東大合格コム

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カテゴリ:大学受験 > 高校数学


こんにちは。いつも図形問題の鮮やかな解答を楽しく拝見しています。
先日、このような図形問題を見つけました。
https://www.youtube.com/watch?v=b8IP5SAOPIU
【中学受験算数】平面図形  超難問の角度の問題 これきちんと説明できる小学生はいるの? 西大和学園中2019年【最難関クラス/偏差値up】
動画では完全に初等幾何の範囲で解いています。
差支えなければ、高校数学ならどのような解法になるのか、主さんの答案をお聞きしたいです。
どうかよろしくお願いします。


算数の入試問題は基本1問5分なので図形はたいていとんち問題です。
点D,E,Mの3点を固定して考えると破綻します。←答えがなくなることがあるから。
A,D,E,Mのうちどれか2点を固定して考えるのがこの問題のポイント。
それに気づけば、算数も高校数学も同じ答えです。

【算数】
BM=MCを保ちながら蟹の爪を閉じていくとMで閉じ切って44°

【算数】
MとAが固定のとき与えられた条件はBM=MC=0でも成り立つので∠DME=22°+22°=44°
Aがどこにあっても同様なので∠DME=44°

【算数】
MとEが固定のとき△AMEは一意的に決まる。そのとき△AMDも一意的に決まり、△AME≡△AMDとなる。したがって∠DME=∠AME+∠AMD=22°+22°=44°
このことは平面上の同一でない任意の2点において成り立つ。

私は数学は自由だと思っているので、「~~禁止」とか「~~で解け」とかは嫌いです。
普通は高校生でも上と同じ解き方をするでしょう。
でも今回は特別に高校数学版別解も考えました。
どんなやりかたもできますが計算が一番楽なのは複素数平面かなと思います。

【高校数学】
複素数平面で考える。
A(0) C(z) また α=cos22°+(sin22°)i とすると sin22°=(α-α*)/(2i)
EはAを90°-22°=68°回転してsin22°を掛けたものなので
E((α-α*)z/(2α)=(1-α*^2)z/2)
またB(kz*) とすると同様に D(k(1-α^2)z*/2)
∴M((z+kz*)/2)
∴∠DME=Arg(((-k(α^2)z*-z)/2)/(-(α*^2)z-kz*)/2))
=Arg((k(α^2)z*+z)/((α*^2)z+kz*))=Arg(α^2)=44°

でもその後のリスポンスがない...

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30
https://www.youtube.com/watch?v=Z2qDqtCK_WA&t=36s
complex plane D(0) C(x) ω^3=1 ω^2+ω+1=0 as ω means 120° and -ω^2 means 60° rotation A(8ω) B(-ω^2(8ω-x)+x)=(-8-ωx)
BD=|-8-ωx|=13 |-8-ωx|^2=(-8-ωx)(-8-ω*x)=(-8-ωx)(-8-ω^2x)=64-8x+x^2=169 ∴x^2-8x-105=(x-15)(x+7)=0 ∴x=15

01
https://www.youtube.com/watch?v=Zm2F47kN4NU&t=1s
sine rule xsinα/sin(60-α)=2xsin75/sin45=2x(cos30+sin30)=x(2cos30+1) sinα=2cos30sin(60-α)+sin(60-α) 2cos30sin(60-α)=sinα-sin(60-α)=2cos30sin(α-30) ∴sin(60-α)=sin(α-30) ∴α=45

02
https://www.youtube.com/watch?v=GvOk3OcO0bM&t=3s
AC=6sin75/sin(105-α) △ABC=(1/2)6(6sin75/sin(75+α))sinα=9 sin(75+α)=2sinαcos15=sin(α+15)+sin(α-15) ∴sin(α-15)=sin(75+α)-sin(α+15)=2cos(45+α)sin30=sin(45-α) ∴2α=60 ∴α=30

03
https://www.youtube.com/watch?v=RwAc93Wrwbs&t=2s
AD=AP=1 PD=2cos80 PB=sin30/sin50=1/(2sin50) PC=sin30/(2sin50sin70)=1/(4sin50sin70)
PC=PD ⇔ 2cos80=1/(4sin50sin70)⇔8sin50sin70cos80=4sin70(sin130-sin30)=4cos20sin50-2sin70=2sin70+2sin30-2sin70=1

