt(3)/57;<br /> E=(z,0);<br /> X=(0,0);<br /> D=(sqrt(3)*7/6,-7/8);<br /> M=(sqrt(3)*7/6,0);<br /> B=z/2*dir(60);<br /> C=z/2*dir(300);<br /> A=(y,-8/7);<br /> F=(x,-sqrt(3)*x/4);<br /> R=circumcenter(A,B,C);<br /> I=circumcenter(M,E,F);<br /> draw(E--X);<br /> draw(A--E);<br /> draw(A--B);<br /> draw(A--C);<br /> draw(B--C);<br /> draw(A--F);<br /> draw(X--F);<br /> draw(E--F);<br /> draw(circumcircle(A,B,C));<br /> draw(circumcircle(M,F,E));<br /> dot(D);<br /> dot(F);<br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(E);<br /> dot(X);<br /> dot(R);<br /> dot(I);<br /> label(&quot;$A$&quot;,A,dir(220));<br /> label(&quot;$B$&quot;,B,dir(110));<br /> label(&quot;$C$&quot;,C,dir(250));<br /> label(&quot;$D$&quot;,D,dir(60));<br /> label(&quot;$E$&quot;,E,dir(0));<br /> label(&quot;$F$&quot;,F,dir(315));<br /> label(&quot;$X$&quot;,X,dir(180));<br /> &lt;/asy&gt;<br /> First of all, use the [[Angle Bisector Theorem]] to find that &lt;math&gt;BD=35/8&lt;/math&gt; and &lt;math&gt;CD=21/8&lt;/math&gt;, and use [[Stewart's Theorem]] to find that &lt;math&gt;AD=15/8&lt;/math&gt;. Then use [[Power of a Point Theorem|Power of a Point]] to find that &lt;math&gt;DE=49/8&lt;/math&gt;. Then use the [[Circumradius|circumradius of a triangle]] formula to find that the length of the circumradius of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;\frac{7\sqrt{3}}{3}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;DE&lt;/math&gt; is the diameter of circle &lt;math&gt;\gamma&lt;/math&gt;, &lt;math&gt;\angle DFE&lt;/math&gt; is &lt;math&gt;90^\circ&lt;/math&gt;. Extending &lt;math&gt;DF&lt;/math&gt; to intersect circle &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;X&lt;/math&gt;, we find that &lt;math&gt;XE&lt;/math&gt; is the diameter of the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt; (since &lt;math&gt;\angle DFE&lt;/math&gt; is &lt;math&gt;90^\circ&lt;/math&gt;). Therefore, &lt;math&gt;XE=\frac{14\sqrt{3}}{3}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;EF=x&lt;/math&gt;, &lt;math&gt;XD=a&lt;/math&gt;, and &lt;math&gt;DF=b&lt;/math&gt;. Then, by the [[Pythagorean Theorem]],<br /> <br /> &lt;cmath&gt;x^2+b^2=\left(\frac{49}{8}\right)^2=\frac{2401}{64}&lt;/cmath&gt;<br /> <br /> and<br /> <br /> &lt;cmath&gt;x^2+(a+b)^2=\left(\frac{14\sqrt{3}}{3}\right)^2=\frac{196}{3}.&lt;/cmath&gt;<br /> <br /> Subtracting the first equation from the second, the &lt;math&gt;x^2&lt;/math&gt; term cancels out and we obtain:<br /> <br /> &lt;cmath&gt;(a+b)^2-b^2=\frac{196}{3}-\frac{2401}{64}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;a^2+2ab = \frac{5341}{192}.&lt;/cmath&gt;<br /> <br /> By Power of a Point, &lt;math&gt;ab=BD \cdot DC=735/64=2205/192&lt;/math&gt;, so<br /> <br /> &lt;cmath&gt;a^2+2 \cdot \frac{2205}{192}=\frac{5341}{192}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;a^2=\frac{931}{192}.&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;a=XD&lt;/math&gt;, &lt;math&gt;XD=\frac{7\sqrt{19}}{8\sqrt{3}}&lt;/math&gt;.<br /> <br /> Because &lt;math&gt;\angle EXF&lt;/math&gt; and &lt;math&gt;\angle EAF&lt;/math&gt; intercept the same arc in circle &lt;math&gt;\omega&lt;/math&gt; and the same goes for &lt;math&gt;\angle XFA&lt;/math&gt; and &lt;math&gt;\angle XEA&lt;/math&gt;, &lt;math&gt;\angle EXF\cong\angle EAF&lt;/math&gt; and &lt;math&gt;\angle XFA\cong\angle XEA&lt;/math&gt;. Therefore, &lt;math&gt;\triangle XDE\sim\triangle ADF&lt;/math&gt; by [[Similarity (geometry)|AA Similarity]]. Since side lengths in similar triangles are proportional,<br /> <br /> &lt;cmath&gt;\frac{AF}{\frac{15}{8}}=\frac{\frac{14\sqrt{3}}{3}}{\frac{7\sqrt{19}}{8\sqrt{3}}}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\frac{AF}{\frac{15}{8}}=\frac{16}{\sqrt{19}}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;AF \cdot \sqrt{19} = 30&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;AF = \frac{30}{\sqrt{19}}.&lt;/cmath&gt;<br /> <br /> However, the problem asks for &lt;math&gt;AF^2&lt;/math&gt;, so &lt;math&gt;AF^2 = \frac{900}{19}\implies 900 + 19 = \boxed{919}&lt;/math&gt;.<br /> <br /> '''-Solution by TheBoomBox77'''<br /> <br /> ==Solution 4==<br /> It can be verified with law of cosines that &lt;math&gt;\angle BAC=120^\circ.&lt;/math&gt; Also, &lt;math&gt;E&lt;/math&gt; is the midpoint of major arc &lt;math&gt;BC&lt;/math&gt; so &lt;math&gt;BE=CE,&lt;/math&gt; and &lt;math&gt;\angle BEC=60.&lt;/math&gt; Thus &lt;math&gt;CBE&lt;/math&gt; is equilateral. Notice now that &lt;math&gt;\angle BFC=\angle BFE= 60.&lt;/math&gt; But &lt;math&gt;\angle DFE=90&lt;/math&gt; so &lt;math&gt;FD&lt;/math&gt; bisects &lt;math&gt;\angle BFC.&lt;/math&gt; Thus, &lt;math&gt;\frac{BF}{CF}=\frac{BD}{CD}=\frac{BA}{CA}=\frac{5}{3}.&lt;/math&gt; <br /> <br /> Let &lt;math&gt;BF=5a, CF=3a.&lt;/math&gt; By law of cosines on &lt;math&gt;BFC&lt;/math&gt; we find &lt;math&gt;a\sqrt{5^2+3^2-5*3}=a\sqrt{19}=7.&lt;/math&gt; But by ptolemy on &lt;math&gt;BFCA&lt;/math&gt;, &lt;math&gt;15a+15a=7*AF,&lt;/math&gt; so &lt;math&gt;AF= \frac{30}{\sqrt{19}},&lt;/math&gt; so &lt;math&gt;AF^2=\frac{900}{19}&lt;/math&gt; and the answer is &lt;math&gt;900+19=\boxed{919}&lt;/math&gt;<br /> <br /> ~abacadaea<br /> <br /> == See Also ==<br /> {{AIME box|year=2012|n=II|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Jj ca888 https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_15&diff=103830 2012 AIME II Problems/Problem 15 2019-02-26T01:08:27Z <p>Jj ca888: /* Solution 1 */</p> <hr /> <div>== Problem 15 ==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is inscribed in circle &lt;math&gt;\omega&lt;/math&gt; with &lt;math&gt;AB=5&lt;/math&gt;, &lt;math&gt;BC=7&lt;/math&gt;, and &lt;math&gt;AC=3&lt;/math&gt;. The bisector of angle &lt;math&gt;A&lt;/math&gt; meets side &lt;math&gt;\overline{BC}&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and circle &lt;math&gt;\omega&lt;/math&gt; at a second point &lt;math&gt;E&lt;/math&gt;. Let &lt;math&gt;\gamma&lt;/math&gt; be the circle with diameter &lt;math&gt;\overline{DE}&lt;/math&gt;. Circles &lt;math&gt;\omega&lt;/math&gt; and &lt;math&gt;\gamma&lt;/math&gt; meet at &lt;math&gt;E&lt;/math&gt; and a second point &lt;math&gt;F&lt;/math&gt;. Then &lt;math&gt;AF^2 = \frac mn&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 Using Only Elementary Geometry Methods ==<br /> Let &lt;math&gt;P&lt;/math&gt; be the unique point on circle &lt;math&gt;\omega&lt;/math&gt; such that &lt;math&gt;\bigtriangleup BCP&lt;/math&gt; is equilateral. Point &lt;math&gt;E&lt;/math&gt; must be the point on &lt;math&gt;\omega&lt;/math&gt; such that &lt;math&gt;\angle PED&lt;/math&gt; is right. We extend line &lt;math&gt;ED&lt;/math&gt; to hit circle &lt;math&gt;\omega&lt;/math&gt; again at &lt;math&gt;X&lt;/math&gt;.<br /> <br /> Claim 1: &lt;math&gt;X&lt;/math&gt; is the midpoint of minor arc &lt;math&gt;\overarc{BC}&lt;/math&gt;. <br /> Proof: Notice that &lt;math&gt;\angle PEX&lt;/math&gt; and &lt;math&gt;\angle PAX&lt;/math&gt; are subtended by the same arc, thus they are both equal to 90 degrees.<br /> <br /> Now it is obvious that &lt;math&gt;\bigtriangleup PED \sim \bigtriangleup ADX&lt;/math&gt;. After this, we length chase with similar triangles and Pythag, which yields &lt;math&gt;AP = 8&lt;/math&gt;, &lt;math&gt;EX = \frac{56}{\sqrt{57}}&lt;/math&gt;, &lt;math&gt;AX = \frac{2}{\sqrt{3}}&lt;/math&gt;, &lt;math&gt;PE = \frac{14}{\sqrt{19}}&lt;/math&gt;, and &lt;math&gt;PX = \frac{14}{\sqrt{3}}&lt;/math&gt;. We can tie this all together with Ptolemy's theorem for cyclic quadrilaterals, and we find that &lt;math&gt;AE = \frac{30}{\sqrt{19}}&lt;/math&gt;. Thus, &lt;math&gt;AE^2 = \frac{900}{19}&lt;/math&gt;, and the answer is &lt;math&gt;\boxed{919}&lt;/math&gt;.<br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> == Solution 1==<br /> Use the angle bisector theorem to find &lt;math&gt;CD=\frac{21}{8}&lt;/math&gt;, &lt;math&gt;BD=\frac{35}{8}&lt;/math&gt;, and use the Stewart's Theorem to find &lt;math&gt;AD=\frac{15}{8}&lt;/math&gt;. Use Power of the Point to find &lt;math&gt;DE=\frac{49}{8}&lt;/math&gt;, and so &lt;math&gt;AE=8&lt;/math&gt;. Use law of cosines to find &lt;math&gt;\angle CAD = \frac{\pi} {3}&lt;/math&gt;, hence &lt;math&gt;\angle BAD = \frac{\pi}{3}&lt;/math&gt; as well, and &lt;math&gt;\triangle BCE&lt;/math&gt; is equilateral, so &lt;math&gt;BC=CE=BE=7&lt;/math&gt;.<br /> <br /> I'm sure there is a more elegant solution from here, but instead we'll do some hairy law of cosines:<br /> <br /> &lt;math&gt;AE^2 = AF^2 + EF^2 - 2 \cdot AF \cdot EF \cdot \cos \angle AFE.&lt;/math&gt; (1)<br /> <br /> &lt;math&gt;AF^2 = AE^2 + EF^2 - 2 \cdot AE \cdot EF \cdot \cos \angle AEF.&lt;/math&gt; Adding these two and simplifying we get:<br /> <br /> &lt;math&gt;EF = AF \cdot \cos \angle AFE + AE \cdot \cos \angle AEF&lt;/math&gt; (2). Ah, but &lt;math&gt;\angle AFE = \angle ACE&lt;/math&gt; (since &lt;math&gt;F&lt;/math&gt; lies on &lt;math&gt;\omega&lt;/math&gt;), and we can find &lt;math&gt;cos \angle ACE&lt;/math&gt; using the law of cosines:<br /> <br /> &lt;math&gt;AE^2 = AC^2 + CE^2 - 2 \cdot AC \cdot CE \cdot \cos \angle ACE&lt;/math&gt;, and plugging in &lt;math&gt;AE = 8, AC = 3, BE = BC = 7,&lt;/math&gt; we get &lt;math&gt;\cos \angle ACE = -1/7 = \cos \angle AFE&lt;/math&gt;.<br /> <br /> Also, &lt;math&gt;\angle AEF = \angle DEF&lt;/math&gt;, and &lt;math&gt;\angle DFE = \pi/2&lt;/math&gt; (since &lt;math&gt;F&lt;/math&gt; is on the circle &lt;math&gt;\gamma&lt;/math&gt; with diameter &lt;math&gt;DE&lt;/math&gt;), so &lt;math&gt;\cos \angle AEF = EF/DE = 8 \cdot EF/49&lt;/math&gt;. <br /> <br /> Plugging in all our values into equation (2), we get:<br /> <br /> &lt;math&gt;EF = -\frac{AF}{7} + 8 \cdot \frac{8EF}{49}&lt;/math&gt;, or &lt;math&gt;EF = \frac{7}{15} \cdot AF&lt;/math&gt;.<br /> <br /> Finally, we plug this into equation (1), yielding:<br /> <br /> &lt;math&gt;8^2 = AF^2 + \frac{49}{225} \cdot AF^2 - 2 \cdot AF \cdot \frac{7AF}{15} \cdot \frac{-1}{7}&lt;/math&gt;. Thus,<br /> <br /> &lt;math&gt;64 = \frac{AF^2}{225} \cdot (225+49+30),&lt;/math&gt; or &lt;math&gt;AF^2 = \frac{900}{19}.&lt;/math&gt; The answer is &lt;math&gt;\boxed{919}&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> <br /> Let &lt;math&gt;a = BC&lt;/math&gt;, &lt;math&gt;b = CA&lt;/math&gt;, &lt;math&gt;c = AB&lt;/math&gt; for convenience. We claim that &lt;math&gt;AF&lt;/math&gt; is a symmedian. Indeed, let &lt;math&gt;M&lt;/math&gt; be the midpoint of segment &lt;math&gt;BC&lt;/math&gt;. Since &lt;math&gt;\angle EAB=\angle EAC&lt;/math&gt;, it follows that &lt;math&gt;EB = EC&lt;/math&gt; and consequently &lt;math&gt;EM\perp BC&lt;/math&gt;. Therefore, &lt;math&gt;M\in \gamma&lt;/math&gt;. Now let &lt;math&gt;G = FD\cap \omega&lt;/math&gt;. Since &lt;math&gt;EG&lt;/math&gt; is a diameter, &lt;math&gt;G&lt;/math&gt; lies on the perpendicular bisector of &lt;math&gt;BC&lt;/math&gt;; hence &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;G&lt;/math&gt; are collinear. From &lt;math&gt;\angle DAG = \angle DMG = 90&lt;/math&gt;, it immediately follows that quadrilateral &lt;math&gt;ADMG&lt;/math&gt; is cyclic. Therefore, &lt;math&gt;\angle MAD = \angle MGD=\angle EAF&lt;/math&gt;, implying that &lt;math&gt;AF&lt;/math&gt; is a symmedian, as claimed.<br /> <br /> The rest is standard; here's a quick way to finish. From above, quadrilateral &lt;math&gt;ABFC&lt;/math&gt; is harmonic, so &lt;math&gt;\dfrac{FB}{FC}=\dfrac{AB}{BC}=\dfrac{c}{a}&lt;/math&gt;. In conjunction with &lt;math&gt;\triangle ABF\sim\triangle AMC&lt;/math&gt;, it follows that &lt;math&gt;AF^2=\dfrac{b^2c^2}{AM^2}=\dfrac{4b^2c^2}{2b^2+2c^2-a^2}&lt;/math&gt;. (Notice that this holds for all triangles &lt;math&gt;ABC&lt;/math&gt;.) To finish, substitute &lt;math&gt;a = 7&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt;, &lt;math&gt;c=5&lt;/math&gt; to obtain &lt;math&gt;AF^2=\dfrac{900}{19}\implies 900+19=\boxed{919}&lt;/math&gt; as before.<br /> <br /> '''-Solution by thecmd999'''<br /> <br /> ==Solution 3==<br /> &lt;asy&gt;<br /> size(6cm);<br /> pair E,X,B,C,A,D,M,F,R,I;<br /> real z=sqrt(3)*14/3;<br /> real y=2*sqrt(3)/21;<br /> real x=224*sqrt(3)/57;<br /> E=(z,0);<br /> X=(0,0);<br /> D=(sqrt(3)*7/6,-7/8);<br /> M=(sqrt(3)*7/6,0);<br /> B=z/2*dir(60);<br /> C=z/2*dir(300);<br /> A=(y,-8/7);<br /> F=(x,-sqrt(3)*x/4);<br /> R=circumcenter(A,B,C);<br /> I=circumcenter(M,E,F);<br /> draw(E--X);<br /> draw(A--E);<br /> draw(A--B);<br /> draw(A--C);<br /> draw(B--C);<br /> draw(A--F);<br /> draw(X--F);<br /> draw(E--F);<br /> draw(circumcircle(A,B,C));<br /> draw(circumcircle(M,F,E));<br /> dot(D);<br /> dot(F);<br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(E);<br /> dot(X);<br /> dot(R);<br /> dot(I);<br /> label(&quot;$A$&quot;,A,dir(220));<br /> label(&quot;$B$&quot;,B,dir(110));<br /> label(&quot;$C$&quot;,C,dir(250));<br /> label(&quot;$D$&quot;,D,dir(60));<br /> label(&quot;$E$&quot;,E,dir(0));<br /> label(&quot;$F$&quot;,F,dir(315));<br /> label(&quot;$X$&quot;,X,dir(180));<br /> &lt;/asy&gt;<br /> First of all, use the [[Angle Bisector Theorem]] to find that &lt;math&gt;BD=35/8&lt;/math&gt; and &lt;math&gt;CD=21/8&lt;/math&gt;, and use [[Stewart's Theorem]] to find that &lt;math&gt;AD=15/8&lt;/math&gt;. Then use [[Power of a Point Theorem|Power of a Point]] to find that &lt;math&gt;DE=49/8&lt;/math&gt;. Then use the [[Circumradius|circumradius of a triangle]] formula to find that the length of the circumradius of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;\frac{7\sqrt{3}}{3}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;DE&lt;/math&gt; is the diameter of circle &lt;math&gt;\gamma&lt;/math&gt;, &lt;math&gt;\angle DFE&lt;/math&gt; is &lt;math&gt;90^\circ&lt;/math&gt;. Extending &lt;math&gt;DF&lt;/math&gt; to intersect circle &lt;math&gt;\omega&lt;/math&gt; at &lt;math&gt;X&lt;/math&gt;, we find that &lt;math&gt;XE&lt;/math&gt; is the diameter of the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt; (since &lt;math&gt;\angle DFE&lt;/math&gt; is &lt;math&gt;90^\circ&lt;/math&gt;). Therefore, &lt;math&gt;XE=\frac{14\sqrt{3}}{3}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;EF=x&lt;/math&gt;, &lt;math&gt;XD=a&lt;/math&gt;, and &lt;math&gt;DF=b&lt;/math&gt;. Then, by the [[Pythagorean Theorem]],<br /> <br /> &lt;cmath&gt;x^2+b^2=\left(\frac{49}{8}\right)^2=\frac{2401}{64}&lt;/cmath&gt;<br /> <br /> and<br /> <br /> &lt;cmath&gt;x^2+(a+b)^2=\left(\frac{14\sqrt{3}}{3}\right)^2=\frac{196}{3}.&lt;/cmath&gt;<br /> <br /> Subtracting the first equation from the second, the &lt;math&gt;x^2&lt;/math&gt; term cancels out and we obtain:<br /> <br /> &lt;cmath&gt;(a+b)^2-b^2=\frac{196}{3}-\frac{2401}{64}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;a^2+2ab = \frac{5341}{192}.&lt;/cmath&gt;<br /> <br /> By Power of a Point, &lt;math&gt;ab=BD \cdot DC=735/64=2205/192&lt;/math&gt;, so<br /> <br /> &lt;cmath&gt;a^2+2 \cdot \frac{2205}{192}=\frac{5341}{192}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;a^2=\frac{931}{192}.&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;a=XD&lt;/math&gt;, &lt;math&gt;XD=\frac{7\sqrt{19}}{8\sqrt{3}}&lt;/math&gt;.<br /> <br /> Because &lt;math&gt;\angle EXF&lt;/math&gt; and &lt;math&gt;\angle EAF&lt;/math&gt; intercept the same arc in circle &lt;math&gt;\omega&lt;/math&gt; and the same goes for &lt;math&gt;\angle XFA&lt;/math&gt; and &lt;math&gt;\angle XEA&lt;/math&gt;, &lt;math&gt;\angle EXF\cong\angle EAF&lt;/math&gt; and &lt;math&gt;\angle XFA\cong\angle XEA&lt;/math&gt;. Therefore, &lt;math&gt;\triangle XDE\sim\triangle ADF&lt;/math&gt; by [[Similarity (geometry)|AA Similarity]]. Since side lengths in similar triangles are proportional,<br /> <br /> &lt;cmath&gt;\frac{AF}{\frac{15}{8}}=\frac{\frac{14\sqrt{3}}{3}}{\frac{7\sqrt{19}}{8\sqrt{3}}}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\frac{AF}{\frac{15}{8}}=\frac{16}{\sqrt{19}}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;AF \cdot \sqrt{19} = 30&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;AF = \frac{30}{\sqrt{19}}.&lt;/cmath&gt;<br /> <br /> However, the problem asks for &lt;math&gt;AF^2&lt;/math&gt;, so &lt;math&gt;AF^2 = \frac{900}{19}\implies 900 + 19 = \boxed{919}&lt;/math&gt;.<br /> <br /> '''-Solution by TheBoomBox77'''<br /> <br /> ==Solution 4==<br /> It can be verified with law of cosines that &lt;math&gt;\angle BAC=120^\circ.&lt;/math&gt; Also, &lt;math&gt;E&lt;/math&gt; is the midpoint of major arc &lt;math&gt;BC&lt;/math&gt; so &lt;math&gt;BE=CE,&lt;/math&gt; and &lt;math&gt;\angle BEC=60.