https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Always+correct&feedformat=atom AoPS Wiki - User contributions [en] 2020-10-24T23:49:42Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=Euler_line&diff=86816 Euler line 2017-08-04T23:56:31Z <p>Always correct: </p> <hr /> <div>In any [[triangle]] &lt;math&gt;\triangle ABC&lt;/math&gt;, the '''Euler line''' is a [[line]] which passes through the [[orthocenter]] &lt;math&gt;H&lt;/math&gt;, [[centroid]] &lt;math&gt;G&lt;/math&gt;, [[circumcenter]] &lt;math&gt;O&lt;/math&gt;, [[nine-point center]] &lt;math&gt;N&lt;/math&gt; and [[De Longchamps point | de Longchamps point]] &lt;math&gt;L&lt;/math&gt;. It is named after [[Leonhard Euler]]. Its existence is a non-trivial fact of Euclidean [[geometry]]. Certain fixed orders and distance [[ratio]]s hold among these points. In particular, &lt;math&gt;\overline{OGNH}&lt;/math&gt; and &lt;math&gt;OG:GN:NH = 2:1:3&lt;/math&gt;<br /> <br /> Given the [[orthic triangle]] &lt;math&gt;\triangle H_AH_BH_C&lt;/math&gt; of &lt;math&gt;\triangle ABC&lt;/math&gt;, the Euler lines of &lt;math&gt;\triangle AH_BH_C&lt;/math&gt;,&lt;math&gt;\triangle BH_CH_A&lt;/math&gt;, and &lt;math&gt;\triangle CH_AH_B&lt;/math&gt; [[concurrence | concur]] at &lt;math&gt;N&lt;/math&gt;, the nine-point circle of &lt;math&gt;\triangle ABC&lt;/math&gt;.<br /> <br /> ==Proof Centroid Lies on Euler Line==<br /> This proof utilizes the concept of [[spiral similarity]], which in this case is a [[rotation]] followed [[homothety]]. Consider the [[medial triangle]] &lt;math&gt;\triangle O_AO_BO_C&lt;/math&gt;. It is similar to &lt;math&gt;\triangle ABC&lt;/math&gt;. Specifically, a rotation of &lt;math&gt;180^\circ&lt;/math&gt; about the midpoint of &lt;math&gt;O_BO_C&lt;/math&gt; followed by a homothety with scale factor &lt;math&gt;2&lt;/math&gt; centered at &lt;math&gt;A&lt;/math&gt; brings &lt;math&gt;\triangle ABC \to \triangle O_AO_BO_C&lt;/math&gt;. Let us examine what else this transformation, which we denote as &lt;math&gt;\mathcal{S}&lt;/math&gt;, will do. <br /> <br /> It turns out &lt;math&gt;O&lt;/math&gt; is the orthocenter, and &lt;math&gt;G&lt;/math&gt; is the centroid of &lt;math&gt;\triangle O_AO_BO_C&lt;/math&gt;. Thus, &lt;math&gt;\mathcal{S}(\{O_A, O, G\}) = \{A, H, G\}&lt;/math&gt;. As a homothety preserves angles, it follows that &lt;math&gt;\measuredangle O_AOG = \measuredangle AHG&lt;/math&gt;. Finally, as &lt;math&gt;\overline{AH} || \overline{O_AO}&lt;/math&gt; it follows that <br /> &lt;cmath&gt;\triangle AHG = \triangle O_AOG&lt;/cmath&gt;<br /> Thus, &lt;math&gt;O, G, H&lt;/math&gt; are collinear, and &lt;math&gt;\frac{OG}{HG} = \frac{1}{2}&lt;/math&gt;.<br /> <br /> ==Proof Nine-Point Center Lies on Euler Line==<br /> Assuming that the [[nine point circle]] exists and that &lt;math&gt;N&lt;/math&gt; is the center, note that a homothety centered at &lt;math&gt;H&lt;/math&gt; with factor &lt;math&gt;2&lt;/math&gt; brings the [[Euler point]]s &lt;math&gt;\{E_A, E_B, E_C\}&lt;/math&gt; onto the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt;. Thus, it brings the nine-point circle to the circumcircle. Additionally, &lt;math&gt;N&lt;/math&gt; should be sent to &lt;math&gt;O&lt;/math&gt;, thus &lt;math&gt;N \in \overline{HO}&lt;/math&gt; and &lt;math&gt;\frac{HN}{ON} = 1&lt;/math&gt;.<br /> <br /> ==Analytic Proof of Existence==<br /> Let the circumcenter be represented by the vector &lt;math&gt;O = (0, 0)&lt;/math&gt;, and let vectors &lt;math&gt;A,B,C&lt;/math&gt; correspond to the vertices of the triangle. It is well known the that the orthocenter is &lt;math&gt;H = A+B+C&lt;/math&gt; and the centroid is &lt;math&gt;G = \frac{A+B+C}{3}&lt;/math&gt;. Thus, &lt;math&gt;O, G, H&lt;/math&gt; are collinear and &lt;math&gt;\frac{OG}{HG} = \frac{1}{2}&lt;/math&gt;<br /> <br /> [[Image:Euler Line.PNG||500px|frame|center]]<br /> <br /> <br /> <br /> <br /> {{stub}}</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=Euler_line&diff=86801 Euler line 2017-08-04T00:35:38Z <p>Always correct: </p> <hr /> <div>In any [[triangle]] &lt;math&gt;\triangle ABC&lt;/math&gt;, the '''Euler line''' is a [[line]] which passes through the [[orthocenter]] &lt;math&gt;H&lt;/math&gt;, [[centroid]] &lt;math&gt;G&lt;/math&gt;, [[circumcenter]] &lt;math&gt;O&lt;/math&gt;, [[nine-point center]] &lt;math&gt;N&lt;/math&gt; and [[De Longchamps point | de Longchamps point]] &lt;math&gt;L&lt;/math&gt;. It is named after [[Leonhard Euler]]. Its existence is a non-trivial fact of Euclidean [[geometry]]. Certain fixed orders and distance [[ratio]]s hold among these points. In particular, &lt;math&gt;\overline{OGNH}&lt;/math&gt; and &lt;math&gt;OG:GN:NH = 2:1:3&lt;/math&gt;<br /> <br /> Given the [[orthic triangle]] &lt;math&gt;\triangle H_AH_BH_C&lt;/math&gt; of &lt;math&gt;\triangle ABC&lt;/math&gt;, the Euler lines of &lt;math&gt;\triangle AH_BH_C&lt;/math&gt;,&lt;math&gt;\triangle BH_CH_A&lt;/math&gt;, and &lt;math&gt;\triangle CH_AH_B&lt;/math&gt; [[concurrence | concur]] at &lt;math&gt;N&lt;/math&gt;, the nine-point circle of &lt;math&gt;\triangle ABC&lt;/math&gt;.<br /> <br /> ==Proof Centroid Lies on Euler Line==<br /> This proof utilizes the concept of [[spiral similarity]], which in this case is a [[rotation]] followed [[homothety]]. Consider the [[medial triangle]] &lt;math&gt;\triangle O_AO_BO_C&lt;/math&gt;. It is similar to &lt;math&gt;\triangle ABC&lt;/math&gt;. Specifically, a rotation of &lt;math&gt;180^\circ&lt;/math&gt; about the midpoint of &lt;math&gt;O_BO_C&lt;/math&gt; followed by a homothety with scale factor &lt;math&gt;2&lt;/math&gt; centered at &lt;math&gt;A&lt;/math&gt; brings &lt;math&gt;\triangle ABC \to \triangle O_AO_BO_C&lt;/math&gt;. Let us examine what else this transformation, which we denote as &lt;math&gt;\mathcal{S}&lt;/math&gt;, will do. <br /> <br /> It turns out &lt;math&gt;O&lt;/math&gt; is the orthocenter, and &lt;math&gt;G&lt;/math&gt; is the centroid of &lt;math&gt;\triangle O_AO_BO_C&lt;/math&gt;. Thus, &lt;math&gt;\mathcal{S}(\{O_A, O, G\}) = \{A, H, G\}&lt;/math&gt;. As a homothety preserves angles, it follows that &lt;math&gt;\measuredangle O_AOG = \measuredangle AHG&lt;/math&gt;. Finally, as &lt;math&gt;\overline{AH} || \overline{O_AO}&lt;/math&gt; it follows that <br /> &lt;cmath&gt;\triangle AHG = \triangle O_AOG&lt;/cmath&gt;<br /> Thus, &lt;math&gt;O, G, H&lt;/math&gt; are collinear, and &lt;math&gt;\frac{OG}{HG} = \frac{1}{2}&lt;/math&gt;.<br /> <br /> ==Proof Nine-Point Center Lies on Euler Line==<br /> Assuming that the [[nine point circle]] exists and that &lt;math&gt;N&lt;/math&gt; is the center, note that a homothety centered at &lt;math&gt;H&lt;/math&gt; with factor &lt;math&gt;2&lt;/math&gt; brings the [[Euler point]]s &lt;math&gt;\{E_A, E_B, E_C\}&lt;/math&gt; onto the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt;. Thus, it brings the nine-point circle to the circumcircle. Additionally, &lt;math&gt;N&lt;/math&gt; should be sent to &lt;math&gt;O&lt;/math&gt;, thus &lt;math&gt;N \in \overline{HO}&lt;/math&gt; and &lt;math&gt;\frac{HN}{ON} = 1&lt;/math&gt;.<br /> <br /> ~always_correct<br /> <br /> <br /> [[Image:Euler Line.PNG||500px|frame|center]]<br /> <br /> <br /> <br /> <br /> {{stub}}</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=Euler_line&diff=86800 Euler line 2017-08-04T00:03:44Z <p>Always correct: </p> <hr /> <div>In any [[triangle]] &lt;math&gt;\triangle ABC&lt;/math&gt;, the '''Euler line''' is a [[line]] which passes through the [[orthocenter]] &lt;math&gt;H&lt;/math&gt;, [[centroid]] &lt;math&gt;G&lt;/math&gt;, [[circumcenter]] &lt;math&gt;O&lt;/math&gt;, [[nine-point center]] &lt;math&gt;N&lt;/math&gt; and [[De Longchamps point | de Longchamps point]] &lt;math&gt;L&lt;/math&gt;. It is named after [[Leonhard Euler]]. Its existence is a non-trivial fact of Euclidean [[geometry]]. Certain fixed orders and distance [[ratio]]s hold among these points. In particular, &lt;math&gt;\overline{OGNH}&lt;/math&gt; and &lt;math&gt;OG:GN:NH = 2:1:3&lt;/math&gt;<br /> <br /> Given the [[orthic triangle]] &lt;math&gt;\triangle H_AH_BH_C&lt;/math&gt; of &lt;math&gt;\triangle ABC&lt;/math&gt;, the Euler lines of &lt;math&gt;\triangle AH_BH_C&lt;/math&gt;,&lt;math&gt;\triangle BH_CH_A&lt;/math&gt;, and &lt;math&gt;\triangle CH_AH_B&lt;/math&gt; [[concurrence | concur]] at &lt;math&gt;N&lt;/math&gt;, the nine-point center of &lt;math&gt;\triangle ABC&lt;/math&gt;. &lt;span style=&quot;color:#ff0000&quot;&gt; Is this true? &lt;/span&gt;<br /> <br /> ==Proof Centroid Lies on Euler Line==<br /> This proof utilizes the concept of [[spiral similarity]], which in this case is a [[rotation]] followed [[homothety]]. Consider the [[medial triangle]] &lt;math&gt;\triangle O_AO_BO_C&lt;/math&gt;. It is similar to &lt;math&gt;\triangle ABC&lt;/math&gt;. Specifically, a rotation of &lt;math&gt;180^\circ&lt;/math&gt; about the midpoint of &lt;math&gt;O_BO_C&lt;/math&gt; followed by a homothety with scale factor &lt;math&gt;2&lt;/math&gt; centered at &lt;math&gt;A&lt;/math&gt; brings &lt;math&gt;\triangle ABC \to \triangle O_AO_BO_C&lt;/math&gt;. Let us examine what else this transformation, which we denote as &lt;math&gt;\mathcal{S}&lt;/math&gt;, will do. <br /> <br /> It turns out &lt;math&gt;O&lt;/math&gt; is the orthocenter, and &lt;math&gt;G&lt;/math&gt; is the centroid of &lt;math&gt;\triangle O_AO_BO_C&lt;/math&gt;. Thus, &lt;math&gt;\mathcal{S}(\{O_A, O, G\}) = \{A, H, G\}&lt;/math&gt;. As a homothety preserves angles, it follows that &lt;math&gt;\measuredangle O_AOG = \measuredangle AHG&lt;/math&gt;. Finally, as &lt;math&gt;\overline{AH} || \overline{O_AO}&lt;/math&gt; it follows that <br /> &lt;cmath&gt;\triangle AHG = \triangle O_AOG&lt;/cmath&gt;<br /> Thus, &lt;math&gt;O, G, H&lt;/math&gt; are collinear, and &lt;math&gt;\frac{OG}{HG} = \frac{1}{2}&lt;/math&gt;.<br /> <br /> ==Proof Nine-Point Center Lies on Euler Line==<br /> Assuming that the [[nine point circle]] exists and that &lt;math&gt;N&lt;/math&gt; is the center, note that a homothety centered at &lt;math&gt;H&lt;/math&gt; with factor &lt;math&gt;2&lt;/math&gt; brings the [[Euler point]]s &lt;math&gt;\{E_A, E_B, E_C\}&lt;/math&gt; onto the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt;. Thus, it brings the nine-point circle to the circumcircle. Additionally, &lt;math&gt;N&lt;/math&gt; should be sent to &lt;math&gt;O&lt;/math&gt;, thus &lt;math&gt;N \in \overline{HO}&lt;/math&gt; and &lt;math&gt;\frac{HN}{ON} = 1&lt;/math&gt;.<br /> <br /> ~always_correct<br /> <br /> <br /> [[Image:Euler Line.PNG||500px|frame|center]]<br /> <br /> <br /> <br /> <br /> {{stub}}</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=Euler_line&diff=86799 Euler line 2017-08-03T23:39:39Z <p>Always correct: /* Proof Nine-Point Center Lies on Euler Line */</p> <hr /> <div>In any [[triangle]] &lt;math&gt;\triangle ABC&lt;/math&gt;, the '''Euler line''' is a [[line]] which passes through the [[orthocenter]] &lt;math&gt;H&lt;/math&gt;, [[centroid]] &lt;math&gt;G&lt;/math&gt;, [[circumcenter]] &lt;math&gt;O&lt;/math&gt;, [[nine-point center]] &lt;math&gt;N&lt;/math&gt; and [[De Longchamps point | de Longchamps point]] &lt;math&gt;L&lt;/math&gt;. It is named after [[Leonhard Euler]]. Its existence is a non-trivial fact of Euclidean [[geometry]]. Certain fixed orders and distance [[ratio]]s hold among these points. In particular, &lt;math&gt;\overline{OGNH}&lt;/math&gt; and &lt;math&gt;OG:GN:NH = 2:1:3&lt;/math&gt;<br /> <br /> Given the [[orthic triangle]] &lt;math&gt;\triangle H_AH_BH_C&lt;/math&gt; of &lt;math&gt;\triangle ABC&lt;/math&gt;, the Euler lines of &lt;math&gt;\triangle AH_BH_C&lt;/math&gt;,&lt;math&gt;\triangle BH_CH_A&lt;/math&gt;, and &lt;math&gt;\triangle CH_AH_B&lt;/math&gt; [[concurrence | concur]] at &lt;math&gt;N&lt;/math&gt;, the nine-point center of &lt;math&gt;\triangle ABC&lt;/math&gt;.<br /> <br /> ==Proof Centroid Lies on Euler Line==<br /> This proof utilizes the concept of [[spiral similarity]], which in this case is a [[rotation]] followed [[homothety]]. Consider the [[medial triangle]] &lt;math&gt;\triangle O_AO_BO_C&lt;/math&gt;. It is similar to &lt;math&gt;\triangle ABC&lt;/math&gt;. Specifically, a rotation of &lt;math&gt;180^\circ&lt;/math&gt; about the midpoint of &lt;math&gt;O_BO_C&lt;/math&gt; followed by a homothety with scale factor &lt;math&gt;2&lt;/math&gt; centered at &lt;math&gt;A&lt;/math&gt; brings &lt;math&gt;\triangle ABC \to \triangle O_AO_BO_C&lt;/math&gt;. Let us examine what else this transformation, which we denote as &lt;math&gt;\mathcal{S}&lt;/math&gt;, will do. <br /> <br /> It turns out &lt;math&gt;O&lt;/math&gt; is the orthocenter, and &lt;math&gt;G&lt;/math&gt; is the centroid of &lt;math&gt;\triangle O_AO_BO_C&lt;/math&gt;. Thus, &lt;math&gt;\mathcal{S}(\{O_A, O, G\}) = \{A, H, G\}&lt;/math&gt;. As a homothety preserves angles, it follows that &lt;math&gt;\measuredangle O_AOG = \measuredangle AHG&lt;/math&gt;. Finally, as &lt;math&gt;\overline{AH} || \overline{O_AO}&lt;/math&gt; it follows that <br /> &lt;cmath&gt;\triangle AHG = \triangle O_AOG&lt;/cmath&gt;<br /> Thus, &lt;math&gt;O, G, H&lt;/math&gt; are collinear, and &lt;math&gt;\frac{OG}{HG} = \frac{1}{2}&lt;/math&gt;.<br /> <br /> ==Proof Nine-Point Center Lies on Euler Line==<br /> Assuming that the [[nine point circle]] exists and that &lt;math&gt;N&lt;/math&gt; is the center, note that a homothety centered at &lt;math&gt;H&lt;/math&gt; with factor &lt;math&gt;2&lt;/math&gt; brings the [[Euler point]]s &lt;math&gt;\{E_A, E_B, E_C\}&lt;/math&gt; onto the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt;. Thus, it brings the nine-point circle to the circumcircle. Additionally, &lt;math&gt;N&lt;/math&gt; should be sent to &lt;math&gt;O&lt;/math&gt;, thus &lt;math&gt;N \in \overline{HO}&lt;/math&gt; and &lt;math&gt;\frac{HN}{ON} = 1&lt;/math&gt;.<br /> <br /> ~always_correct<br /> <br /> <br /> [[Image:Euler Line.PNG||500px|frame|center]]<br /> <br /> <br /> <br /> <br /> {{stub}}</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=Euler_line&diff=86798 Euler line 2017-08-03T23:35:14Z <p>Always correct: /* Proof of Existence */</p> <hr /> <div>In any [[triangle]] &lt;math&gt;\triangle ABC&lt;/math&gt;, the '''Euler line''' is a [[line]] which passes through the [[orthocenter]] &lt;math&gt;H&lt;/math&gt;, [[centroid]] &lt;math&gt;G&lt;/math&gt;, [[circumcenter]] &lt;math&gt;O&lt;/math&gt;, [[nine-point center]] &lt;math&gt;N&lt;/math&gt; and [[De Longchamps point | de Longchamps point]] &lt;math&gt;L&lt;/math&gt;. It is named after [[Leonhard Euler]]. Its existence is a non-trivial fact of Euclidean [[geometry]]. Certain fixed orders and distance [[ratio]]s hold among these points. In particular, &lt;math&gt;\overline{OGNH}&lt;/math&gt; and &lt;math&gt;OG:GN:NH = 2:1:3&lt;/math&gt;<br /> <br /> Given the [[orthic triangle]] &lt;math&gt;\triangle H_AH_BH_C&lt;/math&gt; of &lt;math&gt;\triangle ABC&lt;/math&gt;, the Euler lines of &lt;math&gt;\triangle AH_BH_C&lt;/math&gt;,&lt;math&gt;\triangle BH_CH_A&lt;/math&gt;, and &lt;math&gt;\triangle CH_AH_B&lt;/math&gt; [[concurrence | concur]] at &lt;math&gt;N&lt;/math&gt;, the nine-point center of &lt;math&gt;\triangle ABC&lt;/math&gt;.<br /> <br /> ==Proof Centroid Lies on Euler Line==<br /> This proof utilizes the concept of [[spiral similarity]], which in this case is a [[rotation]] followed [[homothety]]. Consider the [[medial triangle]] &lt;math&gt;\triangle O_AO_BO_C&lt;/math&gt;. It is similar to &lt;math&gt;\triangle ABC&lt;/math&gt;. Specifically, a rotation of &lt;math&gt;180^\circ&lt;/math&gt; about the midpoint of &lt;math&gt;O_BO_C&lt;/math&gt; followed by a homothety with scale factor &lt;math&gt;2&lt;/math&gt; centered at &lt;math&gt;A&lt;/math&gt; brings &lt;math&gt;\triangle ABC \to \triangle O_AO_BO_C&lt;/math&gt;. Let us examine what else this transformation, which we denote as &lt;math&gt;\mathcal{S}&lt;/math&gt;, will do. <br /> <br /> It turns out &lt;math&gt;O&lt;/math&gt; is the orthocenter, and &lt;math&gt;G&lt;/math&gt; is the centroid of &lt;math&gt;\triangle O_AO_BO_C&lt;/math&gt;. Thus, &lt;math&gt;\mathcal{S}(\{O_A, O, G\}) = \{A, H, G\}&lt;/math&gt;. As a homothety preserves angles, it follows that &lt;math&gt;\measuredangle O_AOG = \measuredangle AHG&lt;/math&gt;. Finally, as &lt;math&gt;\overline{AH} || \overline{O_AO}&lt;/math&gt; it follows that <br /> &lt;cmath&gt;\triangle AHG = \triangle O_AOG&lt;/cmath&gt;<br /> Thus, &lt;math&gt;O, G, H&lt;/math&gt; are collinear, and &lt;math&gt;\frac{OG}{HG} = \frac{1}{2}&lt;/math&gt;.