https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Twod+horse&feedformat=atom AoPS Wiki - User contributions [en] 2021-08-05T05:56:11Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=Wooga_Looga_Theorem&diff=153524 Wooga Looga Theorem 2021-05-11T14:35:14Z <p>Twod horse: /* Solution 4 */</p> <hr /> <div>=Definition=<br /> If there is &lt;math&gt;\triangle ABC&lt;/math&gt; and points &lt;math&gt;D,E,F&lt;/math&gt; on the sides &lt;math&gt;BC,CA,AB&lt;/math&gt; respectively such that &lt;math&gt;\frac{DB}{DC}=\frac{EC}{EA}=\frac{FA}{FB}=r&lt;/math&gt;, then the ratio &lt;math&gt;\frac{[DEF]}{[ABC]}=\frac{r^2-r+1}{(r+1)^2}&lt;/math&gt;.<br /> <br /> <br /> Created by Foogle and Hoogle of [https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ The Ooga Booga Tribe of The Caveman Society]<br /> <br /> =Proofs=<br /> ==Proof 1==<br /> Proof by Gogobao:<br /> <br /> We have: &lt;math&gt;\frac{DB}{BC} = \frac{r}{r+1}, \frac{DC}{BC} = \frac{1}{r+1}, \frac{EC}{AC} = \frac{r}{r+1}, \frac{EA}{AC} = \frac{1}{r+1}, \frac{FA}{BA} = \frac{r}{r+1}, \frac{FB}{BA} = \frac{1}{r+1} &lt;/math&gt;<br /> <br /> We have: &lt;math&gt;[DEF] = [ABC] - [DCE] - [FAE] - [FBD]&lt;/math&gt;<br /> <br /> &lt;math&gt;[DCE] = [ABC] \cdot \frac{DC}{CB} \cdot \frac{CE}{CA} = [ABC] \cdot \frac{r}{(r+1)^2}&lt;/math&gt;<br /> <br /> &lt;math&gt;[EAF] = [ABC] \cdot \frac{EA}{CA} \cdot \frac{FA}{BA} = [ABC] \cdot \frac{r}{(r+1)^2}&lt;/math&gt;<br /> <br /> &lt;math&gt;[FBD] = [ABC] \cdot \frac{FB}{AB} \cdot \frac{BD}{CB} = [ABC] \cdot \frac{r}{(r+1)^2}&lt;/math&gt;<br /> <br /> Therefore &lt;math&gt;[DEF] = [ABC] (1-\frac{3r}{(r+1)^2})&lt;/math&gt;<br /> <br /> So we have &lt;math&gt;\frac{[DEF]}{[ABC]} = \frac{r^2-r+1}{(r+1)^2}&lt;/math&gt;<br /> <br /> ==Proof 2==<br /> Proof by franzliszt<br /> <br /> Apply Barycentrics w.r.t. &lt;math&gt;\triangle ABC&lt;/math&gt;. Then &lt;math&gt;A=(1,0,0),B=(0,1,0),C=(0,0,1)&lt;/math&gt;. We can also find that &lt;math&gt;D=\left(0,\tfrac {1}{r+1},\tfrac {r}{r+1}\right),E=\left(\tfrac {r}{r+1},0,\tfrac {1}{r+1}\right),F=\left(\tfrac {1}{r+1},\tfrac {r}{r+1},0\right)&lt;/math&gt;. In the barycentric coordinate system, the area formula is &lt;math&gt;[XYZ]=\begin{vmatrix} x_{1} &amp;y_{1} &amp;z_{1} \\ x_{2} &amp;y_{2} &amp;z_{2} \\ x_{3}&amp; y_{3} &amp; z_{3} \end{vmatrix}\cdot [ABC]&lt;/math&gt; where &lt;math&gt;\triangle XYZ&lt;/math&gt; is a random triangle and &lt;math&gt;\triangle ABC&lt;/math&gt; is the reference triangle. Using this, we find that &lt;cmath&gt;\frac{[DEF]}{[ABC]}= \begin{vmatrix} 0&amp;\tfrac {1}{r+1}&amp;\tfrac {r}{r+1} \\ \tfrac {r}{r+1}&amp;0&amp;\tfrac {1}{r+1}\\ \tfrac {1}{r+1}&amp;\tfrac {r}{r+1}&amp;0 \end{vmatrix}=\frac{r^2-r+1}{(r+1)^2}.&lt;/cmath&gt;<br /> <br /> ==Proof 3==<br /> Proof by [[User:RedFireTruck|&lt;font color=&quot;#FF0000&quot;&gt;RedFireTruck&lt;/font&gt;]] ([[User talk:RedFireTruck|&lt;font color=&quot;#FF0000&quot;&gt;talk&lt;/font&gt;]]) 12:11, 1 February 2021 (EST):<br /> <br /> WLOG we let &lt;math&gt;A=(0, 0)&lt;/math&gt;, &lt;math&gt;B=(1, 0)&lt;/math&gt;, &lt;math&gt;C=(x, y)&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y\in\mathbb{R}&lt;/math&gt;. We then use Shoelace Forumla to get &lt;math&gt;[ABC]=\frac12|y|&lt;/math&gt;. We then figure out that &lt;math&gt;D=\left(\frac{rx+1}{r+1}, \frac{ry}{r+1}\right)&lt;/math&gt;, &lt;math&gt;E=\left(\frac{x}{r+1}, \frac{y}{r+1}\right)&lt;/math&gt;, and &lt;math&gt;F=\left(\frac{r}{r+1}, 0\right)&lt;/math&gt; so we know that by Shoelace Formula &lt;math&gt;\frac{[DEF]}{[ABC]}=\frac{\frac12\left|\frac{r^2y-ry+y}{(r+1)^2}\right|}{\frac12|y|}=\left|\frac{r^2-r+1}{(r+1)^2}\right|&lt;/math&gt;. We know that &lt;math&gt;\frac{r^2-r+1}{(r+1)^2}\ge0&lt;/math&gt; for all &lt;math&gt;r\in\mathbb{R}&lt;/math&gt; so &lt;math&gt;\left|\frac{r^2-r+1}{(r+1)^2}\right|=\frac{r^2-r+1}{(r+1)^2}&lt;/math&gt;.<br /> <br /> ==Proof 4==<br /> Proof by ishanvannadil2008:<br /> <br /> Just use jayasharmaramankumarguptareddybavarajugopal's lemma. (Thanks to tenebrine)<br /> <br /> =Application 1=<br /> ==Problem==<br /> The Wooga Looga Theorem states that the solution to this problem by franzliszt:<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt; points &lt;math&gt;X,Y,Z&lt;/math&gt; are on sides &lt;math&gt;BC,CA,AB&lt;/math&gt; such that &lt;math&gt;\frac{XB}{XC}=\frac{YC}{YA}=\frac{ZA}{ZB}=\frac 71&lt;/math&gt;. Find the ratio of &lt;math&gt;[XYZ]&lt;/math&gt; to &lt;math&gt;[ABC]&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> One solution is this one by [[User:RedFireTruck|&lt;font color=&quot;#FF0000&quot;&gt;RedFireTruck&lt;/font&gt;]] ([[User talk:RedFireTruck|&lt;font color=&quot;#FF0000&quot;&gt;talk&lt;/font&gt;]]) 12:11, 1 February 2021 (EST):<br /> <br /> WLOG let &lt;math&gt;A=(0, 0)&lt;/math&gt;, &lt;math&gt;B=(1, 0)&lt;/math&gt;, &lt;math&gt;C=(x, y)&lt;/math&gt;. Then &lt;math&gt;[ABC]=\frac12|y|&lt;/math&gt; by Shoelace Theorem and &lt;math&gt;X=\left(\frac{7x+1}{8}, \frac{7y}{8}\right)&lt;/math&gt;, &lt;math&gt;Y=\left(\frac{x}{8}, \frac{y}{8}\right)&lt;/math&gt;, &lt;math&gt;Z=\left(\frac78, 0\right)&lt;/math&gt;. Then &lt;math&gt;[XYZ]=\frac12\left|\frac{43y}{64}\right|&lt;/math&gt; by Shoelace Theorem. Therefore the answer is &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> or this solution by franzliszt:<br /> <br /> We apply Barycentric Coordinates w.r.t. &lt;math&gt;\triangle ABC&lt;/math&gt;. Let &lt;math&gt;A=(1,0,0),B=(0,1,0),C=(0,0,1)&lt;/math&gt;. Then we find that &lt;math&gt;X=\left(0,\tfrac 18,\tfrac 78\right),Y=\left(\tfrac 78,0,\tfrac 18\right),Z=\left(\tfrac18,\tfrac78,0\right)&lt;/math&gt;. In the barycentric coordinate system, the area formula is &lt;math&gt;[XYZ]=\begin{vmatrix}<br /> x_{1} &amp;y_{1} &amp;z_{1} \\<br /> x_{2} &amp;y_{2} &amp;z_{2} \\ <br /> x_{3}&amp; y_{3} &amp; z_{3}<br /> \end{vmatrix}\cdot [ABC]&lt;/math&gt; where &lt;math&gt;\triangle XYZ&lt;/math&gt; is a random triangle and &lt;math&gt;\triangle ABC&lt;/math&gt; is the reference triangle. Using this, we find that &lt;cmath&gt;\frac{[XYZ]}{[ABC]}=\begin{vmatrix}<br /> 0&amp;\tfrac 18&amp;\tfrac 78\\<br /> \tfrac 78&amp;0&amp;\tfrac 18\\<br /> \tfrac18&amp;\tfrac78&amp;0<br /> \end{vmatrix}=\frac{43}{64}.&lt;/cmath&gt; &lt;math&gt;\blacksquare&lt;/math&gt;<br /> ==Solution 3==<br /> or this solution by aaja3427:<br /> <br /> According the the Wooga Looga Theorem, It is &lt;math&gt;\frac{49-7+1}{8^2}&lt;/math&gt;. This is &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> <br /> or this solution by AoPS user ilovepizza2020:<br /> <br /> We use the &lt;math&gt;\mathbf{FUNDAMENTAL~THEOREM~OF~GEOGEBRA}&lt;/math&gt; to instantly get &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;. (Note: You can only use this method when you are not in a contest, as this method is so overpowered that the people behind mathematics examinations decided to ban it.)<br /> <br /> ==Solution 5==<br /> or this solution by eduD_looC:<br /> <br /> This is a perfect application of the Adihaya Jayasharmaramankumarguptareddybavarajugopal's Lemma, which results in the answer being &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;. A very beautiful application, which leaves graders and readers speechless. Great for math contests with proofs.<br /> <br /> ==Solution 6==<br /> or this solution by CoolJupiter:<br /> <br /> Wow. All of your solutions are slow, compared to my sol:<br /> <br /> By math, we have &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;.<br /> <br /> ~CoolJupiter<br /> ^<br /> |<br /> EVERYONE USE THIS SOLUTION IT'S BRILLIANT <br /> ~bsu1<br /> Yes, very BRILLIANT!<br /> ~ TheAoPSLebron<br /> <br /> ==The Best Solution==<br /> <br /> By the &lt;math&gt;1+1=\text{BREAD}&lt;/math&gt; principle, we get &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;. Definitely the best method. When asked, please say that OlympusHero taught you this method. Because he did.<br /> <br /> ==Easiest Solution==<br /> <br /> The answer is clearly &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;. We leave the proof and intermediate steps to the reader as an exercise.<br /> <br /> =Application 2=<br /> ==Problem==<br /> The Wooga Looga Theorem states that the solution to this problem by Matholic:<br /> <br /> The figure below shows a triangle ABC whose area is &lt;math&gt;72 \text{cm}^2&lt;/math&gt;. If &lt;math&gt;\dfrac{AD}{DB}=\dfrac{BE}{EC}=\dfrac{CF}{FA}=\dfrac{1}{5}&lt;/math&gt;, find &lt;math&gt;[DEF].&lt;/math&gt;<br /> <br /> ~LaTeX-ifyed by RP3.1415<br /> <br /> ==Solution 1==<br /> is this solution by franzliszt:<br /> <br /> We apply Barycentric Coordinates w.r.t. &lt;math&gt;\triangle ABC&lt;/math&gt;. Let &lt;math&gt;A=(1,0,0),B=(0,1,0),C=(0,0,1)&lt;/math&gt;. Then we find that &lt;math&gt;D=(\tfrac 56,\tfrac 16,0),E=(0,\tfrac 56,\tfrac 16),F=(\tfrac16,0,\tfrac56)&lt;/math&gt;. In the barycentric coordinate system, the area formula is &lt;math&gt;[XYZ]=\begin{vmatrix}<br /> x_{1} &amp;y_{1} &amp;z_{1} \\ <br /> x_{2} &amp;y_{2} &amp;z_{2} \\ <br /> x_{3}&amp; y_{3} &amp; z_{3}<br /> \end{vmatrix}\cdot [ABC]&lt;/math&gt; where &lt;math&gt;\triangle XYZ&lt;/math&gt; is a random triangle and &lt;math&gt;\triangle ABC&lt;/math&gt; is the reference triangle. Using this, we find that&lt;cmath&gt;\frac{[DEF]}{}=\begin{vmatrix}<br /> \tfrac 56&amp;\tfrac 16&amp;0\\<br /> 0&amp;\tfrac 56&amp;\tfrac 16\\<br /> \tfrac16&amp;0&amp;\tfrac56<br /> \end{vmatrix}=\frac{7}{12}&lt;/cmath&gt; so &lt;math&gt;[DEF]=42&lt;/math&gt;. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> ==Solution 2==<br /> or this solution by [[User:RedFireTruck|&lt;font color=&quot;#FF0000&quot;&gt;RedFireTruck&lt;/font&gt;]] ([[User talk:RedFireTruck|&lt;font color=&quot;#FF0000&quot;&gt;talk&lt;/font&gt;]]) 12:11, 1 February 2021 (EST):<br /> <br /> By the Wooga Looga Theorem, &lt;math&gt;\frac{[DEF]}{[ABC]}=\frac{5^2-5+1}{(5+1)^2}=\frac{21}{36}=\frac{7}{12}&lt;/math&gt;. We are given that &lt;math&gt;[ABC]=72&lt;/math&gt; so &lt;math&gt;[DEF]=\frac{7}{12}\cdot72=\boxed{42}&lt;/math&gt;<br /> <br /> =Application 3=<br /> ==Problem==<br /> The Wooga Looga Theorem states that the solution to this problem by RedFireTruck:<br /> <br /> Find the ratio &lt;math&gt;\frac{[GHI]}{[ABC]}&lt;/math&gt; if &lt;math&gt;\frac{AD}{DB}=\frac{BE}{EC}=\frac{CF}{FA}=\frac12&lt;/math&gt; and &lt;math&gt;\frac{DG}{GE}=\frac{EH}{HF}=\frac{FI}{ID}=1&lt;/math&gt; in the diagram below.&lt;asy&gt;<br /> draw((0, 0)--(6, 0)--(4, 3)--cycle);<br /> draw((2, 0)--(16/3, 1)--(8/3, 2)--cycle);<br /> draw((11/3, 1/2)--(4, 3/2)--(7/3, 1)--cycle);<br /> label(&quot;$A$&quot;, (0, 0), SW);<br /> label(&quot;$B$&quot;, (6, 0), SE);<br /> label(&quot;$C$&quot;, (4, 3), N);<br /> label(&quot;$D$&quot;, (2, 0), S);<br /> label(&quot;$E$&quot;, (16/3, 1), NE);<br /> label(&quot;$F$&quot;, (8/3, 2), NW);<br /> label(&quot;$G$&quot;, (11/3, 1/2), SE);<br /> label(&quot;$H$&quot;, (4, 3/2), NE);<br /> label(&quot;$I$&quot;, (7/3, 1), W);<br /> &lt;/asy&gt;<br /> ==Solution 1==<br /> is this solution by franzliszt:<br /> <br /> By the Wooga Looga Theorem, &lt;math&gt;\frac{[DEF]}{[ABC]}=\frac{2^2-2+1}{(1+2)^2}=\frac 13&lt;/math&gt;. Notice that &lt;math&gt;\triangle GHI&lt;/math&gt; is the medial triangle of '''Wooga Looga Triangle ''' of &lt;math&gt;\triangle ABC&lt;/math&gt;. So &lt;math&gt;\frac{[GHI]}{[DEF]}=\frac 14&lt;/math&gt; and &lt;math&gt;\frac{[GHI]}{[ABD]}=\frac{[DEF]}{[ABC]}\cdot\frac{[GHI]}{[DEF]}=\frac 13 \cdot \frac 14 = \frac {1}{12}&lt;/math&gt; by Chain Rule ideas.<br /> <br /> ==Solution 2==<br /> or this solution by franzliszt:<br /> <br /> Apply Barycentrics w.r.t. &lt;math&gt;\triangle ABC&lt;/math&gt; so that &lt;math&gt;A=(1,0,0),B=(0,1,0),C=(0,0,1)&lt;/math&gt;. Then &lt;math&gt;D=(\tfrac 23,\tfrac 13,0),E=(0,\tfrac 23,\tfrac 13),F=(\tfrac 13,0,\tfrac 23)&lt;/math&gt;. And &lt;math&gt;G=(\tfrac 13,\tfrac 12,\tfrac 16),H=(\tfrac 16,\tfrac 13,\tfrac 12),I=(\tfrac 12,\tfrac 16,\tfrac 13)&lt;/math&gt;.<br /> <br /> In the barycentric coordinate system, the area formula is &lt;math&gt;[XYZ]=\begin{vmatrix} x_{1} &amp;y_{1} &amp;z_{1} \\ x_{2} &amp;y_{2} &amp;z_{2} \\ x_{3}&amp; y_{3} &amp; z_{3} \end{vmatrix}\cdot [ABC]&lt;/math&gt; where &lt;math&gt;\triangle XYZ&lt;/math&gt; is a random triangle and &lt;math&gt;\triangle ABC&lt;/math&gt; is the reference triangle. Using this, we find that&lt;cmath&gt;\frac{[GHI]}{[ABC]}=\begin{vmatrix} \tfrac 13&amp;\tfrac 12&amp;\tfrac 16\\ \tfrac 16&amp;\tfrac 13&amp;\tfrac 12\\ \tfrac 12&amp;\tfrac 16&amp;\tfrac 13 \end{vmatrix}=\frac{1}{12}.