It's much simpler than that! You just need to rotate the triangle APB around point B by 90 deg. counterclockwise. Point P will become P1. Let's draw the segment PP1. Triangle PP1B is an isosceles right triangle. The angle P1PB is 45 deg. It is easy to check that triangle APP1 is right-angled (Pythagoras' theorem). The angle APP1 is 90 deg. The angle we are looking for is: 90 + 45 = 135 !!

Stewarts theorem might fail (or work)but If it's too tough don't crack it

That is probably the first time that I have heard of ever rotating a triangle. And I am not surprised that the Law od Cosines had to be used. When is rotating triangle necessary may I ask? And also theta is an obtuse angle. Would the domain of cosine allow for the existence of acute angles a la Law of Cosines? I think so, but I want to check.

φ = 30°; ∎ABCD → AB = AF + BF = (a - y) + y = BC = BG + CG = x + (a - x) = CD = AD = AE + DE = x + (a - x) = a = EG = EP + GP = (a - y) + y; BP = 1; BPA = θ = ? CP = 3; AP = √7 → (a - y)^2 + x^2 + 2 = (a - x)^2 + y^2 = 9; y = √(1 - x^2) → (a^2 - 6)/2a = √(1 - x^2); x^2 = ((a^2 - 8)/2a)^2 → (a^2 - 6)^2 + (a^2 - 8)^2 = 4a^2 → z ∶= a^2 → z = 8 ± √14 → a ≥ 3 → z = 8 + √14 → z = 8 - 2(√7)cos⁡(θ) → cos⁡(θ) = -√2/2 → θ < 6φ → θ = 3φ + 3φ/2 = 9φ/2 btw: ∆ BCP → BCP = α; BC = a; BP = 1; CP = 3 → cos⁡(α) = (a/150)(57 - 4√14) → α ≈ 16,22° ∆ ABP → PAB = β; AB = a; AP = √7; BP = 1 → cos⁡(β) = (√7/350)(√(8+ √14))(49 - 3√14) → β ≈ 11,9°

We can solve it only using the law of cosine for three angles - theta,angle(ABP)=a and angle(CBP)=b. Let AB=x cos(a)=(1+x^2-sqrt(7)^2)/(2*1*x)=(x^2-6)/(2* x) cos(b)=(1+x^2-3^2)/(2*1*x)=(x^2-8)/(2*x) a+b=pi/2 cos(b)=sin(a) from cos(a)^2+sin(a)^2=1 we can get an equation of x X^4-16*x^2+50=0 from which x^2=8+sqrt(14) . With x known, we can then calculate cos(theta)=(1+sqrt(7)^2-(8+sqrt(14))/(2*1*sqrt(7))=-sqrt(2)/2 Which means theta=3*pi/4 or 145 degrees.

Io ho usato il teorema dei coseno..dopo i calcoli risulta 2√7(cosθ)^3-8(cosθ)^2-√7cosθ+4=0...1 soluzione è cosθ=1/√2,ma non va bene perché θ è>90...altra soluzione è cosθ=((8-√14)-√(78+16√14))/4√7=-1/√2...θ=135..la cubica diventa (cosθ-1/√2)(cosθ+1/√2)(2√7cosθ-8)=0...quindi l'unico angolo ottuso è 135

Cunning

Reflect ∆ADB about the side AB to form ∆AD'B. Similarly reflect ∆ADC about the side AC to form ΔAD''C. AD' =AD" = X and ∠D'AD"=90° and ∠AD'B =∠AD"C=90° Extend D'B and D"C to meet at E with ∠E=90°, consequently forming the Square AD'ED" with side X. The Right ∆BEC formed at the bottom, has sides (X-2, X-3, 5) which gives X=6 Hence area(∆ABC) =½*5*6=15

