China Math Olympiad | You should be able to solve this | 2 Methods 

The problem statement doesn't specify the location of point Q on AD. Let's consider the special case of Q being at the midpoint. Then, PQ is parallel to AB and CD. In that case, length AB becomes 11 and BC becomes 6, which is also the diameter of the circle. The area of the rectangle becomes 11 times 6, or 66. Note that both solution methods break down for this special case because the triangles become straight lines. A better way to state the problem might be to identify the midpoint of AD and specify that Q is between the midpoint and point D. The limiting case of Q being at the midpoint is straightforward to solve, as done here. On a multiple choice test, just select its solution, 66, as the correct answer. Otherwise, use the special case to verify your solution for the general case.

Always enjoyable - your explanations, written statements and constructions are always so clear to follow - and you come up with some very ingenious constructions! Thank you.

Considering the right triangle inside the circle in which diameter PE // to DC is the hypotenuse and chord SP is its major side, ES the minor. And considering a second right triangle drawing the estension of OP until it intersects side AD on point F. These two triangles PES and PFQ are similar because are both right and have angle P in common, then we can write: 6 : DC = 2r : (6+5) Area = DC*2r = 6*11 = 66

Outstanding problem with 2 elegant solutions. Success wjith problems of this caliber require Dedication/Practice, starting with easier ones, gradually increasing the difficulty. Inherent problem solving ability and being able to process at higher levels surely help, but they don't guarantee anything.

Very good explanation!

I love the way you edited this, it's so visually appealing.

Should the statement of the problem not specify that circle is tangent to 3 sides of rectangle, and that point P is a point of tangency? Otherwise we are making all sorts of assumptions to solve.

I think that I now understand: similar to the last problem. The first method involves-similar the previous problem setting up right angles ans with that similar right angled triangles. That is without iterating the Pythagoras theorem. The second method involves corresponding right angles and similar right triangles. Both have to be set up AFTER right angles are indented. I think that the first method is easier to understand. I could be wrong.

The definition of the problem never states that P id the midpoint of BC, you just assume that.

the result does not depend on l3=... see line 20. the graphics will show which one of the results makes sense: 10 print "mathbooster-china math olympiad-you should be able" 20 dim x(1,3),y(1,3):l1=5:l2=6:sw=l1^2/(l1+l2)/10:l3=.5:r=sw:goto 120 30 lb=sqr((l1+l2)^2-(r-l3)^2):dy=(r-l3)*l1/(l1+l2):y=l3+dy 40 dis=sqr(abs(r*r-(y-r)^2)) 50 x=lb-r+vz*dis:dgu1=x*x/l1^2:dgu2=(y-l3)^2/l1^2 60 dg=dgu1+dgu2-1:return 70 gosub 30 80 dg1=dg:r1=r:r=r+sw:r2=r:gosub 30:if dg1*dg>0 then 80 90 r=(r1+r2)/2:gosub 30:if dg1*dg>0 then r1=r else r2=r 100 if abs(dg)>1E-10 then 90 110 print r:ages=lb*2*r:print ages:return 120 vz=1:gosub 70:gosub 140:gosub 180:r=sw:vz=-1:gosub 70:gosub 140:gosub 180 130 end 140 x(0,0)=0:y(0,0)=0:x(0,1)=lb:y(0,1)=0:x(0,2)=x(0,1):y(0,2)=r 150 x(0,3)=0:y(0,3)=l3:x(1,0)=0:y(1,0)=l3:x(1,1)=lb:y(1,1)=r 160 x(1,2)=x(1,1):y(1,2)=2*r:x(1,3)=0:y(1,3)=y(1,2) 170 xm=lb-r:ym=r:return 180 masx=1200/lb:masy=800/2/r 190 if masx

φ = 30°; ∎ABCD → AB = CD = a = QN; BC = AD = 2r = BP + CP = r + r; QP = QS + SM + PM = 5 + 3 + 3; PQN = OMP = δ → cos⁡(δ) = 3/r = a/11 → 2ar = 66

Nada se dice de la pendiente del segmento PQ, entonces podemos decir arbitrariamente que PQ es horizontal y el problema se vuelve trivial. >>> Área ABCD =PQ*BC =(5+6)*6=66. Gracias y saludos.