<br /> <br /> ==Proof Nine-Point Center Lies on Euler Line==<br /> Assuming that the [[nine point circle]] exists and that &lt;math&gt;N&lt;/math&gt; is the center, note that a homothety centered at &lt;math&gt;H&lt;/math&gt; with factor &lt;math&gt;2&lt;/math&gt; brings the [[Euler point]]s &lt;math&gt;{E_A, E_B, E_C\}&lt;/math&gt; onto the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt;. Thus, it brings the nine-point circle to the circumcircle. Additionally, &lt;math&gt;N&lt;/math&gt; should be sent to &lt;math&gt;O&lt;/math&gt;, thus &lt;math&gt;N \in \overline{HO}&lt;/math&gt; and &lt;math&gt;\frac{HN}{ON} = 1&lt;/math&gt;.<br /> <br /> ~always_correct<br /> <br /> <br /> [[Image:Euler Line.PNG||500px|frame|center]]<br /> <br /> <br /> <br /> <br /> {{stub}}</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=Nine_point_circle&diff=86797 Nine point circle 2017-08-03T23:14:50Z <p>Always correct: </p> <hr /> <div>[[Image:Euler Line.PNG|thumb|500px|right| Triangle ''ABC'' with the nine point circle in light orange]]<br /> <br /> The '''nine point circle''' (also known as ''Euler's circle'' or ''Feuerbach's circle'') of a given [[triangle]] is a circle which passes through 9 &quot;significant&quot; points:<br /> * The three feet of the [[altitude]]s of the triangle.<br /> * The three [[midpoint]]s of the [[edge]]s of the triangle.<br /> * The three midpoints of the segments joining the [[vertex | vertices]] of the triangle to its [[orthocenter]]. (These points are sometimes known as the [[Euler point]]s of the triangle.)<br /> <br /> That such a circle exists is a non-trivial theorem of Euclidean [[geometry]].<br /> <br /> The center of the nine point circle is the [[nine-point center]] and is usually denoted &lt;math&gt;N&lt;/math&gt;.<br /> <br /> ==First Proof of Existence==<br /> Since &lt;math&gt;O_c&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E_b&lt;/math&gt; is the midpoint of &lt;math&gt;BH&lt;/math&gt;, &lt;math&gt;O_cE_b&lt;/math&gt; is parallel to &lt;math&gt;AH&lt;/math&gt;. Using similar logic, we see that &lt;math&gt;O_bE_c&lt;/math&gt; is also parallel to &lt;math&gt;AH&lt;/math&gt;. Since &lt;math&gt;E_b&lt;/math&gt; is the midpoint of &lt;math&gt;HB&lt;/math&gt; and &lt;math&gt;E_c&lt;/math&gt; is the midpoint of &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;E_bE_c&lt;/math&gt; is parallel to &lt;math&gt;BC&lt;/math&gt;, which is perpendicular to &lt;math&gt;AH&lt;/math&gt;. Similar logic gives us that &lt;math&gt;O_bO_c&lt;/math&gt; is perpendicular to &lt;math&gt;AH&lt;/math&gt; as well. Therefore &lt;math&gt;O_bO_cE_bE_c&lt;/math&gt; is a rectangle, which is a cyclic figure. The diagonals &lt;math&gt;O_bE_b&lt;/math&gt; and &lt;math&gt;O_cE_c&lt;/math&gt; are diagonals of the circumcircle. Similar logic to the above gives us that &lt;math&gt;O_aO_cE_aE_c&lt;/math&gt; is a rectangle with a common diagonal to &lt;math&gt;O_bO_cE_bE_c&lt;/math&gt;. Therefore the circumcircles of the two rectangles are identical. We can also gain that rectangle &lt;math&gt;O_aO_bE_aE_b&lt;/math&gt; is also on the circle.<br /> <br /> We now have a circle with the points &lt;math&gt;O_a&lt;/math&gt;, &lt;math&gt;O_b&lt;/math&gt;, &lt;math&gt;O_c&lt;/math&gt;, &lt;math&gt;E_a&lt;/math&gt;, &lt;math&gt;E_b&lt;/math&gt;, and &lt;math&gt;E_c&lt;/math&gt; on it, with diameters &lt;math&gt;O_aE_A&lt;/math&gt;, &lt;math&gt;O_bE_b&lt;/math&gt;, and &lt;math&gt;O_cE_c&lt;/math&gt;. We now note that &lt;math&gt;\angle E_aH_aO_a=\angle E_bH_bO_b=\angle E_cH_cO_c=90^{\circ}&lt;/math&gt;. Therefore &lt;math&gt;H_a&lt;/math&gt;, &lt;math&gt;H_b&lt;/math&gt;, and &lt;math&gt;H_c&lt;/math&gt; are also on the circle. We now have a circle with the midpoints of the sides on it, the three midpoints of the segments joining the vertices of the triangle to its orthocenter on it, and the three feet of the altitudes of the triangle on it. Therefore the nine points are on the circle, and the nine-point circle exists.<br /> <br /> ==Second Proof of Existence==<br /> We know that the reflection of the orthocenter about the sides and about the midpoints of the triangle's sides lie on the circumcircle. Thus, consider the homothety centered at &lt;math&gt;H&lt;/math&gt; with ratio &lt;math&gt;-\frac{1}{2}&lt;/math&gt;. It maps the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt; to the nine point circle, and the vertices of the triangle to its [[Euler point]]s.<br /> Hence proved.<br /> <br /> <br /> {{stub}}<br /> [[Category:Definition]]</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=Nine_point_circle&diff=86796 Nine point circle 2017-08-03T23:14:08Z <p>Always correct: /* First Proof of Existence */</p> <hr /> <div>[[Image:Euler Line.PNG|thumb|500px|right|Triangle ''ABC'' with the nine point circle in light orange]]<br /> <br /> The '''nine point circle''' (also known as ''Euler's circle'' or ''Feuerbach's circle'') of a given [[triangle]] is a circle which passes through 9 &quot;significant&quot; points:<br /> * The three feet of the [[altitude]]s of the triangle.<br /> * The three [[midpoint]]s of the [[edge]]s of the triangle.<br /> * The three midpoints of the segments joining the [[vertex | vertices]] of the triangle to its [[orthocenter]]. (These points are sometimes known as the [[Euler point]]s of the triangle.)<br /> <br /> That such a circle exists is a non-trivial theorem of Euclidean [[geometry]].<br /> <br /> The center of the nine point circle is the [[nine-point center]] and is usually denoted &lt;math&gt;N&lt;/math&gt;.<br /> <br /> ==First Proof of Existence==<br /> Since &lt;math&gt;O_c&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E_b&lt;/math&gt; is the midpoint of &lt;math&gt;BH&lt;/math&gt;, &lt;math&gt;O_cE_b&lt;/math&gt; is parallel to &lt;math&gt;AH&lt;/math&gt;. Using similar logic, we see that &lt;math&gt;O_bE_c&lt;/math&gt; is also parallel to &lt;math&gt;AH&lt;/math&gt;. Since &lt;math&gt;E_b&lt;/math&gt; is the midpoint of &lt;math&gt;HB&lt;/math&gt; and &lt;math&gt;E_c&lt;/math&gt; is the midpoint of &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;E_bE_c&lt;/math&gt; is parallel to &lt;math&gt;BC&lt;/math&gt;, which is perpendicular to &lt;math&gt;AH&lt;/math&gt;. Similar logic gives us that &lt;math&gt;O_bO_c&lt;/math&gt; is perpendicular to &lt;math&gt;AH&lt;/math&gt; as well. Therefore &lt;math&gt;O_bO_cE_bE_c&lt;/math&gt; is a rectangle, which is a cyclic figure. The diagonals &lt;math&gt;O_bE_b&lt;/math&gt; and &lt;math&gt;O_cE_c&lt;/math&gt; are diagonals of the circumcircle. Similar logic to the above gives us that &lt;math&gt;O_aO_cE_aE_c&lt;/math&gt; is a rectangle with a common diagonal to &lt;math&gt;O_bO_cE_bE_c&lt;/math&gt;. Therefore the circumcircles of the two rectangles are identical. We can also gain that rectangle &lt;math&gt;O_aO_bE_aE_b&lt;/math&gt; is also on the circle.<br /> <br /> We now have a circle with the points &lt;math&gt;O_a&lt;/math&gt;, &lt;math&gt;O_b&lt;/math&gt;, &lt;math&gt;O_c&lt;/math&gt;, &lt;math&gt;E_a&lt;/math&gt;, &lt;math&gt;E_b&lt;/math&gt;, and &lt;math&gt;E_c&lt;/math&gt; on it, with diameters &lt;math&gt;O_aE_A&lt;/math&gt;, &lt;math&gt;O_bE_b&lt;/math&gt;, and &lt;math&gt;O_cE_c&lt;/math&gt;. We now note that &lt;math&gt;\angle E_aH_aO_a=\angle E_bH_bO_b=\angle E_cH_cO_c=90^{\circ}&lt;/math&gt;. Therefore &lt;math&gt;H_a&lt;/math&gt;, &lt;math&gt;H_b&lt;/math&gt;, and &lt;math&gt;H_c&lt;/math&gt; are also on the circle. We now have a circle with the midpoints of the sides on it, the three midpoints of the segments joining the vertices of the triangle to its orthocenter on it, and the three feet of the altitudes of the triangle on it. Therefore the nine points are on the circle, and the nine-point circle exists.<br /> <br /> ==Second Proof of Existence==<br /> We know that the reflection of the orthocenter about the sides and about the midpoints of the triangle's sides lie on the circumcircle. Thus, consider the homothety centered at &lt;math&gt;H&lt;/math&gt; with ratio &lt;math&gt;-\frac{1}{2}&lt;/math&gt;. It maps the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt; to the nine point circle, and the vertices of the triangle to its [[Euler point]]s.<br /> Hence proved.<br /> <br /> <br /> {{stub}}<br /> [[Category:Definition]]</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=Nine_point_circle&diff=86792 Nine point circle 2017-08-03T23:02:42Z <p>Always correct: </p> <hr /> <div>[[Image:Euler Line.PNG|thumb|500px|right|Triangle ''ABC'' with the nine point circle in light orange]]<br /> <br /> The '''nine point circle''' (also known as ''Euler's circle'' or ''Feuerbach's circle'') of a given [[triangle]] is a circle which passes through 9 &quot;significant&quot; points:<br /> * The three feet of the [[altitude]]s of the triangle.<br /> * The three [[midpoint]]s of the [[edge]]s of the triangle.<br /> * The three midpoints of the segments joining the [[vertex | vertices]] of the triangle to its [[orthocenter]]. (These points are sometimes known as the [[Euler point]]s of the triangle.)<br /> <br /> That such a circle exists is a non-trivial theorem of Euclidean [[geometry]].<br /> <br /> The center of the nine point circle is the [[nine-point center]] and is usually denoted &lt;math&gt;N&lt;/math&gt;.<br /> <br /> ==First Proof of Existence==<br /> Since &lt;math&gt;O_c&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E_b&lt;/math&gt; is the midpoint of &lt;math&gt;BH&lt;/math&gt;, &lt;math&gt;O_cE_b&lt;/math&gt; is parallel to &lt;math&gt;AH&lt;/math&gt;. Using similar logic, we see that &lt;math&gt;O_bE_c&lt;/math&gt; is also parallel to &lt;math&gt;AH&lt;/math&gt;. Since &lt;math&gt;E_b&lt;/math&gt; is the midpoint of &lt;math&gt;HB&lt;/math&gt; and &lt;math&gt;E_c&lt;/math&gt; is the midpoint of &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;E_bE_c&lt;/math&gt; is parallel to &lt;math&gt;BC&lt;/math&gt;, which is perpendicular to &lt;math&gt;AH&lt;/math&gt;. Similar logic gives us that &lt;math&gt;O_bO_c&lt;/math&gt; is perpendicular to &lt;math&gt;AH&lt;/math&gt; as well. Therefore &lt;math&gt;O_bO_cE_bE_c&lt;/math&gt; is a rectangle, which is a cyclic figure. The diagonals &lt;math&gt;O_bE_b&lt;/math&gt; and &lt;math&gt;O_cE_c&lt;/math&gt; are diagonals of the circumcircle. Similar logic to the above gives us that &lt;math&gt;O_aO_cE_aE_c&lt;/math&gt; is a rectangle with a common diagonal to &lt;math&gt;O_bO_cE_bE_c&lt;/math&gt;. Therefore the circumcircles of the two rectangles are identical. We can also gain that rectangle &lt;math&gt;O_aO_bE_aE_b&lt;/math&gt; is also on the circle.<br /> <br /> We now have a circle with the points &lt;math&gt;O_a&lt;/math&gt;, &lt;math&gt;O_b&lt;/math&gt;, &lt;math&gt;O_c&lt;/math&gt;, &lt;math&gt;E_a&lt;/math&gt;, &lt;math&gt;E_b&lt;/math&gt;, and &lt;math&gt;E_c&lt;/math&gt; on it, with diameters &lt;math&gt;O_aE_A&lt;/math&gt;, &lt;math&gt;O_bE_b&lt;/math&gt;, and &lt;math&gt;O_cE_c&lt;/math&gt;. We now note that &lt;math&gt;\angle E_aH_aO_a=\angle E_bH_bO_b=\angle E_cH_cO_c=90^{\circ}&lt;/math&gt;. Therefore &lt;math&gt;H_a&lt;/math&gt;, &lt;math&gt;H_b&lt;/math&gt;, ad &lt;math&gt;H_c&lt;/math&gt; are also on the circle. We now have a circle with the midpoints of the sides on it, the three midpoints of the segments joining the vertices of the triangle to its orthocenter on it, and the three feet of the altitudes of the triangle on it. Therefore the nine points are on the circle, and the nine-point circle exists.<br /> <br /> ==Second Proof of Existence==<br /> We know that the reflection of the orthocenter about the sides and about the midpoints of the triangle's sides lie on the circumcircle. Thus, consider the homothety centered at &lt;math&gt;H&lt;/math&gt; with ratio &lt;math&gt;-\frac{1}{2}&lt;/math&gt;. It maps the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt; to the nine point circle, and the vertices of the triangle to its [[Euler point]]s.<br /> Hence proved.<br /> <br /> <br /> {{stub}}<br /> [[Category:Definition]]</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=Nine_point_circle&diff=86791 Nine point circle 2017-08-03T23:00:46Z <p>Always correct: </p> <hr /> <div>[[Image:Euler Line.PNG|thumb|500px|right|Triangle ''ABC'' with the nine point circle in light orange]]<br /> <br /> The '''nine point circle''' (also known as ''Euler's circle'' or ''Feuerbach's circle'') of a given [[triangle]] is a circle which passes through 9 &quot;significant&quot; points:<br /> * The three feet of the [[altitude]]s of the triangle.<br /> * The three [[midpoint]]s of the [[edge]]s of the triangle.<br /> * The three midpoints of the segments joining the [[vertex | vertices]] of the triangle to its [[orthocenter]]. (These points are sometimes known as the [[Euler point]]s of the triangle.)<br /> <br /> That such a circle exists is a non-trivial theorem of Euclidean [[geometry]].<br /> <br /> The center of the nine point circle is the [[nine-point center]] and is usually denoted &lt;math&gt;N&lt;/math&gt;.<br /> <br /> ==First Proof of Existence==<br /> We know that the reflection of the orthocenter about the sides and about the midpoints of the triangle's sides lie on the circumcircle. Thus, consider the homothety centered at &lt;math&gt;H&lt;/math&gt; with ratio &lt;math&gt;-\frac{1}{2}&lt;/math&gt;. It maps the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt; to the nine point circle.<br /> Hence proved.<br /> <br /> =Second Proof of Existence==<br /> Since &lt;math&gt;O_c&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E_b&lt;/math&gt; is the midpoint of &lt;math&gt;BH&lt;/math&gt;, &lt;math&gt;O_cE_b&lt;/math&gt; is parallel to &lt;math&gt;AH&lt;/math&gt;. Using similar logic, we see that &lt;math&gt;O_bE_c&lt;/math&gt; is also parallel to &lt;math&gt;AH&lt;/math&gt;. Since &lt;math&gt;E_b&lt;/math&gt; is the midpoint of &lt;math&gt;HB&lt;/math&gt; and &lt;math&gt;E_c&lt;/math&gt; is the midpoint of &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;E_bE_c&lt;/math&gt; is parallel to &lt;math&gt;BC&lt;/math&gt;, which is perpendicular to &lt;math&gt;AH&lt;/math&gt;. Similar logic gives us that &lt;math&gt;O_bO_c&lt;/math&gt; is perpendicular to &lt;math&gt;AH&lt;/math&gt; as well. Therefore &lt;math&gt;O_bO_cE_bE_c&lt;/math&gt; is a rectangle, which is a cyclic figure. The diagonals &lt;math&gt;O_bE_b&lt;/math&gt; and &lt;math&gt;O_cE_c&lt;/math&gt; are diagonals of the circumcircle. Similar logic to the above gives us that &lt;math&gt;O_aO_cE_aE_c&lt;/math&gt; is a rectangle with a common diagonal to &lt;math&gt;O_bO_cE_bE_c&lt;/math&gt;. Therefore the circumcircles of the two rectangles are identical. We can also gain that rectangle &lt;math&gt;O_aO_bE_aE_b&lt;/math&gt; is also on the circle.<br /> <br /> We now have a circle with the points &lt;math&gt;O_a&lt;/math&gt;, &lt;math&gt;O_b&lt;/math&gt;, &lt;math&gt;O_c&lt;/math&gt;, &lt;math&gt;E_a&lt;/math&gt;, &lt;math&gt;E_b&lt;/math&gt;, and &lt;math&gt;E_c&lt;/math&gt; on it, with diameters &lt;math&gt;O_aE_A&lt;/math&gt;, &lt;math&gt;O_bE_b&lt;/math&gt;, and &lt;math&gt;O_cE_c&lt;/math&gt;. We now note that &lt;math&gt;\angle E_aH_aO_a=\angle E_bH_bO_b=\angle E_cH_cO_c=90^{\circ}&lt;/math&gt;. Therefore &lt;math&gt;H_a&lt;/math&gt;, &lt;math&gt;H_b&lt;/math&gt;, ad &lt;math&gt;H_c&lt;/math&gt; are also on the circle. We now have a circle with the midpoints of the sides on it, the three midpoints of the segments joining the vertices of the triangle to its orthocenter on it, and the three feet of the altitudes of the triangle on it. Therefore the nine points are on the circle, and the nine-point circle exists.<br /> <br /> {{stub}}<br /> [[Category:Definition]]</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=Nine_point_circle&diff=86790 Nine point circle 2017-08-03T22:50:36Z <p>Always correct: /* Proof of the Existence */</p> <hr /> <div>[[Image:Euler Line.PNG|thumb|500px|right|Triangle ''ABC'' with the nine point circle in light orange]]<br /> <br /> The '''nine point circle''' (also known as ''Euler's circle'' or ''Feuerbach's circle'') of a given [[triangle]] is a circle which passes through 9 &quot;significant&quot; points:<br /> * The three feet of the [[altitude]]s of the triangle.