&lt;/math&gt; Thus &lt;math&gt;CBE&lt;/math&gt; is equilateral. Notice now that &lt;math&gt;\angle BFC=\angle BFE= 60.&lt;/math&gt; But &lt;math&gt;\angle DFE=90&lt;/math&gt; so &lt;math&gt;FD&lt;/math&gt; bisects &lt;math&gt;\angle BFC.&lt;/math&gt; Thus, &lt;math&gt;\frac{BF}{CF}=\frac{BD}{CD}=\frac{BA}{CA}=\frac{5}{3}.&lt;/math&gt; <br /> <br /> Let &lt;math&gt;BF=5a, CF=3a.&lt;/math&gt; By law of cosines on &lt;math&gt;BFC&lt;/math&gt; we find &lt;math&gt;a\sqrt{5^2+3^2-5*3}=a\sqrt{19}=7.&lt;/math&gt; But by ptolemy on &lt;math&gt;BFCA&lt;/math&gt;, &lt;math&gt;15a+15a=7*AF,&lt;/math&gt; so &lt;math&gt;AF= \frac{30}{\sqrt{19}},&lt;/math&gt; so &lt;math&gt;AF^2=\frac{900}{19}&lt;/math&gt; and the answer is &lt;math&gt;900+19=\boxed{919}&lt;/math&gt;<br /> <br /> ~abacadaea<br /> <br /> == See Also ==<br /> {{AIME box|year=2012|n=II|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Jj ca888 https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki:Competition_ratings&diff=98506 AoPS Wiki:Competition ratings 2018-11-05T02:31:58Z <p>Jj ca888: /* HMMT (November) */</p> <hr /> <div>This page contains an approximate estimation of the difficulty level of various [[List of mathematics competitions|competitions]]. It is designed with the intention of introducing contests of similar difficulty levels (but possibly different styles of problems) that readers may like to try to gain more experience.<br /> <br /> Each entry groups the problems into sets of similar difficulty levels and suggests an approximate difficulty rating, on a scale from 1 to 10 (from easiest to hardest). Note that many of these ratings are not directly comparable, because the actual competitions have many different rules; the ratings are generally synchronized with the amount of available time, etc. Also, due to variances within a contest, ranges shown may overlap. A sample problem is provided with each entry, with a link to a solution. <br /> <br /> As you may have guessed with time many competitions got more challenging because many countries got more access to books targeted at olympiad preparation. But especially web site where one can discuss Olympiads such as our very own AoPS!<br /> <br /> If you have some experience with mathematical competitions, we hope that you can help us make the difficulty rankings more accurate. Currently, the system is on a scale from 1 to 10 where 1 is the easiest level, e.g. [http://www.mathlinks.ro/resources.php?c=182&amp;cid=44 early AMC problems] and 10 is hardest level, e.g. [http://www.mathlinks.ro/resources.php?c=37&amp;cid=47 China IMO Team Selection Test.] When considering problem difficulty '''put more emphasis on problem-solving aspects and less so on technical skill requirements'''.<br /> <br /> = Scale =<br /> All levels are estimated and refer to ''averages''. The following is a rough standard based on the USA tier system AMC 8 – AMC 10 – AMC 12 – AIME – USAMO/USAJMO, representing Middle School – Junior High – High School – Challenging High School – Olympiad levels. Other contests can be interpolated against this. <br /> # Problems strictly for beginner, on the easiest elementary school or middle school levels (MOEMS, easy Mathcounts questions, #1-20 on AMC 8s, #1-5 AMC 10s, and others that involve standard techniques introduced up to the middle school level), most traditional middle/high school word problems<br /> # For motivated beginners, harder questions from the previous categories (#21-25 on AMC 8, Challenging Mathcounts questions, #5-20 on AMC 10, #5-10 on AMC 12, the easiest AIME questions, etc), traditional middle/high school word problems with extremely complex problem solving <br /> # Beginner/novice problems that require more creative thinking (MathCounts National, #21-25 on AMC 10, #11-20ish on AMC 12, #1-5 on AIMEs, etc.)<br /> # Intermediate-leveled problems, the most difficult questions on AMC 12s (#21-25s), more difficult AIME-styled questions #6-10<br /> # Difficult AIME problems (#10-13), simple proof-based problems (JBMO), etc<br /> # High-leveled AIME-styled questions (#12-15). Introductory-leveled Olympiad-level questions (#1,4s).<br /> # Tougher Olympiad-level questions, #1,4s that require more technical knowledge than new students to Olympiad-type questions have, easier #2,5s, etc. <br /> # High-level difficult Olympiad-level questions, eg #2,5s on difficult Olympiad contest and easier #3,6s, etc. <br /> # Expert Olympiad-level questions, eg #3,6s on difficult Olympiad contests. <br /> # Super Expert problems, problems occasionally even unsuitable for very hard competitions (like the IMO) due to being exceedingly tedious/long/difficult (e.g. very few students are capable of solving, even on a worldwide basis).<br /> <br /> = Competitions =<br /> <br /> ==Introductory Competitions==<br /> Most middle school and first-stage high school competitions would fall under this category. Problems in these competitions are usually ranked from 1 to 3. A full list is available [https://artofproblemsolving.com/wiki/index.php?title=Special%3ASearch&amp;search=Category%3AIntroductory+mathematics+competitions here].<br /> <br /> === [[MOEMS]] ===<br /> *Division E: '''1'''<br /> *: ''The whole number &lt;math&gt;N&lt;/math&gt; is divisible by &lt;math&gt;7&lt;/math&gt;. &lt;math&gt;N&lt;/math&gt; leaves a remainder of &lt;math&gt;1&lt;/math&gt; when divided by &lt;math&gt;2,3,4,&lt;/math&gt; or &lt;math&gt;5&lt;/math&gt;. What is the smallest value that &lt;math&gt;N&lt;/math&gt; can be?'' ([http://www.moems.org/sample_files/SampleE.pdf Solution])<br /> *Division M: '''1'''<br /> *: ''The value of a two-digit number is &lt;math&gt;10&lt;/math&gt; times more than the sum of its digits. The units digit is 1 more than twice the tens digit. Find the two-digit number.'' ([http://www.moems.org/sample_files/SampleM.pdf Solution])<br /> <br /> === [[AMC 8]] ===<br /> <br /> * Problem 1 - Problem 12: '''1''' <br /> *: ''What is the number of degrees in the smaller angle between the hour hand and the minute hand on a clock that reads seven o'clock?'' ([[1989 AJHSME Problems/Problem 10|Solution]])<br /> * Problem 13 - Problem 25: '''1.5'''<br /> *: ''A fifth number, &lt;math&gt;n&lt;/math&gt;, is added to the set &lt;math&gt;\{ 3,6,9,10 \}&lt;/math&gt; to make the mean of the set of five numbers equal to its median. What is the number of possible values of &lt;math&gt;n&lt;/math&gt;? '' ([[1988 AJHSME Problems/Problem 21|Solution]])<br /> <br /> === [[Mathcounts]] ===<br /> <br /> * Countdown: '''0.5''' (School, Chapter), '''1''' (State, National)<br /> * Sprint: '''1-1.5''' (school), '''1.5''' (Chapter),'''2''' (State), '''2-2.5''' (National)<br /> * Target: '''1.5''' (school), '''2''' (Chapter), '''2-2.5''' (State), '''2.5''' (National)<br /> <br /> === [[AMC 10]] ===<br /> <br /> * Problem 1 - 5: '''1'''<br /> *: ''A rectangular box has integer side lengths in the ratio &lt;math&gt;1: 3: 4&lt;/math&gt;. Which of the following could be the volume of the box?'' ([[2016 AMC 10A Problems/Problem 5|Solution]])<br /> * Problem 6 - 20: '''2'''<br /> *: ''Three runners start running simultaneously from the same point on a 500-meter circular track. They each run clockwise around the course maintaining constant speeds of 4.4, 4.8, and 5.0 meters per second. The runners stop once they are all together again somewhere on the circular course. How many seconds do the runners run?'' ([[2012 AMC 10A Problems/Problem 16|Solution]])<br /> * Problem 21 - 25: '''3'''<br /> *: ''The vertices of an equilateral triangle lie on the hyperbola &lt;math&gt;xy=1&lt;/math&gt;, and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?'' ([[2017 AMC 10B Problems/Problem 24|Solution]])<br /> <br /> ===[[CEMC|CEMC Multiple Choice Tests]]===<br /> This covers the CEMC Gauss, Pascal, Cayley, and Fermat tests.<br /> <br /> * Part A: '''0.5-1.5'''<br /> *: ''How many different 3-digit whole numbers can be formed using the digits 4, 7, and 9, assuming that no digit can be repeated in a number?'' (2015 Gauss 7 Problem 10)<br /> * Part B: '''1-2'''<br /> *: ''Two lines with slopes &lt;math&gt;\tfrac14&lt;/math&gt; and &lt;math&gt;\tfrac54&lt;/math&gt; intersect at &lt;math&gt;(1,1)&lt;/math&gt;. What is the area of the triangle formed by these two lines and the vertical line &lt;math&gt;x = 5&lt;/math&gt;?'' (2017 Cayley Problem 19)<br /> * Part C (Gauss/Pascal): '''2-2.5'''<br /> *: ''Suppose that &lt;math&gt;\tfrac{2009}{2014} + \tfrac{2019}{n} = \tfrac{a}{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;n&lt;/math&gt; are positive integers with &lt;math&gt;\tfrac{a}{b}&lt;/math&gt; in lowest terms. What is the sum of the digits of the smallest positive integer &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;a&lt;/math&gt; is a multiple of 1004?'' (2014 Pascal Problem 25)<br /> * Part C (Cayley/Fermat): '''2.5-3'''<br /> *: ''Wayne has 3 green buckets, 3 red buckets, 3 blue buckets, and 3 yellow buckets. He randomly distributes 4 hockey pucks among the green buckets, with each puck equally likely to be put in each bucket. Similarly, he distributes 3 pucks among the red buckets, 2 pucks among the blue buckets, and 1 puck among the yellow buckets. Once he is ﬁnished, what is the probability that a green bucket contains more pucks than each of the other 11 buckets?'' (2018 Fermat Problem 24)<br /> <br /> ===[[CEMC|CEMC Fryer/Galois/Hypatia]]===<br /> <br /> * Problem 1-2: '''1-2'''<br /> * Problem 3-4 (early parts): '''2-3'''<br /> * Problem 3-4 (later parts): '''3-5'''<br /> <br /> ==Intermediate Competitions==<br /> This category consists of all the non-proof math competitions for the middle stages of high school. The difficulty range would normally be from 3 to 6. A full list is available [https://artofproblemsolving.com/wiki/index.php?title=Special%3ASearch&amp;search=Category%3AIntermediate+mathematics+competitions here].<br /> <br /> === [[AMC 12]] ===<br /> <br /> * Problem 1-10: '''2'''<br /> *: ''A solid box is &lt;math&gt;15&lt;/math&gt; cm by &lt;math&gt;10&lt;/math&gt; cm by &lt;math&gt;8&lt;/math&gt; cm. A new solid is formed by removing a cube &lt;math&gt;3&lt;/math&gt; cm on a side from each corner of this box. What percent of the original volume is removed?'' ([[2003 AMC 12A Problems/Problem 3|Solution]])<br /> * Problem 11-20: '''3'''<br /> *: ''An object in the plane moves from one lattice point to another. At each step, the object may move one unit to the right, one unit to the left, one unit up, or one unit down. If the object starts at the origin and takes a ten-step path, how many different points could be the final point?'' ([[2006 AMC 12B Problems/Problem 18|Solution]])<br /> * Problem 21-25: '''4'''<br /> *: ''Functions &lt;math&gt;f&lt;/math&gt; and &lt;math&gt;g&lt;/math&gt; are quadratic, &lt;math&gt;g(x) = - f(100 - x)&lt;/math&gt;, and the graph of &lt;math&gt;g&lt;/math&gt; contains the vertex of the graph of &lt;math&gt;f&lt;/math&gt;. The four &lt;math&gt;x&lt;/math&gt;-intercepts on the two graphs have &lt;math&gt;x&lt;/math&gt;-coordinates &lt;math&gt;x_1&lt;/math&gt;, &lt;math&gt;x_2&lt;/math&gt;, &lt;math&gt;x_3&lt;/math&gt;, and &lt;math&gt;x_4&lt;/math&gt;, in increasing order, and &lt;math&gt;x_3 - x_2 = 150&lt;/math&gt;. The value of &lt;math&gt;x_4 - x_1&lt;/math&gt; is &lt;math&gt;m + n\sqrt p&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt;, and &lt;math&gt;p&lt;/math&gt; are positive integers, and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. What is &lt;math&gt;m + n + p&lt;/math&gt;?'' ([[2009 AMC 12A Problems/Problem 23|Solution]])<br /> <br /> === [[AIME]] ===<br /> <br /> * Problem 1 - 5: '''3'''<br /> *: Starting at &lt;math&gt;(0,0),&lt;/math&gt; an object moves in the coordinate plane via a sequence of steps, each of length one. Each step is left, right, up, or down, all four equally likely. Let &lt;math&gt;p&lt;/math&gt; be the probability that the object reaches &lt;math&gt;(2,2)&lt;/math&gt; in six or fewer steps. Given that &lt;math&gt;p&lt;/math&gt; can be written in the form &lt;math&gt;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; ([[1995 AIME Problems/Problem 3|Solution]])<br /> * Problem 6 - 10: '''3.75''' <br /> *: ''Triangle &lt;math&gt;ABC&lt;/math&gt; has &lt;math&gt;AB=21&lt;/math&gt;, &lt;math&gt;AC=22&lt;/math&gt; and &lt;math&gt;BC=20&lt;/math&gt;. Points &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; are located on &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{AC}&lt;/math&gt;, respectively, such that &lt;math&gt;\overline{DE}&lt;/math&gt; is [[parallel]] to &lt;math&gt;\overline{BC}&lt;/math&gt; and contains the center of the inscribed circle of triangle &lt;math&gt;ABC&lt;/math&gt;. Then &lt;math&gt;DE=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;.'' ([[2001 AIME I Problems/Problem 7|Solution]])<br /> * Problem 10 - 12: '''4.5'''<br /> *: Let &lt;math&gt;z&lt;/math&gt; be a complex number with &lt;math&gt;|z|=2014&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be the polygon in the complex plane whose vertices are &lt;math&gt;z&lt;/math&gt; and every &lt;math&gt;w&lt;/math&gt; such that &lt;math&gt;\frac{1}{z+w}=\frac{1}{z}+\frac{1}{w}&lt;/math&gt;. Then the area enclosed by &lt;math&gt;P&lt;/math&gt; can be written in the form &lt;math&gt;n\sqrt{3}&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is an integer. Find the remainder when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;. ([[2014 AIME II Problems/Problem 10|Solution]])<br /> * Problem 12 - 15: '''5'''<br /> *: ''Let<br /> <br /> &lt;cmath&gt;P(x) = 24x^{24} + \sum_{j = 1}^{23}(24 - j)(x^{24 - j} + x^{24 + j}).&lt;/cmath&gt;<br /> Let &lt;math&gt;z_{1},z_{2},\ldots,z_{r}&lt;/math&gt; be the distinct zeros of &lt;math&gt;P(x),&lt;/math&gt; and let &lt;math&gt;z_{k}^{2} = a_{k} + b_{k}i&lt;/math&gt; for &lt;math&gt;k = 1,2,\ldots,r,&lt;/math&gt; where &lt;math&gt;i = \sqrt { - 1},&lt;/math&gt; and &lt;math&gt;a_{k}&lt;/math&gt; and &lt;math&gt;b_{k}&lt;/math&gt; are real numbers. Let<br /> <br /> &lt;cmath&gt;\sum_{k = 1}^{r}|b_{k}| = m + n\sqrt {p},&lt;/cmath&gt;<br /> where &lt;math&gt;m,&lt;/math&gt; &lt;math&gt;n,&lt;/math&gt; and &lt;math&gt;p&lt;/math&gt; are integers and &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;m + n + p.&lt;/math&gt;.'' ([[2003 AIME II Problems/Problem 15|Solution]])<br /> <br /> === [[ARML]] ===<br /> <br /> * Individuals, Problem 1: '''2'''<br /> <br /> * Individuals, Problems 3, 5, 7, and 9: '''3'''<br /> <br /> * Individuals, Problems 2 and 4: '''3'''<br /> <br /> * Individuals, Problems 6 and 8: '''4''' <br /> <br /> * Individuals, Problem 10: '''6'''<br /> <br /> * Team/power, Problem 1-5: '''3.5''' <br /> <br /> * Team/power, Problem 6-10: '''5'''<br /> <br /> ===[[HMMT|HMMT (November)]]===<br /> * Individual Round, Problem 6-8: '''4'''<br /> * Individual Round, Problem 10: '''4.5'''<br /> * Team Round: '''5'''<br /> * Guts: '''6'''<br /> <br /> ===[[CEMC|CEMC Euclid]]===<br /> <br /> * Problem 1-6: '''1-3'''<br /> * Problem 7-10: '''3-6'''<br /> <br /> ===[[Purple Comet! Math Meet|Purple Comet]]===<br /> <br /> * Problems 1-10 (MS): '''2-3'''<br /> * Problems 11-20 (MS): '''3-5'''<br /> * Problems 1-10 (HS): '''2-4'''<br /> * Problems 11-20 (HS): '''4-5'''<br /> * Problems 21-30 (HS): '''5-6'''<br /> <br /> === [[Philippine Mathematical Olympiad Qualifying Round]] ===<br /> <br /> * Problem 1-15: '''2'''<br /> * Problem 16-25: '''3'''<br /> * Problem 26-30: '''4'''<br /> <br /> ==Beginner Olympiad Competitions==<br /> This category consists of beginning Olympiad math competitions. Most junior and first stage Olympiads fall under this category. The range from the difficulty scale would be around 4 to 6. A full list is available [http://artofproblemsolving.com/wiki/index.php?title=Special%3ASearch&amp;search=Category%3ABeginner+Olympiad+mathematics+competitions here].