Shorter at the end: x=27/8, Hence 3*11- 2*27/8=z

Here is my solution Let BE⊥AC construction and AE=x ,EC=y. Notice that right triangle ABE is isosceles, so BE=AE=x Pythagoras theorem in right triangle ABE => AB²=x²+x² => AB=x√2 Also Pythagoras theorem in ΔBEC => x²+y²=25 (1) Express AD using Pythagoras theorem in triangles ADC, ABD : AD²=(x+y)²-4 and AD²=2x²-9 so (x+y)²-4=2x²-9 => …… => x²-xy=15 => y=(x²-15)/x. Now substitute y in (1) and you have : x²+((x²-15)/x)²=25 => ……. 2x⁴-55x²+225=0 Let ω=x² and we have 2ω²-55ω+225=0 => D=1125>0 ω=45/2 or ω=5 (is rejected) ω=45/2 => x²=45/2 We found AD²=2x²-9=> AD²=(2•45)/2-9 =>AD²=36 AD=6 and area of triangle (ABC)=(AD•BC)/2=(6•5)/2=15 s.u *ω=5 is rejected cause ω=5 => x²=5* *but y=(x²-15)/x <0*

X=2√3

S(ABC)=15

Thanks

*Outra solução:* Sejam AB=x, AC=y, AD= h e [ABC]= A. Usando a lei dos cossenos no triângulo ABC e sabendo que sin45°=cos45, temos: *25=x^2 + y^2 - 2xysin45° (1)* Os triângulos ABD e ADC são triângulos retângulos, logo pelo teorema de Pitágoras, temos: x^2=9 + h^2 e y^2=4 + h^2, somando membro a membro ambas expressões, obtemos: x^2+ y^2=13+ 2h^2, substituindo na expressão (1), temos: 25=13 +2h^2- 2xysin45° => *12=2h^2- 2xysin45° (2).* Observe que: A=xysin45°/2 => *xysin45°=2A* e h.5/2=A => *h=2A/5.* Substituindo tais expressões na expressão (2), temos o seguinte resultado: 12=16A^2/25 - 4A ÷(4) => 3=A^2/25 - A => 2A^2 - 25A-75=0. Daí, ∆=1225 => √∆=35, como A>0 então a única solução possível, segundo a fórmula de Bhaskara é: A=(25 + 35)/4 => *A=15.*

Sine law twice should be enough |AD|/sin(20) = 6/sin(10) |AD| = 6sin(20)/sin(10) |AD| = 12cos(10) x/sin(30) = 12cos(10)/sin(80) x/sin(30) = 12sin(80)/sin(80) 2x = 12 x = 6

Nice second solution by Math booster . I solved it, using only Pythagoras theorem. But the system was a nightmare. I will post it tomorrow cause i am tired now 😊

Right now I think that the second method is easier to understand and easier to intuit. Well tbf the first method was easy to understand and the second method is easier bc congruent and corresponding angles with a common corresponding and congruent angle result in a pair of triangles with the same letters as congruent corresponding angles being congruent. Easier to understand than the comments!

Without Pythagoras: 5*sqrt 2 is a diagonal of a Square, Hence 5 is BD and likewise DC!

Note that sin <PBA=sin(270-<PBC)=-cos<PBC which can be obtained from triangle PBC and cos<PBA is obtained from triangle PBA . Then squaring the sine and cosine of <PBA and adding (=1) gives the quadratic eqnuation for the square of X.

Hi! I have constructed a circumscribed circle of this triangle. Its radius is 2.5*sqrt(2). Further, it is not difficult to find the height of the triangle drawn from the vertex of this angle. It is equal to 6. Then the area of the triangle is 15. THANK YOU FOR THE BEAUTIFUL TASK!

It occured to me that the 45 degree angle at the apex might be divided by the same ratio as the two elements of the base stand relative to the total length of BC. That is to say 3 is 60% of the base, while 2 is 40% of the base. Applying that ratio to the 45 degree angle at the apex and you have 27 degrees for angle BAD and 18 degrees for angle CAD. Anyway, I get an area of 14.92. Which is pretty close to the 15 that the author has put out. Of course this could be sheer luck on my part, but I think there's a theorem somewhere that states this principle. Just can't remember it.