&lt;/cmath&gt;<br /> <br /> =Application 4=<br /> ==Problem==<br /> <br /> Let &lt;math&gt;ABC&lt;/math&gt; be a triangle and &lt;math&gt;D,E,F&lt;/math&gt; be points on sides &lt;math&gt;BC,AC,&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt; respectively. We have that &lt;math&gt;\frac{BD}{DC} = 3&lt;/math&gt; and similar for the other sides. If the area of triangle &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;16&lt;/math&gt;, then what is the area of triangle &lt;math&gt;DEF&lt;/math&gt;? (By ilovepizza2020)<br /> <br /> ==Solution 1==<br /> <br /> By Franzliszt<br /> <br /> By Wooga Looga, &lt;math&gt;\frac{[DEF]}{16} = \frac{3^2-3+1}{(3+1)^2}=\frac{7}{16}&lt;/math&gt; so the answer is &lt;math&gt;\boxed7&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> By franzliszt<br /> <br /> Apply Barycentrics w.r.t. &lt;math&gt;\triangle ABC&lt;/math&gt;. Then &lt;math&gt;A=(1,0,0),B=(0,1,0),C=(0,0,1)&lt;/math&gt;. We can also find that &lt;math&gt;D=(0,\tfrac 14,\tfrac 34),E=(\tfrac 34,0,\tfrac 14),F=(\tfrac 14,\tfrac 34,0)&lt;/math&gt;. In the barycentric coordinate system, the area formula is &lt;math&gt;[XYZ]=\begin{vmatrix} x_{1} &amp;y_{1} &amp;z_{1} \\ x_{2} &amp;y_{2} &amp;z_{2} \\ x_{3}&amp; y_{3} &amp; z_{3} \end{vmatrix}\cdot [ABC]&lt;/math&gt; where &lt;math&gt;\triangle XYZ&lt;/math&gt; is a random triangle and &lt;math&gt;\triangle ABC&lt;/math&gt; is the reference triangle. Using this, we find that&lt;cmath&gt;\frac{[DEF]}{[ABC]}=\begin{vmatrix} 0&amp;\tfrac 14&amp;\tfrac 34\\ \tfrac 34&amp;0&amp;\tfrac 14\\ \tfrac 14&amp;\tfrac 34&amp;0 \end{vmatrix}=\frac{7}{16}.&lt;/cmath&gt;So the answer is &lt;math&gt;\boxed{7}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> A long story short, the answer must be &lt;math&gt;\boxed{7}&lt;/math&gt; by the inverse of the Inverse Wooga Looga Theorem<br /> <br /> ==Solution 4==<br /> <br /> By TwoD_Horse<br /> <br /> Another solution involves using the proof of contradiction.<br /> <br /> We first let a real number &lt;math&gt;x&lt;/math&gt; be the solution of the problem, which we set &lt;math&gt;x\neq 7&lt;/math&gt;. By using induction, we can find out that such &lt;math&gt;x&lt;/math&gt; does not exist. Because the process of induction involves the use of calculus, fractals, and chaos theory, it is omitted in order to not make this page too long (to threaten the position of the great Gmaas). However, you can find details of the proof [https://www.youtube.com/watch?v=dQw4w9WgXcQ here].<br /> <br /> Since such &lt;math&gt;x&lt;/math&gt; does not exist, it is clearly a contradiction. The solution will be all real numbers that are not in the domain of &lt;math&gt;x&lt;/math&gt;, which is &lt;math&gt;\boxed{7}&lt;/math&gt;.<br /> <br /> =Testimonials=<br /> <br /> Pogpr0 = wooga looga - Ladka13<br /> The Wooga Looga Theorem can be used to prove many problems and should be a part of any geometry textbook.<br /> ~ilp2020<br /> <br /> Thanks for rediscovering our theorem RedFireTruck - Foogle and Hoogle of [https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ The Ooga Booga Tribe of The Caveman Society]<br /> <br /> Franzlist is wooga looga howsopro - volkie boy<br /> <br /> this is in fact a pretty sensible theorem. Nothing to be so excited about, though. ~DofL<br /> <br /> The Wooga Looga Theorem is EPIC POGGERS WHOLESOME 100 KEANU CHUNGUS AMAZING SKILL THEOREM!!!!!1!!!111111 -centslordm<br /> <br /> The Wooga Looga Theorem is amazing and can be applied to so many problems and should be taught in every school. - [[User:RedFireTruck|&lt;font color=&quot;#FF0000&quot;&gt;RedFireTruck&lt;/font&gt;]] ([[User talk:RedFireTruck|&lt;font color=&quot;#FF0000&quot;&gt;talk&lt;/font&gt;]]) 11:00, 1 February 2021 (EST)<br /> <br /> The Wooga Looga Theorem is the best. -aaja3427<br /> <br /> The Wooga Looga Theorem is needed for everything and it is great-hi..<br /> <br /> The Wooga Looga Theorem was made by the author of the 5th Testimonial, RedFireTruck, which means they are the ooga booga tribe... proof: go to https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ and click &quot;about&quot;. now copy and paste the aops URL. you got RedFireTruck! Great Job! now go check out his thread for post milestones, https://artofproblemsolving.com/community/c3h2319596, and give him a friend request! -FPT<br /> <br /> This theorem has helped me with school and I am no longer failing my math class. -mchang<br /> <br /> I, who [u]rarely[\u] edits the AoPS Wiki, has edited this to show how amazing this theorem is! The Wooga Looga theorem has actually helped me on a school test, a math competition, and more! My teacher got upset at me for not doing it the way I she taught it though - ChrisalonaLiverspur<br /> <br /> &quot;I can't believe AoPS books don't have this amazing theorem. If you need help with math, you can depend on caveman.&quot; ~CoolJupiter<br /> <br /> Before the Wooga Looga Theorem, I had NO IDEA how to solve any hard geo. But, now that I've learned it, I can solve hard geo in 7 seconds ~ ilp2020 (2nd testimonial by me)<br /> <br /> Too powerful... ~franzliszt<br /> <br /> The Wooga Looga Theorem is so pro ~ ac142931<br /> <br /> It is so epic and awesome that it will blow the minds of people if they saw this ~ ac142931(2nd testimonial by me)<br /> <br /> This theorem changed my life... ~ samrocksnature<br /> <br /> Math competitions need to ban the use of the Wooga Looga Theorem, it's just too good. ~ jasperE3<br /> <br /> It actually can be. I never thought I'd say this, but the Wooga Looga theorem is a legit theorem. ~ jasperE3<br /> <br /> This is franzliszt and I endorse this theorem. ~franzliszt<br /> <br /> This theorem is too OP. ~bestzack66<br /> <br /> This is amazing! However much it looks like a joke, it is a legitimate - and powerful - theorem. -Supernova283<br /> <br /> Wooga Looga Theorem is extremely useful. Someone needs to make a handout on this so everyone can obtain the power of Wooga Looga ~RP3.1415<br /> <br /> The Wooga Looga cavemen were way ahead of their time. Good job (dead) guys! -HIA2020<br /> <br /> It's like the Ooga Booga Theorem (also OP), but better!!! - BobDBuilder321<br /> <br /> The Wooga Looga Theorem is a special case of [https://en.wikipedia.org/wiki/Routh%27s_theorem Routh's Theorem.] So this wiki article is DEFINITELY needed. -peace<br /> <br /> I actually thought this was a joke theorem until I read this page - HumanCalculator9<br /> <br /> I endorse the Wooga Looga theorem for its utter usefulness and seriousness. -HamstPan38825<br /> <br /> &lt;s&gt;I ReAlLy don't get it - Senguamar&lt;/s&gt; HOW DARE YOU!!!!<br /> <br /> The Wooga Looga Theorem is the base of all geometry. It is so OP that even I don't understand how to use it.<br /> <br /> You know what, this is jayasharmaramankumarguptareddybavarajugopal's lemma - Ishan<br /> <br /> If only I knew this on some contests that I had done previously... - JacobJB<br /> <br /> The Wooga Looga Theorem is so pr0 that it needs to be nerfed. - rocketsri<br /> <br /> &quot;The Wooga Looga Theorem should be used in contests and should be part of geometry books.&quot; ~ [[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]] ([[User talk:Aops-g5-gethsemanea2|talk]]) 21:56, 21 December 2020 (EST)<br /> <br /> The Wooga Looga Theorem is so OP BRUH<br /> <br /> thank for the theorem it is trivial by 1/2 ab sin(C) formula but very helpful I have used it zero times so far in competitions so it is of great use thank - bussie<br /> <br /> I have no idea what is going on here - awesomeguy856<br /> <br /> fuzimiao2013 waz hear<br /> <br /> this theorem is bad<br /> <br /> poggers theorem - awesomeming327<br /> <br /> The Wooga Looga theorem is very OP and not to be frowned upon - Yelly314<br /> <br /> person who invented Wooga Looga theorem is orz orz orz wooga looga theorem OP, citing it on ANY olympiad test = instant full marks -awesomeness_in_a_bun<br /> <br /> Wooga Looga Theorem is TRASH.<br /> <br /> HOW DARE YOU @above and @5above DISRESPECT THE WOOGA LOOGA THEOREM. THIS IS THE MOST OP THEOREM EVER AND CAN BE USED TO SOLVE EVERY PROBLEM. BECAUSE OF THIS THEOREM, I GOT A 187 ON AMC 10A AND 10B, a 665 on the AIME, AND A 420 ON THE USAJMO THIS YEAR. I HAVE ALSO USED THIS TO PROVE THE GOLDBACH CONJECTURE, THE TWIN PRIME CONJECTURE, EVERY SINGLE IMO PROBLEM, AND I HAVE PROVED THE RIEMANN HYPOTHESIS. I ALSO INVENTED HAND SANITIZER THAT KILLS 100% OF BACTERIA, MILK THAT IS 0% MILK, AND A TIME MACHINE WITH THIS THEOREM. THANK YOU SO MUCH @RedFireTruck FOR THIS LEGENDARY THEOREM. I WILL ALWAYS BE INDEBTED TO YOU.</div> Twod horse https://artofproblemsolving.com/wiki/index.php?title=Wooga_Looga_Theorem&diff=153521 Wooga Looga Theorem 2021-05-11T14:28:50Z <p>Twod horse: /* Application 4 */</p> <hr /> <div>=Definition=<br /> If there is &lt;math&gt;\triangle ABC&lt;/math&gt; and points &lt;math&gt;D,E,F&lt;/math&gt; on the sides &lt;math&gt;BC,CA,AB&lt;/math&gt; respectively such that &lt;math&gt;\frac{DB}{DC}=\frac{EC}{EA}=\frac{FA}{FB}=r&lt;/math&gt;, then the ratio &lt;math&gt;\frac{[DEF]}{[ABC]}=\frac{r^2-r+1}{(r+1)^2}&lt;/math&gt;.<br /> <br /> <br /> Created by Foogle and Hoogle of [https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ The Ooga Booga Tribe of The Caveman Society]<br /> <br /> =Proofs=<br /> ==Proof 1==<br /> Proof by Gogobao:<br /> <br /> We have: &lt;math&gt;\frac{DB}{BC} = \frac{r}{r+1}, \frac{DC}{BC} = \frac{1}{r+1}, \frac{EC}{AC} = \frac{r}{r+1}, \frac{EA}{AC} = \frac{1}{r+1}, \frac{FA}{BA} = \frac{r}{r+1}, \frac{FB}{BA} = \frac{1}{r+1} &lt;/math&gt;<br /> <br /> We have: &lt;math&gt;[DEF] = [ABC] - [DCE] - [FAE] - [FBD]&lt;/math&gt;<br /> <br /> &lt;math&gt;[DCE] = [ABC] \cdot \frac{DC}{CB} \cdot \frac{CE}{CA} = [ABC] \cdot \frac{r}{(r+1)^2}&lt;/math&gt;<br /> <br /> &lt;math&gt;[EAF] = [ABC] \cdot \frac{EA}{CA} \cdot \frac{FA}{BA} = [ABC] \cdot \frac{r}{(r+1)^2}&lt;/math&gt;<br /> <br /> &lt;math&gt;[FBD] = [ABC] \cdot \frac{FB}{AB} \cdot \frac{BD}{CB} = [ABC] \cdot \frac{r}{(r+1)^2}&lt;/math&gt;<br /> <br /> Therefore &lt;math&gt;[DEF] = [ABC] (1-\frac{3r}{(r+1)^2})&lt;/math&gt;<br /> <br /> So we have &lt;math&gt;\frac{[DEF]}{[ABC]} = \frac{r^2-r+1}{(r+1)^2}&lt;/math&gt;<br /> <br /> ==Proof 2==<br /> Proof by franzliszt<br /> <br /> Apply Barycentrics w.r.t. &lt;math&gt;\triangle ABC&lt;/math&gt;. Then &lt;math&gt;A=(1,0,0),B=(0,1,0),C=(0,0,1)&lt;/math&gt;. We can also find that &lt;math&gt;D=\left(0,\tfrac {1}{r+1},\tfrac {r}{r+1}\right),E=\left(\tfrac {r}{r+1},0,\tfrac {1}{r+1}\right),F=\left(\tfrac {1}{r+1},\tfrac {r}{r+1},0\right)&lt;/math&gt;. In the barycentric coordinate system, the area formula is &lt;math&gt;[XYZ]=\begin{vmatrix} x_{1} &amp;y_{1} &amp;z_{1} \\ x_{2} &amp;y_{2} &amp;z_{2} \\ x_{3}&amp; y_{3} &amp; z_{3} \end{vmatrix}\cdot [ABC]&lt;/math&gt; where &lt;math&gt;\triangle XYZ&lt;/math&gt; is a random triangle and &lt;math&gt;\triangle ABC&lt;/math&gt; is the reference triangle. Using this, we find that &lt;cmath&gt;\frac{[DEF]}{[ABC]}= \begin{vmatrix} 0&amp;\tfrac {1}{r+1}&amp;\tfrac {r}{r+1} \\ \tfrac {r}{r+1}&amp;0&amp;\tfrac {1}{r+1}\\ \tfrac {1}{r+1}&amp;\tfrac {r}{r+1}&amp;0 \end{vmatrix}=\frac{r^2-r+1}{(r+1)^2}.&lt;/cmath&gt;<br /> <br /> ==Proof 3==<br /> Proof by [[User:RedFireTruck|&lt;font color=&quot;#FF0000&quot;&gt;RedFireTruck&lt;/font&gt;]] ([[User talk:RedFireTruck|&lt;font color=&quot;#FF0000&quot;&gt;talk&lt;/font&gt;]]) 12:11, 1 February 2021 (EST):<br /> <br /> WLOG we let &lt;math&gt;A=(0, 0)&lt;/math&gt;, &lt;math&gt;B=(1, 0)&lt;/math&gt;, &lt;math&gt;C=(x, y)&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y\in\mathbb{R}&lt;/math&gt;. We then use Shoelace Forumla to get &lt;math&gt;[ABC]=\frac12|y|&lt;/math&gt;. We then figure out that &lt;math&gt;D=\left(\frac{rx+1}{r+1}, \frac{ry}{r+1}\right)&lt;/math&gt;, &lt;math&gt;E=\left(\frac{x}{r+1}, \frac{y}{r+1}\right)&lt;/math&gt;, and &lt;math&gt;F=\left(\frac{r}{r+1}, 0\right)&lt;/math&gt; so we know that by Shoelace Formula &lt;math&gt;\frac{[DEF]}{[ABC]}=\frac{\frac12\left|\frac{r^2y-ry+y}{(r+1)^2}\right|}{\frac12|y|}=\left|\frac{r^2-r+1}{(r+1)^2}\right|&lt;/math&gt;. We know that &lt;math&gt;\frac{r^2-r+1}{(r+1)^2}\ge0&lt;/math&gt; for all &lt;math&gt;r\in\mathbb{R}&lt;/math&gt; so &lt;math&gt;\left|\frac{r^2-r+1}{(r+1)^2}\right|=\frac{r^2-r+1}{(r+1)^2}&lt;/math&gt;.