<br /> <br /> === [[USAMTS]] ===<br /> USAMTS generally has a different feel to it than olympiads, and is mainly for proofwriting practice instead of olympiad practice depending on how one takes the test. USAMTS allows an entire month to solve problems, with internet resources and books being allowed. However, the ultimate gap is that it permits computer programs to be used, and that Problem 1 is not a proof problem. However, it can still be roughly put to this rating scale:<br /> * Problem 1-2: '''3-4'''<br /> *: ''Find three isosceles triangles, no two of which are congruent, with integer sides, such that each triangle’s area is numerically equal to 6 times its perimeter.'' ([http://usamts.org/Solutions/Solution2_3_16.pdf Solution])<br /> * Problem 3-5: '''5-6'''<br /> *: ''Call a positive real number groovy if it can be written in the form &lt;math&gt;\sqrt{n} + \sqrt{n + 1}&lt;/math&gt; for some positive integer &lt;math&gt;n&lt;/math&gt;. Show that if &lt;math&gt;x&lt;/math&gt; is groovy, then for any positive integer &lt;math&gt;r&lt;/math&gt;, the number &lt;math&gt;x^r&lt;/math&gt; is groovy as well.'' ([http://usamts.org/Solutions/Solutions_20_1.pdf Solution])<br /> <br /> === [[Indonesia Mathematical Olympiad|Indonesia MO]] ===<br /> * Problem 1/5: '''3.5'''<br /> *: '' In a drawer, there are at most &lt;math&gt;2009&lt;/math&gt; balls, some of them are white, the rest are blue, which are randomly distributed. If two balls were taken at the same time, then the probability that the balls are both blue or both white is &lt;math&gt;\frac12&lt;/math&gt;. Determine the maximum amount of white balls in the drawer, such that the probability statement is true?'' &lt;url&gt;viewtopic.php?t=294065 (Solution)&lt;/url&gt;<br /> * Problem 2/6: '''4.5'''<br /> *: ''Find the lowest possible values from the function<br /> &lt;math&gt;f(x) = x^{2008} - 2x^{2007} + 3x^{2006} - 4x^{2005} + 5x^{2004} - \cdots - 2006x^3 + 2007x^2 - 2008x + 2009&lt;/math&gt;<br /> <br /> for any real numbers &lt;math&gt;x&lt;/math&gt;.''&lt;url&gt;viewtopic.php?t=294067 (Solution)&lt;/url&gt;<br /> * Problem 3/7: '''5'''<br /> *: ''A pair of integers &lt;math&gt;(m,n)&lt;/math&gt; is called ''good'' if<br /> &lt;math&gt;m\mid n^2 + n \ \text{and} \ n\mid m^2 + m&lt;/math&gt;<br /> <br /> Given 2 positive integers &lt;math&gt;a,b &gt; 1&lt;/math&gt; which are relatively prime, prove that there exists a ''good'' pair &lt;math&gt;(m,n)&lt;/math&gt; with &lt;math&gt;a\mid m&lt;/math&gt; and &lt;math&gt;b\mid n&lt;/math&gt;, but &lt;math&gt;a\nmid n&lt;/math&gt; and &lt;math&gt;b\nmid m&lt;/math&gt;.'' &lt;url&gt;viewtopic.php?t=294068 (Solution)&lt;/url&gt;<br /> * Problem 4/8: '''6'''<br /> *: ''Given an acute triangle &lt;math&gt;ABC&lt;/math&gt;. The incircle of triangle &lt;math&gt;ABC&lt;/math&gt; touches &lt;math&gt;BC,CA,AB&lt;/math&gt; respectively at &lt;math&gt;D,E,F&lt;/math&gt;. The angle bisector of &lt;math&gt;\angle A&lt;/math&gt; cuts &lt;math&gt;DE&lt;/math&gt; and &lt;math&gt;DF&lt;/math&gt; respectively at &lt;math&gt;K&lt;/math&gt; and &lt;math&gt;L&lt;/math&gt;. Suppose &lt;math&gt;AA_1&lt;/math&gt; is one of the altitudes of triangle &lt;math&gt;ABC&lt;/math&gt;, and &lt;math&gt;M&lt;/math&gt; be the midpoint of &lt;math&gt;BC&lt;/math&gt;.<br /> <br /> (a) Prove that &lt;math&gt;BK&lt;/math&gt; and &lt;math&gt;CL&lt;/math&gt; are perpendicular with the angle bisector of &lt;math&gt;\angle BAC&lt;/math&gt;.<br /> <br /> (b) Show that &lt;math&gt;A_1KML&lt;/math&gt; is a cyclic quadrilateral.'' &lt;url&gt;viewtopic.php?t=294069 (Solution)&lt;/url&gt;<br /> <br /> === [[Central American Olympiad]] ===<br /> * Problem 1: '''4'''<br /> *: ''Find all three-digit numbers &lt;math&gt;abc&lt;/math&gt; (with &lt;math&gt;a \neq 0&lt;/math&gt;) such that &lt;math&gt;a^{2} + b^{2} + c^{2}&lt;/math&gt; is a divisor of 26.'' (&lt;url&gt;viewtopic.php?p=903856#903856 Solution&lt;/url&gt;)<br /> * Problem 2,4,5: '''5-6'''<br /> *: ''Show that the equation &lt;math&gt;a^{2}b^{2} + b^{2}c^{2} + 3b^{2} - c^{2} - a^{2} = 2005&lt;/math&gt; has no integer solutions.'' (&lt;url&gt;viewtopic.php?p=291301#291301 Solution&lt;/url&gt;)<br /> * Problem 3/6: '''6.5''' <br /> *: ''Let &lt;math&gt;ABCD&lt;/math&gt; be a convex quadrilateral. &lt;math&gt;I = AC\cap BD&lt;/math&gt;, and &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;H&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;G&lt;/math&gt; are points on &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;CD&lt;/math&gt; and &lt;math&gt;DA&lt;/math&gt; respectively, such that &lt;math&gt;EF \cap GH = I&lt;/math&gt;. If &lt;math&gt;M = EG \cap AC&lt;/math&gt;, &lt;math&gt;N = HF \cap AC&lt;/math&gt;, show that &lt;math&gt;\frac {AM}{IM}\cdot \frac {IN}{CN} = \frac {IA}{IC}&lt;/math&gt;.'' (&lt;url&gt;viewtopic.php?p=828841#p828841 Solution&lt;/url&gt;<br /> <br /> === [[JBMO]] ===<br /> <br /> * Problem 1: '''4'''<br /> *: ''Find all real numbers &lt;math&gt;a,b,c,d&lt;/math&gt; such that <br /> &lt;cmath&gt; \left\{\begin{array}{cc}a+b+c+d = 20,\\ ab+ac+ad+bc+bd+cd = 150.\end{array}\right. &lt;/cmath&gt;''<br /> * Problem 2: '''5'''<br /> *: ''Let &lt;math&gt;ABCD&lt;/math&gt; be a convex quadrilateral with &lt;math&gt;\angle DAC=\angle BDC=36^\circ&lt;/math&gt;, &lt;math&gt;\angle CBD=18^\circ&lt;/math&gt; and &lt;math&gt;\angle BAC=72^\circ&lt;/math&gt;. The diagonals intersect at point &lt;math&gt;P&lt;/math&gt;. Determine the measure of &lt;math&gt;\angle APD&lt;/math&gt;.''<br /> * Problem 3: '''5'''<br /> *: ''Find all prime numbers &lt;math&gt;p,q,r&lt;/math&gt;, such that &lt;math&gt;\frac pq-\frac4{r+1}=1&lt;/math&gt;.''<br /> * Problem 4: '''6'''<br /> *: ''A &lt;math&gt;4\times4&lt;/math&gt; table is divided into &lt;math&gt;16&lt;/math&gt; white unit square cells. Two cells are called neighbors if they share a common side. A '''move''' consists in choosing a cell and changing the colors of neighbors from white to black or from black to white. After exactly &lt;math&gt;n&lt;/math&gt; moves all the &lt;math&gt;16&lt;/math&gt; cells were black. Find all possible values of &lt;math&gt;n&lt;/math&gt;.''<br /> <br /> ==Olympiad Competitions==<br /> This category consists of standard Olympiad competitions, usually ones from national Olympiads. Average difficulty is from 6 to 8. A full list is available [http://artofproblemsolving.com/wiki/index.php?title=Special%3ASearch&amp;search=Category%3AOlympiad+mathematics+competitions here].<br /> <br /> === [[USAJMO]] ===<br /> * Problem 1/4: ''3'''<br /> * Problem 2/5: '''5.5'''<br /> * Problem 3/6: '''10'''<br /> <br /> ===[[HMMT|HMMT (February)]]===<br /> * Individual Round, Problem 1-5: '''5'''<br /> * Individual Round, Problem 6-10: '''6'''<br /> * Team Round: '''7.5'''<br /> * HMIC: '''8'''<br /> <br /> === [[Canadian MO]] ===<br /> <br /> * Problem 1: '''5.5'''<br /> * Problem 2: '''6'''<br /> * Problem 3: '''6.5''' <br /> * Problem 4: '''7-7.5'''<br /> * Problem 5: '''7.5-8'''<br /> <br /> === Austrian MO ===<br /> <br /> * Regional Competition for Advanced Students, Problems 1-4: '''5''' <br /> * Federal Competition for Advanced Students, Part 1. Problems 1-4: '''6''' <br /> * Federal Competition for Advanced Students, Part 2, Problems 1-6: '''7'''<br /> <br /> === [[Ibero American Olympiad]] ===<br /> <br /> * Problem 1/4: '''5.5'''<br /> * Problem 2/5: '''6.5'''<br /> * Problem 3/6: '''7.5'''<br /> <br /> === [[APMO]] ===<br /> *Problem 1: '''6'''<br /> *Problem 2: '''7'''<br /> *Problem 3: '''7'''<br /> *Problem 4: '''7.5'''<br /> *Problem 5: '''8'''<br /> <br /> === Balkan MO ===<br /> <br /> * Problem 1: '''6'''<br /> *: '' Solve the equation &lt;math&gt;3^x - 5^y = z^2&lt;/math&gt; in positive integers. '' <br /> * Problem 2: '''6.5'''<br /> *: '' Let &lt;math&gt;MN&lt;/math&gt; be a line parallel to the side &lt;math&gt;BC&lt;/math&gt; of a triangle &lt;math&gt;ABC&lt;/math&gt;, with &lt;math&gt;M&lt;/math&gt; on the side &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; on the side &lt;math&gt;AC&lt;/math&gt;. The lines &lt;math&gt;BN&lt;/math&gt; and &lt;math&gt;CM&lt;/math&gt; meet at point &lt;math&gt;P&lt;/math&gt;. The circumcircles of triangles &lt;math&gt;BMP&lt;/math&gt; and &lt;math&gt;CNP&lt;/math&gt; meet at two distinct points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;. Prove that &lt;math&gt;\angle BAQ = \angle CAP&lt;/math&gt;. ''<br /> * Problem 3: '''7.5'''<br /> *: '' A &lt;math&gt;9 \times 12&lt;/math&gt; rectangle is partitioned into unit squares. The centers of all the unit squares, except for the four corner squares and eight squares sharing a common side with one of them, are coloured red. Is it possible to label these red centres &lt;math&gt;C_1,C_2...,C_{96}&lt;/math&gt; in such way that the following to conditions are both fulfilled<br /> <br /> &lt;math&gt;(i)&lt;/math&gt; the distances &lt;math&gt;C_1C_2,...C_{95}C_{96}, C_{96}C_{1}&lt;/math&gt; are all equal to &lt;math&gt;\sqrt {13}&lt;/math&gt;<br /> <br /> &lt;math&gt;(ii)&lt;/math&gt; the closed broken line &lt;math&gt;C_1C_2...C_{96}C_1&lt;/math&gt; has a centre of symmetry? ''<br /> * Problem 4: '''8'''<br /> *: '' Denote by &lt;math&gt;S&lt;/math&gt; the set of all positive integers. Find all functions &lt;math&gt;f: S \rightarrow S&lt;/math&gt; such that<br /> <br /> &lt;math&gt;f \bigg(f^2(m) + 2f^2(n)\bigg) = m^2 + 2 n^2&lt;/math&gt; for all &lt;math&gt;m,n \in S&lt;/math&gt;. '<br /> <br /> ==Hard Olympiad Competitions==<br /> This category consists of harder Olympiad contests. Difficulty is usually from 7 to 10. A full list is available [https://artofproblemsolving.com/wiki/index.php?title=Special%3ASearch&amp;search=Category%3AHard+Olympiad+mathematics+competitions here].<br /> <br /> === [[USAMO]] ===<br /> * Problem 1/4: '''7'''<br /> *: ''Let &lt;math&gt;\mathcal{P}&lt;/math&gt; be a convex polygon with &lt;math&gt;n&lt;/math&gt; sides, &lt;math&gt;n\ge3&lt;/math&gt;. Any set of &lt;math&gt;n - 3&lt;/math&gt; diagonals of &lt;math&gt;\mathcal{P}&lt;/math&gt; that do not intersect in the interior of the polygon determine a ''triangulation'' of &lt;math&gt;\mathcal{P}&lt;/math&gt; into &lt;math&gt;n - 2&lt;/math&gt; triangles. If &lt;math&gt;\mathcal{P}&lt;/math&gt; is regular and there is a triangulation of &lt;math&gt;\mathcal{P}&lt;/math&gt; consisting of only isosceles triangles, find all the possible values of &lt;math&gt;n&lt;/math&gt;.'' ([[2008 USAMO Problems/Problem 4|Solution]]) <br /> * Problem 2/5: '''8'''<br /> *: ''Three nonnegative real numbers &lt;math&gt;r_1&lt;/math&gt;, &lt;math&gt;r_2&lt;/math&gt;, &lt;math&gt;r_3&lt;/math&gt; are written on a blackboard. These numbers have the property that there exist integers &lt;math&gt;a_1&lt;/math&gt;, &lt;math&gt;a_2&lt;/math&gt;, &lt;math&gt;a_3&lt;/math&gt;, not all zero, satisfying &lt;math&gt;a_1r_1 + a_2r_2 + a_3r_3 = 0&lt;/math&gt;. We are permitted to perform the following operation: find two numbers &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; on the blackboard with &lt;math&gt;x \le y&lt;/math&gt;, then erase &lt;math&gt;y&lt;/math&gt; and write &lt;math&gt;y - x&lt;/math&gt; in its place. Prove that after a finite number of such operations, we can end up with at least one &lt;math&gt;0&lt;/math&gt; on the blackboard.'' ([[2008 USAMO Problems/Problem 5|Solution]])<br /> * Problem 3/6: '''9'''<br /> *: ''Prove that any monic polynomial (a polynomial with leading coefficient 1) of degree &lt;math&gt;n &lt;/math&gt; with real coefficients is the average of two monic polynomials of degree &lt;math&gt;n &lt;/math&gt; with &lt;math&gt;n &lt;/math&gt; real roots.'' ([[2002 USAMO Problems/Problem 3|Solution]])<br /> <br /> === [[USA TST]] ===<br /> <br /> (seems to vary more than other contests; estimates based on 08 and 09)<br /> <br /> * Problem 1/4/7: '''7'''<br /> * Problem 2/5/8: '''8'''<br /> * Problem 3/6/9: '''9.5'''<br /> <br /> === [[Putnam]] ===<br /> <br /> * Problem A/B,1-2: '''7'''<br /> *: ''Find the least possible area of a convex set in the plane that intersects both branches of the hyperbola &lt;math&gt;xy = 1&lt;/math&gt; and both branches of the hyperbola &lt;math&gt;xy = - 1.&lt;/math&gt; (A set &lt;math&gt;S&lt;/math&gt; in the plane is called ''convex'' if for any two points in &lt;math&gt;S&lt;/math&gt; the line segment connecting them is contained in &lt;math&gt;S.&lt;/math&gt;)'' ([https://artofproblemsolving.com/community/c7h177227p978383 Solution])<br /> * Problem A/B,3-4: '''8'''<br /> *: ''Let &lt;math&gt;H&lt;/math&gt; be an &lt;math&gt;n\times n&lt;/math&gt; matrix all of whose entries are &lt;math&gt;\pm1&lt;/math&gt; and whose rows are mutually orthogonal. Suppose &lt;math&gt;H&lt;/math&gt; has an &lt;math&gt;a\times b&lt;/math&gt; submatrix whose entries are all &lt;math&gt;1.&lt;/math&gt; Show that &lt;math&gt;ab\le n&lt;/math&gt;.'' ([https://artofproblemsolving.com/community/c7h64435p383280 Solution])<br /> * Problem A/B,5-6: '''9'''<br /> *: ''For any &lt;math&gt;a &gt; 0&lt;/math&gt;, define the set &lt;math&gt;S(a) = \{[an]|n = 1,2,3,...\}&lt;/math&gt;. Show that there are no three positive reals &lt;math&gt;a,b,c&lt;/math&gt; such that &lt;math&gt;S(a)\cap S(b) = S(b)\cap S(c) = S(c)\cap S(a) = \emptyset,S(a)\cup S(b)\cup S(c) = \{1,2,3,...\}&lt;/math&gt;.'' (&lt;url&gt;viewtopic.php?t=127810 Solution&lt;/url&gt;)<br /> <br /> === [[China TST]] ===<br /> <br /> * Problem 1/4: '''7''' <br /> *: ''Given an integer &lt;math&gt;m,&lt;/math&gt; prove that there exist odd integers &lt;math&gt;a,b&lt;/math&gt; and a positive integer &lt;math&gt;k&lt;/math&gt; such that &lt;cmath&gt;2m=a^{19}+b^{99}+k*2^{1000}.&lt;/cmath&gt;''<br /> * Problem 2/5: '''8.5''' <br /> *: ''Given a positive integer &lt;math&gt;n&gt;1&lt;/math&gt; and real numbers &lt;math&gt;a_1 &lt; a_2 &lt; \ldots &lt; a_n,&lt;/math&gt; such that &lt;math&gt;\dfrac{1}{a_1} + \dfrac{1}{a_2} + \ldots + \dfrac{1}{a_n} \le 1,&lt;/math&gt; prove that for any positive real number &lt;math&gt;x,&lt;/math&gt; &lt;cmath&gt;\left(\dfrac{1}{a_1^2+x} + \dfrac{1}{a_2^2+x} + \ldots + \dfrac{1}{a_n^2+x}\right)^2 \ge \dfrac{1}{2a_1(a_1-1)+2x}.&lt;/cmath&gt;''<br /> * Problem 3/6: '''10'''<br /> *: ''Let &lt;math&gt;n&gt;1&lt;/math&gt; be an integer and let &lt;math&gt;a_0,a_1,\ldots,a_n&lt;/math&gt; be non-negative real numbers. Define &lt;math&gt;S_k=\sum_{i=0}^k \binom{k}{i}a_i&lt;/math&gt; for &lt;math&gt;k=0,1,\ldots,n&lt;/math&gt;. Prove that&lt;cmath&gt;\frac{1}{n} \sum_{k=0}^{n-1} S_k^2-\frac{1}{n^2}\left(\sum_{k=0}^{n} S_k\right)^2\le \frac{4}{45} (S_n-S_0)^2.&lt;/cmath&gt;''<br /> <br /> === [[IMO]] ===<br /> <br /> * Problem 1/4: '''6.5'''<br /> *: ''Find all functions &lt;math&gt;f: (0, \infty) \mapsto (0, \infty)&lt;/math&gt; (so that &lt;math&gt;f&lt;/math&gt; is a function from the positive real numbers) such that<br /> &lt;center&gt;&lt;math&gt;\frac {\left( f(w) \right)^2 + \left( f(x) \right)^2}{f(y^2) + f(z^2) } = \frac {w^2 + x^2}{y^2 + z^2}&lt;/math&gt;&lt;/center&gt; for all positive real numbers &lt;math&gt;w,x,y,z,&lt;/math&gt; satisfying &lt;math&gt;wx = yz.&lt;/math&gt; ([[2008 IMO Problems/Problem 4|Solution]])<br /> ''<br /> * Problem 2/5: '''7.5-8'''<br /> *: ''Let &lt;math&gt;P(x)&lt;/math&gt; be a polynomial of degree &lt;math&gt;n&gt;1&lt;/math&gt; with integer coefficients, and let &lt;math&gt;k&lt;/math&gt; be a positive integer. Consider the polynomial &lt;math&gt;Q(x) = P( P ( \ldots P(P(x)) \ldots ))&lt;/math&gt;, where &lt;math&gt;P&lt;/math&gt; occurs &lt;math&gt;k&lt;/math&gt; times. Prove that there are at most &lt;math&gt;n&lt;/math&gt; integers &lt;math&gt;t&lt;/math&gt; such that &lt;math&gt;Q(t)=t&lt;/math&gt;.'' ([[2006 IMO Problems/Problem 5|Solution]])<br /> * Problem 3/6: '''9.5'''<br /> *: ''Assign to each side &lt;math&gt;b&lt;/math&gt; of a convex polygon &lt;math&gt;P&lt;/math&gt; the maximum area of a triangle that has &lt;math&gt;b&lt;/math&gt; as a side and is contained in &lt;math&gt;P&lt;/math&gt;. Show that the sum of the areas assigned to the sides of &lt;math&gt;P&lt;/math&gt; is at least twice the area of &lt;math&gt;P&lt;/math&gt;.'' (&lt;url&gt;viewtopic.php?p=572824#572824 Solution&lt;/url&gt;)<br /> <br /> === [[IMO Shortlist]] ===<br /> <br /> * Problem 1-2: '''5.5-7'''<br /> * Problem 3-4: '''7-8'''<br /> * Problem 5+: '''8-10'''<br /> <br /> [[Category:Mathematics competitions]]</div> Jj ca888 https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=91842 AMC historical results 2018-02-17T20:08:49Z <p>Jj ca888: /* AMC 10A */</p> <hr /> <div>&lt;!