04
https://www.youtube.com/watch?v=BAlG2QUN3Eg&t=4s
CE=1 DC=2cos(90-α)=2sinα ∴sinα=(2sinα+1)/(4sinα) sinα=(1+√5)/4 ∴α=54

05
https://www.youtube.com/watch?v=WZsKsbKfKBM&t=49s
AE=CD=1 sine rule AD=sin2β/sin(90-3β)=sin4β/sin(90-β) ∴cosβ=2cos2βcos3β=cos5β+cosβ ∴cos5β=0 ∴5β=90 ∴2α=180-6β=180-108=72°

AE=CD=1 sine rule AD=sin2β/sin(90-3β)=sin4β/sin(90-β) ∴sin7β+sinβ=sin3β+sinβ ∴5β=90 ∴2α=180-6β=180-108=72°

06
https://www.youtube.com/watch?v=NH1ivIdrCxU&t=21s
let AR=1 then from sine rule for △ARC RC=sin(180-α-γ)/sinγ and for △RBP RC=sin(α-γ)/sinγ+1 ∴sinγ=sin(α+γ)-sin(α-γ)=2cosαsinγ ∴cosα=1/2 ∴α=60° ∴?=120°

07
https://www.youtube.com/watch?v=lalr4H4dj_k
sine rule for △BDE 8/sin3α=4/sinα ∴sinα=1/2 α=30 as similler DEB/ABC=(DB/BC)^2=(2/3)^2=4/9

08
https://www.youtube.com/watch?v=dUWeiF45ykY&t=4s
sine law for △ADE (9-x)/sinβ=6/sin(α+β) sine law for △ADC 6/sinβ=(14-x)/sin(α+β) ∴(9-x)/6=6/(14-x) ∴x^2-23x+90=(x-5)(x-18)=0 ∴x=5

09
https://www.youtube.com/watch?v=XsVKEXmsCBE&t=63s
vector a=6√2(cos60,sin60)=(3√2,3√6) h=(12cos15,0)=(3√6+3√2,0) KL=|k-l|=|(a+b+d)/3-(b+d+h)/3|=|(a-h)/3|=|(√6,√6)|=2√3

B(0,0) D(12cos15, 12sin15) then L(8cos15, 4sin15) K is 15° rotation of (6,2) then K(6cos15-2sin15, 6sin15+2cos15) ∴KL^2=(2cos15+2sin15)^2+(2cos15+2sin15)^2=8+8sin30=12 ∴KL=2√3

10
https://www.youtube.com/watch?v=BhGoEVhFbfM
OB=r then sine law BC/sin58=2r cosα=BC/(2r)=sin58=cos32 ∴α=32

AngleOAB=OBA OAC=OCA ∴OBA+OCA=58 OBC=OCB=α ∴2α+58+58=180 ∴α=32

11
https://www.youtube.com/watch?v=Oq33wRucl5o&t=3s
AngleBCD=θ then CF=2/sinθ CD=4/sinθ sine law (4/sinθ)/sin150=DB/sinθ ∴DB=4/sin150=8

12
https://www.youtube.com/watch?v=mT8MSMImLx4
Set E on AB//AE AD=AE ∴∠ADE=∠AED=15° ∴△DAC≡△EAC ∠DAC=∠EAC=75° ∠DCE=70° ∴∠DAC=35°

AD=1 AB=sin110/sin40 BC=sin110sin(140-α)/(sin40sinα)=1+sin110/sin40 sin110(sin(40+α)-sinα)=sin40sinα cos(20+α)=sinα α=35°

13
https://www.youtube.com/watch?v=arfsBGGsxQ0&t=2s
AN=a BC=b AD=CD=c then from cosine law DB^2=cc+aa-ac=cc+bb-bc ∴aa-bb=c(a+b) ∴a=b+c

ABCD is inscribed in a circle sine law AB/sin80=BC/sin20=CD/sin40=2R BC+CD=2R(sin20+sin40)=2R(2sin30cos10)=2Rcos10=2Rsin80=AB

14
https://www.youtube.com/watch?v=zWXqJOHcKwk&t=1s
The center of the circle is O the foot of the perpendicular drawn from O to AB is H
81(BH-AH)=(BH+AH)(BH-AH)=BH^2-AH^2=(BH^2+OH^2)-(AH^2+OH^2)=OB^2-OA^2=(OB^2-OE^2)-(OA^2-OE^2)=BE^2-AD^2=27^2-18^2=9・45=81・5 ∴BH-AH=5
GH=FH ∴BG-FA=(BG+GH)-(HF+FA)=BH-AH=5