<br /> * The three [[midpoint]]s of the [[edge]]s of the triangle.<br /> * The three midpoints of the segments joining the [[vertex | vertices]] of the triangle to its [[orthocenter]]. (These points are sometimes known as the [[Euler point]]s of the triangle.)<br /> <br /> That such a circle exists is a non-trivial theorem of Euclidean [[geometry]].<br /> <br /> The center of the nine point circle is the [[nine-point center]] and is usually denoted &lt;math&gt;N&lt;/math&gt;.<br /> <br /> ==Proof of Existence==<br /> Since &lt;math&gt;O_c&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E_b&lt;/math&gt; is the midpoint of &lt;math&gt;BH&lt;/math&gt;, &lt;math&gt;O_cE_b&lt;/math&gt; is parallel to &lt;math&gt;AH&lt;/math&gt;. Using similar logic, we see that &lt;math&gt;O_bE_c&lt;/math&gt; is also parallel to &lt;math&gt;AH&lt;/math&gt;. Since &lt;math&gt;E_b&lt;/math&gt; is the midpoint of &lt;math&gt;HB&lt;/math&gt; and &lt;math&gt;E_c&lt;/math&gt; is the midpoint of &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;E_bE_c&lt;/math&gt; is parallel to &lt;math&gt;BC&lt;/math&gt;, which is perpendicular to &lt;math&gt;AH&lt;/math&gt;. Similar logic gives us that &lt;math&gt;O_bO_c&lt;/math&gt; is perpendicular to &lt;math&gt;AH&lt;/math&gt; as well. Therefore &lt;math&gt;O_bO_cE_bE_c&lt;/math&gt; is a rectangle, which is a cyclic figure. The diagonals &lt;math&gt;O_bE_b&lt;/math&gt; and &lt;math&gt;O_cE_c&lt;/math&gt; are diagonals of the circumcircle. Similar logic to the above gives us that &lt;math&gt;O_aO_cE_aE_c&lt;/math&gt; is a rectangle with a common diagonal to &lt;math&gt;O_bO_cE_bE_c&lt;/math&gt;. Therefore the circumcircles of the two rectangles are identical. We can also gain that rectangle &lt;math&gt;O_aO_bE_aE_b&lt;/math&gt; is also on the circle.<br /> <br /> We now have a circle with the points &lt;math&gt;O_a&lt;/math&gt;, &lt;math&gt;O_b&lt;/math&gt;, &lt;math&gt;O_c&lt;/math&gt;, &lt;math&gt;E_a&lt;/math&gt;, &lt;math&gt;E_b&lt;/math&gt;, and &lt;math&gt;E_c&lt;/math&gt; on it, with diameters &lt;math&gt;O_aE_A&lt;/math&gt;, &lt;math&gt;O_bE_b&lt;/math&gt;, and &lt;math&gt;O_cE_c&lt;/math&gt;. We now note that &lt;math&gt;\angle E_aH_aO_a=\angle E_bH_bO_b=\angle E_cH_cO_c=90^{\circ}&lt;/math&gt;. Therefore &lt;math&gt;H_a&lt;/math&gt;, &lt;math&gt;H_b&lt;/math&gt;, ad &lt;math&gt;H_c&lt;/math&gt; are also on the circle. We now have a circle with the midpoints of the sides on it, the three midpoints of the segments joining the vertices of the triangle to its orthocenter on it, and the three feet of the altitudes of the triangle on it. Therefore the nine points are on the circle, and the nine-point circle exists.<br /> &lt;math&gt;Another&lt;/math&gt;&lt;math&gt; proof.&lt;/math&gt;<br /> We know that the reflection of the orthocenter about the Triangle's sides and about the mid points of the triangle's sides lie on the circumcircle.<br /> Thus consider the homothety centred at &lt;math&gt;H&lt;/math&gt; with ratio &lt;math&gt;-1/2&lt;/math&gt;.It maps the circumcircle to the nine point circle.<br /> Hence proved.<br /> {{stub}}<br /> [[Category:Definition]]</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=Nine_point_circle&diff=86789 Nine point circle 2017-08-03T22:50:23Z <p>Always correct: /* Proof of the Nine-Point circle */</p> <hr /> <div>[[Image:Euler Line.PNG|thumb|500px|right|Triangle ''ABC'' with the nine point circle in light orange]]<br /> <br /> The '''nine point circle''' (also known as ''Euler's circle'' or ''Feuerbach's circle'') of a given [[triangle]] is a circle which passes through 9 &quot;significant&quot; points:<br /> * The three feet of the [[altitude]]s of the triangle.<br /> * The three [[midpoint]]s of the [[edge]]s of the triangle.<br /> * The three midpoints of the segments joining the [[vertex | vertices]] of the triangle to its [[orthocenter]]. (These points are sometimes known as the [[Euler point]]s of the triangle.)<br /> <br /> That such a circle exists is a non-trivial theorem of Euclidean [[geometry]].<br /> <br /> The center of the nine point circle is the [[nine-point center]] and is usually denoted &lt;math&gt;N&lt;/math&gt;.<br /> <br /> ==Proof of the Existence==<br /> Since &lt;math&gt;O_c&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E_b&lt;/math&gt; is the midpoint of &lt;math&gt;BH&lt;/math&gt;, &lt;math&gt;O_cE_b&lt;/math&gt; is parallel to &lt;math&gt;AH&lt;/math&gt;. Using similar logic, we see that &lt;math&gt;O_bE_c&lt;/math&gt; is also parallel to &lt;math&gt;AH&lt;/math&gt;. Since &lt;math&gt;E_b&lt;/math&gt; is the midpoint of &lt;math&gt;HB&lt;/math&gt; and &lt;math&gt;E_c&lt;/math&gt; is the midpoint of &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;E_bE_c&lt;/math&gt; is parallel to &lt;math&gt;BC&lt;/math&gt;, which is perpendicular to &lt;math&gt;AH&lt;/math&gt;. Similar logic gives us that &lt;math&gt;O_bO_c&lt;/math&gt; is perpendicular to &lt;math&gt;AH&lt;/math&gt; as well. Therefore &lt;math&gt;O_bO_cE_bE_c&lt;/math&gt; is a rectangle, which is a cyclic figure. The diagonals &lt;math&gt;O_bE_b&lt;/math&gt; and &lt;math&gt;O_cE_c&lt;/math&gt; are diagonals of the circumcircle. Similar logic to the above gives us that &lt;math&gt;O_aO_cE_aE_c&lt;/math&gt; is a rectangle with a common diagonal to &lt;math&gt;O_bO_cE_bE_c&lt;/math&gt;. Therefore the circumcircles of the two rectangles are identical. We can also gain that rectangle &lt;math&gt;O_aO_bE_aE_b&lt;/math&gt; is also on the circle.<br /> <br /> We now have a circle with the points &lt;math&gt;O_a&lt;/math&gt;, &lt;math&gt;O_b&lt;/math&gt;, &lt;math&gt;O_c&lt;/math&gt;, &lt;math&gt;E_a&lt;/math&gt;, &lt;math&gt;E_b&lt;/math&gt;, and &lt;math&gt;E_c&lt;/math&gt; on it, with diameters &lt;math&gt;O_aE_A&lt;/math&gt;, &lt;math&gt;O_bE_b&lt;/math&gt;, and &lt;math&gt;O_cE_c&lt;/math&gt;. We now note that &lt;math&gt;\angle E_aH_aO_a=\angle E_bH_bO_b=\angle E_cH_cO_c=90^{\circ}&lt;/math&gt;. Therefore &lt;math&gt;H_a&lt;/math&gt;, &lt;math&gt;H_b&lt;/math&gt;, ad &lt;math&gt;H_c&lt;/math&gt; are also on the circle. We now have a circle with the midpoints of the sides on it, the three midpoints of the segments joining the vertices of the triangle to its orthocenter on it, and the three feet of the altitudes of the triangle on it. Therefore the nine points are on the circle, and the nine-point circle exists.<br /> &lt;math&gt;Another&lt;/math&gt;&lt;math&gt; proof.&lt;/math&gt;<br /> We know that the reflection of the orthocenter about the Triangle's sides and about the mid points of the triangle's sides lie on the circumcircle.<br /> Thus consider the homothety centred at &lt;math&gt;H&lt;/math&gt; with ratio &lt;math&gt;-1/2&lt;/math&gt;.It maps the circumcircle to the nine point circle.<br /> Hence proved.<br /> {{stub}}<br /> [[Category:Definition]]</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=Euler_line&diff=86788 Euler line 2017-08-03T22:47:59Z <p>Always correct: </p> <hr /> <div>In any [[triangle]] &lt;math&gt;\triangle ABC&lt;/math&gt;, the '''Euler line''' is a [[line]] which passes through the [[orthocenter]] &lt;math&gt;H&lt;/math&gt;, [[centroid]] &lt;math&gt;G&lt;/math&gt;, [[circumcenter]] &lt;math&gt;O&lt;/math&gt;, [[nine-point center]] &lt;math&gt;N&lt;/math&gt; and [[De Longchamps point | de Longchamps point]] &lt;math&gt;L&lt;/math&gt;. It is named after [[Leonhard Euler]]. Its existence is a non-trivial fact of Euclidean [[geometry]]. Certain fixed orders and distance [[ratio]]s hold among these points. In particular, &lt;math&gt;\overline{OGNH}&lt;/math&gt; and &lt;math&gt;OG:GN:NH = 2:1:3&lt;/math&gt;<br /> <br /> Given the [[orthic triangle]] &lt;math&gt;\triangle H_AH_BH_C&lt;/math&gt; of &lt;math&gt;\triangle ABC&lt;/math&gt;, the Euler lines of &lt;math&gt;\triangle AH_BH_C&lt;/math&gt;,&lt;math&gt;\triangle BH_CH_A&lt;/math&gt;, and &lt;math&gt;\triangle CH_AH_B&lt;/math&gt; [[concurrence | concur]] at &lt;math&gt;N&lt;/math&gt;, the nine-point center of &lt;math&gt;\triangle ABC&lt;/math&gt;.<br /> <br /> ==Proof of Existence==<br /> This proof utilizes the concept of [[spiral similarity]], which in this case is a [[rotation]] followed [[homothety]]. Consider the [[medial triangle]] &lt;math&gt;\triangle O_AO_BO_C&lt;/math&gt;. It is similar to &lt;math&gt;\triangle ABC&lt;/math&gt;. Specifically, a rotation of &lt;math&gt;180^\circ&lt;/math&gt; about the midpoint of &lt;math&gt;O_BO_C&lt;/math&gt; followed by a homothety with scale factor &lt;math&gt;2&lt;/math&gt; centered at &lt;math&gt;A&lt;/math&gt; brings &lt;math&gt;\triangle ABC \to \triangle O_AO_BO_C&lt;/math&gt;. Let us examine what else this transformation, which we denote as &lt;math&gt;\mathcal{S}&lt;/math&gt;, will do. <br /> <br /> It turns out &lt;math&gt;O&lt;/math&gt; is the orthocenter, and &lt;math&gt;G&lt;/math&gt; is the centroid of &lt;math&gt;\triangle O_AO_BO_C&lt;/math&gt;. Thus, &lt;math&gt;\mathcal{S}(\{O_A, O, G\}) = \{A, H, G\}&lt;/math&gt;. As a homothety preserves angles, it follows that &lt;math&gt;\measuredangle O_AOG = \measuredangle AHG&lt;/math&gt;. Finally, as &lt;math&gt;\overline{AH} || \overline{O_AO}&lt;/math&gt; it follows that <br /> &lt;cmath&gt;\triangle AHG = \triangle O_AOG&lt;/cmath&gt;<br /> Thus, &lt;math&gt;O, G, H&lt;/math&gt; are collinear, and &lt;math&gt;\frac{OG}{HG} = \frac{1}{2}&lt;/math&gt;.<br /> <br /> ~always_correct<br /> <br /> <br /> [[Image:Euler Line.PNG||500px|frame|center]]<br /> <br /> <br /> <br /> <br /> {{stub}}</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=Euler_line&diff=86787 Euler line 2017-08-03T19:17:17Z <p>Always correct: </p> <hr /> <div>In any [[triangle]] &lt;math&gt;\triangle ABC&lt;/math&gt;, the '''Euler line''' is a [[line]] which passes through the [[orthocenter]] &lt;math&gt;H&lt;/math&gt;, [[centroid]] &lt;math&gt;G&lt;/math&gt;, [[circumcenter]] &lt;math&gt;O&lt;/math&gt;, [[nine-point center]] &lt;math&gt;N&lt;/math&gt; and [[De Longchamps point]] &lt;math&gt;L&lt;/math&gt;. It is named after [[Leonhard Euler]]. Its existence is a non-trivial fact of Euclidean [[geometry]]. Certain fixed orders and distance [[ratio]]s hold among these points. In particular, &lt;math&gt;\overline{OGNH}&lt;/math&gt; and &lt;math&gt;OG:GN:NH = 2:1:3&lt;/math&gt;<br /> <br /> Given the [[orthic triangle]] &lt;math&gt;\triangle H_AH_BH_C&lt;/math&gt; of &lt;math&gt;\triangle ABC&lt;/math&gt;, the Euler lines of &lt;math&gt;\triangle AH_BH_C&lt;/math&gt;,&lt;math&gt;\triangle BH_CH_A&lt;/math&gt;, and &lt;math&gt;\triangle CH_AH_B&lt;/math&gt; [[concurrence | concur]] at &lt;math&gt;N&lt;/math&gt;, the nine-point center of &lt;math&gt;\triangle ABC&lt;/math&gt;.<br /> <br /> ==Proof of Existence==<br /> This proof utilizes the concept of [[spiral similarity]], which in this case is a [[rotation]] followed [[homothety]]. Consider the [[medial triangle]] &lt;math&gt;\triangle O_AO_BO_C&lt;/math&gt;. It is similar to &lt;math&gt;\triangle ABC&lt;/math&gt;. Specifically, a rotation of &lt;math&gt;180^\circ&lt;/math&gt; about the midpoint of &lt;math&gt;O_BO_C&lt;/math&gt; followed by a homothety with scale factor &lt;math&gt;2&lt;/math&gt; centered at &lt;math&gt;A&lt;/math&gt; brings &lt;math&gt;\triangle ABC \to \triangle O_AO_BO_C&lt;/math&gt;. Let us examine what else this transformation, which we denote as &lt;math&gt;\mathcal{S}&lt;/math&gt;, will do. <br /> <br /> It turns out &lt;math&gt;O&lt;/math&gt; is the orthocenter, and &lt;math&gt;G&lt;/math&gt; is the centroid of &lt;math&gt;\triangle O_AO_BO_C&lt;/math&gt;. Thus, &lt;math&gt;\mathcal{S}(\{O_A, O, G\}) = \{A, H, G\}&lt;/math&gt;. As a homothety preserves angles, it follows that &lt;math&gt;\measuredangle O_AOG = \measuredangle AHG&lt;/math&gt;. Finally, as &lt;math&gt;\overline{AH} || \overline{O_AO}&lt;/math&gt; it follows that <br /> &lt;cmath&gt;\triangle AHG = \triangle O_AOG&lt;/cmath&gt;<br /> Thus, &lt;math&gt;O, G, H&lt;/math&gt; are collinear, and &lt;math&gt;\frac{OG}{HG} = \frac{1}{2}&lt;/math&gt;.<br /> <br /> ~always_correct<br /> <br /> <br /> [[Image:Euler Line.PNG||500px|frame|center]]<br /> <br /> <br /> <br /> <br /> {{stub}}</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=Euler_line&diff=86786 Euler line 2017-08-03T19:15:26Z <p>Always correct: </p> <hr /> <div>In any [[triangle]] &lt;math&gt;\triangle ABC&lt;/math&gt;, the '''Euler line''' is a [[line]] which passes through the [[orthocenter]] &lt;math&gt;H&lt;/math&gt;, [[centroid]] &lt;math&gt;G&lt;/math&gt;, [[circumcenter]] &lt;math&gt;O&lt;/math&gt;, [[nine-point center]] &lt;math&gt;N&lt;/math&gt; and [[De Longchamps point]] &lt;math&gt;L&lt;/math&gt;. It is named after [[Leonhard Euler]]. Its existence is a non-trivial fact of Euclidean [[geometry]]. Certain fixed orders and distance [[ratio]]s hold among these points. In particular, &lt;math&gt;\overline{OGNH}&lt;/math&gt; and &lt;math&gt;OG:GN:NH = 2:1:3&lt;/math&gt;<br /> <br /> Given the [[orthic triangle]]&lt;math&gt;\triangle H_AH_BH_C&lt;/math&gt; of &lt;math&gt;\triangle ABC&lt;/math&gt;, the Euler lines of &lt;math&gt;\triangle AH_BH_C&lt;/math&gt;,&lt;math&gt;\triangle BH_CH_A&lt;/math&gt;, and &lt;math&gt;\triangle CH_AH_B&lt;/math&gt; [[concurrence | concur]] at &lt;math&gt;N&lt;/math&gt;, the nine-point center of &lt;math&gt;\triangle ABC&lt;/math&gt;.<br /> <br /> ==Proof of Existence==<br /> This proof utilizes the concept of [[spiral similarity]], which in this case is a [[rotation]] followed [[homothety]]. Consider the [[medial triangle]] &lt;math&gt;\triangle O_AO_BO_C&lt;/math&gt;. It is similar to &lt;math&gt;\triangle ABC&lt;/math&gt;. Specifically, a rotation of &lt;math&gt;180^\circ&lt;/math&gt; about the midpoint of &lt;math&gt;O_BO_C&lt;/math&gt; followed by a homothety with scale factor &lt;math&gt;2&lt;/math&gt; centered at &lt;math&gt;A&lt;/math&gt; brings &lt;math&gt;\triangle ABC \to \triangle O_AO_BO_C&lt;/math&gt;. Let us examine what else this transformation, which we denote as &lt;math&gt;\mathcal{S}&lt;/math&gt;, will do. <br /> <br /> It turns out &lt;math&gt;O&lt;/math&gt; is the orthocenter, and &lt;math&gt;G&lt;/math&gt; is the centroid of &lt;math&gt;\triangle O_AO_BO_C&lt;/math&gt;. Thus, &lt;math&gt;\mathcal{S}(\{O_A, O, G\}) = \{A, H, G\}&lt;/math&gt;. As a homothety preserves angles, it follows that &lt;math&gt;\measuredangle O_AOG = \measuredangle AHG&lt;/math&gt;. Finally, as &lt;math&gt;\overline{AH} || \overline{O_AO}&lt;/math&gt; it follows that <br /> &lt;cmath&gt;\triangle AHG = \triangle O_AOG&lt;/cmath&gt;<br /> Thus, &lt;math&gt;O, G, H&lt;/math&gt; are collinear, and &lt;math&gt;\frac{OG}{HG} = \frac{1}{2}&lt;/math&gt;.<br /> <br /> ~always_correct<br /> <br /> <br /> [[Image:Euler Line.PNG||500px|frame|center]]<br /> <br /> <br /> <br /> <br /> {{stub}}</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_II_Problems/Problem_15&diff=86722 2009 AIME II Problems/Problem 15 2017-07-31T16:29:34Z <p>Always correct: /* Solution 1 (Quick) */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;\overline{MN}&lt;/math&gt; be a diameter of a circle with diameter 1. Let &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; be points on one of the semicircular arcs determined by &lt;math&gt;\overline{MN}&lt;/math&gt; such that &lt;math&gt;A&lt;/math&gt; is the midpoint of the semicircle and &lt;math&gt;MB=\frac{3}5&lt;/math&gt;. Point &lt;math&gt;C&lt;/math&gt; lies on the other semicircular arc. Let &lt;math&gt;d&lt;/math&gt; be the length of the line segment whose endpoints are the intersections of diameter &lt;math&gt;\overline{MN}&lt;/math&gt; with chords &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt;. The largest possible value of &lt;math&gt;d&lt;/math&gt; can be written in the form &lt;math&gt; r-s\sqrt{t} &lt;/math&gt;, where &lt;math&gt;r, s&lt;/math&gt; and &lt;math&gt;t&lt;/math&gt; are positive integers and &lt;math&gt;t&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;r+s+t&lt;/math&gt;.