I prefer the first method using trigonometric ratio and trigonometric identity with no construction needed. It can be more concisely presented as: 1. Let angle BAD be B and angle CAD be C such that B + C = 45 as given. Let height of triangle ABC i.e. AD be H. 2. Use definition of tangent ratio: tanB = 3/H and tanC = 2/H 3. Use compound angle identity for tangent ratio: tan(B+C) = (tanB + tanC)/(1 - tanB tanC) = tan 45 = 1 Hence [(3/H) + (2/H)]/[1 + (3/H)(2/H)] = 1 H^2 - 5H - 6 = 0 H = 6 or H = -1 (negative length rejected) 4. Area of triangle ABC = (1/2)(3+2)(6) = 15. The second method using construction of isosceles triangle which is similar to the original triangle is not that intuitive and it is more complicated to find the height AD from equations formulated by Pythagoras theorem and proportionality of corresponding sides of similar triangle.

With trigonometry it would be very easy

I extended BC and made your angle "alfa" 45 degrees. A big triangle area minus a newly constructed triangle area = the area under question. From the point B I dropped two perpendiculars-one on the CE, another on AC. The BCF triangle is a 5-4-3 straight triangle. Using the similarity of thre triangles I found all sides and hights. But my result was not 15 but 14. I did not check again the calculations, sorry. @ 9:54, sorry, I think the ratio should anclude not EC but EA. I wish anyone can check this, it is not obvious from the drawing.

there are 2 different ways to solve this- the direct product or the thales circle: 10 print "mathbooster-romania math olympiad-very nice geometry":nu=35 20 dim x(2),y(2):l1=6:r1=l1:r2=l1/2:sw=l1/10:xd=sw:goto 50 30 yd=sqr(r2^2-(xd-r2)^2):dgu1=(r2-xd)*xd/l1^2:dgu2=(l1-yd)*yd/l1^2 40 dg=dgu1+dgu2:return 50 gosub 30 60 dg1=dg:xd1=xd:xd=xd+sw:xd2=xd:gosub 30:if dg1*dg>0 then 60 70 xd=(xd1+xd2)/2:gosub 30:if dg1*dg>0 then xd1=xd else xd2=xd 80 if abs(dg)>1E-10 then 70 print xd,yd 90 lad=sqr(xd^2+(yd-l1)^2):l=sw:dx=xd/lad:dy=(yd-l1)/lad :goto 120 100 dxl=l*dx:dyl=l*dy:xc=dxl:yc=l1+dyl:dgu1=xc^2/l1^2:dgu2=yc^2/l1^2 110 dg=dgu1+dgu2-1:return 120 gosub 100 130 dg1=dg:lu1=l:l=l+sw:lu2=l:gosub 100:if dg1*dg>0 then 130 140 l=(lu1+lu2)/2:gosub 100:if dg1*dg>0 then lu1=l else lu2=l 150 if abs(dg)>1E-10 then 140 160 print "die gesuchte entfernung=";l 170 mass=850/l1:goto 190 180 xbu=x*mass:ybu=y*mass:return 190 xmu=0:ymu=0:ru=l1:we=pi/2:gosub 200:goto 240 200 x=ru:x=x+xmu:y=ymu:gosub 180:xba=xbu:yba=ybu:for a=1 to nu:wa=we*a/nu 210 x=ru*cos(wa):x=x+xmu:y=ru*sin(wa):y=y+ymu:gosub 180:xbn=xbu:ybn=ybu:goto 230 220 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return 230 gosub 220:next a:return 240 ru=l1/2:xmu=ru:we=pi:gosub 200:x(0)=0:y(0)=l1:x(1)=xd:y(1)=yd:x(2)=xc:y(2)=yc 250 x=x(0):y=y(0):gosub 180:xba=xbu:yba=ybu:for a=1 to 2:x=x(a):y=y(a):gosub 180 260 xbn=xbu:ybn=ybu:gcola:gosub 220:next a 270 move 0,50:gcol8:print "run in bbc basic sdl and hit ctrl tab to copy from the results window" mathbooster-romania math olympiad-very nice geometry 4.8 2.4 die gesuchte entfernung=7.2 run in bbc basic sdl and hit ctrl tab to copy from the results window >