<br /> <br /> ==Proof 4==<br /> Proof by ishanvannadil2008:<br /> <br /> Just use jayasharmaramankumarguptareddybavarajugopal's lemma. (Thanks to tenebrine)<br /> <br /> =Application 1=<br /> ==Problem==<br /> The Wooga Looga Theorem states that the solution to this problem by franzliszt:<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt; points &lt;math&gt;X,Y,Z&lt;/math&gt; are on sides &lt;math&gt;BC,CA,AB&lt;/math&gt; such that &lt;math&gt;\frac{XB}{XC}=\frac{YC}{YA}=\frac{ZA}{ZB}=\frac 71&lt;/math&gt;. Find the ratio of &lt;math&gt;[XYZ]&lt;/math&gt; to &lt;math&gt;[ABC]&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> One solution is this one by [[User:RedFireTruck|&lt;font color=&quot;#FF0000&quot;&gt;RedFireTruck&lt;/font&gt;]] ([[User talk:RedFireTruck|&lt;font color=&quot;#FF0000&quot;&gt;talk&lt;/font&gt;]]) 12:11, 1 February 2021 (EST):<br /> <br /> WLOG let &lt;math&gt;A=(0, 0)&lt;/math&gt;, &lt;math&gt;B=(1, 0)&lt;/math&gt;, &lt;math&gt;C=(x, y)&lt;/math&gt;. Then &lt;math&gt;[ABC]=\frac12|y|&lt;/math&gt; by Shoelace Theorem and &lt;math&gt;X=\left(\frac{7x+1}{8}, \frac{7y}{8}\right)&lt;/math&gt;, &lt;math&gt;Y=\left(\frac{x}{8}, \frac{y}{8}\right)&lt;/math&gt;, &lt;math&gt;Z=\left(\frac78, 0\right)&lt;/math&gt;. Then &lt;math&gt;[XYZ]=\frac12\left|\frac{43y}{64}\right|&lt;/math&gt; by Shoelace Theorem. Therefore the answer is &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> or this solution by franzliszt:<br /> <br /> We apply Barycentric Coordinates w.r.t. &lt;math&gt;\triangle ABC&lt;/math&gt;. Let &lt;math&gt;A=(1,0,0),B=(0,1,0),C=(0,0,1)&lt;/math&gt;. Then we find that &lt;math&gt;X=\left(0,\tfrac 18,\tfrac 78\right),Y=\left(\tfrac 78,0,\tfrac 18\right),Z=\left(\tfrac18,\tfrac78,0\right)&lt;/math&gt;. In the barycentric coordinate system, the area formula is &lt;math&gt;[XYZ]=\begin{vmatrix}<br /> x_{1} &amp;y_{1} &amp;z_{1} \\<br /> x_{2} &amp;y_{2} &amp;z_{2} \\ <br /> x_{3}&amp; y_{3} &amp; z_{3}<br /> \end{vmatrix}\cdot [ABC]&lt;/math&gt; where &lt;math&gt;\triangle XYZ&lt;/math&gt; is a random triangle and &lt;math&gt;\triangle ABC&lt;/math&gt; is the reference triangle. Using this, we find that &lt;cmath&gt;\frac{[XYZ]}{[ABC]}=\begin{vmatrix}<br /> 0&amp;\tfrac 18&amp;\tfrac 78\\<br /> \tfrac 78&amp;0&amp;\tfrac 18\\<br /> \tfrac18&amp;\tfrac78&amp;0<br /> \end{vmatrix}=\frac{43}{64}.&lt;/cmath&gt; &lt;math&gt;\blacksquare&lt;/math&gt;<br /> ==Solution 3==<br /> or this solution by aaja3427:<br /> <br /> According the the Wooga Looga Theorem, It is &lt;math&gt;\frac{49-7+1}{8^2}&lt;/math&gt;. This is &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> <br /> or this solution by AoPS user ilovepizza2020:<br /> <br /> We use the &lt;math&gt;\mathbf{FUNDAMENTAL~THEOREM~OF~GEOGEBRA}&lt;/math&gt; to instantly get &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;. (Note: You can only use this method when you are not in a contest, as this method is so overpowered that the people behind mathematics examinations decided to ban it.)<br /> <br /> ==Solution 5==<br /> or this solution by eduD_looC:<br /> <br /> This is a perfect application of the Adihaya Jayasharmaramankumarguptareddybavarajugopal's Lemma, which results in the answer being &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;. A very beautiful application, which leaves graders and readers speechless. Great for math contests with proofs.<br /> <br /> ==Solution 6==<br /> or this solution by CoolJupiter:<br /> <br /> Wow. All of your solutions are slow, compared to my sol:<br /> <br /> By math, we have &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;.<br /> <br /> ~CoolJupiter<br /> ^<br /> |<br /> EVERYONE USE THIS SOLUTION IT'S BRILLIANT <br /> ~bsu1<br /> Yes, very BRILLIANT!<br /> ~ TheAoPSLebron<br /> <br /> ==The Best Solution==<br /> <br /> By the &lt;math&gt;1+1=\text{BREAD}&lt;/math&gt; principle, we get &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;. Definitely the best method. When asked, please say that OlympusHero taught you this method. Because he did.<br /> <br /> ==Easiest Solution==<br /> <br /> The answer is clearly &lt;math&gt;\boxed{\frac{43}{64}}&lt;/math&gt;. We leave the proof and intermediate steps to the reader as an exercise.<br /> <br /> =Application 2=<br /> ==Problem==<br /> The Wooga Looga Theorem states that the solution to this problem by Matholic:<br /> <br /> The figure below shows a triangle ABC whose area is &lt;math&gt;72 \text{cm}^2&lt;/math&gt;. If &lt;math&gt;\dfrac{AD}{DB}=\dfrac{BE}{EC}=\dfrac{CF}{FA}=\dfrac{1}{5}&lt;/math&gt;, find &lt;math&gt;[DEF].&lt;/math&gt;<br /> <br /> ~LaTeX-ifyed by RP3.1415<br /> <br /> ==Solution 1==<br /> is this solution by franzliszt:<br /> <br /> We apply Barycentric Coordinates w.r.t. &lt;math&gt;\triangle ABC&lt;/math&gt;. Let &lt;math&gt;A=(1,0,0),B=(0,1,0),C=(0,0,1)&lt;/math&gt;. Then we find that &lt;math&gt;D=(\tfrac 56,\tfrac 16,0),E=(0,\tfrac 56,\tfrac 16),F=(\tfrac16,0,\tfrac56)&lt;/math&gt;. In the barycentric coordinate system, the area formula is &lt;math&gt;[XYZ]=\begin{vmatrix}<br /> x_{1} &amp;y_{1} &amp;z_{1} \\ <br /> x_{2} &amp;y_{2} &amp;z_{2} \\ <br /> x_{3}&amp; y_{3} &amp; z_{3}<br /> \end{vmatrix}\cdot [ABC]&lt;/math&gt; where &lt;math&gt;\triangle XYZ&lt;/math&gt; is a random triangle and &lt;math&gt;\triangle ABC&lt;/math&gt; is the reference triangle. Using this, we find that&lt;cmath&gt;\frac{[DEF]}{}=\begin{vmatrix}<br /> \tfrac 56&amp;\tfrac 16&amp;0\\<br /> 0&amp;\tfrac 56&amp;\tfrac 16\\<br /> \tfrac16&amp;0&amp;\tfrac56<br /> \end{vmatrix}=\frac{7}{12}&lt;/cmath&gt; so &lt;math&gt;[DEF]=42&lt;/math&gt;. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> ==Solution 2==<br /> or this solution by [[User:RedFireTruck|&lt;font color=&quot;#FF0000&quot;&gt;RedFireTruck&lt;/font&gt;]] ([[User talk:RedFireTruck|&lt;font color=&quot;#FF0000&quot;&gt;talk&lt;/font&gt;]]) 12:11, 1 February 2021 (EST):<br /> <br /> By the Wooga Looga Theorem, &lt;math&gt;\frac{[DEF]}{[ABC]}=\frac{5^2-5+1}{(5+1)^2}=\frac{21}{36}=\frac{7}{12}&lt;/math&gt;. We are given that &lt;math&gt;[ABC]=72&lt;/math&gt; so &lt;math&gt;[DEF]=\frac{7}{12}\cdot72=\boxed{42}&lt;/math&gt;<br /> <br /> =Application 3=<br /> ==Problem==<br /> The Wooga Looga Theorem states that the solution to this problem by RedFireTruck:<br /> <br /> Find the ratio &lt;math&gt;\frac{[GHI]}{[ABC]}&lt;/math&gt; if &lt;math&gt;\frac{AD}{DB}=\frac{BE}{EC}=\frac{CF}{FA}=\frac12&lt;/math&gt; and &lt;math&gt;\frac{DG}{GE}=\frac{EH}{HF}=\frac{FI}{ID}=1&lt;/math&gt; in the diagram below.&lt;asy&gt;<br /> draw((0, 0)--(6, 0)--(4, 3)--cycle);<br /> draw((2, 0)--(16/3, 1)--(8/3, 2)--cycle);<br /> draw((11/3, 1/2)--(4, 3/2)--(7/3, 1)--cycle);<br /> label(&quot;$A$&quot;, (0, 0), SW);<br /> label(&quot;$B$&quot;, (6, 0), SE);<br /> label(&quot;$C$&quot;, (4, 3), N);<br /> label(&quot;$D$&quot;, (2, 0), S);<br /> label(&quot;$E$&quot;, (16/3, 1), NE);<br /> label(&quot;$F$&quot;, (8/3, 2), NW);<br /> label(&quot;$G$&quot;, (11/3, 1/2), SE);<br /> label(&quot;$H$&quot;, (4, 3/2), NE);<br /> label(&quot;$I$&quot;, (7/3, 1), W);<br /> &lt;/asy&gt;<br /> ==Solution 1==<br /> is this solution by franzliszt:<br /> <br /> By the Wooga Looga Theorem, &lt;math&gt;\frac{[DEF]}{[ABC]}=\frac{2^2-2+1}{(1+2)^2}=\frac 13&lt;/math&gt;. Notice that &lt;math&gt;\triangle GHI&lt;/math&gt; is the medial triangle of '''Wooga Looga Triangle ''' of &lt;math&gt;\triangle ABC&lt;/math&gt;. So &lt;math&gt;\frac{[GHI]}{[DEF]}=\frac 14&lt;/math&gt; and &lt;math&gt;\frac{[GHI]}{[ABD]}=\frac{[DEF]}{[ABC]}\cdot\frac{[GHI]}{[DEF]}=\frac 13 \cdot \frac 14 = \frac {1}{12}&lt;/math&gt; by Chain Rule ideas.<br /> <br /> ==Solution 2==<br /> or this solution by franzliszt:<br /> <br /> Apply Barycentrics w.r.t. &lt;math&gt;\triangle ABC&lt;/math&gt; so that &lt;math&gt;A=(1,0,0),B=(0,1,0),C=(0,0,1)&lt;/math&gt;. Then &lt;math&gt;D=(\tfrac 23,\tfrac 13,0),E=(0,\tfrac 23,\tfrac 13),F=(\tfrac 13,0,\tfrac 23)&lt;/math&gt;. And &lt;math&gt;G=(\tfrac 13,\tfrac 12,\tfrac 16),H=(\tfrac 16,\tfrac 13,\tfrac 12),I=(\tfrac 12,\tfrac 16,\tfrac 13)&lt;/math&gt;.<br /> <br /> In the barycentric coordinate system, the area formula is &lt;math&gt;[XYZ]=\begin{vmatrix} x_{1} &amp;y_{1} &amp;z_{1} \\ x_{2} &amp;y_{2} &amp;z_{2} \\ x_{3}&amp; y_{3} &amp; z_{3} \end{vmatrix}\cdot [ABC]&lt;/math&gt; where &lt;math&gt;\triangle XYZ&lt;/math&gt; is a random triangle and &lt;math&gt;\triangle ABC&lt;/math&gt; is the reference triangle. Using this, we find that&lt;cmath&gt;\frac{[GHI]}{[ABC]}=\begin{vmatrix} \tfrac 13&amp;\tfrac 12&amp;\tfrac 16\\ \tfrac 16&amp;\tfrac 13&amp;\tfrac 12\\ \tfrac 12&amp;\tfrac 16&amp;\tfrac 13 \end{vmatrix}=\frac{1}{12}.&lt;/cmath&gt;<br /> <br /> =Application 4=<br /> ==Problem==<br /> <br /> Let &lt;math&gt;ABC&lt;/math&gt; be a triangle and &lt;math&gt;D,E,F&lt;/math&gt; be points on sides &lt;math&gt;BC,AC,&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt; respectively. We have that &lt;math&gt;\frac{BD}{DC} = 3&lt;/math&gt; and similar for the other sides. If the area of triangle &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;16&lt;/math&gt;, then what is the area of triangle &lt;math&gt;DEF&lt;/math&gt;? (By ilovepizza2020)<br /> <br /> ==Solution 1==<br /> <br /> By Franzliszt<br /> <br /> By Wooga Looga, &lt;math&gt;\frac{[DEF]}{16} = \frac{3^2-3+1}{(3+1)^2}=\frac{7}{16}&lt;/math&gt; so the answer is &lt;math&gt;\boxed7&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> By franzliszt<br /> <br /> Apply Barycentrics w.r.t. &lt;math&gt;\triangle ABC&lt;/math&gt;. Then &lt;math&gt;A=(1,0,0),B=(0,1,0),C=(0,0,1)&lt;/math&gt;. We can also find that &lt;math&gt;D=(0,\tfrac 14,\tfrac 34),E=(\tfrac 34,0,\tfrac 14),F=(\tfrac 14,\tfrac 34,0)&lt;/math&gt;. In the barycentric coordinate system, the area formula is &lt;math&gt;[XYZ]=\begin{vmatrix} x_{1} &amp;y_{1} &amp;z_{1} \\ x_{2} &amp;y_{2} &amp;z_{2} \\ x_{3}&amp; y_{3} &amp; z_{3} \end{vmatrix}\cdot [ABC]&lt;/math&gt; where &lt;math&gt;\triangle XYZ&lt;/math&gt; is a random triangle and &lt;math&gt;\triangle ABC&lt;/math&gt; is the reference triangle. Using this, we find that&lt;cmath&gt;\frac{[DEF]}{[ABC]}=\begin{vmatrix} 0&amp;\tfrac 14&amp;\tfrac 34\\ \tfrac 34&amp;0&amp;\tfrac 14\\ \tfrac 14&amp;\tfrac 34&amp;0 \end{vmatrix}=\frac{7}{16}.&lt;/cmath&gt;So the answer is &lt;math&gt;\boxed{7}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> A long story short, the answer must be &lt;math&gt;\boxed{7}&lt;/math&gt; by the inverse of the Inverse Wooga Looga Theorem<br /> <br /> ==Solution 4==<br /> <br /> By TwoD_Horse<br /> <br /> Another solution involves using the proof of contradiction.<br /> <br /> We first let a real number &lt;math&gt;x&lt;/math&gt; be the solution of the problem, which we set &lt;math&gt;x\neq 7&lt;/math&gt;. By using induction, we can find out that such &lt;math&gt;x&lt;/math&gt; does not exist. Because the process of induction involves the use of calculus, fractals, and chaos theory, it is omitted in order to not make this page too long (to threaten the position of the great Gmass). However, you can find details of the proof [https://www.youtube.com/watch?v=dQw4w9WgXcQ here].<br /> <br /> Since such &lt;math&gt;x&lt;/math&gt; does not exist, it is clearly a contradiction. The solution will be all real numbers that are not in the domain of &lt;math&gt;x&lt;/math&gt;, which is &lt;math&gt;\boxed{7}&lt;/math&gt;.<br /> <br /> =Testimonials=<br /> <br /> Pogpr0 = wooga looga - Ladka13<br /> The Wooga Looga Theorem can be used to prove many problems and should be a part of any geometry textbook.<br /> ~ilp2020<br /> <br /> Thanks for rediscovering our theorem RedFireTruck - Foogle and Hoogle of [https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ The Ooga Booga Tribe of The Caveman Society]<br /> <br /> Franzlist is wooga looga howsopro - volkie boy<br /> <br /> this is in fact a pretty sensible theorem. Nothing to be so excited about, though. ~DofL<br /> <br /> The Wooga Looga Theorem is EPIC POGGERS WHOLESOME 100 KEANU CHUNGUS AMAZING SKILL THEOREM!!!!!1!!!111111 -centslordm<br /> <br /> The Wooga Looga Theorem is amazing and can be applied to so many problems and should be taught in every school. - [[User:RedFireTruck|&lt;font color=&quot;#FF0000&quot;&gt;RedFireTruck&lt;/font&gt;]] ([[User talk:RedFireTruck|&lt;font color=&quot;#FF0000&quot;&gt;talk&lt;/font&gt;]]) 11:00, 1 February 2021 (EST)<br /> <br /> The Wooga Looga Theorem is the best. -aaja3427<br /> <br /> The Wooga Looga Theorem is needed for everything and it is great-hi..<br /> <br /> The Wooga Looga Theorem was made by the author of the 5th Testimonial, RedFireTruck, which means they are the ooga booga tribe... proof: go to https://www.youtube.com/channel/UC50E9TuLIMWbOPUX45xZPaQ and click &quot;about&quot;. now copy and paste the aops URL. you got RedFireTruck! Great Job! now go check out his thread for post milestones, https://artofproblemsolving.com/community/c3h2319596, and give him a friend request! -FPT<br /> <br /> This theorem has helped me with school and I am no longer failing my math class. -mchang<br /> <br /> I, who [u]rarely[\u] edits the AoPS Wiki, has edited this to show how amazing this theorem is! The Wooga Looga theorem has actually helped me on a school test, a math competition, and more! My teacher got upset at me for not doing it the way I she taught it though - ChrisalonaLiverspur<br /> <br /> &quot;I can't believe AoPS books don't have this amazing theorem. If you need help with math, you can depend on caveman.