<br /> <br /> == Solutions ==<br /> <br /> ===Solution 1 (Quick)===<br /> Let &lt;math&gt;V = \overline{NM} \cap \overline{AC}&lt;/math&gt; and &lt;math&gt;W = \overline{NM} \cap \overline{BC}&lt;/math&gt;. Further more let &lt;math&gt;\angle MNC = \alpha&lt;/math&gt; and &lt;math&gt;\angle NMC = 90^\circ - \alpha&lt;/math&gt;. Angle chasing reveals &lt;math&gt;\angle NBC = \angle NAC = \alpha&lt;/math&gt; and &lt;math&gt;\angle MBC = \angle MAC = 90^\circ - \alpha&lt;/math&gt;. Additionally &lt;math&gt;NB = \frac{4}{5}&lt;/math&gt; and &lt;math&gt;AN = AM&lt;/math&gt; by the Pythagorean Theorem.<br /> <br /> By the Angle Bisector Formula,<br /> &lt;cmath&gt;\frac{NV}{MV} = \frac{\sin (\alpha)}{\sin (90^\circ - \alpha)} = \tan (\alpha)&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{MW}{NW} = \frac{3\sin (90^\circ - \alpha)}{4\sin (\alpha)} = \frac{3}{4} \cot (\alpha)&lt;/cmath&gt;<br /> <br /> As &lt;math&gt;NV + MV =MW + NW = 1&lt;/math&gt; we compute &lt;math&gt;NW = \frac{1}{1+\frac{3}{4}\cot(\alpha)}&lt;/math&gt; and &lt;math&gt;MV = \frac{1}{1+\tan (\alpha)}&lt;/math&gt;, and finally &lt;math&gt;VW = NW + MV - 1 = \frac{1}{1+\frac{3}{4}\cot(\alpha)} + \frac{1}{1+\tan (\alpha)} - 1&lt;/math&gt;. Taking the derivative of &lt;math&gt;VW&lt;/math&gt; with respect to &lt;math&gt;\alpha&lt;/math&gt;, we arrive at<br /> &lt;cmath&gt;VW' = \frac{7\cos^2 (\alpha) - 4}{(\sin(\alpha) + \cos(\alpha))^2(4\sin(\alpha)+3\cos(\alpha))^2}&lt;/cmath&gt;<br /> Clearly the maximum occurs when &lt;math&gt;\alpha = \cos^{-1}\left(\frac{2}{\sqrt{7}}\right)&lt;/math&gt;. Plugging this back in, using the fact that &lt;math&gt;\tan(\cos^{-1}(x)) = \frac{\sqrt{1-x^2}}{x}&lt;/math&gt; and &lt;math&gt;\cot(\cos^{-1}(x)) = \frac{x}{\sqrt{1-x^2}}&lt;/math&gt;, we get<br /> <br /> &lt;math&gt;VW = 7 - 4\sqrt{3}&lt;/math&gt;<br /> with &lt;math&gt;7 + 4 + 3 = \boxed{014}&lt;/math&gt;<br /> <br /> ~always_correct<br /> <br /> ===Solution 2===<br /> Let &lt;math&gt;O&lt;/math&gt; be the center of the circle. Define &lt;math&gt;\angle{MOC}=t&lt;/math&gt;, &lt;math&gt;\angle{BOA}=2a&lt;/math&gt;, and let &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt; intersect &lt;math&gt;MN&lt;/math&gt; at points &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt;, respectively. We will express the length of &lt;math&gt;XY&lt;/math&gt; as a function of &lt;math&gt;t&lt;/math&gt; and maximize that function in the interval &lt;math&gt;[0, \pi]&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;C'&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;MN&lt;/math&gt;. We compute &lt;math&gt;XY&lt;/math&gt; as follows.<br /> <br /> (a) By the Extended Law of Sines in triangle &lt;math&gt;ABC&lt;/math&gt;, we have<br /> <br /> &lt;cmath&gt;CA&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \sin\angle{ABC}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \sin\left(\frac{\widehat{AN} + \widehat{NC}}{2}\right)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \sin\left(\frac{\frac{\pi}{2} + (\pi-t)}{2}\right)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \sin\left(\frac{3\pi}{4} - \frac{t}{2}\right)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \sin\left(\frac{\pi}{4} + \frac{t}{2}\right)&lt;/cmath&gt;<br /> <br /> (b) Note that &lt;math&gt;CC' = CO\sin(t) = \left(\frac{1}{2}\right)\sin(t)&lt;/math&gt; and &lt;math&gt;AO = \frac{1}{2}&lt;/math&gt;. Since &lt;math&gt;CC'Y&lt;/math&gt; and &lt;math&gt;AOY&lt;/math&gt; are similar right triangles, we have &lt;math&gt;CY/AY = CC'/AO = \sin(t)&lt;/math&gt;, and hence,<br /> <br /> &lt;cmath&gt;CY/CA&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \frac{CY}{CY + AY}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \frac{\sin(t)}{1 + \sin(t)}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \frac{\sin(t)}{\sin\left(\frac{\pi}{2}\right) + \sin(t)}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \frac{\sin(t)}{2\sin\left(\frac{\pi}{4} + \frac{t}{2}\right)\cos\left(\frac{\pi}{4} - \frac{t}{2}\right)}&lt;/cmath&gt;<br /> <br /> (c) We have &lt;math&gt;\angle{XCY} = \frac{\widehat{AB}}{2}=a&lt;/math&gt; and &lt;math&gt;\angle{CXY} = \frac{\widehat{MB}+\widehat{CN}}{2} = \frac{\left(\frac{\pi}{2} - 2a\right) + (\pi - t)}{2} = \frac{3\pi}{4} - a - \frac{t}{2}&lt;/math&gt;, and hence by the Law of Sines,<br /> <br /> &lt;cmath&gt;XY/CY&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \frac{\sin\angle{XCY}}{\sin\angle{CXY}}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \frac{\sin(a)}{\sin\left(\frac{3\pi}{4} - a - \frac{t}{2}\right)}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \frac{\sin(a)}{\sin\left(\frac{\pi}{4} + a + \frac{t}{2}\right)}&lt;/cmath&gt;<br /> <br /> (d) Multiplying (a), (b), and (c), we have<br /> <br /> &lt;cmath&gt;XY&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= CA * (CY/CA) * (XY/CY)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \frac{\sin(t)\sin(a)}{2\cos\left(\frac{\pi}{4} - \frac{t}{2}\right)\sin\left(\frac{\pi}{4} + a + \frac{t}{2}\right)}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \frac{\sin(t)\sin(a)}{\sin\left(\frac{\pi}{2} + a\right) + \sin(a + t)}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \sin(a)\times\frac{\sin(t)}{\sin(t + a) + \cos(a)}&lt;/cmath&gt;,<br /> <br /> which is a function of &lt;math&gt;t&lt;/math&gt; (and the constant &lt;math&gt;a&lt;/math&gt;). Differentiating this with respect to &lt;math&gt;t&lt;/math&gt; yields<br /> <br /> &lt;cmath&gt;\sin(a)\times\frac{\cos(t)(\sin(t + a) + \cos(a)) - \sin(t)\cos(t + a)}{(\sin(t + a) + \cos(a))^2}&lt;/cmath&gt;,<br /> <br /> and the numerator of this is<br /> <br /> &lt;cmath&gt;\sin(a) \times(\sin(t + a)\cos(t) - \cos(t + a)\sin(t) + \cos(a)\cos(t))&lt;/cmath&gt; <br /> &lt;cmath&gt;= \sin(a) \times (\sin(a) + \cos(a)\cos(t))&lt;/cmath&gt;,<br /> <br /> which vanishes when &lt;math&gt;\sin(a) + \cos(a)\cos(t) = 0&lt;/math&gt;. Therefore, the length of &lt;math&gt;XY&lt;/math&gt; is maximized when &lt;math&gt;t=t'&lt;/math&gt;, where &lt;math&gt;t'&lt;/math&gt; is the value in &lt;math&gt;[0, \pi]&lt;/math&gt; that satisfies &lt;math&gt;\cos(t') = -\tan(a)&lt;/math&gt;.<br /> <br /> Note that<br /> <br /> &lt;cmath&gt;\frac{1 - \tan(a)}{1 + \tan(a)} = \tan\left(\frac{\pi}{4} - a\right) = \tan((\widehat{MB})/2) = \tan\angle{MNB} = \frac{3}{4}&lt;/cmath&gt;,<br /> <br /> so &lt;math&gt;\tan(a) = \frac{1}{7}&lt;/math&gt;. We compute<br /> <br /> &lt;cmath&gt;\sin(a) = \frac{\sqrt{2}}{10}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\cos(a) = \frac{7\sqrt{2}}{10}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\cos(t') = -\tan(a) = -\frac{1}{7}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\sin(t') = \frac{4\sqrt{3}}{7}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\sin(t' + a)=\sin(t')\cos(a) + \cos(t')\sin(a) = \frac{28\sqrt{6} - \sqrt{2}}{70}&lt;/cmath&gt;,<br /> <br /> so the maximum length of &lt;math&gt;XY&lt;/math&gt; is &lt;math&gt;\sin(a)\times\frac{\sin(t')}{\sin(t' + a) + \cos(a)} = 7 - 4\sqrt{3}&lt;/math&gt;, and the answer is &lt;math&gt;7 + 4 + 3 = \boxed{014}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> &lt;asy&gt;<br /> unitsize(144);<br /> pair A, B, C, M, n;<br /> A = (0,1); B = (-7/25, 24/25); C=(1/7,-4*sqrt(3)/7); M = (-1,0); n = (1,0);<br /> pair [] D = intersectionpoints(A--C,M--n); pair [] e = intersectionpoints(B--C,M--n);<br /> <br /> draw(circle((0,0),1));<br /> draw(M--n--B--M--A--n--C--A--B--C--cycle);<br /> <br /> label(&quot;$A$&quot;,A,N); label(&quot;$B$&quot;,B,NNW); label(&quot;$M$&quot;,M,W); label(&quot;$C$&quot;,C,SSE); label(&quot;$N$&quot;,n,E);<br /> label(&quot;$D$&quot;,D,SE); label(&quot;$E$&quot;,e,SW);<br /> label(&quot;$x$&quot;,(M+C)/2,SW); label(&quot;$y$&quot;,(n+C)/2,SE);<br /> &lt;/asy&gt;<br /> <br /> Suppose &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt; intersect &lt;math&gt;\overline{MN}&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt;, respectively, and let &lt;math&gt;MC = x&lt;/math&gt; and &lt;math&gt;NC = y&lt;/math&gt;. Since &lt;math&gt;A&lt;/math&gt; is the midpoint of arc &lt;math&gt;MN&lt;/math&gt;, &lt;math&gt;\overline{CA}&lt;/math&gt; bisects &lt;math&gt;\angle MCN&lt;/math&gt;, and we get<br /> &lt;cmath&gt;\frac{MC}{MD} = \frac{NC}{ND}\Rightarrow MD = \frac{x}{x + y}.&lt;/cmath&gt;<br /> To find &lt;math&gt;ME&lt;/math&gt;, we note that &lt;math&gt;\triangle BNE\sim\triangle MCE&lt;/math&gt; and &lt;math&gt;\triangle BME\sim\triangle NCE&lt;/math&gt;, so<br /> &lt;cmath&gt;\begin{align*}<br /> \frac{BN}{NE} &amp;= \frac{MC}{CE} \\<br /> \frac{ME}{BM} &amp;= \frac{CE}{NC}.<br /> \end{align*}&lt;/cmath&gt;<br /> Writing &lt;math&gt;NE = 1 - ME&lt;/math&gt;, we can substitute known values and multiply the equations to get<br /> &lt;cmath&gt;\frac{4(ME)}{3 - 3(ME)} = \frac{x}{y}\Rightarrow ME = \frac{3x}{3x + 4y}.&lt;/cmath&gt;<br /> The value we wish to maximize is<br /> &lt;cmath&gt;\begin{align*}<br /> DE &amp;= MD - ME \\<br /> &amp;= \frac{x}{x + y} - \frac{3x}{3x + 4y} \\<br /> &amp;= \frac{xy}{3x^2 + 7xy + 4y^2} \\<br /> &amp;= \frac{1}{3(x/y) + 4(y/x) + 7}.<br /> \end{align*}&lt;/cmath&gt;<br /> By the AM-GM inequality, &lt;math&gt;3(x/y) + 4(y/x)\geq 2\sqrt{12} = 4\sqrt{3}&lt;/math&gt;, so<br /> &lt;cmath&gt;DE\leq \frac{1}{4\sqrt{3} + 7} = 7 - 4\sqrt{3},&lt;/cmath&gt;<br /> giving the answer of &lt;math&gt;7 + 4 + 3 = \boxed{014}&lt;/math&gt;. Equality is achieved when &lt;math&gt;3(x/y) = 4(y/x)&lt;/math&gt; subject to the condition &lt;math&gt;x^2 + y^2 = 1&lt;/math&gt;, which occurs for &lt;math&gt;x = \frac{2\sqrt{7}}{7}&lt;/math&gt; and &lt;math&gt;y = \frac{\sqrt{21}}{7}&lt;/math&gt;.<br /> <br /> ===Solution 4 (Projective)===<br /> By Pythagoras in &lt;math&gt;\triangle BMN,&lt;/math&gt; we get &lt;math&gt;BN=\dfrac{4}{5}.&lt;/math&gt;<br /> <br /> Since cross ratios are preserved upon projecting, note that &lt;math&gt;(M,Y;X,N)\stackrel{C}{=}(M,B;A,N).&lt;/math&gt; By definition of a cross ratio, this becomes &lt;cmath&gt;\dfrac{XM}{NY}:\dfrac{NM}{NY}=\dfrac{AM}{AB}:\dfrac{MN}{NB}.&lt;/cmath&gt; Let &lt;math&gt;MY=a,YX=b,XN=c&lt;/math&gt; such that &lt;math&gt;a+b+c=1.&lt;/math&gt; We know that &lt;math&gt;XM=a+b,XY=b,NM=1,NY=b+c,&lt;/math&gt; so the LHS becomes &lt;math&gt;\dfrac{(a+b)(b+c)}{b}.&lt;/math&gt;<br /> <br /> In the RHS, we are given every value except for &lt;math&gt;AB.&lt;/math&gt; However, Ptolemy's Theorem on &lt;math&gt;MBAN&lt;/math&gt; gives &lt;math&gt;AB\cdot MN+AN\cdot BM=AM\cdot BN\implies AB+\dfrac{3}{5\sqrt{2}}=\dfrac{4}{5\sqrt{2}}\implies AB=\dfrac{1}{5\sqrt{2}}.&lt;/math&gt; Substituting, we get &lt;math&gt;\dfrac{(a+b)(b+c)}{b}=4\implies b(a+b+c)+ac=4b, b=\dfrac{ac}{3}&lt;/math&gt; where we use &lt;math&gt;a+b+c=1.&lt;/math&gt;<br /> <br /> Again using &lt;math&gt;a+b+c=1,&lt;/math&gt; we have &lt;math&gt;a+b+c=1\implies a+\dfrac{ac}{3}+c=1\implies a=3\dfrac{1-c}{c+3}.&lt;/math&gt; Then &lt;math&gt;b=\dfrac{ac}{3}=\dfrac{c-c^2}{c+3}.&lt;/math&gt; Since this is a function in &lt;math&gt;c,&lt;/math&gt; we differentiate WRT &lt;math&gt;c&lt;/math&gt; to find its maximum. By quotient rule, it suffices to solve &lt;cmath&gt;(-2c+1)(c+3)-(c-c^2)=0 \implies c^2+6c-3,c=-3+2\sqrt{3}.&lt;/cmath&gt; Substituting back yields &lt;math&gt;b=7-4\sqrt{3},&lt;/math&gt; so &lt;math&gt;7+4+3=\boxed{014}&lt;/math&gt; is the answer.<br /> <br /> ~Generic_Username<br /> <br /> ==See Also==<br /> {{AIME box|year=2009|n=II|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_II_Problems/Problem_15&diff=86721 2009 AIME II Problems/Problem 15 2017-07-31T16:29:16Z <p>Always correct: /* Solutions */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;\overline{MN}&lt;/math&gt; be a diameter of a circle with diameter 1. Let &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; be points on one of the semicircular arcs determined by &lt;math&gt;\overline{MN}&lt;/math&gt; such that &lt;math&gt;A&lt;/math&gt; is the midpoint of the semicircle and &lt;math&gt;MB=\frac{3}5&lt;/math&gt;. Point &lt;math&gt;C&lt;/math&gt; lies on the other semicircular arc. Let &lt;math&gt;d&lt;/math&gt; be the length of the line segment whose endpoints are the intersections of diameter &lt;math&gt;\overline{MN}&lt;/math&gt; with chords &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt;. The largest possible value of &lt;math&gt;d&lt;/math&gt; can be written in the form &lt;math&gt; r-s\sqrt{t} &lt;/math&gt;, where &lt;math&gt;r, s&lt;/math&gt; and &lt;math&gt;t&lt;/math&gt; are positive integers and &lt;math&gt;t&lt;/math&gt; is not divisible by the square of any prime. Find &lt;math&gt;r+s+t&lt;/math&gt;.<br /> <br /> == Solutions ==<br /> <br /> ===Solution 1 (Quick)===<br /> Let &lt;math&gt;V = \overline{NM} \cap \overline{AC}&lt;/math&gt; and &lt;math&gt;W = \overline{NM} \cap \overline{BC}&lt;/math&gt;. Further more let &lt;math&gt;\angle MNC = \alpha&lt;/math&gt; and &lt;math&gt;\angle NMC = 90^\circ - \alpha&lt;/math&gt;. Angle chasing reveals &lt;math&gt;\angle NBC = \angle NAC = \alpha&lt;/math&gt; and &lt;math&gt;\angle MBC = \angle MAC = 90^\circ - \alpha&lt;/math&gt;. Additionally &lt;math&gt;NB = \frac{4}{5}&lt;/math&gt; and &lt;math&gt;AN = AM&lt;/math&gt; by the Pythagorean Theorem.<br /> <br /> By the Angle Bisector Formula,<br /> &lt;cmath&gt;\frac{NV}{MV} = \frac{\sin (\alpha)}{\sin (90^\circ - \alpha)} = \tan (\alpha)&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{MW}{NW} = \frac{3\sin (90^\circ - \alpha)}{4\sin (\alpha)} = \frac{3}{4} \cot (\alpha)&lt;/cmath&gt;<br /> <br /> As &lt;math&gt;NV + MV =MW + NW = 1&lt;/math&gt; we compute &lt;math&gt;NW = \frac{1}{1+\frac{3}{4}\cot(\alpha)}&lt;/math&gt; and &lt;math&gt;MV = \frac{1}{1+\tan (\alpha)}&lt;/math&gt;, and finally &lt;math&gt;VW = NW + MV - 1 = \frac{1}{1+\frac{3}{4}\cot(\alpha)} + \frac{1}{1+\tan (\alpha)} - 1&lt;/math&gt;. Taking the derivative of &lt;math&gt;VW&lt;/math&gt; with respect to &lt;math&gt;\alpha&lt;/math&gt;, we arrive at<br /> &lt;cmath&gt;VW' = \frac{7\cos^2 (\alpha) - 4}{(\sin(\alpha) + \cos(\alpha))^2(4\sin(\alpha)+3\cos(\alpha))^2}&lt;/cmath&gt;<br /> Clearly the maximum occurs when &lt;math&gt;\alpha = \cos^{-1}\left(\frac{2}{\sqrt{7}}\right)&lt;/math&gt;. Plugging this back in, using the fact that &lt;math&gt;\tan(\cos^{-1}(x)) = \frac{\sqrt{1-x^2}}{x}&lt;/math&gt; and &lt;math&gt;\cot(\cos^{-1}(x)) = \frac{x}{\sqrt{1-x^2}}&lt;/math&gt;, we get<br /> <br /> &lt;math&gt;VW = 7 - 4\sqrt{3}&lt;/math&gt;<br /> with &lt;math&gt;7 + 4 + 3 = \boxed{014}&lt;/math&gt;<br /> ===Solution 2===<br /> Let &lt;math&gt;O&lt;/math&gt; be the center of the circle. Define &lt;math&gt;\angle{MOC}=t&lt;/math&gt;, &lt;math&gt;\angle{BOA}=2a&lt;/math&gt;, and let &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt; intersect &lt;math&gt;MN&lt;/math&gt; at points &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt;, respectively. We will express the length of &lt;math&gt;XY&lt;/math&gt; as a function of &lt;math&gt;t&lt;/math&gt; and maximize that function in the interval &lt;math&gt;[0, \pi]&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;C'&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;MN&lt;/math&gt;. We compute &lt;math&gt;XY&lt;/math&gt; as follows.