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --&gt;<br /> This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br /> <br /> ==2018==<br /> ===AMC 10A===<br /> *Average Score: 150<br /> *AIME floor: 111<br /> *Distinguished Honor Roll: 0<br /> <br /> ===AMC 10B===<br /> *Average score: TBD<br /> *AIME floor: TBD<br /> *Distinguished Honor Roll: TBD<br /> <br /> ===AMC 12A===<br /> *Average score: 59.8<br /> *AIME floor: 96<br /> *Distinguished Honor Roll floor: 117<br /> <br /> ===AMC 12B===<br /> *Average score: TBD<br /> *AIME floor: TBD<br /> *Distinguished Honor Roll: TBD<br /> <br /> ==2017==<br /> ===AMC 10A===<br /> *Average score: 59.33<br /> *AIME floor: 112.5<br /> *DHR: 127.5<br /> <br /> ===AMC 10B===<br /> *Average score: 66.56<br /> *AIME floor: 120<br /> *DHR: 136.5<br /> <br /> ===AMC 12A===<br /> *Average score: 60.32<br /> *AIME floor: 96<br /> *DHR: 115.5<br /> <br /> ===AMC 12B===<br /> *Average score: 58.35<br /> *AIME floor: 100.5<br /> *DHR: 129<br /> <br /> ===AIME I===<br /> *Average score: 5.69<br /> *Median score: 6<br /> *USAMO cutoff: 225(AMC 12A), 235(AMC 12B)<br /> *USAJMO cutoff: 224.5(AMC 10A), 233(AMC 10B)<br /> <br /> ===AIME II===<br /> *Average score: 5.64<br /> *Median score: 6<br /> *USAMO cutoff: 221(AMC 12A), 230.5(AMC 12B)<br /> *USAJMO cutoff: 219(AMC 10A), 225(AMC 10B)<br /> <br /> ==2016==<br /> ===AMC 10A===<br /> *Average score: 65.3<br /> *AIME floor: 110<br /> *DHR: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 65.4<br /> *AIME floor: 110<br /> *DHR: 124.5<br /> <br /> ===AMC 12A===<br /> *Average score: 59.06<br /> *AIME floor: 92<br /> *DHR: 110<br /> <br /> ===AMC 12B===<br /> *Average score: 67.96<br /> *AIME floor: 100<br /> *DHR: 127.5<br /> <br /> ===AIME I===<br /> *Average score: 5.83<br /> *Median score: 6<br /> *USAMO cutoff: 220<br /> *USAJMO cutoff: 210.5<br /> <br /> ===AIME II===<br /> *Average score: 4.33<br /> *Median score: 4<br /> *USAMO cutoff: 205<br /> *USAJMO cutoff: 200<br /> <br /> ==2015==<br /> ===AMC 10A===<br /> *Average score: 73.39<br /> *AIME floor: 106.5<br /> *DHR: 115.5<br /> <br /> ===AMC 10B===<br /> *Average score: 76.10<br /> *AIME floor: 120<br /> *DHR: 132<br /> <br /> ===AMC 12A===<br /> *Average score: 69.90<br /> *AIME floor: 99<br /> *DHR: 117<br /> <br /> ===AMC 12B===<br /> *Average score: 66.92<br /> *AIME floor: 100<br /> *DHR: 126<br /> <br /> ===AIME I===<br /> *Average score: 5.29<br /> *Median score: 5<br /> *USAMO cutoff: 219.0<br /> *USAJMO cutoff: 213.0<br /> <br /> ===AIME II===<br /> *Average score: 6.63<br /> *Median score: 6<br /> *USAMO cutoff: 229.0<br /> *USAJMO cutoff: 223.5<br /> <br /> ==2014==<br /> ===AMC 10A===<br /> *Average score: 63.83<br /> *AIME floor: 120<br /> *DHR: 132<br /> <br /> ===AMC 10B===<br /> *Average score: 71.44<br /> *AIME floor: 120<br /> *DHR: 132<br /> <br /> ===AMC 12A===<br /> *Average score: 64.01<br /> *AIME floor: 93<br /> *DHR: 109.5<br /> <br /> ===AMC 12B===<br /> *Average score: 68.11<br /> *AIME floor: 100<br /> *DHR: 121.5<br /> <br /> ===AIME I===<br /> *Average score: 4.88<br /> *Median score: 5<br /> *USAMO cutoff: 211.5<br /> *USAJMO cutoff: 211<br /> <br /> ===AIME II===<br /> *Average score: 5.49<br /> *Median score: 5<br /> *USAMO cutoff: 211.5<br /> *USAJMO cutoff: 211<br /> <br /> ==2013==<br /> ===AMC 10A===<br /> *Average score: 72.50<br /> *AIME floor: 108<br /> *DHR: 117<br /> <br /> ===AMC 10B===<br /> *Average score: 72.62<br /> *AIME floor: 120<br /> *DHR: 129<br /> <br /> ===AMC 12A===<br /> *Average score: 65.06<br /> *AIME floor: 88.5<br /> *DHR: 106.5<br /> <br /> ===AMC 12B===<br /> *Average score: 64.21<br /> *AIME floor: 93<br /> *DHR: 108<br /> <br /> ===AIME I===<br /> *Average score: 4.69<br /> *Median score: 4<br /> *USAMO cutoff: 209<br /> *USAJMO cutoff: 210.5<br /> <br /> ===AIME II===<br /> *Average score: 6.56<br /> *Median score: 6<br /> *USAMO cutoff: 209<br /> *USAJMO cutoff: 210.5<br /> <br /> ==2012==<br /> ===AMC 10A===<br /> *Average score: 72.51<br /> *AIME floor: 115.5<br /> <br /> ===AMC 10B===<br /> *Average score: 76.59<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 64.62<br /> *AIME floor: 94.5<br /> <br /> ===AMC 12B===<br /> *Average score: 70.08<br /> *AIME floor: 99<br /> <br /> ===AIME I===<br /> *Average score: 5.13<br /> *Median score: <br /> *USAMO cutoff: 204.5<br /> *USAJMO cutoff: 204<br /> <br /> ===AIME II===<br /> *Average score: 4.94<br /> *Median score: <br /> *USAMO cutoff: 204.5<br /> *USAJMO cutoff: 204<br /> <br /> ==2011==<br /> ===AMC 10A===<br /> *Average score: 67.61<br /> *AIME floor: 117<br /> <br /> ===AMC 10B===<br /> *Average score: 71.78<br /> *AIME floor: 117<br /> <br /> ===AMC 12A===<br /> *Average score: 66.77<br /> *AIME floor: 93<br /> <br /> ===AMC 12B===<br /> *Average score: 64.71<br /> *AIME floor: 97.5<br /> <br /> ===AIME I===<br /> *Average score: 5.47<br /> *Median score: <br /> *USAMO cutoff: 215.5<br /> *USAJMO cutoff: 196.5<br /> <br /> ===AIME II===<br /> *Average score: 2.23<br /> *Median score: <br /> *USAMO cutoff: 188<br /> *USAJMO cutoff: 179<br /> <br /> ==2010==<br /> ===AMC 10A===<br /> *Average score: 68.11<br /> *AIME floor: 115.5<br /> <br /> ===AMC 10B===<br /> *Average score: 68.57<br /> *AIME floor: 118.5<br /> <br /> ===AMC 12A===<br /> *Average score: 61.02<br /> *AIME floor: 88.5<br /> <br /> ===AMC 12B===<br /> *Average score: 59.58<br /> *AIME floor: 88.5<br /> <br /> ===AIME I===<br /> *Average score: 5.90<br /> *Median score: <br /> *USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br /> *USAJMO cutoff: 188.5<br /> <br /> ===AIME II===<br /> *Average score: 3.39<br /> *Median score: <br /> *USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br /> *USAJMO cutoff: 188.5<br /> <br /> ==2009==<br /> ===AMC 10A===<br /> *Average score: 67.41<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 74.73<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 66.37<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 71.88<br /> *AIME floor: 100 (Top 5% (1.00))<br /> <br /> ===AIME I===<br /> *Average score: 4.17<br /> *Median score: 4<br /> *USAMO floor: <br /> <br /> ===AIME II===<br /> *Average score: 3.27<br /> *Median score: 3<br /> *USAMO floor:<br /> <br /> ==2008==<br /> ===AMC 10A===<br /> *Average score: <br /> *AIME floor: <br /> <br /> ===AMC 10B===<br /> *Average score: <br /> *AIME floor: <br /> <br /> ===AMC 12A===<br /> *Average score: 65.6<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 68.9<br /> *AIME floor: 97.5<br /> <br /> ===AIME I===<br /> *Average score: <br /> *Median score: <br /> *USAMO floor: <br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2007==<br /> <br /> ===AMC 10A===<br /> *Average score: 67.9<br /> *AIME floor: 117<br /> <br /> ===AMC 10B===<br /> *Average score: 61.5<br /> *AIME floor: 115.5<br /> <br /> ===AMC 12A===<br /> *Average score: 66.8<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 73.1<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score: 5<br /> *Median score: 3<br /> *USAMO floor: 6<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2006==<br /> ===AMC 10A===<br /> *Average score: 79.0<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 68.5<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 85.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 85.5<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2005==<br /> ===AMC 10A===<br /> *Average score: 74.0<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 79.0<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 78.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 83.4<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2004==<br /> ===AMC 10A===<br /> *Average score: 69.1<br /> *AIME floor: 110<br /> <br /> ===AMC 10B===<br /> *Average score: 80.4<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 73.9<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 84.5<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2003==<br /> ===AMC 10A===<br /> *Average score: 74.4<br /> *AIME floor: 119<br /> <br /> ===AMC 10B===<br /> *Average score: 79.6<br /> *AIME floor: 121<br /> <br /> ===AMC 12A===<br /> *Average score: 77.8<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 76.6<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2002==<br /> ===AMC 10A===<br /> *Average score: 68.5<br /> *AIME floor: 115<br /> <br /> ===AMC 10B===<br /> *Average score: 74.9<br /> *AIME floor: 118<br /> <br /> ===AMC 12A===<br /> *Average score: 72.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 80.8<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2001==<br /> ===AMC 10===<br /> *Average score: 67.8<br /> *AIME floor: 116<br /> <br /> ===AMC 12===<br /> *Average score: 56.6<br /> *AIME floor: 84<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2000==<br /> ===AMC 10===<br /> *Average score: 64.2<br /> *AIME floor: <br /> <br /> ===AMC 12===<br /> *Average score: 64.9<br /> *AIME floor: <br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==1999==<br /> ===AHSME===<br /> *Average score: 68.8<br /> *AIME floor:<br /> <br /> ===AIME===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==1998==<br /> none<br /> <br /> ==1997==<br /> ==1996==<br /> ==1995==<br /> ==1994==<br /> ==1993==<br /> ==1992==<br /> ==1991==<br /> ==1990==<br /> ==1989==<br /> ==1988==<br /> ==1987==<br /> ==1986==<br /> ==1985==<br /> ==1984==<br /> ==1983==<br /> ==1982==<br /> ==1981==<br /> ==1980==<br /> ==1979==<br /> ==1978==<br /> ==1977==<br /> ==1976==<br /> ==1975==<br /> ==1974==<br /> ==1973==<br /> ==1972==<br /> ==1971==<br /> ==1970==<br /> ==1969==<br /> ==1968==<br /> ==1967==<br /> ==1966==<br /> ==1965==<br /> ==1964==<br /> ==1963==<br /> ==1962==<br /> ==1961==<br /> ==1960==<br /> <br /> ==1959==</div> Jj ca888 https://artofproblemsolving.com/wiki/index.php?title=2002_AIME_II_Problems/Problem_11&diff=88011 2002 AIME II Problems/Problem 11 2017-10-29T02:31:13Z <p>Jj ca888: /* Solution */</p> <hr /> <div>== Problem ==<br /> Two distinct, real, infinite geometric series each have a sum of &lt;math&gt;1&lt;/math&gt; and have the same second term. The third term of one of the series is &lt;math&gt;1/8&lt;/math&gt;, and the second term of both series can be written in the form &lt;math&gt;\frac{\sqrt{m}-n}p&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt;, and &lt;math&gt;p&lt;/math&gt; are positive integers and &lt;math&gt;m&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;100m+10n+p&lt;/math&gt;.<br /> <br /> == Solution 1==<br /> Let the second term of each series be &lt;math&gt;x&lt;/math&gt;. Then, the common ratio is &lt;math&gt;\frac{1}{8x}&lt;/math&gt;, and the first term is &lt;math&gt;8x^2&lt;/math&gt;. <br /> <br /> So, the sum is &lt;math&gt;\frac{8x^2}{1-\frac{1}{8x}}=1&lt;/math&gt;. Thus, &lt;math&gt;64x^3-8x+1 = (4x-1)(16x^2+4x-1) = 0 \Rightarrow x = \frac{1}{4}, \frac{-1 \pm \sqrt{5}}{8}&lt;/math&gt;. <br /> <br /> The only solution in the appropriate form is &lt;math&gt;x = \frac{\sqrt{5}-1}{8}&lt;/math&gt;. Therefore, &lt;math&gt;100m+10n+p = \boxed{518}&lt;/math&gt;.<br /> <br /> <br /> == Solution 2 ==<br /> <br /> Let the two sequences be &lt;math&gt;a, ar, ar^2 ... \text{ }an^2&lt;/math&gt; and &lt;math&gt;x, xy, xy^2 ... \text{ }xy^n&lt;/math&gt;. We know for a fact that &lt;math&gt;ar = xy&lt;/math&gt;. We also know that the sum of the first sequence = &lt;math&gt;\frac{a}{1-r} = 1&lt;/math&gt;, and the sum of the second sequence = &lt;math&gt;\frac{x}{1-y} = 1&lt;/math&gt;. Therefore we have &lt;cmath&gt;a+r = 1&lt;/cmath&gt;&lt;cmath&gt;x+y = 1&lt;/cmath&gt;&lt;cmath&gt;ar=xy&lt;/cmath&gt; We can then replace &lt;math&gt;r = \frac{xy}{a}&lt;/math&gt; and &lt;math&gt;y = \frac{ar}{x}&lt;/math&gt;. We plug them into the two equations &lt;math&gt;a+r = 1&lt;/math&gt; and &lt;math&gt;x+y = 1&lt;/math&gt;. We then get &lt;cmath&gt;x^2 + ar = x&lt;/cmath&gt;&lt;cmath&gt;a^2 + xy = a&lt;/cmath&gt;We subtract these equations, getting &lt;cmath&gt;x^2 - a^2 + ar - xy = x-a&lt;/cmath&gt;Remember &lt;cmath&gt;ar=xy&lt;/cmath&gt;, so &lt;cmath&gt;(x-a)(x+a-1) = 0&lt;/cmath&gt;Then considering cases, we have either &lt;math&gt;x=a&lt;/math&gt; or &lt;math&gt;y=a&lt;/math&gt;. This suggests that the second sequence is in the form &lt;math&gt;r, ra, ra^2...&lt;/math&gt;, while the first sequence is in the form &lt;math&gt;a, ar, ar^2...&lt;/math&gt; Now we have that &lt;math&gt;ar^2 = \frac18&lt;/math&gt; and we also have that &lt;math&gt;a+r = 1&lt;/math&gt;. We can solve for &lt;math&gt;r&lt;/math&gt; and the only appropriate value for &lt;math&gt;r&lt;/math&gt; is &lt;math&gt;\frac{1+\sqrt{5}}{4}&lt;/math&gt;. All we want is the second term, which is &lt;math&gt;ar = \frac{ar^2}{r} = \frac{\frac18}{\frac{1+\sqrt{5}}{4}} = \frac{\sqrt{5} - 1}{8}&lt;/math&gt;<br /> solution by jj_ca888<br /> <br /> == See also ==<br /> {{AIME box|year=2002|n=II|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Jj ca888 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_13&diff=86112 2006 AMC 12A Problems/Problem 13 2017-06-20T00:14:49Z <p>Jj ca888: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> [[Image:2006 AMC 12A Problem 13.gif]]<br /> <br /> The [[vertex|vertices]] of a &lt;math&gt;3-4-5&lt;/math&gt; [[right triangle]] are the centers of three mutually externally tangent [[circle]]s, as shown. What is the sum of the areas of the three circles?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 12\pi\qquad \mathrm{(B) \ } \frac{25\pi}{2}\qquad \mathrm{(C) \ } 13\pi\qquad \mathrm{(D) \ } \frac{27\pi}{2}\qquad\mathrm{(E) \ } 14\pi&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Let the radius of the smallest circle be &lt;math&gt;r_A&lt;/math&gt;, the radius of the second largest circle be &lt;math&gt;r_B&lt;/math&gt;, and the radius of the largest circle be &lt;math&gt;r_C&lt;/math&gt;. <br /> &lt;cmath&gt;r_A + r_B = 3&lt;/cmath&gt;<br /> &lt;cmath&gt;r_A + r_C = 4&lt;/cmath&gt;<br /> &lt;cmath&gt;r_ B + r_C = 5&lt;/cmath&gt;<br /> <br /> Adding up all these equations and then dividing both sides by 2, we get,<br /> <br /> &lt;cmath&gt;r_A + r_B + r_C = 6&lt;/cmath&gt;<br /> <br /> Then, we get &lt;math&gt;r_A = 1&lt;/math&gt;, &lt;math&gt;r_B = 2&lt;/math&gt;, and &lt;math&gt;r_C = 3&lt;/math&gt; Then we get &lt;math&gt;1^2 \pi + 2^2 \pi + 3^2 \pi = 14 \pi \iff\mathrm{(E)}&lt;/math&gt;<br /> <br /> == See also ==<br /> * [[2006 AMC 12A Problems]]<br /> <br /> {{AMC12 box|year=2006|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Jj ca888 https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12A_Problems/Problem_22&diff=86111 2005 AMC 12A Problems/Problem 22 2017-06-20T00:05:46Z <p>Jj ca888: /* Solution */</p> <hr /> <div>== Problem ==<br /> A rectangular box &lt;math&gt; P &lt;/math&gt; is [[inscribe]]d in a [[sphere]] of [[radius]] &lt;math&gt;r&lt;/math&gt;. The [[surface area]] of &lt;math&gt;P&lt;/math&gt; is 384, and the sum of the lengths of its 12 edges is 112. What is &lt;math&gt;r&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 8\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 14\qquad \mathrm{(E)}\ 16&lt;/math&gt;<br /> <br /> == Solution==<br /> Box P has dimensions &lt;math&gt;l&lt;/math&gt;, &lt;math&gt;w&lt;/math&gt;, and &lt;math&gt;h&lt;/math&gt;. <br /> Surface area = &lt;cmath&gt;2lw+2lh+2wl=384&lt;/cmath&gt;<br /> Sum of all edges = &lt;cmath&gt;4l+4w+4h=112 \Longrightarrow l + w + h = 28&lt;/cmath&gt;<br /> <br /> The diameter of the sphere is the space diagonal of the prism, which is &lt;cmath&gt;\sqrt{l^2 + w^2 +h^2}&lt;/cmath&gt;<br /> &lt;cmath&gt;(l + w + h)^2 - (2lw + 2lh + 2wh) = l^2 + w^2 + h^2 = 784 - 384 = 400&lt;/cmath&gt;<br /> &lt;cmath&gt;\sqrt{l^2 + w^2 +h^2} = 20 = diameter&lt;/cmath&gt;<br /> &lt;cmath&gt;r=\frac{20}{2} = 10&lt;/cmath&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2005|num-b=21|num-a=23|ab=A}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> [[Category:3D Geometry Problems]]<br /> {{MAA Notice}}</div> Jj ca888 https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12A_Problems/Problem_22&diff=86110 2005 AMC 12A Problems/Problem 22 2017-06-20T00:05:27Z <p>Jj ca888: /* Solution */</p> <hr /> <div>== Problem ==<br /> A rectangular box &lt;math&gt; P &lt;/math&gt; is [[inscribe]]d in a [[sphere]] of [[radius]] &lt;math&gt;r&lt;/math&gt;. The [[surface area]] of &lt;math&gt;P&lt;/math&gt; is 384, and the sum of the lengths of its 12 edges is 112. What is &lt;math&gt;r&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 8\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 14\qquad \mathrm{(E)}\ 16&lt;/math&gt;<br /> <br /> == Solution==<br /> Box P has dimensions &lt;math&gt;l&lt;/math&gt;, &lt;math&gt;w&lt;/math&gt;, and &lt;math&gt;h&lt;/math&gt;. <br /> Surface area = &lt;cmath&gt;2lw+2lh+2wl=384&lt;/cmath&gt;<br /> Sum of all edges = &lt;cmath&gt;4l+4w+4h=112 \Longrightarrow l + w + h = 28&lt;/cmath&gt;<br /> <br /> The diameter of the sphere is the space diagonal of the prism, which is &lt;cmath&gt;\sqrt{l^2 + w^2 +h^2}&lt;/cmath&gt;<br /> &lt;cmath&gt;(l + w + h)^2 - (2lw + 2lh + 2wh) = l^2 + w^2 + h^2 = 784 - 384 = 400&lt;/cmath&gt;<br /> &lt;cmath&gt;\sqrt{l^2 + w^2 +h^2} = 20 = diameter&lt;/cmath&gt;<br /> &lt;math&gt;r=\frac{20}{2} = 10&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2005|num-b=21|num-a=23|ab=A}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> [[Category:3D Geometry Problems]]<br /> {{MAA Notice}}</div> Jj ca888 https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12A_Problems/Problem_22&diff=86109 2005 AMC 12A Problems/Problem 22 2017-06-20T00:05:06Z <p>Jj ca888: /* Solution */</p> <hr /> <div>== Problem ==<br /> A rectangular box &lt;math&gt; P &lt;/math&gt; is [[inscribe]]d in a [[sphere]] of [[radius]] &lt;math&gt;r&lt;/math&gt;. The [[surface area]] of &lt;math&gt;P&lt;/math&gt; is 384, and the sum of the lengths of its 12 edges is 112. What is &lt;math&gt;r&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 8\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 14\qquad \mathrm{(E)}\ 16&lt;/math&gt;<br /> <br /> == Solution==<br /> Box P has dimensions &lt;math&gt;l&lt;/math&gt;, &lt;math&gt;w&lt;/math&gt;, and &lt;math&gt;h&lt;/math&gt;. <br /> Surface area = &lt;cmath&gt;2lw+2lh+2wl=384&lt;/cmath&gt;<br /> Sum of all edges = &lt;cmath&gt;4l+4w+4h=112 \Longrightarrow l + w + h = 28&lt;/cmath&gt;<br /> <br /> The diameter of the sphere is the space diagonal of the prism, which is &lt;cmath&gt;\sqrt{l^2 + w^2 +h^2}&lt;/cmath&gt;<br /> &lt;cmath&gt;(l + w + h)^2 - (2lw + 2lh + 2wh) = l^2 + w^2 + h^2 = 784 - 384 = 400&lt;/cmath&gt;<br /> &lt;cmath&gt;\sqrt{l^2 + w^2 +h^2} = 20 = diameter&lt;/cmath&gt;<br /> &lt;math&gt;r=\frac{20}{2} = 10&lt;/math&gt;$<br /> <br /> == See also ==<br /> {{AMC12 box|year=2005|num-b=21|num-a=23|ab=A}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> [[Category:3D Geometry Problems]]<br /> {{MAA Notice}}</div> Jj ca888 https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems/Problem_25&diff=85454 2011 AMC 8 Problems/Problem 25 2017-04-29T01:23:23Z <p>Jj ca888: /* Problem */</p> <hr /> <div>==Problem==<br /> A circle with radius &lt;math&gt;1&lt;/math&gt; is inscribed in a square and circumscribed about another square as shown. Which fraction is closest to the ratio of the circle's shaded area to the area between the two squares?