Napoleon and Pythagoras with Geometry Expressions


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1 Npoleon nd Pythgors with eometry xpressions NPOLON N PYTORS WIT OMTRY XPRSSIONS... 1 INTROUTION... xmple 1: Npoleon s Theorem... 3 xmple : n unexpeted tringle from Pythgorslike digrm... 5 xmple 3: Penequilterl Tringle... 7 xmple 4: nother Penequilterl Tringle xmple 5: Qudrlterl Theorem
2 IRLS N TRINLS WIT OMTRY XPRSSIONS Introdution eometry xpressions utomtilly genertes lgeri expressions from geometri figures. or exmple in the digrm elow, the user hs speified tht the tringle is right nd hs short sides length nd. The system hs lulted n expression for the length of the ltitude: + We present olletion of worked exmples using eometry xpressions. These exmples explore digrms similr to tht of Pythgors Theorem, nd Npoleon s Theorem: in whih similr figures re onstruted on eh side of n originl tringle (or in one se qudrlterl).
3 xmple 1: Npoleon s Theorem Npoleon s Theorem sttes tht if you tke generl tringle nd drw n equilterl tringle on eh side, then the tringle formed y joining the inenters of these new tringles is equilterl. P O R Q T S n you prove the theorem from this digrm y inspetion of the expression for the length of QS? 3
4 IRLS N TRINLS WIT OMTRY XPRSSIONS Now we need to prove the expression in the digrm. One wy of doing this is to remove the onstrint on the length of, nd reple it y onstrint on the ngle K M θ π 6 J L I sin(θ)  os(θ) os(θ) n you use the osine rule to show tht the expression for LJ displyed ove is orret? iven tht n you prove the expression in terms of lengths,,? 4
5 xmple : n unexpeted tringle from Pythgorslike digrm Regrdless of the originl tringle the resulting tringle from this digrm is right ngledisoseles: J π K xmintion of the length J shows tht it is symmetri in nd, nd hene identil to K. 5
6 IRLS N TRINLS WIT OMTRY XPRSSIONS This result n lso e proved y repling the distne with n ngle: +  os(θ) + + sin(θ) os(θ) θ +  os(θ) iven the expression ove for, n you derive the formul in terms of,,? n you derive the expression for? (int: whih ngle of the originl tringle is the sme s? 6
7 xmple 3: Penequilterl Tringle Strting with tringle whose sides re length,,, we onstrut squres on eh side, join the orners of the squres, then join the midpoints of these lines to rete tringle: J K L I 7
8 IRLS N TRINLS WIT OMTRY XPRSSIONS The tringle looks to the nked eye s if it is equilterl. Try drgging the originl points nd oserve tht the new tringle still looks equilterl: J L K I 8
9 Is it in ft? re JKL J K L I Not quite, we oserve the sides re gurnteed to e lose in size, ut not identil unless the originl tringle is isoseles. In ft, the differene in squres of the sides of the new tringle is. 4 4 Notie we n repet the proess drwing squres on JK, KL nd JL. The differene in squres of the ensuing sides will e
10 IRLS N TRINLS WIT OMTRY XPRSSIONS y repeting this proess, we n rete tringle s lose s we like to n equilterl tringle, ut still not extly one. O 6.36 S P M J 6.34 L U N K I T Q R
11 Symolilly: Q n you find reurrene reltion for the side lengths nd the re? 11
12 IRLS N TRINLS WIT OMTRY XPRSSIONS ow out further reursion (this my tke some time to ompute): Q
13 xmple 4: nother Penequilterl Tringle We n do similr onstrution sed on equilterl tringles drwn on the sides of n originl tringle: Side Length re I We see tht the differene in squres of the sides of the new tringle orresponding to nd is. If we repet the proess, this tringle will eventully eome equilterl, ut not s quikly s the previous tringle. 13