<br /> <br /> (a) By the Extended Law of Sines in triangle &lt;math&gt;ABC&lt;/math&gt;, we have<br /> <br /> &lt;cmath&gt;CA&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \sin\angle{ABC}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \sin\left(\frac{\widehat{AN} + \widehat{NC}}{2}\right)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \sin\left(\frac{\frac{\pi}{2} + (\pi-t)}{2}\right)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \sin\left(\frac{3\pi}{4} - \frac{t}{2}\right)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \sin\left(\frac{\pi}{4} + \frac{t}{2}\right)&lt;/cmath&gt;<br /> <br /> (b) Note that &lt;math&gt;CC' = CO\sin(t) = \left(\frac{1}{2}\right)\sin(t)&lt;/math&gt; and &lt;math&gt;AO = \frac{1}{2}&lt;/math&gt;. Since &lt;math&gt;CC'Y&lt;/math&gt; and &lt;math&gt;AOY&lt;/math&gt; are similar right triangles, we have &lt;math&gt;CY/AY = CC'/AO = \sin(t)&lt;/math&gt;, and hence,<br /> <br /> &lt;cmath&gt;CY/CA&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \frac{CY}{CY + AY}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \frac{\sin(t)}{1 + \sin(t)}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \frac{\sin(t)}{\sin\left(\frac{\pi}{2}\right) + \sin(t)}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \frac{\sin(t)}{2\sin\left(\frac{\pi}{4} + \frac{t}{2}\right)\cos\left(\frac{\pi}{4} - \frac{t}{2}\right)}&lt;/cmath&gt;<br /> <br /> (c) We have &lt;math&gt;\angle{XCY} = \frac{\widehat{AB}}{2}=a&lt;/math&gt; and &lt;math&gt;\angle{CXY} = \frac{\widehat{MB}+\widehat{CN}}{2} = \frac{\left(\frac{\pi}{2} - 2a\right) + (\pi - t)}{2} = \frac{3\pi}{4} - a - \frac{t}{2}&lt;/math&gt;, and hence by the Law of Sines,<br /> <br /> &lt;cmath&gt;XY/CY&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \frac{\sin\angle{XCY}}{\sin\angle{CXY}}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \frac{\sin(a)}{\sin\left(\frac{3\pi}{4} - a - \frac{t}{2}\right)}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \frac{\sin(a)}{\sin\left(\frac{\pi}{4} + a + \frac{t}{2}\right)}&lt;/cmath&gt;<br /> <br /> (d) Multiplying (a), (b), and (c), we have<br /> <br /> &lt;cmath&gt;XY&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= CA * (CY/CA) * (XY/CY)&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \frac{\sin(t)\sin(a)}{2\cos\left(\frac{\pi}{4} - \frac{t}{2}\right)\sin\left(\frac{\pi}{4} + a + \frac{t}{2}\right)}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \frac{\sin(t)\sin(a)}{\sin\left(\frac{\pi}{2} + a\right) + \sin(a + t)}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;= \sin(a)\times\frac{\sin(t)}{\sin(t + a) + \cos(a)}&lt;/cmath&gt;,<br /> <br /> which is a function of &lt;math&gt;t&lt;/math&gt; (and the constant &lt;math&gt;a&lt;/math&gt;). Differentiating this with respect to &lt;math&gt;t&lt;/math&gt; yields<br /> <br /> &lt;cmath&gt;\sin(a)\times\frac{\cos(t)(\sin(t + a) + \cos(a)) - \sin(t)\cos(t + a)}{(\sin(t + a) + \cos(a))^2}&lt;/cmath&gt;,<br /> <br /> and the numerator of this is<br /> <br /> &lt;cmath&gt;\sin(a) \times(\sin(t + a)\cos(t) - \cos(t + a)\sin(t) + \cos(a)\cos(t))&lt;/cmath&gt; <br /> &lt;cmath&gt;= \sin(a) \times (\sin(a) + \cos(a)\cos(t))&lt;/cmath&gt;,<br /> <br /> which vanishes when &lt;math&gt;\sin(a) + \cos(a)\cos(t) = 0&lt;/math&gt;. Therefore, the length of &lt;math&gt;XY&lt;/math&gt; is maximized when &lt;math&gt;t=t'&lt;/math&gt;, where &lt;math&gt;t'&lt;/math&gt; is the value in &lt;math&gt;[0, \pi]&lt;/math&gt; that satisfies &lt;math&gt;\cos(t') = -\tan(a)&lt;/math&gt;.<br /> <br /> Note that<br /> <br /> &lt;cmath&gt;\frac{1 - \tan(a)}{1 + \tan(a)} = \tan\left(\frac{\pi}{4} - a\right) = \tan((\widehat{MB})/2) = \tan\angle{MNB} = \frac{3}{4}&lt;/cmath&gt;,<br /> <br /> so &lt;math&gt;\tan(a) = \frac{1}{7}&lt;/math&gt;. We compute<br /> <br /> &lt;cmath&gt;\sin(a) = \frac{\sqrt{2}}{10}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\cos(a) = \frac{7\sqrt{2}}{10}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\cos(t') = -\tan(a) = -\frac{1}{7}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\sin(t') = \frac{4\sqrt{3}}{7}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\sin(t' + a)=\sin(t')\cos(a) + \cos(t')\sin(a) = \frac{28\sqrt{6} - \sqrt{2}}{70}&lt;/cmath&gt;,<br /> <br /> so the maximum length of &lt;math&gt;XY&lt;/math&gt; is &lt;math&gt;\sin(a)\times\frac{\sin(t')}{\sin(t' + a) + \cos(a)} = 7 - 4\sqrt{3}&lt;/math&gt;, and the answer is &lt;math&gt;7 + 4 + 3 = \boxed{014}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> &lt;asy&gt;<br /> unitsize(144);<br /> pair A, B, C, M, n;<br /> A = (0,1); B = (-7/25, 24/25); C=(1/7,-4*sqrt(3)/7); M = (-1,0); n = (1,0);<br /> pair [] D = intersectionpoints(A--C,M--n); pair [] e = intersectionpoints(B--C,M--n);<br /> <br /> draw(circle((0,0),1));<br /> draw(M--n--B--M--A--n--C--A--B--C--cycle);<br /> <br /> label(&quot;$A$&quot;,A,N); label(&quot;$B$&quot;,B,NNW); label(&quot;$M$&quot;,M,W); label(&quot;$C$&quot;,C,SSE); label(&quot;$N$&quot;,n,E);<br /> label(&quot;$D$&quot;,D,SE); label(&quot;$E$&quot;,e,SW);<br /> label(&quot;$x$&quot;,(M+C)/2,SW); label(&quot;$y$&quot;,(n+C)/2,SE);<br /> &lt;/asy&gt;<br /> <br /> Suppose &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt; intersect &lt;math&gt;\overline{MN}&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt;, respectively, and let &lt;math&gt;MC = x&lt;/math&gt; and &lt;math&gt;NC = y&lt;/math&gt;. Since &lt;math&gt;A&lt;/math&gt; is the midpoint of arc &lt;math&gt;MN&lt;/math&gt;, &lt;math&gt;\overline{CA}&lt;/math&gt; bisects &lt;math&gt;\angle MCN&lt;/math&gt;, and we get<br /> &lt;cmath&gt;\frac{MC}{MD} = \frac{NC}{ND}\Rightarrow MD = \frac{x}{x + y}.&lt;/cmath&gt;<br /> To find &lt;math&gt;ME&lt;/math&gt;, we note that &lt;math&gt;\triangle BNE\sim\triangle MCE&lt;/math&gt; and &lt;math&gt;\triangle BME\sim\triangle NCE&lt;/math&gt;, so<br /> &lt;cmath&gt;\begin{align*}<br /> \frac{BN}{NE} &amp;= \frac{MC}{CE} \\<br /> \frac{ME}{BM} &amp;= \frac{CE}{NC}.<br /> \end{align*}&lt;/cmath&gt;<br /> Writing &lt;math&gt;NE = 1 - ME&lt;/math&gt;, we can substitute known values and multiply the equations to get<br /> &lt;cmath&gt;\frac{4(ME)}{3 - 3(ME)} = \frac{x}{y}\Rightarrow ME = \frac{3x}{3x + 4y}.&lt;/cmath&gt;<br /> The value we wish to maximize is<br /> &lt;cmath&gt;\begin{align*}<br /> DE &amp;= MD - ME \\<br /> &amp;= \frac{x}{x + y} - \frac{3x}{3x + 4y} \\<br /> &amp;= \frac{xy}{3x^2 + 7xy + 4y^2} \\<br /> &amp;= \frac{1}{3(x/y) + 4(y/x) + 7}.<br /> \end{align*}&lt;/cmath&gt;<br /> By the AM-GM inequality, &lt;math&gt;3(x/y) + 4(y/x)\geq 2\sqrt{12} = 4\sqrt{3}&lt;/math&gt;, so<br /> &lt;cmath&gt;DE\leq \frac{1}{4\sqrt{3} + 7} = 7 - 4\sqrt{3},&lt;/cmath&gt;<br /> giving the answer of &lt;math&gt;7 + 4 + 3 = \boxed{014}&lt;/math&gt;. Equality is achieved when &lt;math&gt;3(x/y) = 4(y/x)&lt;/math&gt; subject to the condition &lt;math&gt;x^2 + y^2 = 1&lt;/math&gt;, which occurs for &lt;math&gt;x = \frac{2\sqrt{7}}{7}&lt;/math&gt; and &lt;math&gt;y = \frac{\sqrt{21}}{7}&lt;/math&gt;.<br /> <br /> ===Solution 4 (Projective)===<br /> By Pythagoras in &lt;math&gt;\triangle BMN,&lt;/math&gt; we get &lt;math&gt;BN=\dfrac{4}{5}.&lt;/math&gt;<br /> <br /> Since cross ratios are preserved upon projecting, note that &lt;math&gt;(M,Y;X,N)\stackrel{C}{=}(M,B;A,N).&lt;/math&gt; By definition of a cross ratio, this becomes &lt;cmath&gt;\dfrac{XM}{NY}:\dfrac{NM}{NY}=\dfrac{AM}{AB}:\dfrac{MN}{NB}.&lt;/cmath&gt; Let &lt;math&gt;MY=a,YX=b,XN=c&lt;/math&gt; such that &lt;math&gt;a+b+c=1.&lt;/math&gt; We know that &lt;math&gt;XM=a+b,XY=b,NM=1,NY=b+c,&lt;/math&gt; so the LHS becomes &lt;math&gt;\dfrac{(a+b)(b+c)}{b}.&lt;/math&gt;<br /> <br /> In the RHS, we are given every value except for &lt;math&gt;AB.&lt;/math&gt; However, Ptolemy's Theorem on &lt;math&gt;MBAN&lt;/math&gt; gives &lt;math&gt;AB\cdot MN+AN\cdot BM=AM\cdot BN\implies AB+\dfrac{3}{5\sqrt{2}}=\dfrac{4}{5\sqrt{2}}\implies AB=\dfrac{1}{5\sqrt{2}}.&lt;/math&gt; Substituting, we get &lt;math&gt;\dfrac{(a+b)(b+c)}{b}=4\implies b(a+b+c)+ac=4b, b=\dfrac{ac}{3}&lt;/math&gt; where we use &lt;math&gt;a+b+c=1.&lt;/math&gt;<br /> <br /> Again using &lt;math&gt;a+b+c=1,&lt;/math&gt; we have &lt;math&gt;a+b+c=1\implies a+\dfrac{ac}{3}+c=1\implies a=3\dfrac{1-c}{c+3}.&lt;/math&gt; Then &lt;math&gt;b=\dfrac{ac}{3}=\dfrac{c-c^2}{c+3}.&lt;/math&gt; Since this is a function in &lt;math&gt;c,&lt;/math&gt; we differentiate WRT &lt;math&gt;c&lt;/math&gt; to find its maximum. By quotient rule, it suffices to solve &lt;cmath&gt;(-2c+1)(c+3)-(c-c^2)=0 \implies c^2+6c-3,c=-3+2\sqrt{3}.&lt;/cmath&gt; Substituting back yields &lt;math&gt;b=7-4\sqrt{3},&lt;/math&gt; so &lt;math&gt;7+4+3=\boxed{014}&lt;/math&gt; is the answer.<br /> <br /> ~Generic_Username<br /> <br /> ==See Also==<br /> {{AIME box|year=2009|n=II|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=1995_IMO_Problems/Problem_1&diff=86450 1995 IMO Problems/Problem 1 2017-07-19T12:38:53Z <p>Always correct: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;A,B,C,D&lt;/math&gt; be four distinct points on a line, in that order. The circles with diameters &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;BD&lt;/math&gt; intersect at &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt;. The line &lt;math&gt;XY&lt;/math&gt; meets &lt;math&gt;BC&lt;/math&gt; at &lt;math&gt;Z&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be a point on the line &lt;math&gt;XY&lt;/math&gt; other than &lt;math&gt;Z&lt;/math&gt;. The line &lt;math&gt;CP&lt;/math&gt; intersects the circle with diameter &lt;math&gt;AC&lt;/math&gt; at &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;M&lt;/math&gt;, and the line &lt;math&gt;BP&lt;/math&gt; intersects the circle with diameter &lt;math&gt;BD&lt;/math&gt; at &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt;. Prove that the lines &lt;math&gt;AM,DN,XY&lt;/math&gt; are concurrent. <br /> <br /> <br /> == Hint ==<br /> Think about Radical Axis, Power of a Point and Radical Center.<br /> <br /> == Solution ==<br /> Since &lt;math&gt;M&lt;/math&gt; is on the circle with diameter &lt;math&gt;AC&lt;/math&gt;, we have &lt;math&gt;\angle AMC=90&lt;/math&gt; and so &lt;math&gt;\angle MCA=90-A&lt;/math&gt;. We simlarly find that &lt;math&gt;\angle BND=90&lt;/math&gt;. Also, notice that the line &lt;math&gt;XY&lt;/math&gt; is the radical axis of the two circles with diameters &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;BD&lt;/math&gt;. Thus, since &lt;math&gt;P&lt;/math&gt; is on &lt;math&gt;XY&lt;/math&gt;, we have &lt;math&gt;PN\cdot PB=PM\cdot PC&lt;/math&gt; and so by the converse of Power of a Point, the quadrilateral &lt;math&gt;MNBC&lt;/math&gt; is cyclic. Thus, &lt;math&gt;90-A=\angle MCA=\angle BNM&lt;/math&gt;. Thus, &lt;math&gt;\angle MND=180-A&lt;/math&gt; and so quadrilateral &lt;math&gt;AMND&lt;/math&gt; is cyclic. Let the circle which contains the points &lt;math&gt;AMND&lt;/math&gt; be cirle &lt;math&gt;O&lt;/math&gt;. Then, the radical axis of &lt;math&gt;O&lt;/math&gt; and the circle with diameter &lt;math&gt;AC&lt;/math&gt; is line &lt;math&gt;AM&lt;/math&gt;. Also, the radical axis of &lt;math&gt;O&lt;/math&gt; and the circle with diameter &lt;math&gt;BD&lt;/math&gt; is line &lt;math&gt;DN&lt;/math&gt;. Since the pairwise radical axes of 3 circles are concurrent, we have &lt;math&gt;AM,DN,XY&lt;/math&gt; are concurrent as desired.<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;AM&lt;/math&gt; and &lt;math&gt;PT&lt;/math&gt; (a subsegment of &lt;math&gt;XY&lt;/math&gt;) intersect at &lt;math&gt;Z&lt;/math&gt;. Now, assume that &lt;math&gt;Z, N, P&lt;/math&gt; are not collinear. In that case, let &lt;math&gt;ZD&lt;/math&gt; intersect the circle with diameter &lt;math&gt;BD&lt;/math&gt; at &lt;math&gt;N'&lt;/math&gt; and the circle through &lt;math&gt;D, P, T&lt;/math&gt; at &lt;math&gt;N''&lt;/math&gt;.<br /> <br /> We know that &lt;math&gt;\angle AMC = \angle BND = \angle ATP = 90^\circ&lt;/math&gt; via standard formulae, so quadrilaterals &lt;math&gt;AMPT&lt;/math&gt; and &lt;math&gt;DNPT&lt;/math&gt; are cyclic. Thus, &lt;math&gt;N'&lt;/math&gt; and &lt;math&gt;N''&lt;/math&gt; are distinct, as none of them is &lt;math&gt;N&lt;/math&gt;. Hence, by Power of a Point, &lt;cmath&gt;ZM * ZA = ZP * ZT = ZN'' * ZD.&lt;/cmath&gt; However, because &lt;math&gt;Z&lt;/math&gt; lies on radical axis &lt;math&gt;TP&lt;/math&gt; of the two circles, we have &lt;cmath&gt;ZM * ZA = ZN' * ZD.&lt;/cmath&gt; Hence, &lt;math&gt;ZN'' = ZN'&lt;/math&gt;, a contradiction since &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;D'&lt;/math&gt; are distinct. We therefore conclude that &lt;math&gt;Z, N, D&lt;/math&gt; are collinear, which gives the concurrency of &lt;math&gt;AM, XY&lt;/math&gt;, and &lt;math&gt;DN&lt;/math&gt;. This completes the problem.<br /> <br /> == Solution 3==<br /> Let &lt;math&gt;AM&lt;/math&gt; and &lt;math&gt;XY&lt;/math&gt; intersect at &lt;math&gt;Z&lt;/math&gt;. Because &lt;math&gt;\angle AMC = \angle BND = \angle APT = 90^\circ&lt;/math&gt;, we have quadrilaterals &lt;math&gt;AMPT&lt;/math&gt; and &lt;math&gt;DNPT&lt;/math&gt; cyclic. Therefore, &lt;math&gt;Z&lt;/math&gt; lies on the radical axis of the two circumcircles of these quadrilaterals. But &lt;math&gt;Z&lt;/math&gt; also lies on radical axis &lt;math&gt;XY&lt;/math&gt; of the original two circles, so the power of &lt;math&gt;Z&lt;/math&gt; with respect to each of the four circles is all equal to &lt;math&gt;ZM * ZA&lt;/math&gt;. Hence, &lt;math&gt;Z&lt;/math&gt; lies on the radical axis &lt;math&gt;DN&lt;/math&gt; of the two circles passing through &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt;, as desired.<br /> <br /> ==Discussion==<br /> ''Lemma:'' The radical axis of two pairs of circles &lt;math&gt;O_1&lt;/math&gt;, &lt;math&gt;O_2&lt;/math&gt; and &lt;math&gt;O_3&lt;/math&gt;, &lt;math&gt;O_4&lt;/math&gt; are the same line &lt;math&gt;l&lt;/math&gt;. Furthermore, &lt;math&gt;O_1&lt;/math&gt; and &lt;math&gt;O_2&lt;/math&gt; intersect at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;O_3&lt;/math&gt; and &lt;math&gt;O_4&lt;/math&gt; intersect at &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;. Then &lt;math&gt;A, B, C,&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; are concyclic.<br /> <br /> The proof of this lemma is trivial using the argument in Solution 3 and applying the converse of Power of a Point.<br /> <br /> Note that this Problem 1 is a corollary of this lemma. This lemma is an effective way to relate four circles, just as the radical center can relate three circles.<br /> <br /> Solution 1 also gives a trivial lemma that can also be useful:<br /> <br /> ''Lemma 2:'' Chords &lt;math&gt;AB&lt;/math&gt; of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt; of &lt;math&gt;\omega_2&lt;/math&gt; intersect on the segment &lt;math&gt;XY&lt;/math&gt; formed from the intersections of the two circles. Then &lt;math&gt;A, B, C, D&lt;/math&gt; are concyclic.<br /> <br /> Two ways to solve a problem, two different insights into circle geometry. That is cool, but more RADICAL!<br /> <br /> ==See also==<br /> <br /> [[Category:Olympiad Geometry Problems]]</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=1995_IMO_Problems/Problem_1&diff=86449 1995 IMO Problems/Problem 1 2017-07-19T12:38:25Z <p>Always correct: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;A,B,C,D&lt;/math&gt; be four distinct points on a line, in that order. The circles with diameters &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;BD&lt;/math&gt; intersect at &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt;. The line &lt;math&gt;XY&lt;/math&gt; meets &lt;math&gt;BC&lt;/math&gt; at &lt;math&gt;Z&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be a point on the line &lt;math&gt;XY&lt;/math&gt; other than &lt;math&gt;Z&lt;/math&gt;. The line &lt;math&gt;CP&lt;/math&gt; intersects the circle with diameter &lt;math&gt;AC&lt;/math&gt; at &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;M&lt;/math&gt;, and the line &lt;math&gt;BP&lt;/math&gt; intersects the circle with diameter &lt;math&gt;BD&lt;/math&gt; at &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt;. Prove that the lines &lt;math&gt;AM,DN,XY&lt;/math&gt; are concurrent. <br /> <br /> <br /> == Hint ==<br /> Think about Radical Axis, Power of a Point and Radical Center.<br /> <br /> == Solution ==<br /> Since &lt;math&gt;M&lt;/math&gt; is on the circle with diameter &lt;math&gt;AC&lt;/math&gt;, we have &lt;math&gt;\angle AMC=90&lt;/math&gt; and so &lt;math&gt;\angle MCA=90-A&lt;/math&gt;. We simlarly find that &lt;math&gt;\angle BND=90&lt;/math&gt;. Also, notice that the line &lt;math&gt;XY&lt;/math&gt; is the radical axis of the two circles with diameters &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;BD&lt;/math&gt;. Thus, since &lt;math&gt;P&lt;/math&gt; is on &lt;math&gt;XY&lt;/math&gt;, we have &lt;math&gt;PN\cdot PB=PM\cdot PC&lt;/math&gt; and so by the converse of Power of a Point, the quadrilateral &lt;math&gt;MNBC&lt;/math&gt; is cyclic. Thus, &lt;math&gt;90-A=\angle MCA=\angle BNM&lt;/math&gt;. Thus, &lt;math&gt;\angle MND=180-A&lt;/math&gt; and so quadrilateral &lt;math&gt;AMND&lt;/math&gt; is cyclic. Let the circle which contains the points &lt;math&gt;AMND&lt;/math&gt; be cirle &lt;math&gt;O&lt;/math&gt;. Then, the radical axis of &lt;math&gt;O&lt;/math&gt; and the circle with diameter &lt;math&gt;AC&lt;/math&gt; is line &lt;math&gt;AM&lt;/math&gt;. Also, the radical axis of &lt;math&gt;O&lt;/math&gt; and the circle with diameter &lt;math&gt;BD&lt;/math&gt; is line &lt;math&gt;DN&lt;/math&gt;. Since the pairwise radical axes of 3 circles are concurrent, we have &lt;math&gt;AM,DN,XY&lt;/math&gt; are concurrent as desired.