<br /> <br /> &lt;asy&gt;<br /> filldraw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle,mediumgray,black);<br /> filldraw(Circle((0,0),1), mediumgray,black);<br /> filldraw((-1,0)--(0,1)--(1,0)--(0,-1)--cycle,white,black);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}2\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ \frac{3}2\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{5}2 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> The area of the smaller square is the one half of the product of its diagonals. Note that the distance from a corner of the smaller square to the center is equivalent to the circle's radius so the diagonal is equal to the diameter: &lt;math&gt;2*2*1/2=2.&lt;/math&gt;<br /> <br /> The circle's shaded area is the area of the smaller square subtracted from the area of the circle: &lt;math&gt;\pi - 2.&lt;/math&gt;<br /> <br /> If you draw the diagonals of the smaller square, you will see that the larger square is split &lt;math&gt;4&lt;/math&gt; congruent half-shaded squares. The area between the squares is equal to the area of the smaller square: &lt;math&gt;2.&lt;/math&gt;<br /> <br /> Approximating &lt;math&gt;\pi&lt;/math&gt; to &lt;math&gt;3.14,&lt;/math&gt; the ratio of the circle's shaded area to the area between the two squares is about<br /> <br /> &lt;cmath&gt;\frac{\pi-2}{2} \approx \frac{3.14-2}{2} = \frac{1.14}{2} \approx \boxed{\textbf{(A)}\ \frac12}&lt;/cmath&gt;<br /> <br /> ==Solution 2==<br /> For the ratio of the circle's shaded area to the area between the squares to be &lt;math&gt;1,&lt;/math&gt; they would have to be approximately the same size. For any ratio larger than that, the circle's shaded area must be greater. However, we can clearly see that the circle's shaded area is part of the area between the squares, and is approximately &lt;math&gt;\boxed{\textbf{(A)}\ \frac12}&lt;/math&gt;.<br /> <br /> Note that this solution is not rigorous, because we still should show that the ratio is less than &lt;math&gt;\frac{3}{4}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2011|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Jj ca888 https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems/Problem_25&diff=85453 2011 AMC 8 Problems/Problem 25 2017-04-29T01:22:19Z <p>Jj ca888: /* Problem */</p> <hr /> <div>==Problem==<br /> A circle with radius &lt;math&gt;1&lt;/math&gt; is inscribed in a square and circumscribed about another square as shown. Which fraction is closest to the ratio of the circle's shaded area to the area between the two squares?<br /> <br /> &lt;asy&gt;<br /> filldraw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle,white,black);<br /> filldraw(Circle((0,0),1), mediumgray,black);<br /> filldraw((-1,0)--(0,1)--(1,0)--(0,-1)--cycle,white,black);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}2\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ \frac{3}2\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{5}2 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> The area of the smaller square is the one half of the product of its diagonals. Note that the distance from a corner of the smaller square to the center is equivalent to the circle's radius so the diagonal is equal to the diameter: &lt;math&gt;2*2*1/2=2.&lt;/math&gt;<br /> <br /> The circle's shaded area is the area of the smaller square subtracted from the area of the circle: &lt;math&gt;\pi - 2.&lt;/math&gt;<br /> <br /> If you draw the diagonals of the smaller square, you will see that the larger square is split &lt;math&gt;4&lt;/math&gt; congruent half-shaded squares. The area between the squares is equal to the area of the smaller square: &lt;math&gt;2.&lt;/math&gt;<br /> <br /> Approximating &lt;math&gt;\pi&lt;/math&gt; to &lt;math&gt;3.14,&lt;/math&gt; the ratio of the circle's shaded area to the area between the two squares is about<br /> <br /> &lt;cmath&gt;\frac{\pi-2}{2} \approx \frac{3.14-2}{2} = \frac{1.14}{2} \approx \boxed{\textbf{(A)}\ \frac12}&lt;/cmath&gt;<br /> <br /> ==Solution 2==<br /> For the ratio of the circle's shaded area to the area between the squares to be &lt;math&gt;1,&lt;/math&gt; they would have to be approximately the same size. For any ratio larger than that, the circle's shaded area must be greater. However, we can clearly see that the circle's shaded area is part of the area between the squares, and is approximately &lt;math&gt;\boxed{\textbf{(A)}\ \frac12}&lt;/math&gt;.<br /> <br /> Note that this solution is not rigorous, because we still should show that the ratio is less than &lt;math&gt;\frac{3}{4}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2011|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Jj ca888 https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems/Problem_25&diff=85452 2011 AMC 8 Problems/Problem 25 2017-04-29T01:21:35Z <p>Jj ca888: /* Problem */</p> <hr /> <div>==Problem==<br /> A circle with radius &lt;math&gt;1&lt;/math&gt; is inscribed in a square and circumscribed about another square as shown. Which fraction is closest to the ratio of the circle's shaded area to the area between the two squares?<br /> <br /> &lt;asy&gt;<br /> filldraw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle,white,black);<br /> filldraw(Circle((0,0),1), mediumgray,black);<br /> filldraw((-1,0)--(0,1)--(1,0)--(0,-1)--cycle,mediumgray,black);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}2\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ \frac{3}2\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{5}2 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> The area of the smaller square is the one half of the product of its diagonals. Note that the distance from a corner of the smaller square to the center is equivalent to the circle's radius so the diagonal is equal to the diameter: &lt;math&gt;2*2*1/2=2.&lt;/math&gt;<br /> <br /> The circle's shaded area is the area of the smaller square subtracted from the area of the circle: &lt;math&gt;\pi - 2.&lt;/math&gt;<br /> <br /> If you draw the diagonals of the smaller square, you will see that the larger square is split &lt;math&gt;4&lt;/math&gt; congruent half-shaded squares. The area between the squares is equal to the area of the smaller square: &lt;math&gt;2.&lt;/math&gt;<br /> <br /> Approximating &lt;math&gt;\pi&lt;/math&gt; to &lt;math&gt;3.14,&lt;/math&gt; the ratio of the circle's shaded area to the area between the two squares is about<br /> <br /> &lt;cmath&gt;\frac{\pi-2}{2} \approx \frac{3.14-2}{2} = \frac{1.14}{2} \approx \boxed{\textbf{(A)}\ \frac12}&lt;/cmath&gt;<br /> <br /> ==Solution 2==<br /> For the ratio of the circle's shaded area to the area between the squares to be &lt;math&gt;1,&lt;/math&gt; they would have to be approximately the same size. For any ratio larger than that, the circle's shaded area must be greater. However, we can clearly see that the circle's shaded area is part of the area between the squares, and is approximately &lt;math&gt;\boxed{\textbf{(A)}\ \frac12}&lt;/math&gt;.<br /> <br /> Note that this solution is not rigorous, because we still should show that the ratio is less than &lt;math&gt;\frac{3}{4}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2011|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Jj ca888 https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems/Problem_25&diff=85451 2011 AMC 8 Problems/Problem 25 2017-04-29T01:20:49Z <p>Jj ca888: /* Problem */</p> <hr /> <div>==Problem==<br /> A circle with radius &lt;math&gt;1&lt;/math&gt; is inscribed in a square and circumscribed about another square as shown. Which fraction is closest to the ratio of the circle's shaded area to the area between the two squares?<br /> <br /> &lt;asy&gt;<br /> filldraw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle,white,black);<br /> filldraw(Circle((0,0),1), mediumgray,black);<br /> filldraw((-1,0)--(0,1)--(1,0)--(0,-1)--cycle,white,black);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}2\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ \frac{3}2\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{5}2 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> The area of the smaller square is the one half of the product of its diagonals. Note that the distance from a corner of the smaller square to the center is equivalent to the circle's radius so the diagonal is equal to the diameter: &lt;math&gt;2*2*1/2=2.&lt;/math&gt;<br /> <br /> The circle's shaded area is the area of the smaller square subtracted from the area of the circle: &lt;math&gt;\pi - 2.&lt;/math&gt;<br /> <br /> If you draw the diagonals of the smaller square, you will see that the larger square is split &lt;math&gt;4&lt;/math&gt; congruent half-shaded squares. The area between the squares is equal to the area of the smaller square: &lt;math&gt;2.&lt;/math&gt;<br /> <br /> Approximating &lt;math&gt;\pi&lt;/math&gt; to &lt;math&gt;3.14,&lt;/math&gt; the ratio of the circle's shaded area to the area between the two squares is about<br /> <br /> &lt;cmath&gt;\frac{\pi-2}{2} \approx \frac{3.14-2}{2} = \frac{1.14}{2} \approx \boxed{\textbf{(A)}\ \frac12}&lt;/cmath&gt;<br /> <br /> ==Solution 2==<br /> For the ratio of the circle's shaded area to the area between the squares to be &lt;math&gt;1,&lt;/math&gt; they would have to be approximately the same size. For any ratio larger than that, the circle's shaded area must be greater. However, we can clearly see that the circle's shaded area is part of the area between the squares, and is approximately &lt;math&gt;\boxed{\textbf{(A)}\ \frac12}&lt;/math&gt;.<br /> <br /> Note that this solution is not rigorous, because we still should show that the ratio is less than &lt;math&gt;\frac{3}{4}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2011|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Jj ca888 https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=84333 AMC historical results 2017-03-02T01:13:32Z <p>Jj ca888: /* AMC 12B */</p> <hr /> <div>&lt;!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --&gt;<br /> This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br /> <br /> ==2017==<br /> ===AMC 10A===<br /> *Average score: <br /> *AIME floor: 112.5<br /> <br /> ===AMC 10B===<br /> *Average score: <br /> *AIME floor:<br /> <br /> ===AMC 12A===<br /> *Average score:<br /> *AIME floor: 96<br /> <br /> ===AMC 12B===<br /> *Average score:<br /> *AIME floor:<br /> <br /> ===AIME I===<br /> *Average score: <br /> *Median score: <br /> *USAMO cutoff: <br /> *USAJMO cutoff: <br /> <br /> ===AIME II===<br /> *Average score: <br /> *Median score: <br /> *USAMO cutoff: <br /> *USAJMO cutoff:<br /> <br /> ==2015==<br /> ===AMC 10A===<br /> *Average score: 73.39<br /> *AIME floor: 106.5<br /> <br /> ===AMC 10B===<br /> *Average score: 76.10<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 69.90<br /> *AIME floor: 99<br /> <br /> ===AMC 12B===<br /> *Average score: 66.92<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score: 5.29<br /> *Median score: 5<br /> *USAMO cutoff: 219.0<br /> *USAJMO cutoff: 213.0<br /> <br /> ===AIME II===<br /> *Average score: 6.63<br /> *Median score: 6<br /> *USAMO cutoff: 229.0<br /> *USAJMO cutoff: 223.5<br /> <br /> ==2014==<br /> ===AMC 10A===<br /> *Average score: 63.83<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 71.44<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 64.01<br /> *AIME floor: 93<br /> <br /> ===AMC 12B===<br /> *Average score: 68.11<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score: 4.88<br /> *Median score: 5<br /> *USAMO cutoff: 211.5<br /> *USAJMO cutoff: 211<br /> <br /> ===AIME II===<br /> *Average score: 5.49<br /> *Median score: 5<br /> *USAMO cutoff: 211.5<br /> *USAJMO cutoff: 211<br /> <br /> ==2013==<br /> ===AMC 10A===<br /> *Average score: 72.50<br /> *AIME floor: 108<br /> <br /> ===AMC 10B===<br /> *Average score: 72.62<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 65.06<br /> *AIME floor:88.5<br /> <br /> ===AMC 12B===<br /> *Average score: 64.21<br /> *AIME floor: 93<br /> <br /> ===AIME I===<br /> *Average score: 4.69<br /> *Median score: <br /> *USAMO cutoff: 209<br /> *USAJMO cutoff: 210.5<br /> <br /> ===AIME II===<br /> *Average score: 6.56<br /> *Median score: <br /> *USAMO cutoff: 209<br /> *USAJMO cutoff: 210.5<br /> <br /> ==2012==<br /> ===AMC 10A===<br /> *Average score: 72.51<br /> *AIME floor: 115.5<br /> <br /> ===AMC 10B===<br /> *Average score: 76.59<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 64.62<br /> *AIME floor: 94.5<br /> <br /> ===AMC 12B===<br /> *Average score: 70.08<br /> *AIME floor: 99<br /> <br /> ===AIME I===<br /> *Average score: 5.13<br /> *Median score: <br /> *USAMO cutoff: 204.5<br /> *USAJMO cutoff: 204<br /> <br /> ===AIME II===<br /> *Average score: 4.94<br /> *Median score: <br /> *USAMO cutoff: 204.5<br /> *USAJMO cutoff: 204<br /> <br /> ==2011==<br /> ===AMC 10A===<br /> *Average score: 67.61<br /> *AIME floor: 117<br /> <br /> ===AMC 10B===<br /> *Average score: 71.78<br /> *AIME floor: 117<br /> <br /> ===AMC 12A===<br /> *Average score: 66.77<br /> *AIME floor: 93<br /> <br /> ===AMC 12B===<br /> *Average score: 64.71<br /> *AIME floor: 97.5<br /> <br /> ===AIME I===<br /> *Average score: 5.47<br /> *Median score: <br /> *USAMO cutoff: 215.5<br /> *USAJMO cutoff: 196.5<br /> <br /> ===AIME II===<br /> *Average score: 2.23<br /> *Median score: <br /> *USAMO cutoff: 188<br /> *USAJMO cutoff: 179<br /> <br /> ==2010==<br /> ===AMC 10A===<br /> *Average score: 68.11<br /> *AIME floor: 115.5<br /> <br /> ===AMC 10B===<br /> *Average score: 68.57<br /> *AIME floor: 118.5<br /> <br /> ===AMC 12A===<br /> *Average score: 61.02<br /> *AIME floor: 88.5<br /> <br /> ===AMC 12B===<br /> *Average score: 59.58<br /> *AIME floor: 88.5<br /> <br /> ===AIME I===<br /> *Average score: 5.90<br /> *Median score: <br /> *USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br /> *USAJMO cutoff: 188.5<br /> <br /> ===AIME II===<br /> *Average score: 3.39<br /> *Median score: <br /> *USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br /> *USAJMO cutoff: 188.5<br /> <br /> ==2009==<br /> ===AMC 10A===<br /> *Average score: 67.41<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 74.73<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 66.37<br /> *AIME floor: <br /> <br /> ===AMC 12B===<br /> *Average score: 71.88<br /> *AIME floor: <br /> <br /> ===AIME I===<br /> *Average score: <br /> *Median score: <br /> *USAMO floor: <br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2008==<br /> ===AMC 10A===<br /> *Average score: <br /> *AIME floor: <br /> <br /> ===AMC 10B===<br /> *Average score: <br /> *AIME floor: <br /> <br /> ===AMC 12A===<br /> *Average score: 65.6<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 68.9<br /> *AIME floor: 97.5<br /> <br /> ===AIME I===<br /> *Average score: <br /> *Median score: <br /> *USAMO floor: <br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2007==<br /> <br /> ===AMC 10A===<br /> *Average score: 67.9<br /> *AIME floor: 117<br /> <br /> ===AMC 10B===<br /> *Average score: 61.5<br /> *AIME floor: 115.5<br /> <br /> ===AMC 12A===<br /> *Average score: 66.8<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 73.1<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score: 5<br /> *Median score: 3<br /> *USAMO floor: 6<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2006==<br /> ===AMC 10A===<br /> *Average score: 79.0<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 68.5<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 85.