I solved it by the 1st method....😊

Nice solution! It is so nice solution that seems to be simple. Congrat

tan(theta) = 2/h tan(45-theta) = 3/h (1-tan(theta))/(1+tan(theta)) = 3/h (1-2/h)/(1+2/h) = 3/h (h-2)/(h+2) - 3/h = 0 (h(h-2)-3(h+2))/(h(h+2)) = 0 (h^2 - 5h - 6)/(h(h+2)) = 0 h^2 - 5h - 6 = 0 (h - 6)(h + 1) = 0 h = 6 Area = 1/2*5*6 Area = 15

I did it using b*h formula + 1/2absin c

Your method and solution are so intresting!! Wish you the best 🙂

OB😂

BP = 2a BO = R√2 OP = BP - BO OP² + a² = R²

No hace falta aclarar que O es el centro de la circunferencia?

As ∠BDA is an exterior angle to ∆ADC at D, ∠BDA = ∠CAD+∠DCA = 10°+10° = 20°. As ∠CAD = ∠DCA = 10°, ∆ADC is an isosceles triangle and AD = DC. Method 1: Drop a perpendicular from D up to CA at E. As ∆ADC is an isosceles triangle, by rule, DE bisects ∆ADC and forms two congruent right triangles ∆CED and ∆DEA. As ∠EAD = ∠DCE = 10°, ∠ADE = ∠EDC = 90°-10° = 80°. As ∠ABD = 60° and ∠BDA = 20°, ∠DAB = 180°-(60°+30°) = 100°. Extend BA to F and draw DF such that FA = PD, ∠DFA = 90°, and, as ∠FAD = 180°-100° = 80°, ∠ADF = 90°-80° = 10°. As DA is common, by construction, ∆DFA is congruent with ∆DEA (and by extension ∆CED. As ∠ADF = 10° and ∠BDA = 20°, ∠BDF = 30° and ∆BDF is a 30-60-90 special right triangle, such that FB = BD/2 = 1 and DF = √3FB = √3. As FD = √3 and CE = EA = FD, then x = CA = CE + EA = √3 + √3 = 2√3. x = 2√3 units. Method 2: As ∠BDC is a straight angle, ∠ADC = 180°-20° = 160°. As ∠ADC is an exterior angle to ∆BDA at D, ∠DAB = ∠ADC-∠ABD = 160°-60° = 100°. By the law of sines: BD/sin(100°) = DA/sin(60°) DA = BDsin(60°)/sin(100°) DA = 2(√3/2)/sin(100°) = √3/sin(100°) DC = DA = √3/sin(100°) DC = √3/cos(10°) DC/sin(10°) = CA/sin(160°) CA = DCsin(160°)/sin(10°) x = √3sin(180°-20°)/sin(10°)cos(10°) x = √3sin(20°)/sin(10°)cos(10°) x = √3(2sin(10°)cos(10°))/sin(10°)cos(10°) x = 2√3 units