<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;AM&lt;/math&gt; and &lt;math&gt;PT&lt;/math&gt; (a subsegment of &lt;math&gt;XY&lt;/math&gt;) intersect at &lt;math&gt;Z&lt;/math&gt;. Now, assume that &lt;math&gt;Z, N, P&lt;/math&gt; are not collinear. In that case, let &lt;math&gt;ZD&lt;/math&gt; intersect the circle with diameter &lt;math&gt;BD&lt;/math&gt; at &lt;math&gt;N'&lt;/math&gt; and the circle through &lt;math&gt;D, P, T&lt;/math&gt; at &lt;math&gt;N''&lt;/math&gt;.<br /> <br /> We know that &lt;math&gt;\angle AMC = \angle BND = \angle ATP = 90^\circ&lt;/math&gt; via standard formulae, so quadrilaterals &lt;math&gt;AMPT&lt;/math&gt; and &lt;math&gt;DNPT&lt;/math&gt; are cyclic. Thus, &lt;math&gt;N'&lt;/math&gt; and &lt;math&gt;N''&lt;/math&gt; are distinct, as none of them is &lt;math&gt;N&lt;/math&gt;. Hence, by Power of a Point, &lt;cmath&gt;ZM * ZA = ZP * ZT = ZN'' * ZD.&lt;/cmath&gt; However, because &lt;math&gt;Z&lt;/math&gt; lies on radical axis &lt;math&gt;TP&lt;/math&gt; of the two circles, we have &lt;cmath&gt;ZM * ZA = ZN' * ZD.&lt;/cmath&gt; Hence, &lt;math&gt;ZN'' = ZN'&lt;/math&gt;, a contradiction since &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;D'&lt;/math&gt; are distinct. We therefore conclude that &lt;math&gt;Z, N, D&lt;/math&gt; are collinear, which gives the concurrency of &lt;math&gt;AM, XY&lt;/math&gt;, and &lt;math&gt;DN&lt;/math&gt;. This completes the problem.<br /> <br /> == Solution 3==<br /> Let &lt;math&gt;AM&lt;/math&gt; and &lt;math&gt;XY&lt;/math&gt; intersect at &lt;math&gt;Z&lt;/math&gt;. Because &lt;math&gt;&lt;AMC = &lt;BND = &lt;APT = 90^\circ&lt;/math&gt;, we have quadrilaterals &lt;math&gt;AMPT&lt;/math&gt; and &lt;math&gt;DNPT&lt;/math&gt; cyclic. Therefore, &lt;math&gt;Z&lt;/math&gt; lies on the radical axis of the two circumcircles of these quadrilaterals. But &lt;math&gt;Z&lt;/math&gt; also lies on radical axis &lt;math&gt;XY&lt;/math&gt; of the original two circles, so the power of &lt;math&gt;Z&lt;/math&gt; with respect to each of the four circles is all equal to &lt;math&gt;ZM * ZA&lt;/math&gt;. Hence, &lt;math&gt;Z&lt;/math&gt; lies on the radical axis &lt;math&gt;DN&lt;/math&gt; of the two circles passing through &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt;, as desired.<br /> <br /> ==Discussion==<br /> ''Lemma:'' The radical axis of two pairs of circles &lt;math&gt;O_1&lt;/math&gt;, &lt;math&gt;O_2&lt;/math&gt; and &lt;math&gt;O_3&lt;/math&gt;, &lt;math&gt;O_4&lt;/math&gt; are the same line &lt;math&gt;l&lt;/math&gt;. Furthermore, &lt;math&gt;O_1&lt;/math&gt; and &lt;math&gt;O_2&lt;/math&gt; intersect at &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;O_3&lt;/math&gt; and &lt;math&gt;O_4&lt;/math&gt; intersect at &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;. Then &lt;math&gt;A, B, C,&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; are concyclic.<br /> <br /> The proof of this lemma is trivial using the argument in Solution 3 and applying the converse of Power of a Point.<br /> <br /> Note that this Problem 1 is a corollary of this lemma. This lemma is an effective way to relate four circles, just as the radical center can relate three circles.<br /> <br /> Solution 1 also gives a trivial lemma that can also be useful:<br /> <br /> ''Lemma 2:'' Chords &lt;math&gt;AB&lt;/math&gt; of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt; of &lt;math&gt;\omega_2&lt;/math&gt; intersect on the segment &lt;math&gt;XY&lt;/math&gt; formed from the intersections of the two circles. Then &lt;math&gt;A, B, C, D&lt;/math&gt; are concyclic.<br /> <br /> Two ways to solve a problem, two different insights into circle geometry. That is cool, but more RADICAL!<br /> <br /> ==See also==<br /> <br /> [[Category:Olympiad Geometry Problems]]</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=Radical_axis&diff=86444 Radical axis 2017-07-19T01:00:36Z <p>Always correct: /* Exercises */</p> <hr /> <div>==Introduction==<br /> The theory of radical axis is a priceless geometric tool that can solve formidable geometric problems fairly readily. Problems involving it can be found on many major math olympiad competitions, including the prestigious [[USAMO]]. Therefore, any aspiring math olympian should peruse this material carefully, as it may contain the keys to one's future success.<br /> <br /> Not all theorems will be fully proven in this text. The objective of this document is to introduce you to some key concepts, and then give you a chance to derive some of the beautiful results on your own. In that way, you will understand and retain the information in here much more solidly. Finally, your newfound knowledge will be tested on a few challenging problems that are exemplary examples on how radical axis theory can be used and why it pertains to that situation. I hope after you read this text, you will become a better math student, armed with another tool to solve difficult problems. But, anyway, good luck.<br /> <br /> ==Definitions==<br /> The '''power''' of point &lt;math&gt;P&lt;/math&gt; with respect to circle &lt;math&gt;\omega&lt;/math&gt; (with radius &lt;math&gt;r&lt;/math&gt; and center &lt;math&gt;O&lt;/math&gt;, which shall thereafter be dubbed &lt;math&gt;pow(P, \omega)&lt;/math&gt;, is defined to equal &lt;math&gt;OP^2 - r^2&lt;/math&gt;.<br /> <br /> Note that the power of a point is negative if the point is inside the circle.<br /> <br /> The '''radical axis''' of two circles &lt;math&gt;\omega_1, \omega_2&lt;/math&gt; is defined as the locus of the points &lt;math&gt;P&lt;/math&gt; such that the [[power of a point|power]] of &lt;math&gt;P&lt;/math&gt; with respect to &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; are equal. In other words, if &lt;math&gt;O_i, r_i&lt;/math&gt; are the center and radius of &lt;math&gt;\omega_i&lt;/math&gt;, then a point &lt;math&gt;P&lt;/math&gt; is on the radical axis if and only if &lt;cmath&gt;PO_1^2 - r_1^2 = PO_2^2 - r_2^2&lt;/cmath&gt;<br /> <br /> ==Results==<br /> '''Theorem 1: (Power of a Point)'''<br /> If a line drawn through point P intersects circle &lt;math&gt;\omega&lt;/math&gt; at points A and B, then &lt;math&gt;|pow(P, \omega)| = PA * PB&lt;/math&gt;.<br /> <br /> '''Theorem 2: (Radical Axis Theorem)'''<br /> <br /> a. The radical axis is a line perpendicular to the line connecting the circles' centers (line &lt;math&gt;l&lt;/math&gt;).<br /> <br /> b. If the two circles intersect at two common points, their radical axis is the line through these two points.<br /> <br /> c. If they intersect at one point, their radical axis is the common internal tangent.<br /> <br /> d. If the circles do not intersect, and if one does not fully contain the other, their radical axis is the perpendicular to &lt;math&gt;l&lt;/math&gt; through point A, the unique point on &lt;math&gt;l&lt;/math&gt; such that &lt;math&gt;pow(A, \omega_1) = pow(A, \omega_2)&lt;/math&gt;.<br /> <br /> '''Theorem 3: (Radical Axis Concurrence Theorem)'''<br /> The three pairwise radical axes of three circles concur at a point, called the '''radical center'''.<br /> <br /> ==Proofs==<br /> Theorem 1 is trivial Power of a Point, and thus is left to the reader as an exercise. (Hint: Draw a line through P and the center.)<br /> <br /> Theorem 2 shall be proved here. Assume the circles are &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; with centers &lt;math&gt;O_1&lt;/math&gt; and &lt;math&gt;O_2&lt;/math&gt; and radii &lt;math&gt;r_1&lt;/math&gt; and &lt;math&gt;r_2&lt;/math&gt;, respectively. (It may be a good idea for you to draw some circles here.)<br /> <br /> First, we tackle part (b). Suppose the circles intersect at points &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt; and point P lies on &lt;math&gt;XY&lt;/math&gt;. Then by Theorem 1 the powers of P with respect to both circles are equal to &lt;math&gt;PX * PY&lt;/math&gt;, and hence by transitive &lt;math&gt;pow(P, \omega_1) = pow(P, \omega_2)&lt;/math&gt;. Thus, if point P lies on &lt;math&gt;XY&lt;/math&gt;, then the powers of P with respect to both circles are equal.<br /> <br /> Now, we prove the inverse of the statement just proved; because the inverse is equivalent to the converse, the if and only if would then be proven. Suppose that P does not lie on &lt;math&gt;XY&lt;/math&gt;. In particular, line &lt;math&gt;PY&lt;/math&gt; does not intersect X. Then &lt;math&gt;PY&lt;/math&gt; intersects circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; a second time at distinct points &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt;, respectively. (If &lt;math&gt;PY&lt;/math&gt; is tangent to &lt;math&gt;\omega_1&lt;/math&gt;, for example, we adopt the convention that &lt;math&gt;P = M&lt;/math&gt;; similar conventions hold for &lt;math&gt;\omega_2&lt;/math&gt;. Power of a Point still holds in this case. Also, notice that &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; cannot both equal &lt;math&gt;P&lt;/math&gt;, as &lt;math&gt;PY&lt;/math&gt; cannot be tangent to both circles.) Because &lt;math&gt;PM&lt;/math&gt; is not equal to &lt;math&gt;PN&lt;/math&gt;, &lt;math&gt;PY \cdot PM&lt;/math&gt; does not equal &lt;math&gt;PY \cdot PN&lt;/math&gt;, and thus by Theorem 1 &lt;math&gt;pow(P, \omega_1)&lt;/math&gt; is not congruent to &lt;math&gt;pow(P, \omega_2)&lt;/math&gt;, as desired. This completes part (b).<br /> <br /> For the remaining parts, we employ a lemma:<br /> <br /> ''Lemma 1:'' Let &lt;math&gt;P&lt;/math&gt; be a point in the plane, and let &lt;math&gt;P'&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;O_1O_2&lt;/math&gt;. Then &lt;math&gt;pow(P, \omega_1) - pow(P, \omega_2) = pow(P', \omega_1) - pow(P', \omega_2)&lt;/math&gt;.<br /> <br /> The proof of the lemma is an easy application of the Pythagorean Theorem and will again be left to the reader as an exercise.<br /> <br /> ''Lemma 2:'' There is an unique point P on line &lt;math&gt;O_1O_2&lt;/math&gt; such that &lt;math&gt;pow(P, O_1) = pow(P, O_2)&lt;/math&gt;.<br /> <br /> Proof: First show that P lies between &lt;math&gt;O_1&lt;/math&gt; and &lt;math&gt;O_2&lt;/math&gt; via proof by contradiction, by using a bit of inequality theory and the fact that &lt;math&gt;O_1O_2 &gt; r_1 + r_2&lt;/math&gt;. Then, use the fact that &lt;math&gt;O_1P + PO_2 = O_1O_2&lt;/math&gt; (a constant) to prove the lemma.<br /> <br /> Lemma 1 shows that every point on the plane can be equivalently mapped to a line on &lt;math&gt;O_1O_2&lt;/math&gt;. Lemma 2 shows that only one point in this mapping satisfies the given condition. Combining these two lemmas shows that the radical axis is a line perpendicular to &lt;math&gt;l&lt;/math&gt;, completing part (a).<br /> <br /> Parts (c) and (d) will be left to the reader as an exercise. (Also, try proving part (b) solely using the lemmas.)<br /> <br /> Now, try to prove Theorem 3 on your own! (Hint: Let P be the intersection of two of the radical axes.)<br /> <br /> ==Exercises==<br /> If you haven't already done so, prove the theorems and lemmas outlined in the proofs section. <br /> ''Note:'' No solutions will be provided to the following problems. If you are stuck, ask on the forum.<br /> <br /> '''Problem 1.''' Two circles P and Q intersect at X and Y. Point A is located on &lt;math&gt;XY&lt;/math&gt; such that AP = 10 and AQ = &lt;s&gt;15&lt;/s&gt; 12. If the radius of Q is 7, find the radius of P. &lt;span style=&quot;color:#ff0000&quot;&gt;Note: An error in this problem previously rendered it unsolvable.&lt;/span&gt;<br /> <br /> '''Problem 2.''' Solve [[2009_USAMO_Problems|2009 USAMO Problem 1]]. If you already know how to solve it.<br /> '''Problem 3.''' Two circles P and Q with radii 1 and 2, respectively, intersect at X and Y. Circle P is to the left of circle Q. Prove that point A is to the left of &lt;math&gt;XY&lt;/math&gt; if and only if &lt;math&gt;AQ^2 - AP^2 &gt; 3&lt;/math&gt;.<br /> <br /> '''Problem 4.''' Solve [[2012_USAJMO_Problems|2012 USAJMO Problem 1]].<br /> <br /> '''Problem 5.''' Does Theorem 2 apply to circles in which one is contained inside the other? How about internally tangent circles? Concentric circles?<br /> <br /> '''Problem 6.''' Construct the radical axis of two circles. What happens if one circle encloses the other?<br /> <br /> '''Problem 7.''' Solve [[1995_IMO_Problems/Problem_1|1995 IMO Problem 1]] in two different ways. Compare your solutions with the solutions provided.</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=Radical_axis&diff=86440 Radical axis 2017-07-18T20:03:02Z <p>Always correct: /* Proofs */</p> <hr /> <div>==Introduction==<br /> The theory of radical axis is a priceless geometric tool that can solve formidable geometric problems fairly readily. Problems involving it can be found on many major math olympiad competitions, including the prestigious [[USAMO]]. Therefore, any aspiring math olympian should peruse this material carefully, as it may contain the keys to one's future success.<br /> <br /> Not all theorems will be fully proven in this text. The objective of this document is to introduce you to some key concepts, and then give you a chance to derive some of the beautiful results on your own. In that way, you will understand and retain the information in here much more solidly. Finally, your newfound knowledge will be tested on a few challenging problems that are exemplary examples on how radical axis theory can be used and why it pertains to that situation. I hope after you read this text, you will become a better math student, armed with another tool to solve difficult problems. But, anyway, good luck.<br /> <br /> ==Definitions==<br /> The '''power''' of point &lt;math&gt;P&lt;/math&gt; with respect to circle &lt;math&gt;\omega&lt;/math&gt; (with radius &lt;math&gt;r&lt;/math&gt; and center &lt;math&gt;O&lt;/math&gt;, which shall thereafter be dubbed &lt;math&gt;pow(P, \omega)&lt;/math&gt;, is defined to equal &lt;math&gt;OP^2 - r^2&lt;/math&gt;.<br /> <br /> Note that the power of a point is negative if the point is inside the circle.<br /> <br /> The '''radical axis''' of two circles &lt;math&gt;\omega_1, \omega_2&lt;/math&gt; is defined as the locus of the points &lt;math&gt;P&lt;/math&gt; such that the [[power of a point|power]] of &lt;math&gt;P&lt;/math&gt; with respect to &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; are equal. In other words, if &lt;math&gt;O_i, r_i&lt;/math&gt; are the center and radius of &lt;math&gt;\omega_i&lt;/math&gt;, then a point &lt;math&gt;P&lt;/math&gt; is on the radical axis if and only if &lt;cmath&gt;PO_1^2 - r_1^2 = PO_2^2 - r_2^2&lt;/cmath&gt;<br /> <br /> ==Results==<br /> '''Theorem 1: (Power of a Point)'''<br /> If a line drawn through point P intersects circle &lt;math&gt;\omega&lt;/math&gt; at points A and B, then &lt;math&gt;|pow(P, \omega)| = PA * PB&lt;/math&gt;.<br /> <br /> '''Theorem 2: (Radical Axis Theorem)'''<br /> <br /> a. The radical axis is a line perpendicular to the line connecting the circles' centers (line &lt;math&gt;l&lt;/math&gt;).<br /> <br /> b. If the two circles intersect at two common points, their radical axis is the line through these two points.<br /> <br /> c. If they intersect at one point, their radical axis is the common internal tangent.<br /> <br /> d. If the circles do not intersect, and if one does not fully contain the other, their radical axis is the perpendicular to &lt;math&gt;l&lt;/math&gt; through point A, the unique point on &lt;math&gt;l&lt;/math&gt; such that &lt;math&gt;pow(A, \omega_1) = pow(A, \omega_2)&lt;/math&gt;.<br /> <br /> '''Theorem 3: (Radical Axis Concurrence Theorem)'''<br /> The three pairwise radical axes of three circles concur at a point, called the '''radical center'''.<br /> <br /> ==Proofs==<br /> Theorem 1 is trivial Power of a Point, and thus is left to the reader as an exercise. (Hint: Draw a line through P and the center.)