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 85.5<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2005==<br /> ===AMC 10A===<br /> *Average score: 74.0<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 79.0<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 78.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 83.4<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2004==<br /> ===AMC 10A===<br /> *Average score: 69.1<br /> *AIME floor: 110<br /> <br /> ===AMC 10B===<br /> *Average score: 80.4<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 73.9<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 84.5<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2003==<br /> ===AMC 10A===<br /> *Average score: 74.4<br /> *AIME floor: 119<br /> <br /> ===AMC 10B===<br /> *Average score: 79.6<br /> *AIME floor: 121<br /> <br /> ===AMC 12A===<br /> *Average score: 77.8<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 76.6<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2002==<br /> ===AMC 10A===<br /> *Average score: 68.5<br /> *AIME floor: 115<br /> <br /> ===AMC 10B===<br /> *Average score: 74.9<br /> *AIME floor: 118<br /> <br /> ===AMC 12A===<br /> *Average score: 72.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 80.8<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2001==<br /> ===AMC 10===<br /> *Average score: 67.8<br /> *AIME floor: 116<br /> <br /> ===AMC 12===<br /> *Average score: 56.6<br /> *AIME floor: 84<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2000==<br /> ===AMC 10===<br /> *Average score: 64.2<br /> *AIME floor: <br /> <br /> ===AMC 12===<br /> *Average score: 64.9<br /> *AIME floor: <br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==1999==<br /> ===AHSME===<br /> *Average score: 68.8<br /> *AIME floor:<br /> <br /> ===AIME===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==1998==<br /> ==1997==<br /> ==1996==<br /> ==1995==<br /> ==1994==<br /> ==1993==<br /> ==1992==<br /> ==1991==<br /> ==1990==<br /> ==1989==<br /> ==1988==<br /> ==1987==<br /> ==1986==<br /> ==1985==<br /> ==1984==<br /> ==1983==<br /> ==1982==<br /> ==1981==<br /> ==1980==<br /> ==1979==<br /> ==1978==<br /> ==1977==<br /> ==1976==<br /> ==1975==<br /> ==1974==<br /> ==1973==<br /> ==1972==<br /> ==1971==<br /> ==1970==<br /> ==1969==<br /> ==1968==<br /> ==1967==<br /> ==1966==<br /> ==1965==<br /> ==1964==<br /> ==1963==<br /> ==1962==<br /> ==1961==<br /> ==1960==<br /> <br /> ==1959==</div> Jj ca888 https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=84332 AMC historical results 2017-03-02T01:13:22Z <p>Jj ca888: /* AMC 10B */</p> <hr /> <div>&lt;!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --&gt;<br /> This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br /> <br /> ==2017==<br /> ===AMC 10A===<br /> *Average score: <br /> *AIME floor: 112.5<br /> <br /> ===AMC 10B===<br /> *Average score: <br /> *AIME floor:<br /> <br /> ===AMC 12A===<br /> *Average score:<br /> *AIME floor: 96<br /> <br /> ===AMC 12B===<br /> *Average score:<br /> *AIME floor: 100.5<br /> <br /> ===AIME I===<br /> *Average score: <br /> *Median score: <br /> *USAMO cutoff: <br /> *USAJMO cutoff: <br /> <br /> ===AIME II===<br /> *Average score: <br /> *Median score: <br /> *USAMO cutoff: <br /> *USAJMO cutoff:<br /> <br /> ==2015==<br /> ===AMC 10A===<br /> *Average score: 73.39<br /> *AIME floor: 106.5<br /> <br /> ===AMC 10B===<br /> *Average score: 76.10<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 69.90<br /> *AIME floor: 99<br /> <br /> ===AMC 12B===<br /> *Average score: 66.92<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score: 5.29<br /> *Median score: 5<br /> *USAMO cutoff: 219.0<br /> *USAJMO cutoff: 213.0<br /> <br /> ===AIME II===<br /> *Average score: 6.63<br /> *Median score: 6<br /> *USAMO cutoff: 229.0<br /> *USAJMO cutoff: 223.5<br /> <br /> ==2014==<br /> ===AMC 10A===<br /> *Average score: 63.83<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 71.44<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 64.01<br /> *AIME floor: 93<br /> <br /> ===AMC 12B===<br /> *Average score: 68.11<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score: 4.88<br /> *Median score: 5<br /> *USAMO cutoff: 211.5<br /> *USAJMO cutoff: 211<br /> <br /> ===AIME II===<br /> *Average score: 5.49<br /> *Median score: 5<br /> *USAMO cutoff: 211.5<br /> *USAJMO cutoff: 211<br /> <br /> ==2013==<br /> ===AMC 10A===<br /> *Average score: 72.50<br /> *AIME floor: 108<br /> <br /> ===AMC 10B===<br /> *Average score: 72.62<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 65.06<br /> *AIME floor:88.5<br /> <br /> ===AMC 12B===<br /> *Average score: 64.21<br /> *AIME floor: 93<br /> <br /> ===AIME I===<br /> *Average score: 4.69<br /> *Median score: <br /> *USAMO cutoff: 209<br /> *USAJMO cutoff: 210.5<br /> <br /> ===AIME II===<br /> *Average score: 6.56<br /> *Median score: <br /> *USAMO cutoff: 209<br /> *USAJMO cutoff: 210.5<br /> <br /> ==2012==<br /> ===AMC 10A===<br /> *Average score: 72.51<br /> *AIME floor: 115.5<br /> <br /> ===AMC 10B===<br /> *Average score: 76.59<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 64.62<br /> *AIME floor: 94.5<br /> <br /> ===AMC 12B===<br /> *Average score: 70.08<br /> *AIME floor: 99<br /> <br /> ===AIME I===<br /> *Average score: 5.13<br /> *Median score: <br /> *USAMO cutoff: 204.5<br /> *USAJMO cutoff: 204<br /> <br /> ===AIME II===<br /> *Average score: 4.94<br /> *Median score: <br /> *USAMO cutoff: 204.5<br /> *USAJMO cutoff: 204<br /> <br /> ==2011==<br /> ===AMC 10A===<br /> *Average score: 67.61<br /> *AIME floor: 117<br /> <br /> ===AMC 10B===<br /> *Average score: 71.78<br /> *AIME floor: 117<br /> <br /> ===AMC 12A===<br /> *Average score: 66.77<br /> *AIME floor: 93<br /> <br /> ===AMC 12B===<br /> *Average score: 64.71<br /> *AIME floor: 97.5<br /> <br /> ===AIME I===<br /> *Average score: 5.47<br /> *Median score: <br /> *USAMO cutoff: 215.5<br /> *USAJMO cutoff: 196.5<br /> <br /> ===AIME II===<br /> *Average score: 2.23<br /> *Median score: <br /> *USAMO cutoff: 188<br /> *USAJMO cutoff: 179<br /> <br /> ==2010==<br /> ===AMC 10A===<br /> *Average score: 68.11<br /> *AIME floor: 115.5<br /> <br /> ===AMC 10B===<br /> *Average score: 68.57<br /> *AIME floor: 118.5<br /> <br /> ===AMC 12A===<br /> *Average score: 61.02<br /> *AIME floor: 88.5<br /> <br /> ===AMC 12B===<br /> *Average score: 59.58<br /> *AIME floor: 88.5<br /> <br /> ===AIME I===<br /> *Average score: 5.90<br /> *Median score: <br /> *USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br /> *USAJMO cutoff: 188.5<br /> <br /> ===AIME II===<br /> *Average score: 3.39<br /> *Median score: <br /> *USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br /> *USAJMO cutoff: 188.5<br /> <br /> ==2009==<br /> ===AMC 10A===<br /> *Average score: 67.41<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 74.73<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 66.37<br /> *AIME floor: <br /> <br /> ===AMC 12B===<br /> *Average score: 71.88<br /> *AIME floor: <br /> <br /> ===AIME I===<br /> *Average score: <br /> *Median score: <br /> *USAMO floor: <br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2008==<br /> ===AMC 10A===<br /> *Average score: <br /> *AIME floor: <br /> <br /> ===AMC 10B===<br /> *Average score: <br /> *AIME floor: <br /> <br /> ===AMC 12A===<br /> *Average score: 65.6<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 68.9<br /> *AIME floor: 97.5<br /> <br /> ===AIME I===<br /> *Average score: <br /> *Median score: <br /> *USAMO floor: <br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2007==<br /> <br /> ===AMC 10A===<br /> *Average score: 67.9<br /> *AIME floor: 117<br /> <br /> ===AMC 10B===<br /> *Average score: 61.5<br /> *AIME floor: 115.5<br /> <br /> ===AMC 12A===<br /> *Average score: 66.8<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 73.1<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score: 5<br /> *Median score: 3<br /> *USAMO floor: 6<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2006==<br /> ===AMC 10A===<br /> *Average score: 79.0<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 68.5<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 85.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 85.5<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2005==<br /> ===AMC 10A===<br /> *Average score: 74.0<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 79.0<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 78.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 83.4<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2004==<br /> ===AMC 10A===<br /> *Average score: 69.1<br /> *AIME floor: 110<br /> <br /> ===AMC 10B===<br /> *Average score: 80.4<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 73.9<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 84.5<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2003==<br /> ===AMC 10A===<br /> *Average score: 74.4<br /> *AIME floor: 119<br /> <br /> ===AMC 10B===<br /> *Average score: 79.6<br /> *AIME floor: 121<br /> <br /> ===AMC 12A===<br /> *Average score: 77.8<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 76.6<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2002==<br /> ===AMC 10A===<br /> *Average score: 68.5<br /> *AIME floor: 115<br /> <br /> ===AMC 10B===<br /> *Average score: 74.9<br /> *AIME floor: 118<br /> <br /> ===AMC 12A===<br /> *Average score: 72.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 80.8<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2001==<br /> ===AMC 10===<br /> *Average score: 67.8<br /> *AIME floor: 116<br /> <br /> ===AMC 12===<br /> *Average score: 56.6<br /> *AIME floor: 84<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2000==<br /> ===AMC 10===<br /> *Average score: 64.2<br /> *AIME floor: <br /> <br /> ===AMC 12===<br /> *Average score: 64.9<br /> *AIME floor: <br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==1999==<br /> ===AHSME===<br /> *Average score: 68.8<br /> *AIME floor:<br /> <br /> ===AIME===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==1998==<br /> ==1997==<br /> ==1996==<br /> ==1995==<br /> ==1994==<br /> ==1993==<br /> ==1992==<br /> ==1991==<br /> ==1990==<br /> ==1989==<br /> ==1988==<br /> ==1987==<br /> ==1986==<br /> ==1985==<br /> ==1984==<br /> ==1983==<br /> ==1982==<br /> ==1981==<br /> ==1980==<br /> ==1979==<br /> ==1978==<br /> ==1977==<br /> ==1976==<br /> ==1975==<br /> ==1974==<br /> ==1973==<br /> ==1972==<br /> ==1971==<br /> ==1970==<br /> ==1969==<br /> ==1968==<br /> ==1967==<br /> ==1966==<br /> ==1965==<br /> ==1964==<br /> ==1963==<br /> ==1962==<br /> ==1961==<br /> ==1960==<br /> <br /> ==1959==</div> Jj ca888 https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=84331 AMC historical results 2017-03-02T01:12:55Z <p>Jj ca888: /* AMC 10B */</p> <hr /> <div>&lt;!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --&gt;<br /> This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br /> <br /> ==2017==<br /> ===AMC 10A===<br /> *Average score: <br /> *AIME floor: 112.5<br /> <br /> ===AMC 10B===<br /> *Average score: <br /> *AIME floor: 100<br /> <br /> ===AMC 12A===<br /> *Average score:<br /> *AIME floor: 96<br /> <br /> ===AMC 12B===<br /> *Average score:<br /> *AIME floor: 100.5<br /> <br /> ===AIME I===<br /> *Average score: <br /> *Median score: <br /> *USAMO cutoff: <br /> *USAJMO cutoff: <br /> <br /> ===AIME II===<br /> *Average score: <br /> *Median score: <br /> *USAMO cutoff: <br /> *USAJMO cutoff:<br /> <br /> ==2015==<br /> ===AMC 10A===<br /> *Average score: 73.39<br /> *AIME floor: 106.5<br /> <br /> ===AMC 10B===<br /> *Average score: 76.10<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 69.90<br /> *AIME floor: 99<br /> <br /> ===AMC 12B===<br /> *Average score: 66.92<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score: 5.29<br /> *Median score: 5<br /> *USAMO cutoff: 219.0<br /> *USAJMO cutoff: 213.0<br /> <br /> ===AIME II===<br /> *Average score: 6.63<br /> *Median score: 6<br /> *USAMO cutoff: 229.0<br /> *USAJMO cutoff: 223.5<br /> <br /> ==2014==<br /> ===AMC 10A===<br /> *Average score: 63.83<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 71.44<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 64.01<br /> *AIME floor: 93<br /> <br /> ===AMC 12B===<br /> *Average score: 68.11<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score: 4.88<br /> *Median score: 5<br /> *USAMO cutoff: 211.5<br /> *USAJMO cutoff: 211<br /> <br /> ===AIME II===<br /> *Average score: 5.49<br /> *Median score: 5<br /> *USAMO cutoff: 211.5<br /> *USAJMO cutoff: 211<br /> <br /> ==2013==<br /> ===AMC 10A===<br /> *Average score: 72.50<br /> *AIME floor: 108<br /> <br /> ===AMC 10B===<br /> *Average score: 72.62<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 65.06<br /> *AIME floor:88.5<br /> <br /> ===AMC 12B===<br /> *Average score: 64.21<br /> *AIME floor: 93<br /> <br /> ===AIME I===<br /> *Average score: 4.69<br /> *Median score: <br /> *USAMO cutoff: 209<br /> *USAJMO cutoff: 210.5<br /> <br /> ===AIME II===<br /> *Average score: 6.56<br /> *Median score: <br /> *USAMO cutoff: 209<br /> *USAJMO cutoff: 210.5<br /> <br /> ==2012==<br /> ===AMC 10A===<br /> *Average score: 72.51<br /> *AIME floor: 115.5<br /> <br /> ===AMC 10B===<br /> *Average score: 76.59<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 64.62<br /> *AIME floor: 94.5<br /> <br /> ===AMC 12B===<br /> *Average score: 70.08<br /> *AIME floor: 99<br /> <br /> ===AIME I===<br /> *Average score: 5.13<br /> *Median score: <br /> *USAMO cutoff: 204.5<br /> *USAJMO cutoff: 204<br /> <br /> ===AIME II===<br /> *Average score: 4.94<br /> *Median score: <br /> *USAMO cutoff: 204.5<br /> *USAJMO cutoff: 204<br /> <br /> ==2011==<br /> ===AMC 10A===<br /> *Average score: 67.61<br /> *AIME floor: 117<br /> <br /> ===AMC 10B===<br /> *Average score: 71.78<br /> *AIME floor: 117<br /> <br /> ===AMC 12A===<br /> *Average score: 66.77<br /> *AIME floor: 93<br /> <br /> ===AMC 12B===<br /> *Average score: 64.71<br /> *AIME floor: 97.5<br /> <br /> ===AIME I===<br /> *Average score: 5.47<br /> *Median score: <br /> *USAMO cutoff: 215.5<br /> *USAJMO cutoff: 196.5<br /> <br /> ===AIME II===<br /> *Average score: 2.23<br /> *Median score: <br /> *USAMO cutoff: 188<br /> *USAJMO cutoff: 179<br /> <br /> ==2010==<br /> ===AMC 10A===<br /> *Average score: 68.11<br /> *AIME floor: 115.5<br /> <br /> ===AMC 10B===<br /> *Average score: 68.57<br /> *AIME floor: 118.5<br /> <br /> ===AMC 12A===<br /> *Average score: 61.02<br /> *AIME floor: 88.5<br /> <br /> ===AMC 12B===<br /> *Average score: 59.58<br /> *AIME floor: 88.5<br /> <br /> ===AIME I===<br /> *Average score: 5.90<br /> *Median score: <br /> *USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br /> *USAJMO cutoff: 188.5<br /> <br /> ===AIME II===<br /> *Average score: 3.39<br /> *Median score: <br /> *USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br /> *USAJMO cutoff: 188.5<br /> <br /> ==2009==<br /> ===AMC 10A===<br /> *Average score: 67.41<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 74.73<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 66.37<br /> *AIME floor: <br /> <br /> ===AMC 12B===<br /> *Average score: 71.88<br /> *AIME floor: <br /> <br /> ===AIME I===<br /> *Average score: <br /> *Median score: <br /> *USAMO floor: <br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2008==<br /> ===AMC 10A===<br /> *Average score: <br /> *AIME floor: <br /> <br /> ===AMC 10B===<br /> *Average score: <br /> *AIME floor: <br /> <br /> ===AMC 12A===<br /> *Average score: 65.6<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 68.9<br /> *AIME floor: 97.5<br /> <br /> ===AIME I===<br /> *Average score: <br /> *Median score: <br /> *USAMO floor: <br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2007==<br /> <br /> ===AMC 10A===<br /> *Average score: 67.9<br /> *AIME floor: 117<br /> <br /> ===AMC 10B===<br /> *Average score: 61.5<br /> *AIME floor: 115.5<br /> <br /> ===AMC 12A===<br /> *Average score: 66.8<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 73.1<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score: 5<br /> *Median score: 3<br /> *USAMO floor: 6<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2006==<br /> ===AMC 10A===<br /> *Average score: 79.0<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 68.5<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 85.