&quot; ~CoolJupiter<br /> <br /> Before the Wooga Looga Theorem, I had NO IDEA how to solve any hard geo. But, now that I've learned it, I can solve hard geo in 7 seconds ~ ilp2020 (2nd testimonial by me)<br /> <br /> Too powerful... ~franzliszt<br /> <br /> The Wooga Looga Theorem is so pro ~ ac142931<br /> <br /> It is so epic and awesome that it will blow the minds of people if they saw this ~ ac142931(2nd testimonial by me)<br /> <br /> This theorem changed my life... ~ samrocksnature<br /> <br /> Math competitions need to ban the use of the Wooga Looga Theorem, it's just too good. ~ jasperE3<br /> <br /> It actually can be. I never thought I'd say this, but the Wooga Looga theorem is a legit theorem. ~ jasperE3<br /> <br /> This is franzliszt and I endorse this theorem. ~franzliszt<br /> <br /> This theorem is too OP. ~bestzack66<br /> <br /> This is amazing! However much it looks like a joke, it is a legitimate - and powerful - theorem. -Supernova283<br /> <br /> Wooga Looga Theorem is extremely useful. Someone needs to make a handout on this so everyone can obtain the power of Wooga Looga ~RP3.1415<br /> <br /> The Wooga Looga cavemen were way ahead of their time. Good job (dead) guys! -HIA2020<br /> <br /> It's like the Ooga Booga Theorem (also OP), but better!!! - BobDBuilder321<br /> <br /> The Wooga Looga Theorem is a special case of [https://en.wikipedia.org/wiki/Routh%27s_theorem Routh's Theorem.] So this wiki article is DEFINITELY needed. -peace<br /> <br /> I actually thought this was a joke theorem until I read this page - HumanCalculator9<br /> <br /> I endorse the Wooga Looga theorem for its utter usefulness and seriousness. -HamstPan38825<br /> <br /> &lt;s&gt;I ReAlLy don't get it - Senguamar&lt;/s&gt; HOW DARE YOU!!!!<br /> <br /> The Wooga Looga Theorem is the base of all geometry. It is so OP that even I don't understand how to use it.<br /> <br /> You know what, this is jayasharmaramankumarguptareddybavarajugopal's lemma - Ishan<br /> <br /> If only I knew this on some contests that I had done previously... - JacobJB<br /> <br /> The Wooga Looga Theorem is so pr0 that it needs to be nerfed. - rocketsri<br /> <br /> &quot;The Wooga Looga Theorem should be used in contests and should be part of geometry books.&quot; ~ [[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]] ([[User talk:Aops-g5-gethsemanea2|talk]]) 21:56, 21 December 2020 (EST)<br /> <br /> The Wooga Looga Theorem is so OP BRUH<br /> <br /> thank for the theorem it is trivial by 1/2 ab sin(C) formula but very helpful I have used it zero times so far in competitions so it is of great use thank - bussie<br /> <br /> I have no idea what is going on here - awesomeguy856<br /> <br /> fuzimiao2013 waz hear<br /> <br /> this theorem is bad<br /> <br /> poggers theorem - awesomeming327<br /> <br /> The Wooga Looga theorem is very OP and not to be frowned upon - Yelly314<br /> <br /> person who invented Wooga Looga theorem is orz orz orz wooga looga theorem OP, citing it on ANY olympiad test = instant full marks -awesomeness_in_a_bun<br /> <br /> Wooga Looga Theorem is TRASH.<br /> <br /> HOW DARE YOU @above and @5above DISRESPECT THE WOOGA LOOGA THEOREM. THIS IS THE MOST OP THEOREM EVER AND CAN BE USED TO SOLVE EVERY PROBLEM. BECAUSE OF THIS THEOREM, I GOT A 187 ON AMC 10A AND 10B, a 665 on the AIME, AND A 420 ON THE USAJMO THIS YEAR. I HAVE ALSO USED THIS TO PROVE THE GOLDBACH CONJECTURE, THE TWIN PRIME CONJECTURE, EVERY SINGLE IMO PROBLEM, AND I HAVE PROVED THE RIEMANN HYPOTHESIS. I ALSO INVENTED HAND SANITIZER THAT KILLS 100% OF BACTERIA, MILK THAT IS 0% MILK, AND A TIME MACHINE WITH THIS THEOREM. THANK YOU SO MUCH @RedFireTruck FOR THIS LEGENDARY THEOREM. I WILL ALWAYS BE INDEBTED TO YOU.</div> Twod horse https://artofproblemsolving.com/wiki/index.php?title=Midpoint&diff=147902 Midpoint 2021-02-25T03:10:48Z <p>Twod horse: /* Medians */</p> <hr /> <div>== Definition ==<br /> In [[Euclidean geometry]], the '''midpoint''' of a [[line segment]] is the [[point]] on the segment equidistant from both endpoints.<br /> <br /> A midpoint [[bisect]]s the line segment that the midpoint lies on. Because of this property, we say that for any line segment &lt;math&gt;\overline{AB}&lt;/math&gt; with midpoint &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;AM=BM=\frac{1}{2}AB&lt;/math&gt;. Alternatively, any point &lt;math&gt;M&lt;/math&gt; on &lt;math&gt;\overline{AB}&lt;/math&gt; such that &lt;math&gt;AM=BM&lt;/math&gt; is the midpoint of the segment.<br /> &lt;asy&gt;<br /> draw((0,0)--(4,0));<br /> dot((0,0));<br /> label(&quot;A&quot;,(0,0),N);<br /> dot((4,0));<br /> label(&quot;B&quot;,(4,0),N);<br /> dot((2,0));<br /> label(&quot;M&quot;,(2,0),N);<br /> label(&quot;Figure 1&quot;,(2,0),4S);<br /> &lt;/asy&gt;<br /> <br /> == Midpoints and Triangles ==<br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,G;<br /> A=(0,0);<br /> B=(4,0);<br /> C=(1,3);<br /> D=(2,0);<br /> E=(2.5,1.5);<br /> F=(0.5,1.5);<br /> G=(5/3,1);<br /> draw(A--B--C--cycle);<br /> draw(D--E--F--cycle,green);<br /> dot(A--B--C--D--E--F--G);<br /> draw(A--E,red);<br /> draw(B--F,red);<br /> draw(C--D,red);<br /> label(&quot;A&quot;,A,S);<br /> label(&quot;B&quot;,B,S);<br /> label(&quot;C&quot;,C,N);<br /> label(&quot;D&quot;,D,S);<br /> label(&quot;E&quot;,E,E);<br /> label(&quot;F&quot;,F,W);<br /> label(&quot;G&quot;,G,NE);<br /> label(&quot;Figure 2&quot;,D,4S);<br /> &lt;/asy&gt;<br /> === Midsegments ===<br /> As shown in Figure 2, &lt;math&gt;\Delta ABC&lt;/math&gt; is a triangle with &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt; midpoints on &lt;math&gt;\overline{AB}&lt;/math&gt;, &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{CA}&lt;/math&gt; respectively. Connect &lt;math&gt;\overline{EF}&lt;/math&gt;, &lt;math&gt;\overline{FD}&lt;/math&gt;, &lt;math&gt;\overline{DE}&lt;/math&gt; (segments highlighted in green). They are midsegments to their corresponding sides. Using SAS Similarity Postulate, we can see that &lt;math&gt;\Delta CFE \sim \Delta CAB&lt;/math&gt; and likewise for &lt;math&gt;\Delta ADF&lt;/math&gt; and &lt;math&gt;\Delta BED&lt;/math&gt;. Because of this, we know that<br /> &lt;cmath&gt;\begin{align*}<br /> AB &amp;= 2FE \\<br /> BC &amp;= 2DE \\<br /> CA &amp;= 2ED \\<br /> \end{align*}&lt;/cmath&gt;<br /> Which is the Triangle Midsegment Theorem. Because we have a relationship between these segment lengths, &lt;math&gt;\Delta ABC \sim \Delta EFD (SSS)&lt;/math&gt; with similar ratio 2:1. The area ratio is then 4:1; this tells us<br /> &lt;cmath&gt;\begin{align*}<br /> [ABC] &amp;= 4[EFD]<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> === Medians ===<br /> The [[median of a triangle]] is defined as one of the three line segments connecting a midpoint to its opposite vertex. As for the case of Figure 2, the medians are &lt;math&gt;\overline{AE}&lt;/math&gt;, &lt;math&gt;\overline{BF}&lt;/math&gt;, and &lt;math&gt;\overline{CD}&lt;/math&gt;, segments highlighted in red.<br /> <br /> These three line segments are [[concurrent]] at point &lt;math&gt;G&lt;/math&gt;, which is otherwise known as the [[centroid]]. This concurrence can be proven through many ways, one of which involves the most simple usage of [[Ceva's Theorem]]. A median is always within its triangle.<br /> <br /> The centroid is one of the points that trisect a median. For a median in any triangle, the ratio of the median's length from vertex to centroid and centroid to the base is always 2:1.<br /> <br /> For right triangles, the median to the hypotenuse always equals to half the length of the hypotenuse.<br /> <br /> For equilateral triangles, its median to one side is the same as the angle bisector and altitude. It can be calculated as &lt;math&gt;\frac{\sqrt3}{2}s&lt;/math&gt;, where &lt;math&gt;s&lt;/math&gt; denotes its side length.<br /> <br /> == Cartesian Plane ==<br /> In the Cartesian Plane, the coordinates of the midpoint &lt;math&gt;M&lt;/math&gt; can be obtained when the two endpoints &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; of the line segment &lt;math&gt;\overline{AB}&lt;/math&gt; is known. Say that &lt;math&gt;A: A(x_A,y_A)&lt;/math&gt; and &lt;math&gt;B: B(x_B,y_B)&lt;/math&gt;. The Midpoint Formula states that the coordinates of &lt;math&gt;M&lt;/math&gt; can be calculated as:<br /> &lt;cmath&gt;\begin{align*}<br /> M(\frac{x_A+x_B}{2}&amp;,\frac{y_A+y_B}{2})<br /> \end{align*}&lt;/cmath&gt;<br /> == See Also ==<br /> * [[Bisect]]<br /> * [[Median of a triangle]]<br /> <br /> {{stub}}<br /> <br /> [[Category:Definition]]<br /> [[Category:Geometry]]</div> Twod horse https://artofproblemsolving.com/wiki/index.php?title=Midpoint&diff=147901 Midpoint 2021-02-25T03:03:20Z <p>Twod horse: /* Midpoints and Triangles */</p> <hr /> <div>== Definition ==<br /> In [[Euclidean geometry]], the '''midpoint''' of a [[line segment]] is the [[point]] on the segment equidistant from both endpoints.<br /> <br /> A midpoint [[bisect]]s the line segment that the midpoint lies on. Because of this property, we say that for any line segment &lt;math&gt;\overline{AB}&lt;/math&gt; with midpoint &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;AM=BM=\frac{1}{2}AB&lt;/math&gt;. Alternatively, any point &lt;math&gt;M&lt;/math&gt; on &lt;math&gt;\overline{AB}&lt;/math&gt; such that &lt;math&gt;AM=BM&lt;/math&gt; is the midpoint of the segment.<br /> &lt;asy&gt;<br /> draw((0,0)--(4,0));<br /> dot((0,0));<br /> label(&quot;A&quot;,(0,0),N);<br /> dot((4,0));<br /> label(&quot;B&quot;,(4,0),N);<br /> dot((2,0));<br /> label(&quot;M&quot;,(2,0),N);<br /> label(&quot;Figure 1&quot;,(2,0),4S);<br /> &lt;/asy&gt;<br /> <br /> == Midpoints and Triangles ==<br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,G;<br /> A=(0,0);<br /> B=(4,0);<br /> C=(1,3);<br /> D=(2,0);<br /> E=(2.5,1.5);<br /> F=(0.5,1.5);<br /> G=(5/3,1);<br /> draw(A--B--C--cycle);<br /> draw(D--E--F--cycle,green);<br /> dot(A--B--C--D--E--F--G);<br /> draw(A--E,red);<br /> draw(B--F,red);<br /> draw(C--D,red);<br /> label(&quot;A&quot;,A,S);<br /> label(&quot;B&quot;,B,S);<br /> label(&quot;C&quot;,C,N);<br /> label(&quot;D&quot;,D,S);<br /> label(&quot;E&quot;,E,E);<br /> label(&quot;F&quot;,F,W);<br /> label(&quot;G&quot;,G,NE);<br /> label(&quot;Figure 2&quot;,D,4S);<br /> &lt;/asy&gt;<br /> === Midsegments ===<br /> As shown in Figure 2, &lt;math&gt;\Delta ABC&lt;/math&gt; is a triangle with &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt; midpoints on &lt;math&gt;\overline{AB}&lt;/math&gt;, &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{CA}&lt;/math&gt; respectively. Connect &lt;math&gt;\overline{EF}&lt;/math&gt;, &lt;math&gt;\overline{FD}&lt;/math&gt;, &lt;math&gt;\overline{DE}&lt;/math&gt; (segments highlighted in green). They are midsegments to their corresponding sides. Using SAS Similarity Postulate, we can see that &lt;math&gt;\Delta CFE \sim \Delta CAB&lt;/math&gt; and likewise for &lt;math&gt;\Delta ADF&lt;/math&gt; and &lt;math&gt;\Delta BED&lt;/math&gt;. Because of this, we know that<br /> &lt;cmath&gt;\begin{align*}<br /> AB &amp;= 2FE \\<br /> BC &amp;= 2DE \\<br /> CA &amp;= 2ED \\<br /> \end{align*}&lt;/cmath&gt;<br /> Which is the Triangle Midsegment Theorem. Because we have a relationship between these segment lengths, &lt;math&gt;\Delta ABC \sim \Delta EFD (SSS)&lt;/math&gt; with similar ratio 2:1. The area ratio is then 4:1; this tells us<br /> &lt;cmath&gt;\begin{align*}<br /> [ABC] &amp;= 4[EFD]<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> === Medians ===<br /> The [[median of a triangle]] is defined as one of the three line segments connecting a midpoint to its opposite vertex. As for the case of Figure 2, the medians are &lt;math&gt;\overline{AE}&lt;/math&gt;, &lt;math&gt;\overline{BF}&lt;/math&gt;, and &lt;math&gt;\overline{CD}&lt;/math&gt;, segments highlighted in red.<br /> <br /> These three line segments are [[concurrent]] at point &lt;math&gt;G&lt;/math&gt;, which is otherwise known as the [[centroid]]. This concurrence can be proven through many ways, one of which involves the most simple usage of [[Ceva's Theorem]] and the properties of a midpoint. A median is always within its triangle.<br /> <br /> The centroid is one of the points that trisect a median. For a median in any triangle, the ratio of the median's length from vertex to centroid and centroid to the base is always 2:1.<br /> <br /> == Cartesian Plane ==<br /> In the Cartesian Plane, the coordinates of the midpoint &lt;math&gt;M&lt;/math&gt; can be obtained when the two endpoints &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; of the line segment &lt;math&gt;\overline{AB}&lt;/math&gt; is known. Say that &lt;math&gt;A: A(x_A,y_A)&lt;/math&gt; and &lt;math&gt;B: B(x_B,y_B)&lt;/math&gt;. The Midpoint Formula states that the coordinates of &lt;math&gt;M&lt;/math&gt; can be calculated as:<br /> &lt;cmath&gt;\begin{align*}<br /> M(\frac{x_A+x_B}{2}&amp;,\frac{y_A+y_B}{2})<br /> \end{align*}&lt;/cmath&gt;<br /> == See Also ==<br /> * [[Bisect]]<br /> * [[Median of a triangle]]<br /> <br /> {{stub}}<br /> <br /> [[Category:Definition]]<br /> [[Category:Geometry]]</div> Twod horse https://artofproblemsolving.com/wiki/index.php?title=Midpoint&diff=147900 Midpoint 2021-02-25T03:02:54Z <p>Twod horse: /* Midpoints and Triangles */</p> <hr /> <div>== Definition ==<br /> In [[Euclidean geometry]], the '''midpoint''' of a [[line segment]] is the [[point]] on the segment equidistant from both endpoints.