<br /> <br /> Theorem 2 shall be proved here. Assume the circles are &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; with centers &lt;math&gt;O_1&lt;/math&gt; and &lt;math&gt;O_2&lt;/math&gt; and radii &lt;math&gt;r_1&lt;/math&gt; and &lt;math&gt;r_2&lt;/math&gt;, respectively. (It may be a good idea for you to draw some circles here.)<br /> <br /> First, we tackle part (b). Suppose the circles intersect at points &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt; and point P lies on &lt;math&gt;XY&lt;/math&gt;. Then by Theorem 1 the powers of P with respect to both circles are equal to &lt;math&gt;PX * PY&lt;/math&gt;, and hence by transitive &lt;math&gt;pow(P, \omega_1) = pow(P, \omega_2)&lt;/math&gt;. Thus, if point P lies on &lt;math&gt;XY&lt;/math&gt;, then the powers of P with respect to both circles are equal.<br /> <br /> Now, we prove the inverse of the statement just proved; because the inverse is equivalent to the converse, the if and only if would then be proven. Suppose that P does not lie on &lt;math&gt;XY&lt;/math&gt;. In particular, line &lt;math&gt;PY&lt;/math&gt; does not intersect X. Then &lt;math&gt;PY&lt;/math&gt; intersects circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; a second time at distinct points &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt;, respectively. (If &lt;math&gt;PY&lt;/math&gt; is tangent to &lt;math&gt;\omega_1&lt;/math&gt;, for example, we adopt the convention that &lt;math&gt;P = M&lt;/math&gt;; similar conventions hold for &lt;math&gt;\omega_2&lt;/math&gt;. Power of a Point still holds in this case. Also, notice that &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; cannot both equal &lt;math&gt;P&lt;/math&gt;, as &lt;math&gt;PY&lt;/math&gt; cannot be tangent to both circles.) Because &lt;math&gt;PM&lt;/math&gt; is not equal to &lt;math&gt;PN&lt;/math&gt;, &lt;math&gt;PY \cdot PM&lt;/math&gt; does not equal &lt;math&gt;PY \cdot PN&lt;/math&gt;, and thus by Theorem 1 &lt;math&gt;pow(P, \omega_1)&lt;/math&gt; is not congruent to &lt;math&gt;pow(P, \omega_2)&lt;/math&gt;, as desired. This completes part (b).<br /> <br /> For the remaining parts, we employ a lemma:<br /> <br /> ''Lemma 1:'' Let &lt;math&gt;P&lt;/math&gt; be a point in the plane, and let &lt;math&gt;P'&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;O_1O_2&lt;/math&gt;. Then &lt;math&gt;pow(P, \omega_1) - pow(P, \omega_2) = pow(P', \omega_1) - pow(P', \omega_2)&lt;/math&gt;.<br /> <br /> The proof of the lemma is an easy application of the Pythagorean Theorem and will again be left to the reader as an exercise.<br /> <br /> ''Lemma 2:'' There is an unique point P on line &lt;math&gt;O_1O_2&lt;/math&gt; such that &lt;math&gt;pow(P, O_1) = pow(P, O_2)&lt;/math&gt;.<br /> <br /> Proof: First show that P lies between &lt;math&gt;O_1&lt;/math&gt; and &lt;math&gt;O_2&lt;/math&gt; via proof by contradiction, by using a bit of inequality theory and the fact that &lt;math&gt;O_1O_2 &gt; r_1 + r_2&lt;/math&gt;. Then, use the fact that &lt;math&gt;O_1P + PO_2 = O_1O_2&lt;/math&gt; (a constant) to prove the lemma.<br /> <br /> Lemma 1 shows that every point on the plane can be equivalently mapped to a line on &lt;math&gt;O_1O_2&lt;/math&gt;. Lemma 2 shows that only one point in this mapping satisfies the given condition. Combining these two lemmas shows that the radical axis is a line perpendicular to &lt;math&gt;l&lt;/math&gt;, completing part (a).<br /> <br /> Parts (c) and (d) will be left to the reader as an exercise. (Also, try proving part (b) solely using the lemmas.)<br /> <br /> Now, try to prove Theorem 3 on your own! (Hint: Let P be the intersection of two of the radical axes.)<br /> <br /> ==Exercises==<br /> If you haven't already done so, prove the theorems and lemmas outlined in the proofs section. <br /> ''Note:'' No solutions will be provided to the following problems. If you are stuck, ask on the forum.<br /> <br /> '''Problem 1.''' Two circles P and Q intersect at X and Y. Point A is located on &lt;math&gt;XY&lt;/math&gt; such that AP = 10 and AQ = 15. If the radius of Q is 7, find the radius of P.<br /> <br /> '''Problem 2.''' Solve [[2009_USAMO_Problems|2009 USAMO Problem 1]]. If you already know how to solve it.<br /> '''Problem 3.''' Two circles P and Q with radii 1 and 2, respectively, intersect at X and Y. Circle P is to the left of circle Q. Prove that point A is to the left of &lt;math&gt;XY&lt;/math&gt; if and only if &lt;math&gt;AQ^2 - AP^2 &gt; 3&lt;/math&gt;.<br /> <br /> '''Problem 4.''' Solve [[2012_USAJMO_Problems|2012 USAJMO Problem 1]].<br /> <br /> '''Problem 5.''' Does Theorem 2 apply to circles in which one is contained inside the other? How about internally tangent circles? Concentric circles?<br /> <br /> '''Problem 6.''' Construct the radical axis of two circles. What happens if one circle encloses the other?<br /> <br /> '''Problem 7.''' Solve [[1995_IMO_Problems/Problem_1|1995 IMO Problem 1]] in two different ways. Compare your solutions with the solutions provided.</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_II_Problems/Problem_3&diff=86305 2007 AIME II Problems/Problem 3 2017-07-10T00:12:43Z <p>Always correct: /* Solution */</p> <hr /> <div>== Problem ==<br /> [[Square]] &lt;math&gt;ABCD&lt;/math&gt; has side length &lt;math&gt;13&lt;/math&gt;, and [[point]]s &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are exterior to the square such that &lt;math&gt;BE=DF=5&lt;/math&gt; and &lt;math&gt;AE=CF=12&lt;/math&gt;. Find &lt;math&gt;EF^{2}&lt;/math&gt;.<br /> &lt;asy&gt;unitsize(0.2 cm);<br /> <br /> pair A, B, C, D, E, F;<br /> <br /> A = (0,13);<br /> B = (13,13);<br /> C = (13,0);<br /> D = (0,0);<br /> E = A + (12*12/13,5*12/13);<br /> F = D + (5*5/13,-5*12/13);<br /> <br /> draw(A--B--C--D--cycle);<br /> draw(A--E--B);<br /> draw(C--F--D);<br /> <br /> dot(&quot;$A$&quot;, A, W);<br /> dot(&quot;$B$&quot;, B, dir(0));<br /> dot(&quot;$C$&quot;, C, dir(0));<br /> dot(&quot;$D$&quot;, D, W);<br /> dot(&quot;$E$&quot;, E, N);<br /> dot(&quot;$F$&quot;, F, S);&lt;/asy&gt;<br /> <br /> __TOC__<br /> <br /> == Solution ==<br /> <br /> ===Solution 1===<br /> <br /> Let &lt;math&gt;\angle FCD = \alpha&lt;/math&gt;, so that &lt;math&gt;FB = \sqrt{12^2 + 13^2 + 2\cdot12\cdot13\sin(\alpha)} = \sqrt{433}&lt;/math&gt;. By the diagonal, &lt;math&gt;DB = 13\sqrt{2}, DB^2 = 338&lt;/math&gt;. <br /> <br /> &lt;b&gt;The sum of the squares of the sides of a parallelogram is the sum of the squares of the diagonals.&lt;/b&gt;<br /> &lt;cmath&gt;EF^2 = 2\cdot(5^2 + 433) - 338 = 578.&lt;/cmath&gt;<br /> <br /> === Solution 2 ===<br /> <br /> Extend &lt;math&gt;\overline{AE}, \overline{DF}&lt;/math&gt; and &lt;math&gt;\overline{BE}, \overline{CF}&lt;/math&gt; to their points of intersection. Since &lt;math&gt;\triangle ABE \cong \triangle CDF&lt;/math&gt; and are both &lt;math&gt;5-12-13&lt;/math&gt; [[right triangle]]s, we can come to the conclusion that the two new triangles are also congruent to these two (use [[ASA]], as we know all the sides are &lt;math&gt;13&lt;/math&gt; and the angles are mostly complementary). Thus, we create a [[square]] with sides &lt;math&gt;5 + 12 = 17&lt;/math&gt;.<br /> <br /> &lt;asy&gt;unitsize(0.25 cm);<br /> <br /> pair A, B, C, D, E, F, G, H;<br /> <br /> A = (0,13);<br /> B = (13,13);<br /> C = (13,0);<br /> D = (0,0);<br /> E = A + (12*12/13,5*12/13);<br /> F = D + (5*5/13,-5*12/13);<br /> G = rotate(90,(A + C)/2)*(E);<br /> H = rotate(90,(A + C)/2)*(F);<br /> <br /> draw(A--B--C--D--cycle);<br /> draw(E--G--F--H--cycle);<br /> <br /> dot(&quot;$A$&quot;, A, N);<br /> dot(&quot;$B$&quot;, B, dir(0));<br /> dot(&quot;$C$&quot;, C, S);<br /> dot(&quot;$D$&quot;, D, W);<br /> dot(&quot;$E$&quot;, E, N);<br /> dot(&quot;$F$&quot;, F, S);<br /> dot(&quot;$G$&quot;, G, W);<br /> dot(&quot;$H$&quot;, H, dir(0));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\overline{EF}&lt;/math&gt; is the diagonal of the square, with length &lt;math&gt;17\sqrt{2}&lt;/math&gt;; the answer is &lt;math&gt;EF^2 = (17\sqrt{2})^2 = 578&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> A slightly more analytic/brute-force approach:<br /> <br /> [[Image:AIME II prob10 bruteforce.PNG ]]<br /> <br /> Drop perpendiculars from &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;I&lt;/math&gt; and &lt;math&gt;J&lt;/math&gt;, respectively; construct right triangle &lt;math&gt;EKF&lt;/math&gt; with right angle at K and &lt;math&gt;EK || BC&lt;/math&gt;. Since &lt;math&gt;2[CDF]=DF*CF=CD*JF&lt;/math&gt;, we have &lt;math&gt;JF=5\times12/13 = \frac{60}{13}&lt;/math&gt;. Similarly, &lt;math&gt;EI=\frac{60}{13}&lt;/math&gt;. Since &lt;math&gt;\triangle DJF \sim \triangle DFC&lt;/math&gt;, we have &lt;math&gt;DJ=\frac{5JF}{12}=\frac{25}{13}&lt;/math&gt;.<br /> <br /> Now, we see that &lt;math&gt;FK=DC-(DJ+IB)=DC-2DJ=13-\frac{50}{13}=\frac{119}{13}&lt;/math&gt;. Also, &lt;math&gt;EK=BC+(JF+IE)=BC+2JF=13+\frac{120}{13}=\frac{289}{13}&lt;/math&gt;. By the Pythagorean Theorem, we have &lt;math&gt;EF=\sqrt{\left(\frac{289}{13}\right)^2+\left(\frac{119}{13} \right)^2}=\frac{\sqrt{(17^2)(17^2+7^2)}}{13}&lt;/math&gt;&lt;math&gt;=\frac{17\sqrt{338}}{13}=\frac{17(13\sqrt{2})}{13}=17\sqrt{2}&lt;/math&gt;. Therefore, &lt;math&gt;EF^2=(17\sqrt{2})^2=578&lt;/math&gt;.<br /> <br /> ===Solution 4===<br /> Based on the symmetry, we know that &lt;math&gt;F&lt;/math&gt; is a reflection of &lt;math&gt;E&lt;/math&gt; across the center of the square, which we will denote as &lt;math&gt;O&lt;/math&gt;. Since &lt;math&gt;\angle BEA&lt;/math&gt; and &lt;math&gt;\angle AOB&lt;/math&gt; are right, we can conclude that figure &lt;math&gt;AOBE&lt;/math&gt; is a cyclic quadrilateral. Pythagorean Theorem yields that &lt;math&gt;BO=AO=\frac{13\sqrt{2}}{2}&lt;/math&gt;. Now, using Ptolemy's Theorem, we get that <br /> &lt;cmath&gt;AO\cdot BE + BO\cdot AE = AB\cdot AO&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{13\sqrt{2}}{2}\cdot 5+\frac{13\sqrt{2}}{2}\cdot 12 = 13\cdot OE&lt;/cmath&gt;<br /> &lt;cmath&gt;OE=\frac{17\sqrt{2}}{2}&lt;/cmath&gt;<br /> Now, since we stated in the first step that &lt;math&gt;F&lt;/math&gt; is a reflection of &lt;math&gt;E&lt;/math&gt; across &lt;math&gt;O&lt;/math&gt;, we can say that &lt;math&gt;EF=2EO=17\sqrt{2}&lt;/math&gt;. This gives that &lt;cmath&gt;EF^2=(17\sqrt{2})^2=578&lt;/cmath&gt; AWD with this bash solution<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=II|num-b=2|num-a=4}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_II_Problems/Problem_3&diff=86304 2007 AIME II Problems/Problem 3 2017-07-10T00:11:50Z <p>Always correct: /* Solution */</p> <hr /> <div>== Problem ==<br /> [[Square]] &lt;math&gt;ABCD&lt;/math&gt; has side length &lt;math&gt;13&lt;/math&gt;, and [[point]]s &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are exterior to the square such that &lt;math&gt;BE=DF=5&lt;/math&gt; and &lt;math&gt;AE=CF=12&lt;/math&gt;. Find &lt;math&gt;EF^{2}&lt;/math&gt;.<br /> &lt;asy&gt;unitsize(0.2 cm);<br /> <br /> pair A, B, C, D, E, F;<br /> <br /> A = (0,13);<br /> B = (13,13);<br /> C = (13,0);<br /> D = (0,0);<br /> E = A + (12*12/13,5*12/13);<br /> F = D + (5*5/13,-5*12/13);<br /> <br /> draw(A--B--C--D--cycle);<br /> draw(A--E--B);<br /> draw(C--F--D);<br /> <br /> dot(&quot;$A$&quot;, A, W);<br /> dot(&quot;$B$&quot;, B, dir(0));<br /> dot(&quot;$C$&quot;, C, dir(0));<br /> dot(&quot;$D$&quot;, D, W);<br /> dot(&quot;$E$&quot;, E, N);<br /> dot(&quot;$F$&quot;, F, S);&lt;/asy&gt;<br /> <br /> __TOC__<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> Extend &lt;math&gt;\overline{AE}, \overline{DF}&lt;/math&gt; and &lt;math&gt;\overline{BE}, \overline{CF}&lt;/math&gt; to their points of intersection. Since &lt;math&gt;\triangle ABE \cong \triangle CDF&lt;/math&gt; and are both &lt;math&gt;5-12-13&lt;/math&gt; [[right triangle]]s, we can come to the conclusion that the two new triangles are also congruent to these two (use [[ASA]], as we know all the sides are &lt;math&gt;13&lt;/math&gt; and the angles are mostly complementary). Thus, we create a [[square]] with sides &lt;math&gt;5 + 12 = 17&lt;/math&gt;.<br /> <br /> &lt;asy&gt;unitsize(0.25 cm);<br /> <br /> pair A, B, C, D, E, F, G, H;<br /> <br /> A = (0,13);<br /> B = (13,13);<br /> C = (13,0);<br /> D = (0,0);<br /> E = A + (12*12/13,5*12/13);<br /> F = D + (5*5/13,-5*12/13);<br /> G = rotate(90,(A + C)/2)*(E);<br /> H = rotate(90,(A + C)/2)*(F);<br /> <br /> draw(A--B--C--D--cycle);<br /> draw(E--G--F--H--cycle);<br /> <br /> dot(&quot;$A$&quot;, A, N);<br /> dot(&quot;$B$&quot;, B, dir(0));<br /> dot(&quot;$C$&quot;, C, S);<br /> dot(&quot;$D$&quot;, D, W);<br /> dot(&quot;$E$&quot;, E, N);<br /> dot(&quot;$F$&quot;, F, S);<br /> dot(&quot;$G$&quot;, G, W);<br /> dot(&quot;$H$&quot;, H, dir(0));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\overline{EF}&lt;/math&gt; is the diagonal of the square, with length &lt;math&gt;17\sqrt{2}&lt;/math&gt;; the answer is &lt;math&gt;EF^2 = (17\sqrt{2})^2 = 578&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> A slightly more analytic/brute-force approach:<br /> <br /> [[Image:AIME II prob10 bruteforce.PNG ]]<br /> <br /> Drop perpendiculars from &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;I&lt;/math&gt; and &lt;math&gt;J&lt;/math&gt;, respectively; construct right triangle &lt;math&gt;EKF&lt;/math&gt; with right angle at K and &lt;math&gt;EK || BC&lt;/math&gt;. Since &lt;math&gt;2[CDF]=DF*CF=CD*JF&lt;/math&gt;, we have &lt;math&gt;JF=5\times12/13 = \frac{60}{13}&lt;/math&gt;. Similarly, &lt;math&gt;EI=\frac{60}{13}&lt;/math&gt;. Since &lt;math&gt;\triangle DJF \sim \triangle DFC&lt;/math&gt;, we have &lt;math&gt;DJ=\frac{5JF}{12}=\frac{25}{13}&lt;/math&gt;.<br /> <br /> Now, we see that &lt;math&gt;FK=DC-(DJ+IB)=DC-2DJ=13-\frac{50}{13}=\frac{119}{13}&lt;/math&gt;. Also, &lt;math&gt;EK=BC+(JF+IE)=BC+2JF=13+\frac{120}{13}=\frac{289}{13}&lt;/math&gt;. By the Pythagorean Theorem, we have &lt;math&gt;EF=\sqrt{\left(\frac{289}{13}\right)^2+\left(\frac{119}{13} \right)^2}=\frac{\sqrt{(17^2)(17^2+7^2)}}{13}&lt;/math&gt;&lt;math&gt;=\frac{17\sqrt{338}}{13}=\frac{17(13\sqrt{2})}{13}=17\sqrt{2}&lt;/math&gt;. Therefore, &lt;math&gt;EF^2=(17\sqrt{2})^2=578&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> Based on the symmetry, we know that &lt;math&gt;F&lt;/math&gt; is a reflection of &lt;math&gt;E&lt;/math&gt; across the center of the square, which we will denote as &lt;math&gt;O&lt;/math&gt;. Since &lt;math&gt;\angle BEA&lt;/math&gt; and &lt;math&gt;\angle AOB&lt;/math&gt; are right, we can conclude that figure &lt;math&gt;AOBE&lt;/math&gt; is a cyclic quadrilateral. Pythagorean Theorem yields that &lt;math&gt;BO=AO=\frac{13\sqrt{2}}{2}&lt;/math&gt;. Now, using Ptolemy's Theorem, we get that <br /> &lt;cmath&gt;AO\cdot BE + BO\cdot AE = AB\cdot AO&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{13\sqrt{2}}{2}\cdot 5+\frac{13\sqrt{2}}{2}\cdot 12 = 13\cdot OE&lt;/cmath&gt;<br /> &lt;cmath&gt;OE=\frac{17\sqrt{2}}{2}&lt;/cmath&gt;<br /> Now, since we stated in the first step that &lt;math&gt;F&lt;/math&gt; is a reflection of &lt;math&gt;E&lt;/math&gt; across &lt;math&gt;O&lt;/math&gt;, we can say that &lt;math&gt;EF=2EO=17\sqrt{2}&lt;/math&gt;. This gives that &lt;cmath&gt;EF^2=(17\sqrt{2})^2=578&lt;/cmath&gt; AWD with this bash solution<br /> <br /> ===Solution 3===<br /> Let &lt;math&gt;\angle FCD = \alpha&lt;/math&gt;, so that &lt;math&gt;FB = \sqrt{12^2 + 13^2 + 2\cdot12\cdot13\sin(\alpha)} = \sqrt{433}&lt;/math&gt;. By the diagonal, &lt;math&gt;DB = 13\sqrt{2}, DB^2 = 338&lt;/math&gt;. <br /> <br /> &lt;b&gt;The sum of the squares of the sides of a parallelogram is the sum of the squares of the diagonals.&lt;/b&gt;<br /> &lt;cmath&gt;EF^2 = 2\cdot(5^2 + 433) - 338 = 578.&lt;/cmath&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=II|num-b=2|num-a=4}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=Law_of_Tangents&diff=84357 Law of Tangents 2017-03-02T21:49:23Z <p>Always correct: /* Proof */</p> <hr /> <div>The '''Law of Tangents''' is a rather obscure [[trigonometric identity]] that is sometimes used in place of its better-known counterparts, the [[law of sines]] and [[law of cosines]], to calculate [[angle]]s or sides in a [[triangle]].