Ⅰ Sques e Letue: iu ng Mtemtis dution oundtion Pesident Wen-Hsien SUN Ⅱ Retngles = Te e of sque of side is Ⅲ Pllelogms = Te e of etngle of sides nd is = Te e of pllelogm is te podut of te lengt of one side nd te oesponding eigt Ⅳ Tingles () Rigt-ngled tingle () ity tingle = =
Ⅴ Tpezis e= ( + ) = m Te e of tpezium is te podut of its mid-line nd its eigt. Te mid-line of tpezium is pllel to te se lines nd lf s long s tei sum. Ⅵ Kites Te e of kite is lf te podut of te lengts of te digonls. () Romus () Kites m m x Ⅶ iles e= e= + = x ( ) + x = Te e of ile is te podut of π nd te sque of te dius d O e= π = π d 4 Ⅷ Seto of ile O θ s θ π e= 360 Te e of te seto depends on te entl ngle. Te lengt of te ounding te seto/ iumfeene of te ile = s
Lemm. If two tingles ve te sme se nd te sme ltitude, ten tey ve te sme e. Speil se: te medin of tingle uts te tingle into two tingles of equl e Lemm. If two tingles ve te sme ltitude, ten te tio of tei e is equl to te tio of tei ses Lemm 3. If two tingles ve te sme ses, ten te tio of tei e is equl to te tio of tei ltitude In tingle, extend te lengt of side y times to point, extend te lengt of side y 3 times to point, extend te lengt of side y 4 times to point. Wt is te tio of e nd? Teoem of Pytgos ² +²=²
Poof Poof Poof 3 Poof 4 Poof 5
Poof 6 Poved y 0t pesident of US Jmes. Gfield (876 ) = + = + = ( + ) Poof 7 G = + = + = + Ⅸ Rtio Sques Te tio of te e of two sques =Te sque of te lengt of its sides =²/² =(/) ² Tingles k k If te side lengts of tingle e enlged k times, ten its e is enlged k times. Retngles k If two etngles ve one side lengt in ommon, ten te tio of tei es equls te tio of tei seond sides: / =/
In te figue, te e of te tee etngles e 4, 30, 40. Wt is te e of te fout etngle? 4 40 30? Tee sques wit te sme side lengt e put in lge sque. Te numes sown in te figue e te e we n see. Wt is te totl e of te sded pts? 30 4 line pllel to tingle s side uts te two ote sides in te sme tio. = = = Speil se: =, =, = Ⅹ Simil tingles Two tingles e simil if () te tion etween tee oesponding sides e equl ; () te tee oesponding ngles e equl ; (3) te two oesponding ngles e equl ; (4) te tio etween two sides of one is equl to te tio etween te oesponding sides of te ote nd te enlosed ngles e equl. Te ltitude on te ypotenuse divides igt-ngled tingle into two tingles tt e ot simil to te oiginl tingle nd ene to e ote. Gvity vey medin is divided in te tio : y te ente of gvity, te longe pt djoining te vetex. =,=,=, G=G,G=G, G=G G G
If two simil tingles ve lengts of oesponding sides in te tio of k, ten tei es e in te tio of k ''/ =''/ =''/ = k, ~ ''' ''/ = ''/=k '''/ = k² Some fts of tio d si Popety: = d =, 0, 0 d ± n d ± n oesponding ddition: = =, 0, 0 oesponding ddition nd suttion: d f n = = = =, 0, 0, e 0,, m 0 e m ± d + f + + v =, + + e + + m 0 + + e + + m XI lge lgei sums e multiplied y multiplying evey tem of one sum into evey tem of te ote nd dding tese poduts. ' ' ' ' (+)=+ d d d (+d) (+) = +d++d y x xy x y xy (x+y)² = x²+xy+y² x y x y x x²-y² =(x+y)(x-y) x xy (x-y) y y y (x-y)² =x²-xy+y² y x x-y y x-y