<br /> <br /> A midpoint [[bisect]]s the line segment that the midpoint lies on. Because of this property, we say that for any line segment &lt;math&gt;\overline{AB}&lt;/math&gt; with midpoint &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;AM=BM=\frac{1}{2}AB&lt;/math&gt;. Alternatively, any point &lt;math&gt;M&lt;/math&gt; on &lt;math&gt;\overline{AB}&lt;/math&gt; such that &lt;math&gt;AM=BM&lt;/math&gt; is the midpoint of the segment.<br /> &lt;asy&gt;<br /> draw((0,0)--(4,0));<br /> dot((0,0));<br /> label(&quot;A&quot;,(0,0),N);<br /> dot((4,0));<br /> label(&quot;B&quot;,(4,0),N);<br /> dot((2,0));<br /> label(&quot;M&quot;,(2,0),N);<br /> label(&quot;Figure 1&quot;,(2,0),4S);<br /> &lt;/asy&gt;<br /> <br /> == Midpoints and Triangles ==<br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,G;<br /> A=(0,0);<br /> B=(4,0);<br /> C=(1,3)<br /> D=(2,0);<br /> E=(2.5,1.5);<br /> F=(0.5,1.5);<br /> G=(5/3,1);<br /> draw(A--B--C--cycle);<br /> draw(D--E--F--cycle,green);<br /> dot(A--B--C--D--E--F--G);<br /> draw(A--E,red);<br /> draw(B--F,red);<br /> draw(C--D,red);<br /> label(&quot;A&quot;,A,S);<br /> label(&quot;B&quot;,B,S);<br /> label(&quot;C&quot;,C,N);<br /> label(&quot;D&quot;,D,S);<br /> label(&quot;E&quot;,E,E);<br /> label(&quot;F&quot;,F,W);<br /> label(&quot;G&quot;,G,NE);<br /> label(&quot;Figure 2&quot;,D,4S);<br /> &lt;/asy&gt;<br /> === Midsegments ===<br /> As shown in Figure 2, &lt;math&gt;\Delta ABC&lt;/math&gt; is a triangle with &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt; midpoints on &lt;math&gt;\overline{AB}&lt;/math&gt;, &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{CA}&lt;/math&gt; respectively. Connect &lt;math&gt;\overline{EF}&lt;/math&gt;, &lt;math&gt;\overline{FD}&lt;/math&gt;, &lt;math&gt;\overline{DE}&lt;/math&gt; (segments highlighted in green). They are midsegments to their corresponding sides. Using SAS Similarity Postulate, we can see that &lt;math&gt;\Delta CFE \sim \Delta CAB&lt;/math&gt; and likewise for &lt;math&gt;\Delta ADF&lt;/math&gt; and &lt;math&gt;\Delta BED&lt;/math&gt;. Because of this, we know that<br /> &lt;cmath&gt;\begin{align*}<br /> AB &amp;= 2FE \\<br /> BC &amp;= 2DE \\<br /> CA &amp;= 2ED \\<br /> \end{align*}&lt;/cmath&gt;<br /> Which is the Triangle Midsegment Theorem. Because we have a relationship between these segment lengths, &lt;math&gt;\Delta ABC \sim \Delta EFD (SSS)&lt;/math&gt; with similar ratio 2:1. The area ratio is then 4:1; this tells us<br /> &lt;cmath&gt;\begin{align*}<br /> [ABC] &amp;= 4[EFD]<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> === Medians ===<br /> The [[median of a triangle]] is defined as one of the three line segments connecting a midpoint to its opposite vertex. As for the case of Figure 2, the medians are &lt;math&gt;\overline{AE}&lt;/math&gt;, &lt;math&gt;\overline{BF}&lt;/math&gt;, and &lt;math&gt;\overline{CD}&lt;/math&gt;, segments highlighted in red.<br /> <br /> These three line segments are [[concurrent]] at point &lt;math&gt;G&lt;/math&gt;, which is otherwise known as the [[centroid]]. This concurrence can be proven through many ways, one of which involves the most simple usage of [[Ceva's Theorem]] and the properties of a midpoint. A median is always within its triangle.<br /> <br /> The centroid is one of the points that trisect a median. For a median in any triangle, the ratio of the median's length from vertex to centroid and centroid to the base is always 2:1.<br /> <br /> == Cartesian Plane ==<br /> In the Cartesian Plane, the coordinates of the midpoint &lt;math&gt;M&lt;/math&gt; can be obtained when the two endpoints &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; of the line segment &lt;math&gt;\overline{AB}&lt;/math&gt; is known. Say that &lt;math&gt;A: A(x_A,y_A)&lt;/math&gt; and &lt;math&gt;B: B(x_B,y_B)&lt;/math&gt;. The Midpoint Formula states that the coordinates of &lt;math&gt;M&lt;/math&gt; can be calculated as:<br /> &lt;cmath&gt;\begin{align*}<br /> M(\frac{x_A+x_B}{2}&amp;,\frac{y_A+y_B}{2})<br /> \end{align*}&lt;/cmath&gt;<br /> == See Also ==<br /> * [[Bisect]]<br /> * [[Median of a triangle]]<br /> <br /> {{stub}}<br /> <br /> [[Category:Definition]]<br /> [[Category:Geometry]]</div> Twod horse https://artofproblemsolving.com/wiki/index.php?title=Midpoint&diff=147899 Midpoint 2021-02-25T03:01:01Z <p>Twod horse: /* Definition */</p> <hr /> <div>== Definition ==<br /> In [[Euclidean geometry]], the '''midpoint''' of a [[line segment]] is the [[point]] on the segment equidistant from both endpoints.<br /> <br /> A midpoint [[bisect]]s the line segment that the midpoint lies on. Because of this property, we say that for any line segment &lt;math&gt;\overline{AB}&lt;/math&gt; with midpoint &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;AM=BM=\frac{1}{2}AB&lt;/math&gt;. Alternatively, any point &lt;math&gt;M&lt;/math&gt; on &lt;math&gt;\overline{AB}&lt;/math&gt; such that &lt;math&gt;AM=BM&lt;/math&gt; is the midpoint of the segment.<br /> &lt;asy&gt;<br /> draw((0,0)--(4,0));<br /> dot((0,0));<br /> label(&quot;A&quot;,(0,0),N);<br /> dot((4,0));<br /> label(&quot;B&quot;,(4,0),N);<br /> dot((2,0));<br /> label(&quot;M&quot;,(2,0),N);<br /> label(&quot;Figure 1&quot;,(2,0),4S);<br /> &lt;/asy&gt;<br /> <br /> == Midpoints and Triangles ==<br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,G;<br /> A=(0,0);<br /> B=(4,0;<br /> C=(1,3)<br /> D=(2,0);<br /> E=(2.5,1.5);<br /> F=(0.5,1.5);<br /> G=(5/3,1);<br /> draw(A--B--C--cycle);<br /> draw(D--E--F--cycle,green);<br /> dot(A--B--C--D--E--F--G);<br /> draw(A--E,red);<br /> draw(B--F,red);<br /> draw(C--D,red);<br /> label(&quot;A&quot;,A,S);<br /> label(&quot;B&quot;,B,S);<br /> label(&quot;C&quot;,C,N);<br /> label(&quot;D&quot;,D,S);<br /> label(&quot;E&quot;,E,E);<br /> label(&quot;F&quot;,F,W);<br /> label(&quot;G&quot;,G,NE);<br /> label(&quot;Figure 2&quot;,D,4S);<br /> &lt;/asy&gt;<br /> === Midsegments ===<br /> As shown in Figure 2, &lt;math&gt;\Delta ABC&lt;/math&gt; is a triangle with &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt; midpoints on &lt;math&gt;\overline{AB}&lt;/math&gt;, &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{CA}&lt;/math&gt; respectively. Connect &lt;math&gt;\overline{EF}&lt;/math&gt;, &lt;math&gt;\overline{FD}&lt;/math&gt;, &lt;math&gt;\overline{DE}&lt;/math&gt; (segments highlighted in green). They are midsegments to their corresponding sides. Using SAS Similarity Postulate, we can see that &lt;math&gt;\Delta CFE \sim \Delta CAB&lt;/math&gt; and likewise for &lt;math&gt;\Delta ADF&lt;/math&gt; and &lt;math&gt;\Delta BED&lt;/math&gt;. Because of this, we know that<br /> &lt;cmath&gt;\begin{align*}<br /> AB &amp;= 2FE \\<br /> BC &amp;= 2DE \\<br /> CA &amp;= 2ED \\<br /> \end{align*}&lt;/cmath&gt;<br /> Which is the Triangle Midsegment Theorem. Because we have a relationship between these segment lengths, &lt;math&gt;\Delta ABC \sim \Delta EFD (SSS)&lt;/math&gt; with similar ratio 2:1. The area ratio is then 4:1; this tells us<br /> &lt;cmath&gt;\begin{align*}<br /> [ABC] &amp;= 4[EFD]<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> === Medians ===<br /> The [[median of a triangle]] is defined as one of the three line segments connecting a midpoint to its opposite vertex. As for the case of Figure 2, the medians are &lt;math&gt;\overline{AE}&lt;/math&gt;, &lt;math&gt;\overline{BF}&lt;/math&gt;, and &lt;math&gt;\overline{CD}&lt;/math&gt;, segments highlighted in red.<br /> <br /> These three line segments are [[concurrent]] at point &lt;math&gt;G&lt;/math&gt;, which is otherwise known as the [[centroid]]. This concurrence can be proven through many ways, one of which involves the most simple usage of [[Ceva's Theorem]] and the properties of a midpoint. A median is always within its triangle.<br /> <br /> The centroid is one of the points that trisect a median. For a median in any triangle, the ratio of the median's length from vertex to centroid and centroid to the base is always 2:1.<br /> <br /> == Cartesian Plane ==<br /> In the Cartesian Plane, the coordinates of the midpoint &lt;math&gt;M&lt;/math&gt; can be obtained when the two endpoints &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; of the line segment &lt;math&gt;\overline{AB}&lt;/math&gt; is known. Say that &lt;math&gt;A: A(x_A,y_A)&lt;/math&gt; and &lt;math&gt;B: B(x_B,y_B)&lt;/math&gt;. The Midpoint Formula states that the coordinates of &lt;math&gt;M&lt;/math&gt; can be calculated as:<br /> &lt;cmath&gt;\begin{align*}<br /> M(\frac{x_A+x_B}{2}&amp;,\frac{y_A+y_B}{2})<br /> \end{align*}&lt;/cmath&gt;<br /> == See Also ==<br /> * [[Bisect]]<br /> * [[Median of a triangle]]<br /> <br /> {{stub}}<br /> <br /> [[Category:Definition]]<br /> [[Category:Geometry]]</div> Twod horse https://artofproblemsolving.com/wiki/index.php?title=Midpoint&diff=147898 Midpoint 2021-02-25T03:00:22Z <p>Twod horse: </p> <hr /> <div>== Definition ==<br /> In '''Euclidean geometry''', the '''midpoint''' of a [[line segment]] is the [[point]] on the segment equidistant from both endpoints.<br /> <br /> A midpoint [[bisect]]s the line segment that the midpoint lies on. Because of this property, we say that for any line segment &lt;math&gt;\overline{AB}&lt;/math&gt; with midpoint &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;AM=BM=\frac{1}{2}AB&lt;/math&gt;. Alternatively, any point &lt;math&gt;M&lt;/math&gt; on &lt;math&gt;\overline{AB}&lt;/math&gt; such that &lt;math&gt;AM=BM&lt;/math&gt; is the midpoint of the segment.<br /> &lt;asy&gt;<br /> draw((0,0)--(4,0));<br /> dot((0,0));<br /> label(&quot;A&quot;,(0,0),N);<br /> dot((4,0));<br /> label(&quot;B&quot;,(4,0),N);<br /> dot((2,0));<br /> label(&quot;M&quot;,(2,0),N);<br /> label(&quot;Figure 1&quot;,(2,0),4S);<br /> &lt;/asy&gt;<br /> == Midpoints and Triangles ==<br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,G;<br /> A=(0,0);<br /> B=(4,0;<br /> C=(1,3)<br /> D=(2,0);<br /> E=(2.5,1.5);<br /> F=(0.5,1.5);<br /> G=(5/3,1);<br /> draw(A--B--C--cycle);<br /> draw(D--E--F--cycle,green);<br /> dot(A--B--C--D--E--F--G);<br /> draw(A--E,red);<br /> draw(B--F,red);<br /> draw(C--D,red);<br /> label(&quot;A&quot;,A,S);<br /> label(&quot;B&quot;,B,S);<br /> label(&quot;C&quot;,C,N);<br /> label(&quot;D&quot;,D,S);<br /> label(&quot;E&quot;,E,E);<br /> label(&quot;F&quot;,F,W);<br /> label(&quot;G&quot;,G,NE);<br /> label(&quot;Figure 2&quot;,D,4S);<br /> &lt;/asy&gt;<br /> === Midsegments ===<br /> As shown in Figure 2, &lt;math&gt;\Delta ABC&lt;/math&gt; is a triangle with &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt; midpoints on &lt;math&gt;\overline{AB}&lt;/math&gt;, &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{CA}&lt;/math&gt; respectively. Connect &lt;math&gt;\overline{EF}&lt;/math&gt;, &lt;math&gt;\overline{FD}&lt;/math&gt;, &lt;math&gt;\overline{DE}&lt;/math&gt; (segments highlighted in green). They are midsegments to their corresponding sides. Using SAS Similarity Postulate, we can see that &lt;math&gt;\Delta CFE \sim \Delta CAB&lt;/math&gt; and likewise for &lt;math&gt;\Delta ADF&lt;/math&gt; and &lt;math&gt;\Delta BED&lt;/math&gt;. Because of this, we know that<br /> &lt;cmath&gt;\begin{align*}<br /> AB &amp;= 2FE \\<br /> BC &amp;= 2DE \\<br /> CA &amp;= 2ED \\<br /> \end{align*}&lt;/cmath&gt;<br /> Which is the Triangle Midsegment Theorem. Because we have a relationship between these segment lengths, &lt;math&gt;\Delta ABC \sim \Delta EFD (SSS)&lt;/math&gt; with similar ratio 2:1. The area ratio is then 4:1; this tells us<br /> &lt;cmath&gt;\begin{align*}<br /> [ABC] &amp;= 4[EFD]<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> === Medians ===<br /> The [[median of a triangle]] is defined as one of the three line segments connecting a midpoint to its opposite vertex. As for the case of Figure 2, the medians are &lt;math&gt;\overline{AE}&lt;/math&gt;, &lt;math&gt;\overline{BF}&lt;/math&gt;, and &lt;math&gt;\overline{CD}&lt;/math&gt;, segments highlighted in red.<br /> <br /> These three line segments are [[concurrent]] at point &lt;math&gt;G&lt;/math&gt;, which is otherwise known as the [[centroid]]. This concurrence can be proven through many ways, one of which involves the most simple usage of [[Ceva's Theorem]] and the properties of a midpoint. A median is always within its triangle.<br /> <br /> The centroid is one of the points that trisect a median. For a median in any triangle, the ratio of the median's length from vertex to centroid and centroid to the base is always 2:1.