Japan College Test Problem | A Very Nice Geometry Challenge

College Test Problem | A Very Nice Geometry Challenge
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[@harikatragadda]

With point G on DC, draw EG parallel to AD and join AG. ∠ADG = ∠AEG =2θ, making a Rhombus ADGE. Since the diagonals of Rhombus AG and DE are perpendicular, AG is parallel to EF. Hence ∆EBF is Similar to ∆AEG and consequently Isosceles. Therefore, EB = 3 and X = 8+3 = 11

[@hongningsuen1348]

The lesson to learn is not to be lured into the Pythagoras theorem trap by a given right angle and answer wanted being a length. It is obviously a trap as all 3 sides of the right-angled triangle can not be derived by given lengths.

[@Irtsak]

We must always keep in mind that by *bringing the perpendicular to the bisector, an isosceles triangle is created.* So if we bring a perpendicular from point C to DE and let P be the point where the perpendicular intersects the extension of DA. then the triangle DCP is isosceles => x=8+AP. Then it suffices to show AP=3 Have a nice day.

[@quigonkenny]

As ABCD is a parallelogram, then ∠CBA = ∠ADC = 2θ and ∠BAD = ∠DCB = (360°-4θ)/2 = 180°-2θ. As a parallelogram, opposite sides are of equal length, so BA = DC = x and AD = CB = 5+3 = 8. Draw EG, where G is the point on DC where EG is parallel to AD and CB. This will create a new parallelogram AEGD. As ED is a diagonal of AEGD and bisects ∠GDA, rhen AEGD is a rhombus and DG = GE = EA = AD = 8. Therefore GC = BE = x-8. Draw AG. As AEGD is a rhombus, AG is perpendicular to and mutually bisects ED (at point H), and AG additionally bisects ∠EAD and ∠DGE. Let ∠EAH = ∠HAD = ∠DGH = ∠HGE = α = (180°-2θ)/2 = 90°-θ. α and θ are thus complementary angles that sum to 90°. As EG is parallel to CB, and as AG is perpendicular to ED and thus parallel to EF, ∠EAG = ∠BEF = α and ∠AGE = ∠EFB = α as well, so ∆FBE and ∆GEA are similar. As ∠BEF = ∠EFB and GE = EA, ∆FBE is an isosceles triangle, and FB = BE = 3. GC = BE x - 8 = 3 x = 3 + 8 x = 11

[@hanswust6972]

Very nice solution!

[@michaeldoerr5810]

I thought that was an interesting application of the definition of parrallel lines. I thought that getting x required the Pythagoras theorem. That was bc I forgot that we were dealing with a parraleogram and parrallel lines justify corresponding angles such as 2theta. Simply using both the definition of parrallel lines and a supplementary angle, the two triangles ADE and BEF are equilateral and equiangular triangles. And side AE is 8 and EB is 3. I could be wrong.

[@comdo777]

asnwer=12cm ah hu

[@Tmwyl]

I got this one!

[@haiduy7627]

❤❤❤❤🎉🎉🎉🎉❤❤❤❤

[@zdrastvutye]

it is just a repeated calculation of 2 equations and it shows that the result does not depend on wsg: 10 print "mathbooster-japen college test problem-a very nice..." 20 dim x(2,2),y(2,2):l1=5:l2=3:wsg=80:ws=wsg*pi/180:yo=(l1+l2)*sin(ws) 30 sw=l1^2/(l1+l2)/10:xg11=0:yg11=0:goto 150 40 xg12=l1*cos(ws/2):yg12=l1*sin(ws/2):dx21=l1*cos(ws):xg21=x+dx21 50 yg21=l1*sin(ws):dx22=l1*cos(pi/2-ws/2):xg22=xg21-dx22:dy22=l1*sin(pi/2-ws/2) 60 yg22=yg21+dy22 70 a11=yg12-yg11:a12=xg11-xg12:a131=xg11*(yg12-yg11):a132=yg11*(xg11-xg12) 80 a21=yg22-yg21:a22=xg21-xg22:a231=xg21*(yg22-yg21):a232=yg21*(xg21-xg22) 90 a13=a131+a132:a23=a231+a232 100 ngl1=a12*a21:ngl2=a22*a11 110 ngl=ngl1-ngl2:if ngl=0 then print "keine loesung":end 120 zx1=a23*a12:zx2=a13*a22:zx=zx1-zx2 130 zy1=a13*a21:zy2=a23*a11:zy=zy1-zy2 140 xl=zx/ngl:yl=zy/ngl:dg=(yo-yl)/l1:return 150 x=sw:gosub 40 160 dg1=dg:x1=x:x=x+sw:x2=x:gosub 40:if dg1*dg>0 then 160 170 x=(x1+x2)/2:gosub 40:if dg1*dg>0 then x1=x else x2=x 180 if abs(dg)>1E-10 then 170 190 print "die gesuchte laenge=";x 200 x(0,0)=0:y(0,0)=0:x(0,1)=x:y(0,1)=0:x(0,2)=xg21:y(0,2)=yg21 210 x(1,0)=xg21:y(1,0)=yg21:x(1,1)=(l1+l2)*cos(ws):x(1,1)=x(1,1)+x:y(1,1)=yo:x(1,2)=xl:y(1,2)=yl 220 x(2,0)=0:y(2,0)=0:x(2,1)=xl:y(2,1)=yl:x(2,2)=x(1,1)-x:y(2,2)=yo 230 masx=1200/x(1,1):masy=850/yo:if masx run in bbc basic sdl and hit ctrl tab to copy from the results windows.