Russian Math Olympiad | A Very Nice Geometry Problem

[@RAG981]

Brilliant! Putting APD and DQC together was a phenomenal move. I have resubscribed.

[@AMOSKLEIN]

Brilliant! Superb! I'm speechless.

[@marioalb9726]

A = 3+4 = 7 cm² ( Solved √ ) Easy!!! Suppose equilateral triangle rotates over point B, with its vertices always over the sides of rectangle. Like an engine cam Then, the ratio of rectangle sides, varies, between horizontal rectangle, square and vertical rectangle Let's search the border conditions: Border condition N° 1) Triangle side BP, vertical : A₃ becomes 0, A becomes equal to A₄ , and A=A₃+A₄=0+A₄= A₄ Border condition N°2) Triangle side BQ horizontal : A₄ becomes 0, A becomes equal to A₃ , and A=A₃+A₄= A₃+0 = A₃ Border condition N°3) Triangle side PQ at 45° The rectangle becomes a square. A₄ becomes equal to A₃ A₃+A₄ = ½s²sin30° = 7 cm² s² = 14/sin30°= 28cm² --> s=5,2915cm A = ½b.h = ½s½s= ¼s²= 7cm² A = A₃+A₄ = 3,5+3,5 = 7 cm² IN ALL 3 border conditions A = A₃ + A₄ Always A = A₃ + A₄ , for any angular position of equilateral triangle

[@petermischler7324]

The sum of the two triangles is always the triangle in the upper right corner. I proved it by substituting Q by (rcos(phi), rsin(phi)) and P by (rcos(phi+60°),(rsin(phi+60°)), where phi is the angle of CBQ and using the trigonometric identities.

[@alainpeugny1146]

Agree. In that case, trigonometry formulas give an immediate result.

[@oscarcastaneda5310]

My God, we did it the very same way : ) The Analysis took me some time but I'm glad I finally got to see the light in joining the two sides of the equilateral triangle like in your solution.

[@jimlocke9320]

It would appear that Θ could take on any value between 0° and 30° and we could solve a special case and apply it to the general case, but that is not the case! Only one value of Θ makes the figure valid. Another approach: At 3:10, x = a tan(Θ) and y = b tan(30° - Θ). From the areas, we generate two equations: For ΔABP: ax/2 = 3, ax = 6, (a)(a tan(Θ) = 6 and a²tan(Θ) = 6. For ΔBCQ, by/2 = 4, by = 8, (b)(b tan(30° - Θ) = 8 and b²tan(30° - Θ) = 8. From the Pythagorean theorem, for ΔABP, a² + (a tan(Θ))² = (PB)², or a²(1 + tan²(Θ))² = (PB)² and, for ΔBCQ, b² + (b tan(30° - Θ))² = (BQ)² or b²(1 + tan²(30° - Θ)) = (BQ)². However, PB = BQ, so a²(1 + tan²(Θ))² = b²(1 + tan²(30° - Θ)). From a²tan(Θ) = 6, a² = 6/tan(Θ) and b² = 8/tan(30° - Θ). These values can be substituted in a²(1 + tan²(Θ))² = b²(1 + tan²(30° - Θ)) for (6/tan(Θ))(1 + tan²(Θ))² = (8/tan(30° - Θ))(1 + tan²(30° - Θ)), which can be solved for Θ. The area of the rectangle is ab. The side of the equilateral ΔBPQ can be calculated and used to compute its area. Area ΔPDQ = area rectangle ABCD - ΔABP - ΔBCQ - ΔBPQ. This is a much more difficult way of solving but it should produce a result!

[@shaharnevo2086]

Solution without sinus: From the center O of the equilateral triangle draw lines to P and Q and also two verticals to the upper and right sides of the rectangle.These 4 lines and these two sides of the rectangle create two right triangles. One is similar to triangle APB, One to BQC and one to PDQ. The ratio of similarity between those to APB and BQC is sqrt(3). From this we get equation that gives the desired 7.

[@markwu2939]

Set BP = x and ∠ABP = θ. For ΔBAP, the area = x²sin(2θ)/4 = 3. For ΔBCQ, the area = x²sin(60°-2θ)/4 = 4. These two equations give that cot(2θ) = 11/3√3. For ΔBCQ, the area = x²sin(60°+2θ)/4 = x²sin(60°)cos(2θ)/4 + x²cos(60°)sin(2θ)/4 = sin(60°)[x²sin(2θ)/4]cot(2θ) + cos(60°)[x²sin(2θ)/4] = (√3/2)(3)(11/3√3) + (1/2)(3) = (11/2)+(3/2) = 7

[@senirudilmith-n8z]

thank you sir.. everyday learn from you.. and improve my math skill..

[@RealQinnMalloryu4]

(3)^2 (4)^2={9+16}=25 360°ABCD/25=7.5 (ABCD ➖ 7ABCD+5).

[@imetroangola17]

Linda solução!!! 🎉🎉🎉

[@math-problem6940]

Are there another way than rotation way that easier ?

[@zdrastvutye]

it is nested like a russian babushka. however, i solved it without angle functions, but i got a slightly different result: 10 print "math booster-russian math olympiad-a very nice geometry problem" 20 a1=3:a2=4:na=a1+a2:sw=sqr(a1+a2)/53:l1=sw:dim x(2,2),y(2,2):goto 130 30 l2=2*a1/l1:l3=2*a2/l1:dg=(l1^2+l2^3-l4^2)/na:return 40 l4=sw:gosub 30 50 l41=l4:dg1=dg:l4=l4+sw:if l4>20*l1 then 90 60 l42=l4:gosub 30:if dg1*dg>0 then 50 70 l4=(l41+l42)/2:gosub 30:if dg1*dg>0 then l41=l4 else l42=l4 80 if abs(dg)>1E-10 then 70 90 return 100 gosub 40:if l4>20*l1 then return 110 l5=sqr(l4^2-l3^2):dfu1=(l1-l3)^2/na:dfu2=(l5-l2)^2/na:dfu3=l4^2/na 120 df=dfu1+dfu2-dfu3:return 130 l1=sw 140 gosub 100:if l4>20*l1 then else 160 150 l1=l1+sw:goto 140 160 l11=l1:df1=df:l1=l1+sw:if l1>10*sqr(na) then stop 170 l12=l1:gosub 100:if df1*df>0 then 160 180 l1=(l11+l12)/2:gosub 100:if df1*df>0 then l11=l1 else l12=l1 190 if abs(df)>1E-10 then 180 else 210 200 xbu=x*mass:ybu=y*mass:return 210 print l1;:ar=(l1-l3)*(l5-l2)/2:print "die flaeche=";ar 220 x(0,0)=0:y(0,0)=0:x(0,1)=l5:y(0,1)=0:x(0,2)=l5:y(0,2)=l3 230 x(1,0)=l5:y(1,0)=l3:x(1,1)=l5:y(1,1)=l1:x(1,2)=l2:y(1,2)=l1 240 x(2,0)=0:y(2,0)=0:x(2,1)=l2:y(2,1)=l1:x(2,2)=0:y(2,2)=l1 250 masx=1200/l5:masy=850/l1:if masx run in bbc basic sdl and hit ctrl tab to copy from the results window. you may add "@zoom%=1.4*@zoom%" for fullscreen graphics

[@blogfilmes1134]

Linda resolução !

[@marcgriselhubert3915]

Fine.

[@shrikrishnagokhale3557]

In triangle APC ,P=150,A=C=15?

[@shaharnevo2086]

Test