<br /> <br /> == Cartesian Plane ==<br /> In the Cartesian Plane, the coordinates of the midpoint &lt;math&gt;M&lt;/math&gt; can be obtained when the two endpoints &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; of the line segment &lt;math&gt;\overline{AB}&lt;/math&gt; is known. Say that &lt;math&gt;A: A(x_A,y_A)&lt;/math&gt; and &lt;math&gt;B: B(x_B,y_B)&lt;/math&gt;. The Midpoint Formula states that the coordinates of &lt;math&gt;M&lt;/math&gt; can be calculated as:<br /> &lt;cmath&gt;\begin{align*}<br /> M(\frac{x_A+x_B}{2}&amp;,\frac{y_A+y_B}{2})<br /> \end{align*}&lt;/cmath&gt;<br /> == See Also ==<br /> * [[Bisect]]<br /> * [[Median of a triangle]]<br /> <br /> {{stub}}<br /> <br /> [[Category:Definition]]<br /> [[Category:Geometry]]</div> Twod horse https://artofproblemsolving.com/wiki/index.php?title=Midpoint&diff=147065 Midpoint 2021-02-16T03:41:50Z <p>Twod horse: </p> <hr /> <div>== Definition ==<br /> The '''midpoint''' of a [[line segment]] is the [[point]] on the segment equidistant from both endpoints.<br /> <br /> A midpoint [[bisect]]s the line segment that the midpoint lies on. Because of this property, we say that for any line segment &lt;math&gt;\overline{AB}&lt;/math&gt; with midpoint &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;AM=BM=\frac{1}{2}AB&lt;/math&gt;. Alternatively, any point &lt;math&gt;M&lt;/math&gt; on &lt;math&gt;\overline{AB}&lt;/math&gt; such that &lt;math&gt;AM=BM&lt;/math&gt; is the midpoint of the segment.<br /> &lt;asy&gt;<br /> draw((0,0)--(4,0));<br /> dot((0,0));<br /> label(&quot;A&quot;,(0,0),N);<br /> dot((4,0));<br /> label(&quot;B&quot;,(4,0),N);<br /> dot((2,0));<br /> label(&quot;M&quot;,(2,0),N);<br /> label(&quot;Figure 1&quot;,(2,0),4S);<br /> &lt;/asy&gt;<br /> == Midpoints and Triangles ==<br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,G;<br /> A=(0,0);<br /> B=(4,0);<br /> C=(1,3);<br /> D=(2,0);<br /> E=(2.5,1.5);<br /> F=(0.5,1.5);<br /> G=(5/3,1);<br /> draw(A--B--C--cycle);<br /> draw(D--E--F--cycle,green);<br /> dot(A--B--C--D--E--F--G);<br /> draw(A--E,red);<br /> draw(B--F,red);<br /> draw(C--D,red);<br /> label(&quot;A&quot;,A,S);<br /> label(&quot;B&quot;,B,S);<br /> label(&quot;C&quot;,C,N);<br /> label(&quot;D&quot;,D,S);<br /> label(&quot;E&quot;,E,E);<br /> label(&quot;F&quot;,F,W);<br /> label(&quot;G&quot;,G,NE);<br /> label(&quot;Figure 2&quot;,D,4S);<br /> &lt;/asy&gt;<br /> === Midsegments ===<br /> As shown in Figure 2, &lt;math&gt;\Delta ABC&lt;/math&gt; is a triangle with &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt; midpoints on &lt;math&gt;\overline{AB}&lt;/math&gt;, &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{CA}&lt;/math&gt; respectively. Connect &lt;math&gt;\overline{EF}&lt;/math&gt;, &lt;math&gt;\overline{FD}&lt;/math&gt;, &lt;math&gt;\overline{DE}&lt;/math&gt; (segments highlighted in green). They are midsegments to their corresponding sides. Using SAS Similarity Postulate, we can see that &lt;math&gt;\Delta CFE \sim \Delta CAB&lt;/math&gt; and likewise for &lt;math&gt;\Delta ADF&lt;/math&gt; and &lt;math&gt;\Delta BED&lt;/math&gt;. Because of this, we know that<br /> &lt;cmath&gt;\begin{align*}<br /> AB &amp;= 2FE \\<br /> BC &amp;= 2DE \\<br /> CA &amp;= 2ED \\<br /> \end{align*}&lt;/cmath&gt;<br /> Which is the Triangle Midsegment Theorem. Because we have a relationship between these segment lengths, &lt;math&gt;\Delta ABC \sim \Delta EFD (SSS)&lt;/math&gt; with similar ratio 2:1. The area ratio is then 4:1; this tells us<br /> &lt;cmath&gt;\begin{align*}<br /> [ABC] &amp;= 4[EFD]<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> === Medians ===<br /> The [[median of a triangle]] is defined as one of the three line segments connecting a midpoint to its opposite vertex. As for the case of Figure 2, the medians are &lt;math&gt;\overline{AE}&lt;/math&gt;, &lt;math&gt;\overline{BF}&lt;/math&gt;, and &lt;math&gt;\overline{CD}&lt;/math&gt;, segments highlighted in red.<br /> <br /> These three line segments are [[concurrent]] at point &lt;math&gt;G&lt;/math&gt;, which is otherwise known as the [[centroid]]. This concurrence can be proven through many ways, one of which involves the most simple usage of [[Ceva's Theorem]] and the properties of a midpoint. A median is always within its triangle.<br /> <br /> The centroid is one of the points that trisect a median. For a median in any triangle, the ratio of the median's length from vertex to centroid and centroid to the base is always 2:1.<br /> <br /> == Cartesian Plane ==<br /> In the Cartesian Plane, the coordinates of the midpoint &lt;math&gt;M&lt;/math&gt; can be obtained when the two endpoints &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; of the line segment &lt;math&gt;\overline{AB}&lt;/math&gt; is known. Say that &lt;math&gt;A: A(x_A,y_A)&lt;/math&gt; and &lt;math&gt;B: B(x_B,y_B)&lt;/math&gt;. The Midpoint Formula states that the coordinates of &lt;math&gt;M&lt;/math&gt; can be calculated as:<br /> &lt;cmath&gt;\begin{align*}<br /> M(\frac{x_A+x_B}{2}&amp;,\frac{y_A+y_B}{2})<br /> \end{align*}&lt;/cmath&gt;<br /> == See Also ==<br /> * [[Bisect]]<br /> * [[Median of a triangle]]<br /> <br /> {{stub}}<br /> <br /> [[Category:Definition]]<br /> [[Category:Geometry]]</div> Twod horse https://artofproblemsolving.com/wiki/index.php?title=Midpoint&diff=146420 Midpoint 2021-02-13T15:07:57Z <p>Twod horse: </p> <hr /> <div>== Definition ==<br /> The '''midpoint''' of a [[line segment]] is the [[point]] on the segment equidistant from both endpoints.<br /> <br /> A midpoint [[bisect]]s the line segment that the midpoint lies on. Because of this property, we say that for any line segment &lt;math&gt;\overline{AB}&lt;/math&gt; with midpoint &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;AM=BM=\frac{1}{2}AB&lt;/math&gt;. Alternatively, any point &lt;math&gt;M&lt;/math&gt; on &lt;math&gt;\overline{AB}&lt;/math&gt; such that &lt;math&gt;AM=BM&lt;/math&gt; is the midpoint of the segment.<br /> &lt;asy&gt;<br /> draw((0,0)--(4,0));<br /> dot((0,0));<br /> label(&quot;A&quot;,(0,0),N);<br /> dot((4,0));<br /> label(&quot;B&quot;,(4,0),N);<br /> dot((2,0));<br /> label(&quot;M&quot;,(2,0),N);<br /> label(&quot;Figure 1&quot;,(2,0),4S);<br /> &lt;/asy&gt;<br /> == Midpoints and Triangles ==<br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,G;<br /> A=(0,0);<br /> B=(4,0);<br /> C=(1,3);<br /> D=(2,0);<br /> E=(2.5,1.5);<br /> F=(0.5,1.5);<br /> G=(5/3,1);<br /> draw(A--B--C--cycle);<br /> draw(D--E--F--cycle,green);<br /> dot(A--B--C--D--E--F--G);<br /> draw(A--E,red);<br /> draw(B--F,red);<br /> draw(C--D,red);<br /> label(&quot;A&quot;,A,S);<br /> label(&quot;B&quot;,B,S);<br /> label(&quot;C&quot;,C,N);<br /> label(&quot;D&quot;,D,S);<br /> label(&quot;E&quot;,E,E);<br /> label(&quot;F&quot;,F,W);<br /> label(&quot;G&quot;,G,NE);<br /> label(&quot;Figure 2&quot;,D,4S);<br /> &lt;/asy&gt;<br /> === Midsegments ===<br /> As shown in Figure 2, &lt;math&gt;\Delta ABC&lt;/math&gt; is a triangle with &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt; midpoints on &lt;math&gt;\overline{AB}&lt;/math&gt;, &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{CA}&lt;/math&gt; respectively. Connect &lt;math&gt;\overline{EF}&lt;/math&gt;, &lt;math&gt;\overline{FD}&lt;/math&gt;, &lt;math&gt;\overline{DE}&lt;/math&gt; (segments highlighted in green). They are midsegments to their corresponding sides. Using SAS Similarity Postulate, we can see that &lt;math&gt;\Delta CFE \sim \Delta CAB&lt;/math&gt; and likewise for &lt;math&gt;\Delta ADF&lt;/math&gt; and &lt;math&gt;\Delta BED&lt;/math&gt;. Because of this, we know that<br /> &lt;cmath&gt;\begin{align*}<br /> AB &amp;= 2FE \\<br /> BC &amp;= 2DE \\<br /> CA &amp;= 2ED \\<br /> \end{align*}&lt;/cmath&gt;<br /> Which is the Triangle Midsegment Theorem. Because we have a relationship between these segment lengths, &lt;math&gt;\Delta ABC \sim \Delta EFD (SSS)&lt;/math&gt; with similar ratio 2:1. The area ratio is then 4:1; this tells us<br /> &lt;cmath&gt;\begin{align*}<br /> [ABC] &amp;= 4[EFD]<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> === Medians ===<br /> The [[median of a triangle]] is defined as one of the three line segments connecting a midpoint to its opposite vertex. As for the case of Figure 2, the medians are &lt;math&gt;\overline{AE}&lt;/math&gt;, &lt;math&gt;\overline{BF}&lt;/math&gt;, and &lt;math&gt;\overline{CD}&lt;/math&gt;, segments highlighted in red.<br /> <br /> These three line segments are [[concurrent]] at point &lt;math&gt;G&lt;/math&gt;, which is otherwise known as the [[centroid]]. This concurrence can be proven through many ways, one of which involves the most simple usage of [[Ceva's Theorem]] and the properties of a midpoint.<br /> <br /> == Cartesian Plane ==<br /> In the Cartesian Plane, the coordinates of the midpoint &lt;math&gt;M&lt;/math&gt; can be obtained when the two endpoints &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; of the line segment &lt;math&gt;\overline{AB}&lt;/math&gt; is known. Say that &lt;math&gt;A: A(x_A,y_A)&lt;/math&gt; and &lt;math&gt;B: B(x_B,y_B)&lt;/math&gt;. The Midpoint Formula states that the coordinates of &lt;math&gt;M&lt;/math&gt; can be calculated as:<br /> &lt;cmath&gt;\begin{align*}<br /> M(\frac{x_A+x_B}{2}&amp;,\frac{y_A+y_B}{2})<br /> \end{align*}&lt;/cmath&gt;<br /> == See Also ==<br /> * [[Bisect]]<br /> * [[Median of a triangle]]<br /> <br /> {{stub}}<br /> <br /> [[Category:Definition]]<br /> [[Category:Geometry]]</div> Twod horse https://artofproblemsolving.com/wiki/index.php?title=Midpoint&diff=145960 Midpoint 2021-02-12T15:02:28Z <p>Twod horse: /* Midsegments */</p> <hr /> <div>== Definition ==<br /> The '''midpoint''' of a [[line segment]] is the [[point]] on the segment equidistant from both endpoints.<br /> <br /> A midpoint [[bisect]]s the line segment that the midpoint lies on. Because of this property, we say that for any line segment &lt;math&gt;\overline{AB}&lt;/math&gt; with midpoint &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;AM=BM=\frac{1}{2}AB&lt;/math&gt;. Alternatively, any point &lt;math&gt;M&lt;/math&gt; on &lt;math&gt;\overline{AB}&lt;/math&gt; such that &lt;math&gt;AM=BM&lt;/math&gt; is the midpoint of the segment.<br /> &lt;asy&gt;<br /> draw((0,0)--(4,0));<br /> dot((0,0));<br /> label(&quot;A&quot;,(0,0),N);<br /> dot((4,0));<br /> label(&quot;B&quot;,(4,0),N);<br /> dot((2,0));<br /> label(&quot;M&quot;,(2,0),N);<br /> label(&quot;Figure 1&quot;,(2,0),4S);<br /> &lt;/asy&gt;<br /> == Midpoints and Triangles ==<br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,G;<br /> A=(0,0);<br /> B=(4,0);<br /> C=(1,3);<br /> D=(2,0);<br /> E=(2.5,1.5);<br /> F=(0.5,1.5);<br /> G=(5/3,1);<br /> draw(A--B--C--cycle);<br /> draw(D--E--F--cycle,green);<br /> dot(A--B--C--D--E--F--G);<br /> draw(A--E,red);<br /> draw(B--F,red);<br /> draw(C--D,red);<br /> label(&quot;A&quot;,A,S);<br /> label(&quot;B&quot;,B,S);<br /> label(&quot;C&quot;,C,N);<br /> label(&quot;D&quot;,D,S);<br /> label(&quot;E&quot;,E,E);<br /> label(&quot;F&quot;,F,W);<br /> label(&quot;G&quot;,G,NE);<br /> label(&quot;Figure 2&quot;,D,4S);<br /> &lt;/asy&gt;<br /> === Midsegments ===<br /> As shown in Figure 2, &lt;math&gt;\Delta ABC&lt;/math&gt; is a triangle with &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt; midpoints on &lt;math&gt;\overline{AB}&lt;/math&gt;, &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{CA}&lt;/math&gt; respectively. Connect &lt;math&gt;\overline{EF}&lt;/math&gt;, &lt;math&gt;\overline{FD}&lt;/math&gt;, &lt;math&gt;\overline{DE}&lt;/math&gt; (segments highlighted in green). They are midsegments to their corresponding sides. Using SAS Similarity Postulate, we can see that &lt;math&gt;\Delta CFE \sim \Delta CAB&lt;/math&gt; and likewise for &lt;math&gt;\Delta ADF&lt;/math&gt; and &lt;math&gt;\Delta BED&lt;/math&gt;. Because of this, we know that<br /> &lt;cmath&gt;\begin{align*}<br /> AB &amp;= 2FE \\<br /> BC &amp;= 2DE \\<br /> CA &amp;= 2ED \\<br /> \end{align*}&lt;/cmath&gt;<br /> Which is the Triangle Midsegment Theorem. Because we have a relationship between these segment lengths, &lt;math&gt;\Delta ABC \sim \Delta EFD (SSS)&lt;/math&gt; with similar ratio 2:1. The area ratio is then 4:1; this tells us<br /> &lt;cmath&gt;\begin{align*}<br /> [ABC] &amp;= 4[EFD]<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> == Cartesian Plane ==<br /> In the Cartesian Plane, the coordinates of the midpoint &lt;math&gt;M&lt;/math&gt; can be obtained when the two endpoints &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; of the line segment &lt;math&gt;\overline{AB}&lt;/math&gt; is known. Say that &lt;math&gt;A: A(x_A,y_A)&lt;/math&gt; and &lt;math&gt;B: B(x_B,y_B)&lt;/math&gt;. The Midpoint Formula states that the coordinates of &lt;math&gt;M&lt;/math&gt; can be calculated as:<br /> &lt;cmath&gt;\begin{align*}<br /> M(\frac{x_A+x_B}{2}&amp;,\frac{y_A+y_B}{2})<br /> \end{align*}&lt;/cmath&gt;<br /> == See Also ==<br /> * [[Bisect]]<br /> <br /> {{stub}}<br /> <br /> [[Category:Definition]]<br /> [[Category:Geometry]]</div> Twod horse https://artofproblemsolving.com/wiki/index.php?title=Multinomial_Theorem&diff=145892 Multinomial Theorem 2021-02-12T07:59:41Z <p>Twod horse: </p> <hr /> <div>The '''Multinomial Theorem''' states that<br /> &lt;cmath&gt;<br /> (a_1+a_2+\cdots+a_k)^n=\sum_{\substack{j_1,j_2,\ldots,j_k \\ 0 \leq j_i \leq n \textrm{ for each } i \\<br /> \textrm{and } j_1 + \ldots + j_k = n}}\binom{n}{j_1; j_2; \ldots ; j_k}a_1^{j_1}a_2^{j_2}\cdots a_k^{j_k}<br /> &lt;/cmath&gt;<br /> where &lt;math&gt;\binom{n}{j_1; j_2; \ldots ; j_k}&lt;/math&gt; is the [[multinomial coefficient]] &lt;math&gt;\binom{n}{j_1; j_2; \ldots ; j_k}=\dfrac{n!