Nice question and nice solition

once the position of the center of the semicircle is known, the line on which the second center point must be on, can be described, see line 140: 10 vdu5:for a=0 to 15:gcola:print a;:next a:vdu3:gcol8 20 @zoom%=@zoom%*1.4:nu=55 30 print "math booster-germany math olympiad challenge-very nice geometry problem" 40 dim x(2),y(2):l1=3:l2=4:sw=l1^2/(l1+l2)/10:ym1=0:r1=sw:yp=ym1 50 xg1=0:yg1=l1:xg2=l2:yg2=0:dx=xg2-xg1:dy=yg2-yg1:goto 90 60 xm1=r1:xp=xm1:zx=dx*(xp-xg1):zy=dy*(yp-yg1):k=(zx+zy)/(dx^2+dy^2) 70 xl=k*dx:xl=xl+xg1:yl=k*dy:yl=yl+yg1:lo=sqr((xp-xl)^2+(yp-yl)^2) 80 dg=r1/lo:dg=dg-1:return 90 gosub 60 100 dg1=dg:r11=r1:r1=r1+sw:r12=r1:gosub 60:if dg1*dg>0 then 100 110 r1=(r11+r12)/2:gosub 60:if dg1*dg>0 then r11=r1 else r12=r1 120 if abs(dg)>1E-10 then 110 130 l=sw:lm=sqr(l1^2+xm1^2):goto 160 140 xm2=xm1/lm*l:r2=xm2:dy=l/lm:ym2=l1*(1-dy):dgu1=(xm2-xm1)^2/l2^2:dgu2=(ym1-ym2)^2/l2^2 150 dgu3=(r1+r2)^2/l2^2:dg=dgu1+dgu2-dgu3:return 160 gosub 140 170 dg1=dg:lu1=l:l=l+sw:lu2=l:gosub 140:if dg1*dg>0 then 170 180 l=(lu1+lu2)/2:gosub 140:if dg1*dg>0 then lu1=l else lu2=l 190 if abs(dg)>1E-10 then 180 200 print l,"%",r2 210 x(0)=0:y(0)=0:x(1)=l2:y(1)=0:x(2)=0:y(2)=l1:masx=1200/l2:masy=850/l1 220 if masx<masy then mass=masx else mass=masy 230 goto 250 240 xbu=x*mass:ybu=y*mass:return 250 x=x(0):y=y(0):gosub 240:xba=xbu:yba=ybu:for a=1 to 3 260 ia=a:if ia=3 then ia=0 270 x=x(ia):y=y(ia):gosub 240:xbn=xbu:ybn=ybu:goto 290 280 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return 290 gosub 280:next a:goto 320 300 x=xmu+ru:y=ymu:gosub 240:xba=xbu:yba=ybu:for a=1 to nu:wa=a/nu*we:x=ru*cos(wa):x=x+xmu 310 y=ru*sin(wa):y=y+ymu:gosub 240:xbn=xbu:ybn=ybu:gosub 280:next a:return 320 x=xm1:y=ym1:xmu=xm1:ymu=ym1:ru=r1:we=pi:gosub 300:gcol9:x=xm2:y=ym2:gosub 240:circle xbu,ybu,r2*mass 330 move 0,50:gcol8:print "run in bbc basic sdl and hit ctrl tab to copy from the results window" > 8 9 10 11 12 13 14 15 math booster-germany math olympiad challenge-very nice geometry problem 1.28115295% 0.572949017 run in bbc basic sdl and hit ctrl tab to copy from the results window 0 1 2 3 4 5 6 7

Nice Video , Keep on going . I sugest to mention if the provlem is solved with pure Geometry , cause sometimes we try to prove the impossible !

Rabbit out of the hat

This is a nice solution because this shows that why and how the simplest solution and just staring at you in plain-sight. It just requires right angles to be identified in order to justify "a", a right-angled square, a certain circle theorem that correlates to that square, and then use the circle theorem in order to justify R, and then simplify in terms of a. I almost forgot to include that cos45 degrees is part of the first step of simplifying in terms of "a" and R. And your explanation makes more sense than the comments as always!

Thank you so much for every video you make! As always, it was a very interesting exercise, having to deal with a problem only with geometry!

I want to have a go at this so I haven't listened to the whole video yet, but... You didn't mention, in the 'givens' preface, whether point O is the center of the circle. Is it?

Solução linda! Parabéns!

OB =√2R. ∆ABC is Isosceles, hence, BP= 2a. If OP= b, then b=2a-√2R In the ∆OPD, R²=b²+a² This gives, 5a²-4√2Ra+R²=0 Cosθ= a/R = 1/(2√2-√3)

Answer: θ ≈ 20.2°

Nice solution❤

From Figure mentioned by Mathbooster length of OE=2*OC Cos70°=10Cos70°=OP(as angle OEP=angle OPE=70°). Area of shaded region=Area of (∆BOD-∆EOP)=1/2*5*5*Sin40°-1/2*10Cos70°*10Cos70*Sin40°=25/2 sin40°(1-4Cos^2 70°); •Let us simplify trigonometric expression sin40°(1-4Cos^2 70°)= 2Sin20°Cos20°(1-4sin^2 20°) and now put 1=cos^2 20°+Sin^2 20° Following expression becomes 2Sin20°Cos20°(Cos^2 20°-3Sin^2 20°)= 2Sin20°(Cos^3 20°-3Cos20° Sin^2 20°)= [ Cos3§=Real parts of(Cos§+i Sin§)^3=Cos^3 §-3Cos§Sin^2 §(From De -Moivre theorem).] 2Sin20°(Cos60°)=Sin20°. Area =25/2 Sin20° Remark : This expression could also be simplified in following way : 2Sin20°Cos20°(1-4sin^2 20) =2Sin20°Cos20(1-4(1-cos^2 20°) =2Sin20°Cos20(4Cos^2 20°-3) =2Sin20(4Cos^3 20°-3Cos20°)[from well known formula of Cos 3§=4Cos^3 §-3Cos§)] =2Sin20°(Cos60°) =Sin20°.