Think of the figure multiplied by 1/9 the center of the circle is O the foot of the perpendicular drawn from O to AB is H ∠ACB=2θ
From the cosine law 2(cosθ)^2-1=cos2θ=(49+64-81)/112=2 /7 ∴cosθ=3/√14 ∴OC=5/cosθ=(5/3)√14
From the cosine law OA^2=49+OC^2-70=OC^2-21 OB^2=64+OC^2-80=OC^2-16
9(BH-AH)=BH^2-AH^2=(OB^2-OH^2)-(OA^2-OH^2)=OB^2-OA^2=(OC^2-16) -(OC^2-21)=5 ∴BH-AH=5/9
GH=FH ∴BG-FA=(BG+GH)-(HF+FA)=BH-AH=5/9 ∴True BG-FA=5

△BEG is similar to △BFE (81-y)/27=27/x △ADF is similar to △AGD (81-x)/18=18/y hence 81(x-y)=27^2-18^2=9•45=81•5 ∴x-y=5

15
https://www.youtube.com/watch?v=9sB9IOpxXKM&t=62s
AngleCAD=CBE=θ AngleOED=ODE=α sinθ=1/3 cosθ=2√2/3 from sine rule for CDE √2/sin(90-α)=(2√2/3)/sin(θ+α)=2rcosα/sin(90-θ) sin(θ+α)=(2/3)cosα sinα=cosα/(2√2) cosα=(2/3)√2 r=2/(3(cosα)^2)=(2/3)(9/8)=3/4

Suppose x(1/√2) plane D(0,0) A(0,2√2) B(-1,0) C(1,0) AC:y=-√2(x-1) BE:y=(1/(2√2))(x+1) E(7/9,(4/9)√2) O(x,y) x^2+y^2=(x-(7/9))^2+(y-((4/9)√2))^2 (9-14x)/4=2√2y=x+1 x=5/18 y=23/(36√2) r=√(x^2+y^2)=3/(4√2) real r=3/4

F is the middle of EC, EDF is the same as CDF DE=√2 angleBAD=CBE=DEB=θ sinθ=1/3 cosθ=2√2/3 DE=2rcosθ=√2 r=3/4

16
https://www.youtube.com/watch?v=8I8Zvd4O7VA&t=15s
Vector B as the starting point A(α) F(b) C(2b) |a|=|b| D((α+b)/2) E((-1/3)(2b)+(4/3)(α+b)/2=(2/3)α) ∴AE=(1/3)|α| EB=2|b| ∴EB/AF=6

△ADB≡△FDB bisector AG:GC=AB:BC=1:2 Ceva EB/AE=(2/1)(1/1)=2 BC/AD=(2a)/(a/3)=6

BF=CF=1 AE=a △BDA≡△BDF AB=1 BE=1-a Menelaus rule (1-a)/a(1/2)(1/1)=1 a=1/3 2/a=6

17
https://www.youtube.com/watch?v=g2jOIDxeCvo&t=12s
B(0,0) A(0,1) C(2,0) D(a,b) because (a,b)•(2,-1)=0 and (a/2,b/2) is on y+x/2=1, r=a=4/5

18
https://www.youtube.com/watch?v=s8jwKqRuqYE
middle of EB is O, AB is F, angleAEF=BED
ABC:EAB:AFC=2:√2:1 EA=1 AF=1/√2 FC=1 EC=3/√2

19
https://www.youtube.com/watch?v=YM0cPFQudTw&t=24s
SR=1 from sine rule 1/sin36°=SD/sin72° SD=sin72°/sin36°=2cos36° let cos36°=c 4c^3-3c=cos108°=-cos72°=1-2c^2 4c^3+2c^2-3c-1=(c+1)(4c^2-2c-1)=0 c=(1+√5)/4 DE=(2c)SD=(2c)^2 as line ratio is (2c)^2 area ratio is (2c)^4=16c^4=(4c^2+2c+2)(4c^2-2c-1)+6c+2=6c+2=(7+3√5)/2

△ACD is similar to △DTC TC=1 CD=AT=x AD=x^2=x+1 x=(1+√5)/2 PQ=1/x area ratio=(x/(1/x))^2=x^4=3x+2=(7+3√5)/2