}{j_1!\cdot j_2!\cdots j_k!}&lt;/math&gt;.<br /> <br /> Note that this is a direct generalization of the [[Binomial Theorem]]: when &lt;math&gt;k = 2&lt;/math&gt; it simplifies to<br /> &lt;cmath&gt;<br /> (a_1 + a_2)^n = \sum_{\substack{0\leq j_1, j_2 \leq n \\ j_1 + j_2 = n}} \binom{n}{j_1; j_2} a_1^{j_1}a_2^{j_2} = \sum_{j = 0}^n \binom{n}{j} a_1^j a_2^{n - j}<br /> &lt;/cmath&gt;<br /> <br /> == Proof ==<br /> ===Proof by Induction===<br /> Proving the Multinomial Theorem by Induction<br /> <br /> For a positive integer &lt;math&gt;k&lt;/math&gt; and a non-negative integer &lt;math&gt;n&lt;/math&gt;,<br /> &lt;cmath&gt;(x_1 + x_2 + x_3 + ... + x_{k-1} + x_k)^n = \sum_{b_1 + b_2 + b_3 + ... + b_{k-1} + b_k= n}{\binom{n}{b_1, b_2, b_3, ..., b_{k-1}, b_k} \prod_{j=1}^{k}{x_j^{b_j}}}&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\bf Proof:&lt;/math&gt;<br /> When &lt;math&gt;k=1&lt;/math&gt; the result is true, and when &lt;math&gt;k=2&lt;/math&gt; the result is the binomial theorem. Assume that &lt;math&gt;k \ge 3&lt;/math&gt; and that the result is true for &lt;math&gt;k=p&lt;/math&gt; When &lt;math&gt;k=p+1&lt;/math&gt;<br /> &lt;cmath&gt;(x_1 + x_2 + x_3 + ... + x_{p-1} + x_p)^n = (x_1 + x_2 + x_3 + ... + x_{p-1} + (x_p +x_{p+1})^n&lt;/cmath&gt;<br /> Treating &lt;math&gt;x_p + x_{p+1}&lt;/math&gt; as a single term and using the induction hypothesis:<br /> &lt;cmath&gt;\sum_{b_1 + b_2 + b_3 + ... + b_{p-1} + B = n}{\binom{n}{b_1, b_2, b_3, ..., b_{p-1}, B} \cdot (x_p + x_{p+1})^B \cdot \prod_{j=1}^{p-1}{x_j^{b_j}}}&lt;/cmath&gt;<br /> By the Binomial Theorem, this becomes:<br /> &lt;cmath&gt;\sum_{b_1 + b_2 + b_3 + ... + b_{p-1} + B = n}{\binom{n}{b_1, b_2, b_3, ..., b_{p-1}, B}} (\prod_{j=1}^{p-1}{x_j^{b_j}}) \cdot \sum_{b_p + b_{p+1} = B}{\binom{B}{b_p} \cdot x_p^{b_p} x_{p+1}^{b_{p+1}}}&lt;/cmath&gt;<br /> Since &lt;math&gt;\binom{n}{b_1, b_2, b_3, ... ,b_p, B}\binom{B}{b_p} = \binom{n}{b_1, b_2, b_3, ... ,b_p, b_{p+1}}&lt;/math&gt;, this can be rewritten as:<br /> &lt;cmath&gt;\sum_{b_1 + b_2 + ... b_p + b_{p+1}= n}{\binom{n}{b_1, b_2, b_3, ..., b_p, b_{p+1}}\prod_{j=1}^{k}{x_j^{b_j}}}&lt;/cmath&gt;<br /> <br /> === Combinatorial proof ===<br /> {{stub}}<br /> <br /> ==Problems==<br /> ===Intermediate===<br /> *The [[expression]]<br /> <br /> &lt;math&gt;(x+y+z)^{2006}+(x-y-z)^{2006}&lt;/math&gt;<br /> <br /> is simplified by expanding it and combining like terms. How many terms are in the simplified expression?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 6018\qquad \mathrm{(B) \ } 671,676\qquad \mathrm{(C) \ } 1,007,514\qquad \mathrm{(D) \ } 1,008,016\qquad\mathrm{(E) \ } 2,015,028&lt;/math&gt;<br /> <br /> (Source: [[2006_AMC_12A_Problems/Problem_24|2006 AMC 12A Problem 24]])<br /> <br /> ===Olympiad===<br /> [[Category:Theorems]]<br /> [[Category:Combinatorics]]<br /> [[Category:Multinomial Theorem]]</div> Twod horse https://artofproblemsolving.com/wiki/index.php?title=Bretschneider%27s_formula&diff=145879 Bretschneider's formula 2021-02-12T07:51:18Z <p>Twod horse: </p> <hr /> <div>Suppose we have a [[quadrilateral]] with [[edge]]s of length &lt;math&gt;a,b,c,d&lt;/math&gt; (in that order) and [[diagonal]]s of length &lt;math&gt;p, q&lt;/math&gt;. '''Bretschneider's formula''' states that the [[area]]<br /> &lt;math&gt;[ABCD]=\frac{1}{4} \cdot \sqrt{4p^2q^2-(b^2+d^2-a^2-c^2)^2}&lt;/math&gt;.<br /> <br /> It can be derived with [[vector]] [[geometry]].<br /> <br /> ==Proof==<br /> <br /> Suppose a quadrilateral has sides &lt;math&gt;\vec{a}, \vec{b}, \vec{c}, \vec{d}&lt;/math&gt; such that &lt;math&gt;\vec{a} + \vec{b} + \vec{c} + \vec{d} = \vec{0}&lt;/math&gt; and that the diagonals of the quadrilateral are &lt;math&gt;\vec{p} = \vec{b} + \vec{c} = -\vec{a} - \vec{d}&lt;/math&gt; and &lt;math&gt;\vec{q} = \vec{a} + \vec{b} = -\vec{c} - \vec{d}&lt;/math&gt;. The area of any such quadrilateral is &lt;math&gt;\frac{1}{2} |\vec{p} \times \vec{q}|&lt;/math&gt;.<br /> <br /> <br /> &lt;math&gt;K = \frac{1}{2} |\vec{p} \times \vec{q}| &lt;/math&gt;<br /> <br /> [[Lagrange's Identity]] states that &lt;math&gt;|\vec{a}|^2|\vec{b}|^2-(\vec{a}\cdot\vec{b})^2=|\vec{a}\times\vec{b}|^2 \implies \sqrt{|\vec{a}|^2|\vec{b}|^2-(\vec{a}\cdot\vec{b})^2}=|\vec{a}\times\vec{b}|&lt;/math&gt;. Therefore:<br /> <br /> &lt;math&gt;K = \frac{1}{2} \sqrt{|\vec{p}|^2|\vec{q}|^2 - (\vec{p} \cdot \vec{q})^2} &lt;/math&gt;<br /> <br /> &lt;math&gt;= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - (2 \vec{p} \cdot \vec{q})^2} &lt;/math&gt;<br /> <br /> &lt;math&gt;= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [2 (\vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b})]^2} &lt;/math&gt;<br /> <br /> &lt;math&gt;= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [2 \vec{b} \cdot (\vec{a} + \vec{b}) + 2 \vec{c} \cdot (\vec{a} + \vec{b})]^2} &lt;/math&gt;<br /> <br /> &lt;math&gt;= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [-2 \vec{b} \cdot (\vec{c} + \vec{d}) + 2 \vec{c} \cdot (\vec{a} + \vec{b})]^2} &lt;/math&gt;<br /> <br /> &lt;math&gt;= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [-2 \vec{b} \cdot \vec{c} - 2 \vec{b} \cdot \vec{d} + 2 \vec{a} \cdot \vec{c} + 2 \vec{b} \cdot \vec{c}]^2}&lt;/math&gt;<br /> <br /> &lt;math&gt;= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [2 \vec{a} \cdot \vec{c} - 2 \vec{b} \cdot \vec{d}]^2} &lt;/math&gt;<br /> <br /> &lt;math&gt;= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - ([(\vec{a} + \vec{c})\cdot(\vec{a} + \vec{c}) - \vec{a}\cdot\vec{a} - \vec{c}\cdot\vec{c}] - [(\vec{b} + \vec{d})\cdot(\vec{b} + \vec{d}) - \vec{b}\cdot\vec{b} - \vec{d}\cdot\vec{d}])^2} &lt;/math&gt;<br /> <br /> &lt;math&gt;= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [|\vec{b}|^2 + |\vec{d}|^2 - |\vec{a}|^2 - |\vec{c}|^2 + (\vec{a} + \vec{c})\cdot(\vec{a} + \vec{c}) - (\vec{b} + \vec{d})\cdot(\vec{b} + \vec{d})]^2} &lt;/math&gt;<br /> <br /> &lt;math&gt;= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [|\vec{b}|^2 + |\vec{d}|^2 - |\vec{a}|^2 - |\vec{c}|^2 + |\vec{a} + \vec{c}|^2 - |\vec{b} + \vec{d}|^2]^2} &lt;/math&gt;<br /> <br /> &lt;math&gt;= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [|\vec{b}|^2 + |\vec{d}|^2 - |\vec{a}|^2 - |\vec{c}|^2 + |\vec{a} + \vec{c}|^2 - |-(\vec{a} + \vec{c})|^2]^2} &lt;/math&gt;<br /> <br /> &lt;math&gt;= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [|\vec{b}|^2 + |\vec{d}|^2 - |\vec{a}|^2 - |\vec{c}|^2]^2} &lt;/math&gt;<br /> <br /> Then if &lt;math&gt;a, b, c, d&lt;/math&gt; represent &lt;math&gt;|\vec{a}|, |\vec{b}|, |\vec{c}|, |\vec{d}|&lt;/math&gt; (and are thus the side lengths) while &lt;math&gt;p, q&lt;/math&gt; represent &lt;math&gt;|\vec{p}|, |\vec{q}|&lt;/math&gt; (and are thus the diagonal lengths), the area of a quadrilateral is:<br /> <br /> &lt;cmath&gt; K = \frac{1}{4} \sqrt{4p^2q^2 - (b^2 + d^2 - a^2 - c^2)^2} &lt;/cmath&gt;<br /> <br /> <br /> ==See Also==<br /> * [[Brahmagupta's formula]]<br /> * [[Geometry]]<br /> <br /> [[Category:Geometry]]<br /> [[Category:Theorems]]</div> Twod horse https://artofproblemsolving.com/wiki/index.php?title=Cohn%27s_criterion&diff=145876 Cohn's criterion 2021-02-12T07:48:06Z <p>Twod horse: </p> <hr /> <div>Let &lt;math&gt;p&lt;/math&gt; be a prime number, and &lt;math&gt;b\geq 2&lt;/math&gt; an integer. If &lt;math&gt;\overline{p_np_{n-1}\cdots p_1p_0}&lt;/math&gt; is the base-&lt;math&gt;b&lt;/math&gt; representation of &lt;math&gt;p&lt;/math&gt;, and &lt;math&gt;0\leq p_i&lt;b&lt;/math&gt;, then<br /> &lt;cmath&gt;f(x)=p_nx^n+p_{n-1}x^{n-1}+\cdots+p_1x+p_0&lt;/cmath&gt;<br /> is irreducible.<br /> <br /> ==Proof==<br /> The following proof is due to M. Ram Murty.<br /> <br /> We start off with a lemma. Let &lt;math&gt;g(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\in \mathbb{Z}[x]&lt;/math&gt;. Suppose &lt;math&gt;a_n\geq 1&lt;/math&gt;, &lt;math&gt;a-{n-1}\geq 0&lt;/math&gt;, and &lt;math&gt;|a_i|\leq H&lt;/math&gt;. Then, any complex root of &lt;math&gt;f(x)&lt;/math&gt;, &lt;math&gt;\phi&lt;/math&gt;, has a non positive real part or satisfies &lt;math&gt;|\phi|&lt;\frac{1+\sqrt{1+4H}}{2}&lt;/math&gt;.<br /> <br /> Proof: If &lt;math&gt;|z|&gt;1&lt;/math&gt; and Re &lt;math&gt;z&gt;0&lt;/math&gt;, note that:<br /> &lt;cmath&gt;|\frac{g(z)}{z^n}|\geq |a_n+\frac{a_{n-1}}{z}|-H(\frac{1}{|z|^2}+\cdots+\frac{1}{|z|^n})&gt;Re(a_n+\frac{a_{n-1}}{z})-\frac{H}{|z|^2-|z|}\geq 1-\frac{H}{|z|^2-|z|}=\frac{|z|^2-|z|-H}{|z|^2-|z|}&lt;/cmath&gt;<br /> This means &lt;math&gt;g(z)&gt;0&lt;/math&gt; if &lt;math&gt;|z|\geq \frac{1+\sqrt{1+4H}}{2}&lt;/math&gt;, so &lt;math&gt;|\phi|&lt;\frac{1+\sqrt{1+4H}}{2}&lt;/math&gt;.<br /> <br /> If &lt;math&gt;b\geq 3&lt;/math&gt;, this implies &lt;math&gt;|b-\phi|&gt;1&lt;/math&gt; if &lt;math&gt;b\geq 3&lt;/math&gt; and &lt;math&gt;f(\phi)=0&lt;/math&gt;. Let &lt;math&gt;f(x)=g(x)h(x)&lt;/math&gt;. Since &lt;math&gt;f(b)=p&lt;/math&gt;, one of &lt;math&gt;|g(b)|&lt;/math&gt; and &lt;math&gt;h(b)&lt;/math&gt; is 1. WLOG, assume &lt;math&gt;g(b)=1&lt;/math&gt;. Let &lt;math&gt;\phi_1, phi_2,\cdots,\phi_r&lt;/math&gt; be the roots of &lt;math&gt;g(x)&lt;/math&gt;. This means that &lt;math&gt;|g(b)|=|b-\phi_1||b-\phi_2|\cdots|b-\phi_r|&gt;1&lt;/math&gt;. Therefore, &lt;math&gt;f(x)&lt;/math&gt; is irreducible.<br /> <br /> If &lt;math&gt;b=2&lt;/math&gt;, we will need to prove another lemma:<br /> <br /> All of the zeroes of &lt;math&gt;f(x)&lt;/math&gt; satisfy Re &lt;math&gt;z&gt;\frac{3}{2}&lt;/math&gt;.<br /> <br /> Proof: If &lt;math&gt;n=1&lt;/math&gt;, then the two polynomials are &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;x\pm 1&lt;/math&gt;, both of which satisfy our constraint. For &lt;math&gt;n=2&lt;/math&gt;, we get the polynomials &lt;math&gt;x^2&lt;/math&gt;, &lt;math&gt;x^2\pm x&lt;/math&gt;, &lt;math&gt;x^2\pm 1&lt;/math&gt;, and &lt;math&gt;x^2\pm x\pm 1&lt;/math&gt;, all of which satisfy the constraint. If &lt;math&gt;n\geq 3&lt;/math&gt;, <br /> &lt;cmath&gt;|\frac{f(z)}{z^n}|\geq |1+\frac{a_{n-1}}{z}+\frac{a_{n-2}}{z^2}|-(\frac{1}{|z|^3}+\cdots+\frac{1}{|z|^n})&gt;|1+\frac{a_{n-1}}{z}+\frac{a_{n-2}}{z^2}|-\frac{1}{|z|^2(|z|-1)}&lt;/cmath&gt;<br /> <br /> If Re &lt;math&gt;z\geq 0&lt;/math&gt;, we have Re &lt;math&gt;\frac{1}{z^2}\geq 0&lt;/math&gt;, and then<br /> &lt;cmath&gt;|\frac{f(z)}{z^n}|&gt;1-\frac{1}{|z|^2(|z|-1)}&lt;/cmath&gt;<br /> For &lt;math&gt;|z|\geq \frac{3}{2}&lt;/math&gt;, then &lt;math&gt;|z|^2(|z|-1)\geq(\frac{3}{2})^2(\frac{1}{2})=\frac{9}{8}&gt;1&lt;/math&gt;. Therefore, &lt;math&gt;z&lt;/math&gt; is not a root of &lt;math&gt;f(x)&lt;/math&gt;.<br /> <br /> However, if Re &lt;math&gt;z&lt;0&lt;/math&gt;, we have from our first lemma, that &lt;math&gt;|z|&lt;\frac{1+\sqrt{5}}{2}&lt;/math&gt;, so Re &lt;math&gt;z&lt;\frac{1+\sqrt{5}}{2\sqrt{2}}&lt;\frac{3}{2}&lt;/math&gt;. Thus we have proved the lemma.<br /> <br /> To finish the proof, let &lt;math&gt;f(x)=g(x)h(x)&lt;/math&gt;. Since &lt;math&gt;f(2)=p&lt;/math&gt;, one of &lt;math&gt;g(2)&lt;/math&gt; and &lt;math&gt;h(2)&lt;/math&gt; is 1. WLOG, assume &lt;math&gt;g(2)=1&lt;/math&gt;. By our lemma, &lt;math&gt;|z-2|&gt;|z-1|&lt;/math&gt;. Thus, if &lt;math&gt;\phi_1, \phi_2,\cdots,\phi_r&lt;/math&gt; are the roots of &lt;math&gt;g(x)&lt;/math&gt;, then &lt;math&gt;|g(2)|=|2-\phi_1||2-\phi_2|\cdots|2-\phi_n|&gt;|1-\phi_1||1-\phi_2|\cdots|1-\phi_n|=|g(1)|\leq 1&lt;/math&gt;. This is a contradiction, so &lt;math&gt;f(x)&lt;/math&gt; is irreducible.<br /> <br /> <br /> [[Category:Algebra]]</div> Twod horse https://artofproblemsolving.com/wiki/index.php?title=Absolute_zero&diff=145869 Absolute zero 2021-02-12T07:44:12Z <p>Twod horse: </p> <hr /> <div>Absolute zero is the theoretical point in [[temperature]] where all kinetic motion of an atom at an atomic scale stops. As said, no object in the observable universe exhibits such temperature; although scientists are capable of cooling the temperature to a very close approximation to 0 Kelvin, no existing technology allows any object to arrive at absolute zero.