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 85.5<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2005==<br /> ===AMC 10A===<br /> *Average score: 74.0<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 79.0<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 78.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 83.4<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2004==<br /> ===AMC 10A===<br /> *Average score: 69.1<br /> *AIME floor: 110<br /> <br /> ===AMC 10B===<br /> *Average score: 80.4<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 73.9<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 84.5<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2003==<br /> ===AMC 10A===<br /> *Average score: 74.4<br /> *AIME floor: 119<br /> <br /> ===AMC 10B===<br /> *Average score: 79.6<br /> *AIME floor: 121<br /> <br /> ===AMC 12A===<br /> *Average score: 77.8<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 76.6<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2002==<br /> ===AMC 10A===<br /> *Average score: 68.5<br /> *AIME floor: 115<br /> <br /> ===AMC 10B===<br /> *Average score: 74.9<br /> *AIME floor: 118<br /> <br /> ===AMC 12A===<br /> *Average score: 72.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 80.8<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2001==<br /> ===AMC 10===<br /> *Average score: 67.8<br /> *AIME floor: 116<br /> <br /> ===AMC 12===<br /> *Average score: 56.6<br /> *AIME floor: 84<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2000==<br /> ===AMC 10===<br /> *Average score: 64.2<br /> *AIME floor: <br /> <br /> ===AMC 12===<br /> *Average score: 64.9<br /> *AIME floor: <br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==1999==<br /> ===AHSME===<br /> *Average score: 68.8<br /> *AIME floor:<br /> <br /> ===AIME===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==1998==<br /> ==1997==<br /> ==1996==<br /> ==1995==<br /> ==1994==<br /> ==1993==<br /> ==1992==<br /> ==1991==<br /> ==1990==<br /> ==1989==<br /> ==1988==<br /> ==1987==<br /> ==1986==<br /> ==1985==<br /> ==1984==<br /> ==1983==<br /> ==1982==<br /> ==1981==<br /> ==1980==<br /> ==1979==<br /> ==1978==<br /> ==1977==<br /> ==1976==<br /> ==1975==<br /> ==1974==<br /> ==1973==<br /> ==1972==<br /> ==1971==<br /> ==1970==<br /> ==1969==<br /> ==1968==<br /> ==1967==<br /> ==1966==<br /> ==1965==<br /> ==1964==<br /> ==1963==<br /> ==1962==<br /> ==1961==<br /> ==1960==<br /> <br /> ==1959==</div> Jj ca888 https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=84093 AMC historical results 2017-02-19T01:57:54Z <p>Jj ca888: /* 1960 */</p> <hr /> <div>&lt;!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --&gt;<br /> This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br /> <br /> ==2017==<br /> ===AMC 10A===<br /> *Average score: TBA<br /> *AIME floor: TBA<br /> <br /> ===AMC 10B===<br /> *Average score: TBA<br /> *AIME floor: TBA<br /> <br /> ===AMC 12A===<br /> *Average score:<br /> *AIME floor:<br /> <br /> ===AMC 12B===<br /> *Average score:<br /> *AIME floor:<br /> <br /> ===AIME I===<br /> *Average score: <br /> *Median score: <br /> *USAMO cutoff: <br /> *USAJMO cutoff: <br /> <br /> ===AIME II===<br /> *Average score: <br /> *Median score: <br /> *USAMO cutoff: <br /> *USAJMO cutoff:<br /> <br /> ==2016==<br /> ===AMC 10A===<br /> *Average score: 60<br /> *AIME floor: 110<br /> <br /> ===AMC 10B===<br /> *Average score: 64.54<br /> *AIME floor: 110<br /> <br /> ===AMC 12A===<br /> *Average score: 59.06<br /> *AIME floor: 92<br /> <br /> ===AMC 12B===<br /> *Average score: 67.96<br /> *AIME floor: 100.0 (Top 5% 106.5)<br /> <br /> ===AIME I===<br /> *Average score: <br /> *Median score: <br /> *USAMO cutoff: 220<br /> *USAJMO cutoff: 210.5<br /> <br /> ===AIME II===<br /> *Average score: <br /> *Median score: <br /> *USAMO cutoff: 205<br /> *USAJMO cutoff: 200<br /> <br /> ==2015==<br /> ===AMC 10A===<br /> *Average score: 73.39<br /> *AIME floor: 106.5<br /> <br /> ===AMC 10B===<br /> *Average score: 76.10<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 69.90<br /> *AIME floor: 99<br /> <br /> ===AMC 12B===<br /> *Average score: 66.92<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score: 5.29<br /> *Median score: 5<br /> *USAMO cutoff: 219.0<br /> *USAJMO cutoff: 213.0<br /> <br /> ===AIME II===<br /> *Average score: 6.63<br /> *Median score: 6<br /> *USAMO cutoff: 229.0<br /> *USAJMO cutoff: 223.5<br /> <br /> ==2014==<br /> ===AMC 10A===<br /> *Average score: 63.83<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 71.44<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 64.01<br /> *AIME floor: 93<br /> <br /> ===AMC 12B===<br /> *Average score: 68.11<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score: 4.88<br /> *Median score: 5<br /> *USAMO cutoff: 211.5<br /> *USAJMO cutoff: 211<br /> <br /> ===AIME II===<br /> *Average score: 5.49<br /> *Median score: 5<br /> *USAMO cutoff: 211.5<br /> *USAJMO cutoff: 211<br /> <br /> ==2013==<br /> ===AMC 10A===<br /> *Average score: 72.50<br /> *AIME floor: 108<br /> <br /> ===AMC 10B===<br /> *Average score: 72.62<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 65.06<br /> *AIME floor:88.5<br /> <br /> ===AMC 12B===<br /> *Average score: 64.21<br /> *AIME floor: 93<br /> <br /> ===AIME I===<br /> *Average score: 4.69<br /> *Median score: <br /> *USAMO cutoff: 209<br /> *USAJMO cutoff: 210.5<br /> <br /> ===AIME II===<br /> *Average score: 6.56<br /> *Median score: <br /> *USAMO cutoff: 209<br /> *USAJMO cutoff: 210.5<br /> <br /> ==2012==<br /> ===AMC 10A===<br /> *Average score: 72.51<br /> *AIME floor: 115.5<br /> <br /> ===AMC 10B===<br /> *Average score: 76.59<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 64.62<br /> *AIME floor: 94.5<br /> <br /> ===AMC 12B===<br /> *Average score: 70.08<br /> *AIME floor: 99<br /> <br /> ===AIME I===<br /> *Average score: 5.13<br /> *Median score: <br /> *USAMO cutoff: 204.5<br /> *USAJMO cutoff: 204<br /> <br /> ===AIME II===<br /> *Average score: 4.94<br /> *Median score: <br /> *USAMO cutoff: 204.5<br /> *USAJMO cutoff: 204<br /> <br /> ==2011==<br /> ===AMC 10A===<br /> *Average score: 67.61<br /> *AIME floor: 117<br /> <br /> ===AMC 10B===<br /> *Average score: 71.78<br /> *AIME floor: 117<br /> <br /> ===AMC 12A===<br /> *Average score: 66.77<br /> *AIME floor: 93<br /> <br /> ===AMC 12B===<br /> *Average score: 64.71<br /> *AIME floor: 97.5<br /> <br /> ===AIME I===<br /> *Average score: 5.47<br /> *Median score: <br /> *USAMO cutoff: 215.5<br /> *USAJMO cutoff: 196.5<br /> <br /> ===AIME II===<br /> *Average score: 2.23<br /> *Median score: <br /> *USAMO cutoff: 188<br /> *USAJMO cutoff: 179<br /> <br /> ==2010==<br /> ===AMC 10A===<br /> *Average score: 68.11<br /> *AIME floor: 115.5<br /> <br /> ===AMC 10B===<br /> *Average score: 68.57<br /> *AIME floor: 118.5<br /> <br /> ===AMC 12A===<br /> *Average score: 61.02<br /> *AIME floor: 88.5<br /> <br /> ===AMC 12B===<br /> *Average score: 59.58<br /> *AIME floor: 88.5<br /> <br /> ===AIME I===<br /> *Average score: 5.90<br /> *Median score: <br /> *USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br /> *USAJMO cutoff: 188.5<br /> <br /> ===AIME II===<br /> *Average score: 3.39<br /> *Median score: <br /> *USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br /> *USAJMO cutoff: 188.5<br /> <br /> ==2009==<br /> ===AMC 10A===<br /> *Average score: 67.41<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 74.73<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 66.37<br /> *AIME floor: <br /> <br /> ===AMC 12B===<br /> *Average score: 71.88<br /> *AIME floor: <br /> <br /> ===AIME I===<br /> *Average score: <br /> *Median score: <br /> *USAMO floor: <br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2008==<br /> ===AMC 10A===<br /> *Average score: <br /> *AIME floor: <br /> <br /> ===AMC 10B===<br /> *Average score: <br /> *AIME floor: <br /> <br /> ===AMC 12A===<br /> *Average score: 65.6<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 68.9<br /> *AIME floor: 97.5<br /> <br /> ===AIME I===<br /> *Average score: <br /> *Median score: <br /> *USAMO floor: <br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2007==<br /> <br /> ===AMC 10A===<br /> *Average score: 67.9<br /> *AIME floor: 117<br /> <br /> ===AMC 10B===<br /> *Average score: 61.5<br /> *AIME floor: 115.5<br /> <br /> ===AMC 12A===<br /> *Average score: 66.8<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 73.1<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score: 5<br /> *Median score: 3<br /> *USAMO floor: 6<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2006==<br /> ===AMC 10A===<br /> *Average score: 79.0<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 68.5<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 85.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 85.5<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2005==<br /> ===AMC 10A===<br /> *Average score: 74.0<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 79.0<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 78.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 83.4<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2004==<br /> ===AMC 10A===<br /> *Average score: 69.1<br /> *AIME floor: 110<br /> <br /> ===AMC 10B===<br /> *Average score: 80.4<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 73.9<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 84.5<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2003==<br /> ===AMC 10A===<br /> *Average score: 74.4<br /> *AIME floor: 119<br /> <br /> ===AMC 10B===<br /> *Average score: 79.6<br /> *AIME floor: 121<br /> <br /> ===AMC 12A===<br /> *Average score: 77.8<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 76.6<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2002==<br /> ===AMC 10A===<br /> *Average score: 68.5<br /> *AIME floor: 115<br /> <br /> ===AMC 10B===<br /> *Average score: 74.9<br /> *AIME floor: 118<br /> <br /> ===AMC 12A===<br /> *Average score: 72.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 80.8<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2001==<br /> ===AMC 10===<br /> *Average score: 67.8<br /> *AIME floor: 116<br /> <br /> ===AMC 12===<br /> *Average score: 56.6<br /> *AIME floor: 84<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2000==<br /> ===AMC 10===<br /> *Average score: 64.2<br /> *AIME floor: <br /> <br /> ===AMC 12===<br /> *Average score: 64.9<br /> *AIME floor: <br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==1999==<br /> ===AHSME===<br /> *Average score: 68.8<br /> *AIME floor:<br /> <br /> ===AIME===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==1998==<br /> ==1997==<br /> ==1996==<br /> ==1995==<br /> ==1994==<br /> ==1993==<br /> ==1992==<br /> ==1991==<br /> ==1990==<br /> ==1989==<br /> ==1988==<br /> ==1987==<br /> ==1986==<br /> ==1985==<br /> ==1984==<br /> ==1983==<br /> ==1982==<br /> ==1981==<br /> ==1980==<br /> ==1979==<br /> ==1978==<br /> ==1977==<br /> ==1976==<br /> ==1975==<br /> ==1974==<br /> ==1973==<br /> ==1972==<br /> ==1971==<br /> ==1970==<br /> ==1969==<br /> ==1968==<br /> ==1967==<br /> ==1966==<br /> ==1965==<br /> ==1964==<br /> ==1963==<br /> ==1962==<br /> ==1961==<br /> ==1960==<br /> <br /> ==1959==</div> Jj ca888 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems&diff=83176 2017 AMC 10A Problems 2017-02-09T01:09:12Z <p>Jj ca888: /* Problem 22 */</p> <hr /> <div>{{AMC10 Problems|year=2017|ab=A}}<br /> <br /> ==Problem 1==<br /> What is the value of &lt;math&gt;(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 70\qquad\textbf{(B)}\ 97\qquad\textbf{(C)}\ 127\qquad\textbf{(D)}\ 159\qquad\textbf{(E)}\ 729&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> Pablo buys popsicles for his friends. The store sells single popsicles for \$1 each, 3-popsicle boxes for \$2 each, and 5-popsicle boxes for \$3. What is the greatest number of popsicles that Pablo can buy with \\$8?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 8\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 15&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Tamara has three rows of two 6-feet by 2-feet flower beds in her garden. The beds are separated and also surrounded by 1-foot-wide walkways, as shown on the diagram. What is the total area of the walkways, in square feet?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 72\qquad\textbf{(B)}\ 78\qquad\textbf{(C)}\ 90\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 150&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> Mia is “helping” her mom pick up &lt;math&gt;30&lt;/math&gt; toys that are strewn on the floor. Mia’s mom manages to put &lt;math&gt;3&lt;/math&gt; toys into the toy box every &lt;math&gt;30&lt;/math&gt; seconds, but each time immediately after those &lt;math&gt;30&lt;/math&gt; seconds have elapsed, Mia takes &lt;math&gt;2&lt;/math&gt; toys out of the box. How much time, in minutes, will it take Mia and her mom to put all &lt;math&gt;30&lt;/math&gt; toys into the box for the first time?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 14.5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 15.5&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> The sum of two nonzero real numbers is 4 times their product. What is the sum of the reciprocals of the two numbers?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Ms. Carroll promised that anyone who got all the multiple choice questions right on the upcoming exam would receive an A on the exam. Which of of these statements necessarily follows logically?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{If Lewis did not receive an A, then he got all of the multiple choice questions wrong.}\\\textbf{(B)}\ \text{If Lewis did not receive an A, then he got at least one of the multiple choice questions wrong.}\\\textbf{(C)}\ \text{If Lewis got at least one of the multiple choice questions wrong, then he did not receive an A. }\\\textbf{(D)}\ \text{If Lewis received an A, then he got all of the multiple choice questions right.}\\\textbf{(E)}\ \text{If Lewis received an A, then he got at least one of the multiple choice questions right.}&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> Jerry and Silvia wanted to go from the southwest corner of a square field to the northeast corner. Jerry walked due east and then due north to reach the goal, but Silvia headed northeast and reached the goal walking in a straight line. Which of the following is closest to how much shorter Silvia's trip was, compared to Jerry's trip?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 30\%\qquad\textbf{(B)}\ 40\%\qquad\textbf{(C)}\ 50\%\qquad\textbf{(D)}\ 60\%\qquad\textbf{(E)}\ 70\%&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> At a gathering of 30 people, there are 20 people who all know each other and 10 people who know no one. People who know each other a hug, and people who do not know each other shake hands. How many handshakes occur?