14 IRLS N TRINLS WIT OMTRY XPRSSIONS xmple 5: Qudrlterl Theorem This theorem sttes tht if you drw squre on eh side of qudrlterl, then onnet the enter of opposite sides, the resulting lines hve the sme length, nd re perpendiulr. ere is the result in eometry xpressions z d  d + e + e + e +d e +e 4 + e +d+e +de d+e +d+e+ e ++e +e +e ++e+ +d+e +de d+e +d+e ++e +e +e ++e e J I π e K d d L d  d + e + e + e +d e e 4 + +d+e +de d+e +d+e ++e +e +e ++e e If we rete the length of the other side we n y reful exmintion see tht the lengths re identil. lterntively, we n do some simplifition. Our onstrints re neessrily symetri eometry xpressions will not let you overonstrin the digrm, nd one digonl is suffiient to define the qudrlterl. owever, we might expet the formul to e simpler if expressed in terms of oth digonls. lose inspetion of the formul for the length shows tht it inorportes the squre of the other digonl of the figure, s well s eron s formul for the res of the tringles 14
15 nd. The following Mthemti worksheet ontins the formuls from eometry xpressions for L the length of the desired line, nd f the length of the other digonl. little mnipultion gives simple formul for L^f^/. This n e simplified further y noting tht the remining terms re e^/ nd twie the re of the qudrlterl: In[6]:= L = 1 e J  J + + d  d + e + e + e + d e +e 4 + e "################### + d +e "################### +d  e "################### d +e "#######################  +d +e + e "################# # + +e "################### +  e "################# #  +e "######################  + +e + "################### + d + e "################### +d e "###################  d + e "#######################  + d + e "################### + + e "################# # + e "################# #  +e "######################  + +e NN 1 Out[6]= e I, I + + d d + e + e + e + d e + e 4 + è!!!!!!!!!!!!!!! + ee è!!!!!!!!!!!!! + e! è!!!!!!!!!!!!!!!!! + + e è!!!!!!!!!!!!!!! + + e + è!!!!!!!!!!!!!!! + d ee è!!!!!!!!!!!!! d + e! è!!!!!!!!!!!!!!!!! + d + e è!!!!!!!!!!!!! + d + e! + è!!!!!!!!!!!!! + e! è!!!!!!!!!!!!!!! + d e è!!!!!!!!!!!!!!! + e è!!!!!!!!!!!!!!!!! + + e è!!!!!!!!!!!!!!! + + e è!!!!!!!!!!!!! d + e! è!!!!!!!!!!!!!!!!! + d + e è!!!!!!!!!!!!! + d + e! MM In[4]:= f = 1 e J "#### Out[4]= In[7]:= L^ Out[7]= In[8]:= f^ Out[8]=  J + + d  d + e + e + e + d e e 4 + "################### + d + e "################### +d  e "################### d +e "#######################  +d +e "################## + + e "################# # + e "##################  + e "###################### e NN 1 è!!! e I, I + + d d + e + e + e + d e e 4 + è!!!!!!!!!!!!!! + e è!!!!!!!!!!!!!!! + d e è!!!!!!!!!!!!!!! + e è!!!!!!!!!!!!!!!!! + + e è!!!!!!!!!!!!!!! + + e è!!!!!!!!!!!!!!! d + e è!!!!!!!!!!!!!!!!! + d + e è!!!!!!!!!!!!!! + d + e MM 1 4e I + + d d + e + e + e + d e + e 4 + è!!!!!!!!!!!!!!! + ee è!!!!!!!!!!!!!!! + e è!!!!!!!!!!!!!!!!! + + e è!!!!!!!!!!!!!!! + + e + è!!!!!!!!!!!!!!! + d ee è!!!!!!!!!!!!!! d + e è!!!!!!!!!!!!!!!!! + d + e è!!!!!!!!!!!!!! + d + e + è!!!!!!!!!!!!!! + e è!!!!!!!!!!!!!!! + d e è!!!!!!!!!!!!!!! + e è!!!!!!!!!!!!!!!!! + + e è!!!!!!!!!!!!!!! + + e è!!!!!!!!!!!!!! d + e è!!!!!!!!!!!!!!!!! + d + e è!!!!!!!!!!!!!! + d + e M 1 e I + + d d + e + e + e + d e e 4 + è!!!!!!!!!!!!!!! + e è!!!!!!!!!!!!!!! + d e è!!!!!!!!!!!!!!! + e è!!!!!!!!!!!!!!!!! + + e è!!!!!!!!!!!!!! + + e è!!!!!!!!!!!!!!! d + e è!!!!!!!!!!!!!!!!! + d + e è!!!!!!!!!!!!!!! + d + e M In[11]:= f^ê Out[11]= 1 Ie + è!!!!!!!!!!!!!!! + e è!!!!!!!!!!!!!!! + e è!!!!!!!!!!!!!!!!! + + e è!!!!!!!!!!!!!! + + e + è!!!!!!!!!!!!!!! + d e è!!!!!!!!!!!!!! d + e è!!!!!!!!!!!!!!!!! + d + e è!!!!!!!!!!!!!!! + d + e M rom whih we n derive tht: e f L =
16 IRLS N TRINLS WIT OMTRY XPRSSIONS res re simpler when expressed in terms of ngles. ere is revision of the digrm with ngles inserted. This gives us more of lue of how to prove the result: + + sin(θ) π 4 θ π 4 I φ π 4 d + d + d sin(φ) + + d + sin(θ)+ d sin(φ) d sin(θ+φ) os(θ) d os(φ) J 16
17 If we lso exmine the equtions for the digonls nd the missing side in terms of these ngles we n see how the expression ove is pieed together: +  os(θ) θ φ + +d  os(θ) d os(φ)+ d os(θ+φ) d +d  d os(φ) 17
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