3 4 3 4 - - 3 + + + 3 + + 3 + 4 3-4 4 + + 3 3 + 4 4 =( + + 3 + 4 ) 4 +( + + 3 )( 3-4 ) +( + )( - 3 ) ommon Side Teoem Let P nd Q ve ommon side, nd ve lines nd PQ meet t P PM point M. Ten = Q QM P P P P M Q G Q M M Q Let te medins nd of inteset t point G, ten G=G : G/ G = / = G/ G = / = G = G = G = /3 / G = / G = 3 Q M Teoem of ev Let point P e ny point not olline wit ny two veties of tingle, nd let te lines P, P nd P inteset te lines, nd t points, nd, espetively. Ten = : P /= P/ P,/= P/ P, /= P/ P, /././ = P/ P. P/ P. P/ P =
ommon ngle teoem If two tingles ve equl (o supplementy) of oesponding ngle, ten tei es e in te tio of te podut of two sides tt fom tis ngle. () qul ngles =' / ''' =( / '') ( ''/ ''') =(/'').( /'') ' () Supplementy ngles +'=80 / ''' =( / '') ( ''/ ''') =(/'').(/'') (') ' ' In, =, so = = / =(.)/(.)= / = ' (') ngle iseto teoem Te ngle iseto of tingle divides te opposite side into two segments wose lengts e in te sme tio s te sides of te ngle: /=/ / =./.=/ / =/ /=/ Let e tingle, nd let points, nd e on te sides,, espetively, su tt =, = nd =. Te lines nd inteset t point P, te lines nd inteset t point R, Te lines nd inteset t point Q. ind te tio of es of PQR nd.
Q P R Q/ Q=/=/ Q/ Q=/= = Q+ Q+ Q =(++/) Q Q =/7 Similly, P= R=/7 PQR = - P- Q- R =/7 If =, =, nd =, ten te tio of es of PQR nd is (-) /(++)(++)(++) Speil se (ev s Teoem):,, e onuent, o PQR=0 - =0 = / / /= ue Te ue s 6 fes, nd e fe is sque. Te segments wee two fes inteset e edges. ue s edges nd ll edges ve te sme lengt. Te points wee tee edges inteset e veties. ue s 8 veties. In te igt digm, Sufe e of te ue =6 =6 Volume of te ue = = 3 Lengt of segment = Lengt of segment = 3 Retngul uoid Te etngul uoid lso s six fes. fe is etngul o sque. etngul uoid lso s edges nd 8 veties. It is lso pism, wit volume = (se e) (eigt). In te igt digm, Sufe e of te uoid = ( + + ) Volume of te uoid = Lengt of segment = + Lengt of segment = + +
xmple: Te figue to te igt is wooden etngul uoid. n nt is t point, nd it wnts to wl to point long te sufe of te uoid. Wt is te sotest pt it n tke? (Te dimensions of te uoid e lengt 5m, widt 4m, eigt 3m.) Solution: Te nt wnts to wl fom point to point. Te uoid is solid, so te nt must wl on te sufe nd not inside te uoid. Tee e infinitely mny pts on te uoid s sufe fom to, ut we n lssify tem into only tee types: (I) pt toug fe, pssing toug te edge nd eing toug fe. Rotte te fe out 90, tus mking te fes nd opln. Sine te sotest pt etween two points is te line joining tem, te sotest pt fom to is te segment. y te Pytgoen Teoem, = = 9 3 ( + ) + = (5 + 4) + 3 + = 8+ 9 = 90 9.487 m (II) pt toug fe, edge, nd fe to point. fte simil ottion, we get y te Pytgoen Teoem = ( + ) + = (3 + 5) + 4 = 8 + 4 = 64 + 6 = 80 8.944m (III) pt toug fe, edge, nd fe to point. Rotte nd pply te Pytgoen Teoem similly: = = 5 7 + ( + ) = 5 + (3 + 4) + = 5 + 49 = 74 8.60m Tee e infinitely mny pts fom to. Te tee types ove e ve fixed minimum nd te tid type s te sotest pt. nswe: Te sotest pt is fo te nt to wl toug fe, edge, nd fe to. Te sotest pt s lengt ppoximtely 8.60m. Polem: n nt wls on te sufe of etngul uoid. Te distne etween two points on te sufe is te lengt of te sotest pt te nt n tke etween te two points. om te point of view of te nt, e te two points futest fom e 5m 3m 4m