If we let BE = x then side length CE we can calculate from Pythagorean identity ABE is similar to EDC by angle measures equality x/2 = sqrt((10-x)^2+5^2)/5 (x/2)^2 - ((10-x)^2+5^2)/25 = 0 x^2/4 - ((10-x)^2)/25 - 1 = 0 x^2/4 - (x^2-20x+100)/25 - 1 = 0 x^2/4 -x^2/25 +4/5x-5 = 0 21x^2 + 80x - 500 =0 80^2-4*21*(-500) = 6400+42000 80^2-4*21*(-500) =48400 x = (-80+220)/42 x = 140/42 x = 10/3 Area = 1/2*10*5 - 1/2*(10-10/3)*5 Area = 5/2 (10-(10-10/3)) Area = 5/2*10/3 Area = 25/3

Side length |AC| from sine law in triangle ACD Side length |AB| from cosine law in triangle ABC Angle measure ABC = theta from cosine law in triangle ABC

Assume Chord as A and B and centre as O, draw AO ane BO, we know that inscribed angles subtended by the same arc (AB) are equal, hence angle AOB will be 60 degeee, and traingle AOB will be equilateral, hence AB would be 5.

to find the angles, draw a line in D that is parallel to line BC: 10 vdu5:for a=0 to 15:gcola:print a;:next a:vdu0:gcol8:@zoom%=@zoom%*1.4 20 print "mathbooster-russian math olympiad-a very nice geometry problem" 30 dim x(1,2),y(1,2):la=1:w1=75: sw=.1:wth=sw:goto 60 40 w2=90-w1:w3=wth+w2:w4=180-w3:lbu=la*sin(rad(wth))/sin(rad(w4)) 50 l2=2*la*cos(rad(w1)):dg=l2/lbu:dg=dg-1:return 60 gosub 40 70 dg1=dg:wth1=wth:wth=wth+sw:gosub 40:wth2=wth:if dg1*dg>0 then 70 80 wth=(wth1+wth2)/2:gosub 40:if dg1*dg>0 then wth1=wth else wth2=wth 90 if abs(dg)>1E-10 then 80 100 print "der gesuchte winkel="; wth;"°" 110 x(0,0)=0:y(0,0)=0:x(0,1)=2*la*sin(rad(w1)):y(0,1)=0:x(0,2)=0:y(0,2)=l2 120 x(1,0)=x(0,1)-l2:y(1,0)=0:x(1,1)=x(0,1):y(1,1)=0:x(1,2)=x(0,1)-la*cos(rad(w2)) 130 y(1,2)=la*sin(rad(w2)):masx=1200/x(0,1):masy=850/l2 140 if masx<masy then mass=masx else mass=masy 150 goto 170 160 xbu=x*mass:ybu=y*mass:return 170 for a=0 to 1:x=x(a,0):y=y(a,0):gosub 160:xba=xbu:yba=ybu:gcol a:for b=1 to 3:ib=b:if ib=3 then ib=0 180 x=x(a,ib):y=y(a,ib):gosub 160:xbn=xbu:ybn=ybu:goto 200 190 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return 200 gosub 190:next b:next a:gcol 8 210 print "run in bbc basic sdl and hit ctrl tab to copy from the results window" hit ctrl tab to copy from the0results window 12 13 14 15 >athbooster-russian math olympiad-a very nice geometry problem der gesuchte winkel=15° run in bbc basic sdl and hit ctrl tab to copy from the results window