<br /> <br /> Because of its unique property, the absolute zero is defined as 0 for the [[Kelvin]] scale -- the base SI unit for temperature.<br /> <br /> <br /> == At absolute zero: ==<br /> <br /> All gases have the same pressure at absolute zero.<br /> <br /> -273.15 degrees [[Celsius]]<br /> <br /> 0 degrees [[Kelvin]] (Celsius increments)<br /> <br /> -459.67 degrees [[Fahrenheit]]<br /> <br /> 0 degrees [[Rankine]] (Fahrenheit increment)<br /> <br /> <br /> [[Category:Physics]]<br /> <br /> {{Stub}}</div> Twod horse https://artofproblemsolving.com/wiki/index.php?title=Midpoint&diff=145866 Midpoint 2021-02-12T07:36:04Z <p>Twod horse: </p> <hr /> <div>== Definition ==<br /> The '''midpoint''' of a [[line segment]] is the [[point]] on the segment equidistant from both endpoints.<br /> <br /> A midpoint [[bisect]]s the line segment that the midpoint lies on. Because of this property, we say that for any line segment &lt;math&gt;\overline{AB}&lt;/math&gt; with midpoint &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;AM=BM=\frac{1}{2}AB&lt;/math&gt;. Alternatively, any point &lt;math&gt;M&lt;/math&gt; on &lt;math&gt;\overline{AB}&lt;/math&gt; such that &lt;math&gt;AM=BM&lt;/math&gt; is the midpoint of the segment.<br /> &lt;asy&gt;<br /> draw((0,0)--(4,0));<br /> dot((0,0));<br /> label(&quot;A&quot;,(0,0),N);<br /> dot((4,0));<br /> label(&quot;B&quot;,(4,0),N);<br /> dot((2,0));<br /> label(&quot;M&quot;,(2,0),N);<br /> label(&quot;Figure 1&quot;,(2,0),4S);<br /> &lt;/asy&gt;<br /> == Midpoints and Triangles ==<br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,G;<br /> A=(0,0);<br /> B=(4,0);<br /> C=(1,3);<br /> D=(2,0);<br /> E=(2.5,1.5);<br /> F=(0.5,1.5);<br /> G=(5/3,1);<br /> draw(A--B--C--cycle);<br /> draw(D--E--F--cycle,green);<br /> dot(A--B--C--D--E--F--G);<br /> draw(A--E,red);<br /> draw(B--F,red);<br /> draw(C--D,red);<br /> label(&quot;A&quot;,A,S);<br /> label(&quot;B&quot;,B,S);<br /> label(&quot;C&quot;,C,N);<br /> label(&quot;D&quot;,D,S);<br /> label(&quot;E&quot;,E,E);<br /> label(&quot;F&quot;,F,W);<br /> label(&quot;G&quot;,G,NE);<br /> label(&quot;Figure 2&quot;,D,4S);<br /> &lt;/asy&gt;<br /> === Midsegments ===<br /> As shown in Figure 2, &lt;math&gt;\Delta ABC&lt;/math&gt; is a triangle with &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt; midpoints on &lt;math&gt;\overline{AB}&lt;/math&gt;, &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{CA}&lt;/math&gt; respectively. Connect &lt;math&gt;\overline{EF}&lt;/math&gt;, &lt;math&gt;\overline{FD}&lt;/math&gt;, &lt;math&gt;\overline{DE}&lt;/math&gt; (segments highlighted in green). They are midsegments to their corresponding sides. Using SAS Similarity Postulate, we can see that &lt;math&gt;\Delta CFE \sim \Delta CAB&lt;/math&gt; and likewise for &lt;math&gt;\Delta ADF&lt;/math&gt; and &lt;math&gt;\Delta BED&lt;/math&gt;. Because of this, we know that<br /> &lt;cmath&gt;\begin{align*}<br /> AB &amp;= 2FE \\<br /> BC &amp;= 2DE \\<br /> CA &amp;= 2ED \\<br /> \end{align*}&lt;/cmath&gt;<br /> Which is the Triangle Midsegment Theorem. Because we have a relationship between these segment lengths, &lt;math&gt;\Delta ABC \sim \Delta EFD (SSS)&lt;/math&gt; with similar ratio 2:1. The area ratio is then 4:1; this tells us<br /> &lt;cmath&gt;\begin{align*}<br /> [ABC] &amp;= 4[EFD]<br /> \end{align*}&lt;/cmath&gt;<br /> == Cartesian Plane ==<br /> In the Cartesian Plane, the coordinates of the midpoint &lt;math&gt;M&lt;/math&gt; can be obtained when the two endpoints &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; of the line segment &lt;math&gt;\overline{AB}&lt;/math&gt; is known. Say that &lt;math&gt;A: A(x_A,y_A)&lt;/math&gt; and &lt;math&gt;B: B(x_B,y_B)&lt;/math&gt;. The Midpoint Formula states that the coordinates of &lt;math&gt;M&lt;/math&gt; can be calculated as:<br /> &lt;cmath&gt;\begin{align*}<br /> M(\frac{x_A+x_B}{2}&amp;,\frac{y_A+y_B}{2})<br /> \end{align*}&lt;/cmath&gt;<br /> == See Also ==<br /> * [[Bisect]]<br /> <br /> {{stub}}<br /> <br /> [[Category:Definition]]<br /> [[Category:Geometry]]</div> Twod horse https://artofproblemsolving.com/wiki/index.php?title=Midpoint&diff=145865 Midpoint 2021-02-12T07:34:30Z <p>Twod horse: </p> <hr /> <div>{{stub}}<br /> == Definition ==<br /> The '''midpoint''' of a [[line segment]] is the [[point]] on the segment equidistant from both endpoints.<br /> <br /> A midpoint [[bisect]]s the line segment that the midpoint lies on. Because of this property, we say that for any line segment &lt;math&gt;\overline{AB}&lt;/math&gt; with midpoint &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;AM=BM=\frac{1}{2}AB&lt;/math&gt;. Alternatively, any point &lt;math&gt;M&lt;/math&gt; on &lt;math&gt;\overline{AB}&lt;/math&gt; such that &lt;math&gt;AM=BM&lt;/math&gt; is the midpoint of the segment.<br /> &lt;asy&gt;<br /> draw((0,0)--(4,0));<br /> dot((0,0));<br /> label(&quot;A&quot;,(0,0),N);<br /> dot((4,0));<br /> label(&quot;B&quot;,(4,0),N);<br /> dot((2,0));<br /> label(&quot;M&quot;,(2,0),N);<br /> label(&quot;Figure 1&quot;,(2,0),4S);<br /> &lt;/asy&gt;<br /> == Midpoints and Triangles ==<br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,G;<br /> A=(0,0);<br /> B=(4,0);<br /> C=(1,3);<br /> D=(2,0);<br /> E=(2.5,1.5);<br /> F=(0.5,1.5);<br /> G=(5/3,1);<br /> draw(A--B--C--cycle);<br /> draw(D--E--F--cycle,green);<br /> dot(A--B--C--D--E--F--G);<br /> draw(A--E,red);<br /> draw(B--F,red);<br /> draw(C--D,red);<br /> label(&quot;A&quot;,A,S);<br /> label(&quot;B&quot;,B,S);<br /> label(&quot;C&quot;,C,N);<br /> label(&quot;D&quot;,D,S);<br /> label(&quot;E&quot;,E,E);<br /> label(&quot;F&quot;,F,W);<br /> label(&quot;G&quot;,G,NE);<br /> label(&quot;Figure 2&quot;,D,4S);<br /> &lt;/asy&gt;<br /> === Midsegments ===<br /> As shown in Figure 2, &lt;math&gt;\Delta ABC&lt;/math&gt; is a triangle with &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt; midpoints on &lt;math&gt;\overline{AB}&lt;/math&gt;, &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{CA}&lt;/math&gt; respectively. Connect &lt;math&gt;\overline{EF}&lt;/math&gt;, &lt;math&gt;\overline{FD}&lt;/math&gt;, &lt;math&gt;\overline{DE}&lt;/math&gt; (segments highlighted in green). They are midsegments to their corresponding sides. Using SAS Similarity Postulate, we can see that &lt;math&gt;\Delta CFE \sim \Delta CAB&lt;/math&gt; and likewise for &lt;math&gt;\Delta ADF&lt;/math&gt; and &lt;math&gt;\Delta BED&lt;/math&gt;. Because of this, we know that<br /> &lt;cmath&gt;\begin{align*}<br /> AB &amp;= 2FE \\<br /> BC &amp;= 2DE \\<br /> CA &amp;= 2ED \\<br /> \end{align*}&lt;/cmath&gt;<br /> Which is the Triangle Midsegment Theorem. Because we have a relationship between these segment lengths, &lt;math&gt;\Delta ABC \sim \Delta EFD (SSS)&lt;/math&gt; with similar ratio 2:1. The area ratio is then 4:1; this tells us<br /> &lt;cmath&gt;\begin{align*}<br /> [ABC] &amp;= 4[EFD]<br /> \end{align*}&lt;/cmath&gt;<br /> == Cartesian Plane ==<br /> In the Cartesian Plane, the coordinates of the midpoint &lt;math&gt;M&lt;/math&gt; can be obtained when the two endpoints &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; of the line segment &lt;math&gt;\overline{AB}&lt;/math&gt; is known. Say that &lt;math&gt;A: A(x_A,y_A)&lt;/math&gt; and &lt;math&gt;B: B(x_B,y_B)&lt;/math&gt;. The Midpoint Formula states that the coordinates of &lt;math&gt;M&lt;/math&gt; can be calculated as:<br /> &lt;cmath&gt;\begin{align*}<br /> M(\frac{x_A+x_B}{2}&amp;,\frac{y_A+y_B}{2})<br /> \end{align*}&lt;/cmath&gt;<br /> == See Also ==<br /> * [[Bisect]]<br /> <br /> [[Category:Definition]]<br /> [[Category:Geometry]]</div> Twod horse https://artofproblemsolving.com/wiki/index.php?title=Midpoint&diff=145861 Midpoint 2021-02-12T07:31:21Z <p>Twod horse: </p> <hr /> <div>{{stub}}<br /> == Definition ==<br /> The '''midpoint''' of a [[line segment]] is the [[point]] on the segment equidistant from both endpoints.<br /> <br /> A midpoint [[bisect]]s the line segment that the midpoint lies on. Because of this property, we say that for any line segment &lt;math&gt;\overline{AB}&lt;/math&gt; with midpoint &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;AM=BM=\frac{1}{2}AB&lt;/math&gt;. Alternatively, any point &lt;math&gt;M&lt;/math&gt; on &lt;math&gt;\overline{AB}&lt;/math&gt; such that &lt;math&gt;AM=BM&lt;/math&gt; is the midpoint of the segment.<br /> &lt;asy&gt;<br /> draw((0,0)--(4,0));<br /> dot((0,0));<br /> label(&quot;A&quot;,(0,0),N);<br /> dot((4,0));<br /> label(&quot;B&quot;,(4,0),N);<br /> dot((2,0));<br /> label(&quot;M&quot;,(2,0),N);<br /> label(&quot;Figure 1&quot;,(2,0),4S);<br /> &lt;/asy&gt;<br /> == Midpoints and Triangles ==<br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,G;<br /> A=(0,0);<br /> B=(4,0);<br /> C=(1,3);<br /> D=(2,0);<br /> E=(2.5,1.5);<br /> F=(0.5,1.5);<br /> G=(5/3,1);<br /> draw(A--B--C--cycle);<br /> draw(D--E--F--cycle,green);<br /> dot(A--B--C--D--E--F--G);<br /> draw(A--E,red);<br /> draw(B--F,red);<br /> draw(C--D,red);<br /> label(&quot;A&quot;,A,S);<br /> label(&quot;B&quot;,B,S);<br /> label(&quot;C&quot;,C,N);<br /> label(&quot;D&quot;,D,S);<br /> label(&quot;E&quot;,E,E);<br /> label(&quot;F&quot;,F,W);<br /> label(&quot;G&quot;,G,NE);<br /> label(&quot;Figure 2&quot;,D,4S);<br /> &lt;/asy&gt;<br /> === Midsegments ===<br /> As shown in Figure 2, &lt;math&gt;\Delta ABC&lt;/math&gt; is a triangle with &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt; midpoints on &lt;math&gt;\overline{AB}&lt;/math&gt;, &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{CA}&lt;/math&gt; respectively. Connect &lt;math&gt;\overline{EF}&lt;/math&gt;, &lt;math&gt;\overline{FD}&lt;/math&gt;, &lt;math&gt;\overline{DE}&lt;/math&gt; (segments highlighted in green). They are midsegments to their corresponding sides. Using SAS Similarity Postulate, we can see that &lt;math&gt;\Delta CFE \sim \Delta CAB&lt;/math&gt; and likewise for &lt;math&gt;\Delta ADF&lt;/math&gt; and &lt;math&gt;\Delta BED&lt;/math&gt;. Because of this, we know that<br /> &lt;cmath&gt;\begin{align*}<br /> AB &amp;= 2FE \\<br /> BC &amp;= 2DE \\<br /> CA &amp;= 2ED \\<br /> \end{align*}&lt;/cmath&gt;<br /> Which is the Triangle Midsegment Theorem. Because we have a relationship between these segment lengths, &lt;math&gt;\Delta ABC \sim \Delta EFD (SSS)&lt;/math&gt; with similar ratio 2:1. The area ratio is then 4:1; this tells us<br /> &lt;cmath&gt;\begin{align*}<br /> [ABC] &amp;= 4[EFD]<br /> \end{align*}&lt;/cmath&gt;<br /> == In Cartesian Plane ==<br /> In the Cartesian Plane, the coordinates of the midpoint &lt;math&gt;M&lt;/math&gt; can be obtained when the two endpoints &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; of the line segment &lt;math&gt;\overline{AB}&lt;/math&gt; is known. Say that &lt;math&gt;A: A(x_A,y_A)&lt;/math&gt; and &lt;math&gt;B: B(x_B,y_B)&lt;/math&gt;. The Midpoint Formula states that the coordinates of &lt;math&gt;M&lt;/math&gt; can be calculated as:<br /> &lt;cmath&gt;\begin{align*}<br /> M(\frac{x_A+x_B}{2}&amp;,\frac{y_A+y_B}{2})<br /> \end{align*}&lt;/cmath&gt;<br /> == See Also ==<br /> * [[Bisect]]<br /> <br /> [[Category:Definition]]<br /> [[Category:Geometry]]</div> Twod horse https://artofproblemsolving.com/wiki/index.php?title=Midpoint&diff=145708 Midpoint 2021-02-12T03:42:58Z <p>Twod horse: </p> <hr /> <div>{{stub}}<br /> == Definition ==<br /> The '''midpoint''' of a [[line segment]] is the [[point]] on the segment equidistant from both endpoints.<br /> <br /> A midpoint [[bisect]]s the line segment that the midpoint lies on. Because of this property, we say that for any line segment &lt;math&gt;\overline{AB}&lt;/math&gt; with midpoint &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;AM=BM=\frac{1}{2}AB&lt;/math&gt;. Alternatively, any point &lt;math&gt;M&lt;/math&gt; on &lt;math&gt;\overline{AB}&lt;/math&gt; such that &lt;math&gt;AM=BM&lt;/math&gt; is the midpoint of the segment.<br /> &lt;asy&gt;<br /> draw((0,0)--(4,0));<br /> dot((0,0));<br /> label(&quot;A&quot;,(0,0),N);<br /> dot((4,0));<br /> label(&quot;B&quot;,(4,0),N);<br /> dot((2,0));<br /> label(&quot;M&quot;,(2,0),N);<br /> &lt;/asy&gt;<br /> == In Cartesian Plane ==<br /> In the Cartesian Plane, the coordinates of the midpoint &lt;math&gt;M&lt;/math&gt; can be obtained when the two endpoints &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; of the line segment &lt;math&gt;\overline{AB}&lt;/math&gt; is known. Say that &lt;math&gt;A: A(x_A,y_A)&lt;/math&gt; and &lt;math&gt;B: B(x_B,y_B)&lt;/math&gt;. The Midpoint Formula states that the coordinates of &lt;math&gt;M&lt;/math&gt; can be calculated as:<br /> &lt;cmath&gt;\begin{align*}<br /> M(\frac{x_A+x_B}{2}&amp;,\frac{y_A+y_B}{2})<br /> \end{align*}&lt;/cmath&gt;<br /> == See Also ==<br /> * [[Bisect]]<br /> <br /> [[Category:Definition]]<br /> [[Category:Geometry]]</div> Twod horse