ote lwys two digonlly opposite veties of te uoid (tese two veties e symmeti wit espet to te ente of te uoid)? Solution: Unde tis definition of distne, te nswe is negtive. Suppose points nd e opposite veties of 4 4 8 uoid. Let point e point on te 4 4 fe tt ontins, su tt te distne fom to te two edges of lengt 4 tt ontin e e. Tee e two wys to unfold te uoid. Let s see if te distne etween nd is te longest distne on te uoid. In te left digm, = + 4 = 60, = + 3 = 30 ;in te igt digm, = 8 + 8 = 8, = 9 + 7 = 30. y definition, te distne etween two points on te uoid is te distne te nt must wl to get fom one point to te ote. So te distne fom to is 8, ut te distne fom to is 30. lely, is fte fom tn is. ylinde () iul ylinde Te solid of evolution of etngle out one of its sides is iul ylinde. Volume of ylinde =te e of te se eigt =π Sufe e of ylinde=te e of te top se + te e of te ottom se + te e of te side-fes =π + π + π = π ( + ) () Rigt pism Tke two onguent polygons in pllel plnes nd onnet te oesponding points su tt e line is pependiul to te se plne. Te esulting solid is igt pism. Volume of igt pism =se e eigt= Sufe e of igt pism =Top se e+lowe se e+ltel sufe e =+se peimete
() Olique pism Tke two onguent polygons in pllel plnes nd onnet te oesponding points su tt te lines e not pependiul to te se plne. Te esulting solid is n olique pism. Te distne etween te two plnes is te eigt, wile te distne etween two oesponding points is te slnt eigt. Volume of n olique pism =se e eigt Sufe e of n olique pism =Top se e + Lowe se e + Ltel sufe e one () Rigt one Te solid of evolution of igt tingle out one of its legs is igt one. In te igt digm, is te xis, is te geneto (o te slnt eigt), is te one s pex. Volume of one= se e eigt 3 = 3 π Sufe e of one=se e + ltel e=π + π g g g π lttened out, te ltel e is iul seto wit e π g = π g () Rigt pymid If te segment onneting te ente of te pymid s se nd te pex is pependiul to te se, te pymid is lled igt pymid. s in te digm, is te eigt of te pymid, is te side-edge, nd te ltitude of tingle, ee, is te slnt eigt. Volume of igt pymid = se e eigt 3 Sufe e of igt pymid =se e + ltel sufe e Ltel sufe e of igt pymid is peimete slnt eigt
Regul Tetedon (Regul tingul pymid) Te solid omposed of fou equiltel tingle fes is te egul tetedon. s in te pitue, if =, we ve: Slnt eigt M= 3, Te sufe e is Heigt O= 3 4 = 3, O = 3 Te volume of te egul tetedon is Spee 3 = 3 = 6 3 3 6 = 4 3 4 Te solid of evolution of ile out its dimete is spee. 4 3 Te e of te spee is 3 π. Te sufe e of te spee is 4π. Questions:. In pentgon, te digonl inteset te digonls nd t points nd G espetively. If :=5:4, G:G=:, :G:G =::3 ind te tio of es of nd. (983 Russi Mt Olympid ). In qudiltel, te tio of es of, nd is 3:4:,points M nd N on segments nd espetively su tt M:= N:, nd points, M nd N e olline. Pove tt points M nd N e midpoint of nd espetively. (983 in Mt Olympid) 3. In exgon, te digonls nd e points M nd N divide in tio M:=N:=. If points, M nd N e olline, find. (98 IMO) 4. ind te e of te sded potions. 3 O M π [(4 -π ) 4] (4 -π ) 4