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 490&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> Minnie rides on a flat road at &lt;math&gt;20&lt;/math&gt; kilometers per hour (kph), downhill at &lt;math&gt;30&lt;/math&gt; kph, and uphill at &lt;math&gt;5&lt;/math&gt; kph. Penny rides on a flat road at &lt;math&gt;30&lt;/math&gt; kph, downhill at &lt;math&gt;40&lt;/math&gt; kph, and uphill at &lt;math&gt;10&lt;/math&gt; kph. Minnie goes from town &lt;math&gt;A&lt;/math&gt; to town &lt;math&gt;B&lt;/math&gt;, a distance of &lt;math&gt;10&lt;/math&gt; km all uphill, then from town &lt;math&gt;B&lt;/math&gt; to town &lt;math&gt;C&lt;/math&gt;, a distance of &lt;math&gt;15&lt;/math&gt; km all downhill, and then back to town &lt;math&gt;A&lt;/math&gt;, a distance of &lt;math&gt;20&lt;/math&gt; km on the flat. Penny goes the other way around using the same route. How many more minutes does it take Minnie to complete the &lt;math&gt;45&lt;/math&gt;-km ride than it takes Penny?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 45\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 65\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ 95&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> Joy has &lt;math&gt;30&lt;/math&gt; thin rods, one each of every integer length from &lt;math&gt;1&lt;/math&gt; cm through &lt;math&gt;30&lt;/math&gt; cm. She places the rods with lengths &lt;math&gt;3&lt;/math&gt; cm, &lt;math&gt;7&lt;/math&gt; cm, and &lt;math&gt;15&lt;/math&gt; cm on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area. How many of the remaining rods can she choose as the fourth rod?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 20&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> The region consisting of all points in three-dimensional space within 3 units of line segment &lt;math&gt;\overline{AB}&lt;/math&gt; has volume 216&lt;math&gt;\pi&lt;/math&gt;. What is the length &lt;math&gt;\textit{AB}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 6\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> Let &lt;math&gt;S&lt;/math&gt; be a set of points &lt;math&gt;(x,y)&lt;/math&gt; in the coordinate plane such that two of the three quantities &lt;math&gt;3,~x+2,&lt;/math&gt; and &lt;math&gt;y-4&lt;/math&gt; are equal and the third of the three quantities is no greater than this common value. Which of the following is a correct description for &lt;math&gt;S?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{a single point} \qquad\textbf{(B)}\ \text{two intersecting lines} \\\qquad\textbf{(C)}\ \text{ three lines whose pairwise intersections are three distinct points} \\\qquad\textbf{(D)}\ \text{a triangle} \qquad\textbf{(E)}\ \text{three rays with a common endpoint}&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Define a sequence recursively by &lt;math&gt;F_{0}=0,~F_{1}=1,&lt;/math&gt; and &lt;math&gt;F_{n}=&lt;/math&gt; the remainder when &lt;math&gt;F_{n-1}+F_{n-2}&lt;/math&gt; is divided by &lt;math&gt;3,&lt;/math&gt; for all &lt;math&gt;n\geq 2.&lt;/math&gt; Thus the sequence starts &lt;math&gt;0,1,1,2,0,2,\ldots&lt;/math&gt; What is &lt;math&gt;F_{2017}+F_{2018}+F_{2019}+F_{2020}+F_{2021}+F_{2022}+F_{2023}+F_{2024}?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was &lt;math&gt;A&lt;/math&gt; dollars. The cost of his movie ticket was &lt;math&gt;20\%&lt;/math&gt; of the difference between &lt;math&gt;A&lt;/math&gt; and the cost of his soda, while the cost of his soda was &lt;math&gt;5\%&lt;/math&gt; of the difference between &lt;math&gt;A&lt;/math&gt; and the cost of his movie ticket. To the nearest whole percent, what fraction of &lt;math&gt;A&lt;/math&gt; did Roger pay for his movie ticket and soda?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 9\%\qquad\textbf{(B)}\ 19\%\qquad\textbf{(C)}\ 22\%\qquad\textbf{(D)}\ 23\%\qquad\textbf{(E)}\ 25\%&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Chloé chooses a real number uniformly at random from the interval &lt;math&gt;[0, 2017]&lt;/math&gt;. Independently, Laurent cooses a real number uniformly at random from the interval &lt;math&gt;[0, 4034]&lt;/math&gt;. What is the probability that Laurent's number is greater than Chloé's number?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{2}\qquad\textbf{(B)}\ \frac{2}{3}\qquad\textbf{(C)}\ \frac{3}{4}\qquad\textbf{(D)}\ \frac{5}{6}\qquad\textbf{(E)}\ \frac{7}{8}&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> There are 10 horses, named Horse 1, Horse 2, &lt;math&gt;\ldots&lt;/math&gt;, Horse 10. They get their names from how many minutes it takes them to run one lap around a circular race track: Horse &lt;math&gt;k&lt;/math&gt; runs one lap in exactly &lt;math&gt;k&lt;/math&gt; minutes. At time 0 all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time &lt;math&gt;S&gt;0&lt;/math&gt;, in minutes, at which all 10 horses will again simultaneously be at the starting point is &lt;math&gt;S=2520&lt;/math&gt;. Let &lt;math&gt;T&gt;0&lt;/math&gt; be the least time, in minutes, such that at least 5 of the horses are again at the starting point. What is the sum of the digits of &lt;math&gt;T&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> Distinct points &lt;math&gt;P&lt;/math&gt;, &lt;math&gt;Q&lt;/math&gt;, &lt;math&gt;R&lt;/math&gt;, &lt;math&gt;S&lt;/math&gt; lie on the circle &lt;math&gt;x^2+y^2=25&lt;/math&gt; and have integer coordinates. The distances &lt;math&gt;PQ&lt;/math&gt; and &lt;math&gt;RS&lt;/math&gt; are irrational numbers. What is the greatest possible value of the ratio &lt;math&gt;\frac{PQ}{RS}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 3\sqrt{5}\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 5\sqrt{2}&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> Amelia has a coin that lands heads with probability &lt;math&gt;\frac{1}{3}&lt;/math&gt;, and Blaine has a coin that lands on heads with probability &lt;math&gt;\frac{2}{5}&lt;/math&gt;. Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins. All coin tosses are independent. Amelia goes first. The probability that Amelia wins is &lt;math&gt;\frac{p}{q}&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;q-p&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of 5 chairs under these conditions?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 12\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 40&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> Let &lt;math&gt;S(n)&lt;/math&gt; equal the sum of the digits of positive integer &lt;math&gt;n&lt;/math&gt;. For example, &lt;math&gt;S(1507) = 13&lt;/math&gt;. For a particular positive integer &lt;math&gt;n&lt;/math&gt;, &lt;math&gt;S(n) = 1274&lt;/math&gt;. Which of the following could be the value of &lt;math&gt;S(n+1)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> A square with side length &lt;math&gt;x&lt;/math&gt; is inscribed in a right triangle with sides of length &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, and &lt;math&gt;5&lt;/math&gt; so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length &lt;math&gt;y&lt;/math&gt; is inscribed in another right triangle with sides of length &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, and &lt;math&gt;5&lt;/math&gt; so that one side of the square lies on the hypotenuse of the triangle. What is &lt;math&gt;\tfrac{x}{y}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } \dfrac{12}{13} \qquad \textbf{(B) } \dfrac{35}{37} \qquad \textbf{(C) } 1 \qquad \textbf{(D) } \dfrac{37}{35} \qquad \textbf{(E) } \dfrac{13}{12}&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> Sides &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{AC}&lt;/math&gt; of equilateral triangle &lt;math&gt;ABC&lt;/math&gt; are tangent to a circle at points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; respectively. What fraction of the area of &lt;math&gt;\triangle ABC&lt;/math&gt; lies outside the circle?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{4\sqrt{3}\pi}{27}-\frac{1}{3}\qquad\textbf{(B)}\ \frac{\sqrt{3}}{2}-\frac{\pi}{8}\qquad\textbf{(C)}\ \frac{1}{2}\qquad\textbf{(D)}\ \sqrt{3}-\frac{2\sqrt{3}\pi}{9}\qquad\textbf{(E)}\ \frac{4}{3}-\frac{4\sqrt{3}\pi}{27}&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> How many triangles with positive area have all their vertices at points &lt;math&gt;(i,j)&lt;/math&gt; in the coordinate plane, where &lt;math&gt;i&lt;/math&gt; and &lt;math&gt;j&lt;/math&gt; are integers between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt;, inclusive?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2128 \qquad\textbf{(B)}\ 2148 \qquad\textbf{(C)}\ 2160 \qquad\textbf{(D)}\ 2200 \qquad\textbf{(E)}\ 2300&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> For certain real numbers &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt;, the polynomial &lt;cmath&gt;g(x) = x^3 + ax^2 + x + 10&lt;/cmath&gt;has three distinct roots, and each root of &lt;math&gt;g(x)&lt;/math&gt; is also a root of the polynomial &lt;cmath&gt;f(x) = x^4 + x^3 + bx^2 + 100x + c.&lt;/cmath&gt;What is &lt;math&gt;f(1)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ -9009\qquad\textbf{(B)}\ -8008\qquad\textbf{(C)}\ -7007\qquad\textbf{(D)}\ -6006\qquad\textbf{(E)}\ -5005&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> How many integers between 100 and 999, inclusive, have the property that some permutation of its digits is a multiple of 11 between 100 and 999? For example, both 121 and 211 have this property.<br /> <br /> &lt;math&gt;\textbf{(A)}\ 226\qquad\textbf{(B)}\ 243\qquad\textbf{(C)}\ 270\qquad\textbf{(D)}\ 469\qquad\textbf{(E)}\ 486&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC10 box|year=2017|ab=A|before=[[2016 AMC 10B Problems]]|after=[[2017 AMC 10B Problems]]}}<br /> {{MAA Notice}}</div> Jj ca888 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_25&diff=81190 2013 AMC 10B Problems/Problem 25 2016-11-20T04:20:34Z <p>Jj ca888: /* See also */</p> <hr /> <div>{{duplicate|[[2013 AMC 12B Problems|2013 AMC 12B #23]] and [[2013 AMC 10B Problems|2013 AMC 10B #25]]}}<br /> <br /> ==Problem==<br /> <br /> Bernardo chooses a three-digit positive integer &lt;math&gt;N&lt;/math&gt; and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer &lt;math&gt;S&lt;/math&gt;. For example, if &lt;math&gt;N = 749&lt;/math&gt;, Bernardo writes the numbers &lt;math&gt;10,\!444&lt;/math&gt; and &lt;math&gt;3,\!245&lt;/math&gt;, and LeRoy obtains the sum &lt;math&gt;S = 13,\!689&lt;/math&gt;. For how many choices of &lt;math&gt;N&lt;/math&gt; are the two rightmost digits of &lt;math&gt;S&lt;/math&gt;, in order, the same as those of &lt;math&gt;2N&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 5 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 25&lt;/math&gt;<br /> <br /> ==Solution==<br /> First, we can examine the units digits of the number base 5 and base 6 and eliminate some possibilities.<br /> <br /> Say that &lt;math&gt;N \equiv a \pmod{6}&lt;/math&gt;<br /> <br /> also that &lt;math&gt;N \equiv b \pmod{5}&lt;/math&gt;<br /> <br /> Substituting these equations into the question and setting the units digits of 2N and S equal to each other, it can be seen that &lt;math&gt;a=b&lt;/math&gt;, and &lt;math&gt;b &lt; 5&lt;/math&gt;, (otherwise &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; always have different parities) so<br /> &lt;math&gt;N \equiv a \pmod{6}&lt;/math&gt;,<br /> &lt;math&gt;N \equiv a \pmod{5}&lt;/math&gt;,<br /> &lt;math&gt;\implies N=a \pmod{30}&lt;/math&gt;,<br /> &lt;math&gt;0 \le a \le 4 &lt;/math&gt;<br /> <br /> Therefore, &lt;math&gt;N&lt;/math&gt; can be written as &lt;math&gt;30x+y&lt;/math&gt;<br /> and &lt;math&gt;2N&lt;/math&gt; can be written as &lt;math&gt;60x+2y&lt;/math&gt; <br /> <br /> Keep in mind that &lt;math&gt;y&lt;/math&gt; can be one of five choices: &lt;math&gt;0, 1, 2, 3,&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt;, ;<br /> Also, we have already found which digits of &lt;math&gt;y&lt;/math&gt; will add up into the units digits of &lt;math&gt;2N&lt;/math&gt;.<br /> <br /> Now, examine the tens digit, &lt;math&gt;x&lt;/math&gt; by using &lt;math&gt;\mod{25}&lt;/math&gt; and &lt;math&gt;\mod{36}&lt;/math&gt; to find the tens digit (units digits can be disregarded because &lt;math&gt;y=0,1,2,3,4&lt;/math&gt; will always work)<br /> Then we see that &lt;math&gt;N=30x+y&lt;/math&gt; and take it &lt;math&gt;\mod{25}&lt;/math&gt; and &lt;math&gt;\mod{36}&lt;/math&gt; to find the last two digits in the base &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;6&lt;/math&gt; representation.<br /> &lt;cmath&gt;N \equiv 30x \pmod{36}&lt;/cmath&gt;<br /> &lt;cmath&gt;N \equiv 30x \equiv 5x \pmod{25}&lt;/cmath&gt; <br /> Both of those must add up to <br /> &lt;cmath&gt;2N\equiv60x \pmod{100}&lt;/cmath&gt;<br /> <br /> (&lt;math&gt;33 \ge x \ge 4&lt;/math&gt;)<br /> <br /> Now, since &lt;math&gt;y=0,1,2,3,4&lt;/math&gt; will always work if &lt;math&gt;x&lt;/math&gt; works, then we can treat &lt;math&gt;x&lt;/math&gt; as a units digit instead of a tens digit in the respective bases and decrease the mods so that &lt;math&gt;x&lt;/math&gt; is now the units digit.<br /> &lt;cmath&gt;N \equiv 6x \equiv x \pmod{5}&lt;/cmath&gt; <br /> &lt;cmath&gt;N \equiv 5x \pmod{6}&lt;/cmath&gt; <br /> &lt;cmath&gt;2N\equiv 6x \pmod{10}&lt;/cmath&gt;<br /> <br /> Say that &lt;math&gt;x=5m+n&lt;/math&gt; (m is between 0-6, n is 0-4 because of constraints on x)<br /> Then <br /> <br /> &lt;cmath&gt;N \equiv 5m+n \pmod{5}&lt;/cmath&gt; <br /> &lt;cmath&gt;N \equiv 25m+5n \pmod{6}&lt;/cmath&gt; <br /> &lt;cmath&gt;2N\equiv30m + 6n \pmod{10}&lt;/cmath&gt;<br /> <br /> and this simplifies to <br /> <br /> &lt;cmath&gt;N \equiv n \pmod{5}&lt;/cmath&gt; <br /> &lt;cmath&gt;N \equiv m+5n \pmod{6}&lt;/cmath&gt;<br /> &lt;cmath&gt;2N\equiv 6n \pmod{10}&lt;/cmath&gt;<br /> <br /> From inspection, when<br /> <br /> &lt;math&gt;n=0, m=6&lt;/math&gt;<br /> <br /> &lt;math&gt;n=1, m=6&lt;/math&gt;<br /> <br /> &lt;math&gt;n=2, m=2&lt;/math&gt;<br /> <br /> &lt;math&gt;n=3, m=2&lt;/math&gt;<br /> <br /> &lt;math&gt;n=4, m=4&lt;/math&gt;<br /> <br /> This gives you &lt;math&gt;5&lt;/math&gt; choices for &lt;math&gt;x&lt;/math&gt;, and &lt;math&gt;5&lt;/math&gt; choices for &lt;math&gt;y&lt;/math&gt;, so the answer is <br /> &lt;math&gt;5* 5 = \boxed{\textbf{(E) }25}&lt;/math&gt;<br /> <br /> ==Shortcut==<br /> Notice that there are exactly &lt;math&gt;1000-100=900=5^2\cdot 6^2&lt;/math&gt; possible values of &lt;math&gt;n&lt;/math&gt;. This means, from &lt;math&gt;100\le n\le 999&lt;/math&gt;, every possible combination of &lt;math&gt;2&lt;/math&gt; digits will happen exactly once. We know that &lt;math&gt;n=900,901,902,903,904&lt;/math&gt; works because &lt;math&gt;900\equiv\dots00_5\equiv\dots00_6&lt;/math&gt;.<br /> <br /> We know for sure that the units digit will add perfectly every &lt;math&gt;30&lt;/math&gt; added or subtracted, because &lt;math&gt;\text{lcm }5,6=30&lt;/math&gt;. So we only have to care about cases of &lt;math&gt;n&lt;/math&gt; every &lt;math&gt;30&lt;/math&gt; subtracted. In each case, &lt;math&gt;2n&lt;/math&gt; subtracts &lt;math&gt;6&lt;/math&gt;/adds &lt;math&gt;4&lt;/math&gt;, &lt;math&gt;n_5&lt;/math&gt; subtracts &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;n_6&lt;/math&gt; adds &lt;math&gt;1&lt;/math&gt; for the &lt;math&gt;10&lt;/math&gt;'s digit.<br /> <br /> &lt;cmath&gt;\textbf{5 }\textcolor{red}{\text{ 0}}\text{ 4 3 2 1 0 }\textcolor{red}{\text{4}}\text{ 3 2 1 0 4 3 2 1 0 4 }\textcolor{red}{\text{3 2}}\text{ 1 0 4 3 2 1 0 4 3 2 }\textcolor{red}{\text{1}}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\textbf{6 }\textcolor{red}{\text{ 0}}\text{ 1 2 3 4 5 }\textcolor{red}{\text{0}}\text{ 1 2 3 4 5 0 1 2 3 4 }\textcolor{red}{\text{5 0}}\text{ 1 2 3 4 5 0 1 2 3 4 }\textcolor{red}{\text{5}}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\textbf{10}\textcolor{red}{\text{ 0}}\text{ 4 8 2 6 0 }\textcolor{red}{\text{4}}\text{ 8 2 6 0 4 8 2 6 0 4 }\textcolor{red}{\text{8 2}}\text{ 6 0 4 8 2 6 0 4 8 2 }\textcolor{red}{\text{6}}&lt;/cmath&gt;<br /> <br /> As we can see, there are &lt;math&gt;5&lt;/math&gt; cases, including the original, that work. These are highlighted in &lt;math&gt;\textcolor{red}{\text{red}}&lt;/math&gt;. So, thus, there are &lt;math&gt;5&lt;/math&gt; possibilities for each case, and &lt;math&gt;5\cdot 5=\boxed{\textbf{(E) }25}&lt;/math&gt;.<br /> <br /> == See also == Ciao<br /> {{AMC12 box|year=2013|ab=B|num-b=22|num-a=24}}<br /> <br /> {{AMC10 box|year=2013|ab=B|num-b=24|after=Last Question}}<br /> <br /> {{MAA Notice}}</div> Jj ca888