20
https://www.youtube.com/watch?v=Te1WCris2ao
OAB OBD EBD ECD are all isosceles triangles CE=EB=2 OB=OA=r AC^2-AE^2=(2r)^2+16-((2r)^2+4)=12

21
https://www.youtube.com/watch?v=v7R09Rfhj_U&t=113s
AngleBDC=α from cosine rule for BCD DC^2=196 DC=14
from sine rule for BCD 3√6/sinα=14/sin45° sinα=(3/14)√3 cosα=13/14
from cosine rule for ACD AC^2=100+196-280cos(60°+α)=296-140(cosα-√3sinα)=256 AC=16

E is the intersection of AC and BD, angle AED=BEC=θ then from sine rule 10sin(120-θ)/sinθ+3√6sin(135-θ)/sinθ=13+3√3 hence tanθ=√3 θ=60° AC=10sin60/sinθ+3√6sin45/sinθ=8√3/sinθ=16

22
https://www.youtube.com/watch?v=ZKFKZrv-jl8
sine rule BE=sin40/sinx=sin70/sin(50-x) sin(x+20)+sin(x-20)=sin(100-x)-sinx=2cos(x+50)sin30=sin(40-x) sin(x-20)=sinx-sin(40-x)=2cos20sin(x-20) sin(x-20)=0 ∴x=20

23
https://www.youtube.com/watch?v=YyLwLU3q6aE&t=16s
BD=DC=1 from sine rule AD=sinx/sin(135°-x)=sin30°/sin15°=2cos15° sinx=2sin(45°+x)cos15°=sin(60°+x)+sin(30°+x) sin(30°+x)=sinx-sin(60°+x)=2cos(30°+x)sin(-30°)=-cos(30°+x) tan(30°+x)=-1 30°+x=135° x=105°

24
https://www.youtube.com/watch?v=cT40FC_qB6E&t=4s
E as intersection of AF and BC, angleBCE=BAF=θ sinθ=7/25 cosθ=24/25 AF=25/cosθ+(25-25tanθ)sinθ=25(cosθ+sinθ)=7+24=31

25
https://www.youtube.com/watch?v=__CSN_2iSdg&t=93s
AngleFCB=2θ cosθ=2/√5 as angleFBC=180-4θ FD=(20-20cos2θ+20cos4θ)/cos2θ=20(2cos2θ-1)=20(4(cosθ)^2-3)=20(16/5-3)=4

Angle between DE and CB is θ then BC=10cosθ+(10+10sinθ)tanθ=20 cosθ=4/5 sinθ=3/5 FD=40sinθ-(10+10sinθ)/cosθ=10(12/5-8/4)=4

26
https://www.youtube.com/watch?v=1hG8COtdnk8&t=18s
Same height, DF:FG=1:2 A1:A2=1:2

27
https://www.youtube.com/watch?v=OOWf8IA0xaM
AngleBCD=BAD=θ BC=√(26^2-24^2)=10 HB=10sinθ BD=26sinθ 12^2+(10sinθ)^2=(26sinθ)^2 sinθ=1/2 θ=30°

28
https://www.youtube.com/watch?v=T14gw92KwFA&t=7s
from Heron ACD=36 AC=8 x=√(100-64)=6

AngleACB=θ cosine rule cosθ=(17^2+9^2-10^2)/(2•9•17)=15/17 x=17cosθ-9=15-9=6

29
https://www.youtube.com/watch?v=dHW9UOWCad0&t=27s
CE=x AD=y then AC=y+10 sine law for △ACD 2xsinα/sin(180-4α)=y/sin(3α-90)=(y+10)/sin(90+α)=10/(cosα+cos3α)=5/(cos2αcosα)=x/(2cos2αcosα) ∴x=10

Viviani's theorem
The sum of the distances from any interior point to the sides of an equilateral triangle equals the length of the triangle's altitude.
250px-Viviani_Theorem.svg
u+s+t=h

Viviani's theorem ー Vector Proof
Viviani's theorem ー Complex Plane Proof

img004

(別解)
img003
z-a=lω^2+(1-l)ω
z-bω=m+(1-m)ω^2
z-cω^2=nω+(1-n)

z=(1-2l)ω+(a-l)
z=(b+m-1)ω+(2m-1)
z=(n-c)ω+(1-n-c)

1-2l=b+m-1=n-c
a-l=2m-1=1-n-c
∴a+b+c=3/2

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