Question

3. ABCD is a rectangle, P lies on AD and Q on AB. The
triangles PAQ, QBC and PCD all have the same area,
and BQ = 2. The length of AQ, is :
(A) 3+ 15
(B) 2/3
(C) 5+1
(D) not uniquely determined

Answer 1

Answer:

The points P and Q lie on AD and AB respectively. If the triangles PAQ, QBC and PCD all have the same areas and BQ = 2 then AQ = ? 1+√5.

option c you want to remove+and√ ok answer is 5+1

Answer 2

Answer:

Start with your triangle ABC.

Take E on BA such that BCE isosceles in C.

Then take F on AC, such that CEF be isosceles in E.

Finally take G on AB such that EFG isosceles in F.

The angles are easily computed and you get

BCE = 20 so that ECF = 80 - 20 = 60. Hence ECF is equilateral.

FEG = 180 - 60 - 80 = 40. Hence EGA = 140.

This implies that GFA = 180 - 140 - 20 = 20 and FGA is isosceles in G.

Notice that BC = CE = EF = EG = GA. Hence G = D. Moreover CFG is isosceles in F.

Since CFG = 160, FGC = 10 and AGC = ADC = 140 + 10 = 150.

Edit: solution #4 (#2 and #3 are Duke's)

Take BCE an equilateral triangle with E the top vertex. Then consider a rhombus EFGH pointing upwards on top of BCE with angle GEF = 20 and EF = EB.

Suppose FH and BC point in the same direction.

Angle(BEF) = 180 - 10 - 30 = 140. So the isosceles triangle BEF has angles 20,20,140.

Therefore BF || EH || FG and BFG are aligned. This implies that G = A and H = D and then you know everything.

edit @ Duke, your #5 is actually contained in #4: indeed in #4 CEFH are obviously the sides of a regular polygone with 18 sides since 10 = 180/18 whose center is precisely your P.

Step-by-step explanation:

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