<br /> <br /> == Statement ==<br /> <br /> If &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; are angles in a triangle opposite sides &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; respectively, then<br /> &lt;cmath&gt; \frac{a-b}{a+b}=\frac{\tan [\frac{1}{2}(A-B)]}{\tan [\frac{1}{2}(A+B)]} . &lt;/cmath&gt;<br /> <br /> == Proof ==<br /> <br /> Let &lt;math&gt;s&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; denote &lt;math&gt;(A+B)/2&lt;/math&gt;, &lt;math&gt;(A-B)/2&lt;/math&gt;, respectively. By the [[Law of Sines]],<br /> &lt;cmath&gt; \frac{a-b}{a+b} = \frac{\sin A - \sin B}{\sin A + \sin B} = \frac{ \sin(s+d) - \sin (s-d)}{\sin(s+d) + \sin(s-d)} . &lt;/cmath&gt;<br /> By the angle addition identities,<br /> &lt;cmath&gt; \frac{\sin(s+d) - \sin(s-d)}{\sin(s+d) + \sin(s-d)} = \frac{2\cos s \sin d}{2\sin s \cos d} = \frac{\tan d}{\tan s} = \frac{\tan [\frac{1}{2} (A-B)]}{\tan[ \frac{1}{2} (A+B)]} &lt;/cmath&gt;<br /> as desired. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> ==Problems==<br /> ===Introductory===<br /> {{problem}}<br /> ===Intermediate===<br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, let &lt;math&gt;D&lt;/math&gt; be a point in &lt;math&gt;BC&lt;/math&gt; such that &lt;math&gt;AD&lt;/math&gt; bisects &lt;math&gt;\angle A&lt;/math&gt;. Given that &lt;math&gt;AD=6,BD=4&lt;/math&gt;, and &lt;math&gt;DC=3&lt;/math&gt;, find &lt;math&gt;AB&lt;/math&gt;.<br /> &lt;div align=&quot;right&quot;&gt;([[Mu Alpha Theta]] 1991)&lt;/div&gt;<br /> ===Olympiad===<br /> Show that &lt;math&gt;[ABC]=r^2\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}&lt;/math&gt;.<br /> <br /> &lt;div align=&quot;right&quot;&gt;(AoPS Vol. 2)&lt;/div&gt;<br /> ==See Also==<br /> * [[Trigonometry]]<br /> * [[Trigonometric identities]]<br /> * [[Law of Sines]]<br /> * [[Law of Cosines]]<br /> <br /> [[Category:Theorems]]<br /> [[Category:Trigonometry]]</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=Law_of_Tangents&diff=84356 Law of Tangents 2017-03-02T21:48:53Z <p>Always correct: /* Proof */</p> <hr /> <div>The '''Law of Tangents''' is a rather obscure [[trigonometric identity]] that is sometimes used in place of its better-known counterparts, the [[law of sines]] and [[law of cosines]], to calculate [[angle]]s or sides in a [[triangle]].<br /> <br /> == Statement ==<br /> <br /> If &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; are angles in a triangle opposite sides &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; respectively, then<br /> &lt;cmath&gt; \frac{a-b}{a+b}=\frac{\tan [\frac{1}{2}(A-B)]}{\tan [\frac{1}{2}(A+B)]} . &lt;/cmath&gt;<br /> <br /> == Proof ==<br /> <br /> Let &lt;math&gt;s&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; denote &lt;math&gt;(A+B)/2&lt;/math&gt;, &lt;math&gt;(A-B)/2&lt;/math&gt;, respectively. By the [[Law of Sines]],<br /> &lt;cmath&gt; \frac{a-b}{a+b} = \frac{\sin A - \sin B}{\sin A + \sin B} = \frac{ \sin(s+d) - \sin (s-d)}{\sin(s+d) + \sin(s-d)} . &lt;/cmath&gt;<br /> By the angle addition identities,<br /> &lt;cmath&gt; \frac{\sin(s+d) - \sin(s-d)}{\sin(s+d) + \sin(s-d)} = \frac{2\cos s \sin d}{2\sin s \cos d} = \frac{\tan d}{\tan s} = \frac{\tan \frac{1}{2} (A-B)}{\tan \frac{1}{2} (A+B)} &lt;/cmath&gt;<br /> as desired. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> ==Problems==<br /> ===Introductory===<br /> {{problem}}<br /> ===Intermediate===<br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, let &lt;math&gt;D&lt;/math&gt; be a point in &lt;math&gt;BC&lt;/math&gt; such that &lt;math&gt;AD&lt;/math&gt; bisects &lt;math&gt;\angle A&lt;/math&gt;. Given that &lt;math&gt;AD=6,BD=4&lt;/math&gt;, and &lt;math&gt;DC=3&lt;/math&gt;, find &lt;math&gt;AB&lt;/math&gt;.<br /> &lt;div align=&quot;right&quot;&gt;([[Mu Alpha Theta]] 1991)&lt;/div&gt;<br /> ===Olympiad===<br /> Show that &lt;math&gt;[ABC]=r^2\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}&lt;/math&gt;.<br /> <br /> &lt;div align=&quot;right&quot;&gt;(AoPS Vol. 2)&lt;/div&gt;<br /> ==See Also==<br /> * [[Trigonometry]]<br /> * [[Trigonometric identities]]<br /> * [[Law of Sines]]<br /> * [[Law of Cosines]]<br /> <br /> [[Category:Theorems]]<br /> [[Category:Trigonometry]]</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_1&diff=83063 2017 AMC 10A Problems/Problem 1 2017-02-08T22:52:43Z <p>Always correct: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <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 /> <br /> ==Solution 1==<br /> <br /> Notice this is the term &lt;math&gt;a_6&lt;/math&gt; in a recursive sequence, defined recursively as &lt;math&gt;a_1 = 3, a_n = 2a_{n-1} + 1.&lt;/math&gt; Thus:<br /> &lt;cmath&gt;\begin{split}<br /> a_2 = 3*2 + 1 = 7.\\<br /> a_3 = 7 *2 + 1 = 15.\\<br /> a_4 = 15*2 + 1 = 31.\\<br /> a_5 = 31*2 + 1 = 63.\\<br /> a_6 = 63*2 + 1 = \boxed{\textbf{(C)}\ 127}<br /> \end{split}&lt;/cmath&gt;<br /> <br /> ==Solution 2==<br /> Starting to compute the inner expressions, we see the results are &lt;math&gt;1, 3, 7, 15, \ldots&lt;/math&gt;. This is always &lt;math&gt;1&lt;/math&gt; less than a power of &lt;math&gt;2&lt;/math&gt;. The only admissible answer choice by this rule is thus &lt;math&gt;\boxed{\textbf{(C)}\ 127}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|before=First Problem|num-a=2}}<br /> {{MAA Notice}}</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_1&diff=83062 2017 AMC 10A Problems/Problem 1 2017-02-08T22:52:21Z <p>Always correct: /* Solution */</p> <hr /> <div>==Problem==<br /> <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 /> <br /> ==Solution 1==<br /> <br /> Notice this is the term &lt;math&gt;a_6&lt;/math&gt; in a recursive sequence, defined recursively as &lt;math&gt;a_1 = 3, a_n = 2a_{n-1} + 1.&lt;/math&gt; Thus:<br /> &lt;cmath&gt;\begin{split}<br /> a_2 = 3*2 + 1 = 7.\\<br /> a_3 = 7 *2 + 1 = 15.\\<br /> a_4 = 15*2 + 1 = 31.\\<br /> a_5 = 31*2 + 1 = 63.\\<br /> a_6 = 63*2 + 1 = \boxed{\textbf{(C)}\ 127}<br /> \end{split}&lt;/cmath&gt;<br /> <br /> ==Solution 2==<br /> Starting to compute the inner expressions, we see the results are &lt;math&gt;1, 3, 7, 15, \ldots&lt;/math&gt;. This is always &lt;math&gt;1&lt;/math&gt; less than a power of &lt;math&gt;2&lt;/math&gt;. The only admissible answer choice by this rule is thus \boxed{\textbf{(C)}\ 127}.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|before=First Problem|num-a=2}}<br /> {{MAA Notice}}</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_16&diff=83054 2017 AMC 10A Problems/Problem 16 2017-02-08T22:49:19Z <p>Always correct: </p> <hr /> <div>==Problem==<br /> There are &lt;math&gt;10&lt;/math&gt; horses, named Horse &lt;math&gt;1&lt;/math&gt;, Horse &lt;math&gt;2&lt;/math&gt;, . . . , Horse &lt;math&gt;10&lt;/math&gt;. 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 &lt;math&gt;0&lt;/math&gt; 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 &lt;math&gt;10&lt;/math&gt; 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 /> ==Solution==<br /> If we have horses, &lt;math&gt;a_1, a_2, \ldots, a_n&lt;/math&gt;, then any number that is a multiple of the all those numbers is a time when all horses will meet at the starting point. The least of these numbers is the LCM. To minimize the LCM, we need the smallest primes, and we need to repeat them a lot. By inspection, we find that &lt;math&gt;\text{lcm}(1,2,3,2\cdot2,2\cdot3) = 12&lt;/math&gt;. Finally, &lt;math&gt;1+2 = \boxed{\textbf{(B)}\ 3}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_18&diff=83000 2017 AMC 10A Problems/Problem 18 2017-02-08T22:22:31Z <p>Always correct: /* Solution */</p> <hr /> <div>==Problem==<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 /> ==Solution==<br /> Let &lt;math&gt;P&lt;/math&gt; be the probability Amelia wins. Note that &lt;math&gt;P = \text{chance she wins on her first turn} + \text{chance she gets to her turn again}\cdot P&lt;/math&gt;, as if she gets to her turn again, she is back where she started with probability of winning &lt;math&gt;P&lt;/math&gt;. The chance she wins on her first turn is &lt;math&gt;\frac{1}{3}&lt;/math&gt;. The chance she makes it to her turn again is a combination of her failing to win the first turn—&lt;math&gt;\frac{2}{3}&lt;/math&gt; and Blaine failing to win—&lt;math&gt;\frac{3}{5}&lt;/math&gt;. Multiplying gives us &lt;math&gt;\frac{2}{5}&lt;/math&gt;. Thus,<br /> &lt;cmath&gt;P = \frac{1}{3} + \frac{2}{5} \implies P = \frac{5}{9}&lt;/cmath&gt;<br /> Finally, we do &lt;math&gt;9-5=\boxed{\textbf{(D)}\ 4}&lt;/math&gt;.</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_18&diff=82995 2017 AMC 10A Problems/Problem 18 2017-02-08T22:20:56Z <p>Always correct: /* Solution */</p> <hr /> <div>==Problem==<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 /> ==Solution==<br /> Let &lt;math&gt;P&lt;/math&gt; be the probability Amelia wins. Note that &lt;math&gt;P = \text{chance she wins on her first turn} + \text{chance she gets to her turn again}\cdot P&lt;/math&gt;, as if she gets to her turn again, she is back where she started with probability of winning &lt;math&gt;P&lt;/math&gt;. The chance she wins on her first turn is &lt;math&gt;\frac{1}{3}&lt;/math&gt;. The chance she makes it to her turn again is a combination of her failing to win the first turn—&lt;math&gt;\frac{2}{3}&lt;/math&gt; and Blaine failing to win—&lt;math&gt;\frac{3}{5}&lt;/math&gt;. Multiplying gives us &lt;math&gt;\frac{2}{5}&lt;/math&gt;. Thus,<br /> &lt;cmath&gt;P = \frac{1}{3} + \frac{2}{5} \implies P = \frac{5}{9}&lt;/cmath&gt;<br /> Finally, we do &lt;math&gt;9-5=\boxed{\textbf{(E)}\ 4}&lt;/math&gt;.</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_18&diff=82993 2017 AMC 10A Problems/Problem 18 2017-02-08T22:20:06Z <p>Always correct: </p> <hr /> <div>==Problem==<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 /> ==Solution==<br /> Let &lt;math&gt;P&lt;/math&gt; be the probability Amelia wins. Note that &lt;math&gt;P = \text{chance she wins on her first turn} + \text{chance she gets to her turn again}\cdot P&lt;/math&gt;, as if she gets to her turn again, she is back where she started with probability of winning &lt;math&gt;P&lt;/math&gt;. The chance she wins on her first turn is &lt;math&gt;\frac{1}{3}&lt;/math&gt;, and the chance she makes it to her turn again is a combination of her failing to win the first turn—&lt;math&gt;\frac{2}{3}&lt;/math&gt; and Blaine failing to win—&lt;math&gt;\frac{3}{5}&lt;/math&gt;. Multiplying gives us &lt;math&gt;\frac{2}{5}&lt;/math&gt;. Thus,<br /> &lt;cmath&gt;P = \frac{1}{3} + \frac{2}{5} \implies P = frac{5}{9}&lt;/cmath&gt;.<br /> Finally, we do &lt;math&gt;9-5=\boxed{\textbf{(E)}\ 4}&lt;/math&gt;.</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_20&diff=82956 2017 AMC 10A Problems/Problem 20 2017-02-08T22:07:08Z <p>Always correct: Created page with &quot;==Problem== 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;ma...&quot;</p> <hr /> <div>==Problem==<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 /> ==Solution==<br /> Note that &lt;math&gt;n \equiv S(n) \pmod{9}&lt;/math&gt;. This can be seen from the fact that &lt;math&gt;\sum_{k=0}^{n}10^{k}a_k \equiv \sum_{k=0}^{n}a_k \pmod{9}&lt;/math&gt;. Thus, if &lt;math&gt;S(n) = 1274&lt;/math&gt;, then &lt;math&gt;n \equiv 5 \pmod{9}&lt;/math&gt;, and thus &lt;math&gt;n+1 \equiv S(n+1) \equiv 6 \pmod{9}&lt;/math&gt;. The only answer choice that is &lt;math&gt;6 \pmod{9}&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(D)}\ 1265}&lt;/math&gt;.</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_11&diff=82933 2017 AMC 10A Problems/Problem 11 2017-02-08T21:54:40Z <p>Always correct: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> The region consisting of all point 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 /> ==Solution 1==<br /> In order to solve this problem, we must first visualize what the region contained looks like. We know that, in a three dimensional plane, the region consisting of all points within &lt;math&gt;3&lt;/math&gt; units of a point would be a sphere with radius &lt;math&gt;3&lt;/math&gt;. However, we need to find the region containing all points within 3 units of a segment. It can be seen that our region is a cylinder with two hemispheres on either end. We know the volume of our region, so we set up the following equation (the volume of our cylinder + the volume of our two hemispheres will equal &lt;math&gt;216 \pi&lt;/math&gt;):<br /> <br /> &lt;math&gt;\frac{4 \pi }{3}3^3+9 \pi x=216&lt;/math&gt;<br /> <br /> Where &lt;math&gt;x&lt;/math&gt; is equal to the length of our line segment.<br /> <br /> We isolate &lt;math&gt;x&lt;/math&gt;. This comes out to be &lt;math&gt;\boxed{\textbf{(D)}\ 20}&lt;/math&gt;<br /> <br /> ==Visualizing the Region==<br /> To envision what the region must look like, we simplify the problem to finding all points within &lt;math&gt;3&lt;/math&gt; units from a point. This is a sphere. To account for the line, we drag the sphere's center across the line, sweeping out the desired volume. As stated above, this is a cylinder with two hemispheres on both ends.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_21&diff=82928 2017 AMC 10A Problems/Problem 21 2017-02-08T21:49:34Z <p>Always correct: Created page with &quot;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 squ...&quot;</p> <hr /> <div>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 /> ==Solution==<br /> <br /> Analyze the first right triangle.<br /> <br /> &lt;asy&gt;<br /> pair A,B,C;<br /> pair D, e, F;<br /> A = (0,0);<br /> B = (4,0);<br /> C = (0,3);<br /> <br /> D = (0, 12/7);<br /> e = (12/7 , 12/7);<br /> F = (12/7, 0);<br /> <br /> draw(A--B--C--cycle);<br /> draw(D--e--F);<br /> <br /> label(&quot;$x$&quot;, D/2, W);<br /> label(&quot;$A$&quot;, A, SW);<br /> label(&quot;$B$&quot;, B, SE);<br /> label(&quot;$C$&quot;, C, N);<br /> label(&quot;$D$&quot;, D, W);<br /> label(&quot;$E$&quot;, e, NE);<br /> label(&quot;$F$&quot;, F, S);<br /> &lt;/asy&gt;<br /> <br /> Note that &lt;math&gt;\triangle ABC&lt;/math&gt; and &lt;math&gt;\triangle FBE&lt;/math&gt; are similar, so &lt;math&gt;\frac{BF}{FE} = \frac{AB}{AC}&lt;/math&gt;. This can be written as &lt;math&gt;\frac{4-x}{x}=\frac{4}{3}&lt;/math&gt;. Solving, &lt;math&gt;x = \frac{12}{7}&lt;/math&gt;.<br /> <br /> Now we analyze the second triangle.<br /> <br /> <br /> &lt;asy&gt;<br /> pair A,B,C;<br /> pair q, R, S, T;<br /> A = (0,0);<br /> B = (4,0);<br /> C = (0,3);<br /> <br /> q = (1.297, 0);<br /> R = (2.27 , 1.297);<br /> S = (0.973, 2.27);<br /> T = (0, 0.973);<br /> <br /> draw(A--B--C--cycle);<br /> draw(q--R--S--T--cycle);<br /> <br /> label(&quot;$y$&quot;, (q+R)/2, NW);<br /> label(&quot;$A'$&quot;, A, SW);<br /> label(&quot;$B'$&quot;, B, SE);<br /> label(&quot;$C'$&quot;, C, N);<br /> label(&quot;$Q$&quot;, (q-(0,0.3)));<br /> label(&quot;$R$&quot;, R, NE);<br /> label(&quot;$S$&quot;, S, NE);<br /> label(&quot;$T$&quot;, T, W);<br /> &lt;/asy&gt;<br /> <br /> Similary, &lt;math&gt;\triangle A'B'C'&lt;/math&gt; and &lt;math&gt;\triangle RB'Q&lt;/math&gt; are similar, so &lt;math&gt;RB' = \frac{4}{3}y&lt;/math&gt;, and &lt;math&gt;C'S = \frac{3}{4}y&lt;/math&gt;. Thus, &lt;math&gt;C'B' = C'S + SR + RB' = \frac{4}{3}y + y + \frac{3}{4}y = 5&lt;/math&gt;. Solving for &lt;math&gt;y&lt;/math&gt;, we get &lt;math&gt;y = \frac{60}{37}&lt;/math&gt;. Thus, &lt;math&gt;\frac{x}{y} = \frac{37}{35}&lt;/math&gt;.</div> Always correct https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_8&diff=82862 2017 AMC 10A Problems/Problem 8 2017-02-08T20:41:08Z <p>Always correct: /* Solution */</p> <hr /> <div>==Problem==<br /> <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 /> ==Solution==<br /> Each one of the ten people has to shake hands with all the &lt;math&gt;20&lt;/math&gt; other people they don’t know. So &lt;math&gt;10\cdot20&lt;/math&gt; = 200. From there, we calculate how many handshakes occurred between the people who don’t know each other. This is simply counting how many ways to choose two people to shake hands, or &lt;math&gt;\binom{10}{2} = 45&lt;/math&gt;. Thus the answer is &lt;math&gt;200 + 45 = \boxed{\textbf{(B)} 245}&lt;/math&gt;.</div> Always correct