Try it! Just a little geometry

Geometry Level 3

A B C D ABCD is a square which has an area of 1 1 . E E and F F are points on the square's sides as shown in the figure above. Δ A E F \Delta AEF has a perimeter of 2 2 . Diagonal B D BD divides Δ E C F \Delta ECF into two pieces: a smaller triangle and a quadrilateral. Which has a larger area?

They have the same area The blue quadrilateral The red rectangle

Áron Bán-Szabó by Áron Bán-Szabó , 17, Hungary

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1 solution

In the next figure the E E and F F points are P P and Q Q . The square was rotated by 90 ° 90° .

If A P = x AP=x and A Q = y AQ=y then P B = 1 − x = P ′ B ′ PB=1-x=P'B' , Q D = 1 − y = Q ′ D ′ QD=1-y=Q'D' and P Q = P ′ Q ′ = 2 − x − y PQ=P'Q'=2-x-y . So P Q ′ = P B + B Q ′ = 1 − x + 1 − y = 2 − x − y PQ'=PB+BQ'=1-x+1-y=2-x-y . Since C Q = C Q ′ CQ=CQ' , the P C Q , P ′ C Q ′ , P C Q ′ PCQ, P'CQ', PCQ' triangles are congurent because they have the same sides. So P C Q ∠ = P C Q ′ ∠ = P ′ C Q ′ ∠ = α PCQ\angle =PCQ'\angle =P'CQ'\angle =\alpha , and Q C D ∠ + P ′ C B ′ ∠ = 90 ° − α QCD\angle+P'CB'\angle=90°-\alpha . Now we get D C B ′ ∠ = 180 ° = 3 ∗ α + ( 90 ° − α ) = 2 α + 90 ° DCB'\angle =180°=3*\alpha+(90°-\alpha)=2\alpha+90° . So α = 45 ° \alpha=45° .

You can find the end of the solution here: Solution part 2

1 Helpful 0 Interesting 0 Brilliant 0 Confused

@aron ban With the configuration provided in the problem you can go for pure sides chasing as well.

Vishwash Kumar ΓΞΩ - 4 years ago

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