MathDB

18th Cabri Clubs 2007

Part of Cabri Clubs

Subcontests

(1)
2

18th Cabri Clubs 2007, round 1, level 1, 3 problems, Argentinian geo contest

level 1
p1. Let PP be a point inside triangle ABCABC such that the line APAP is the bisector of the ∠BAC\angle BAC, BPBP ​​is the bisector of ∠ABC\angle ABC and CPCP is the bisector of ∠ACB\angle ACB. If ∠APB=110o\angle APB = 110^o, ∠BPC=120o\angle BPC = 120^o and ∠CPA=130o\angle CPA = 130^o, find the angles of triangle ABCABC. https://cdn.artofproblemsolving.com/attachments/1/c/6969b4781b5c225a1631a29c2c83754d3d70b4.jpg
p2. Let ABCDABCD be a parallelogram. We mark the midpoint of CDCD, which we call MM. We call P the intersection of BCBC with the line AMAM. It is known that AMAM is the bisector of angle ∠BAD\angle BAD and that PB=10PB = 10. Find the perimeter of parallelogram ABCDABCD.
p3. Let ABCDABCD be a square with side 1 1. Let P,Q,RP, Q, R, and SS be points interior to the square, such that ABPABP, BCQBCQ, CDRCDR, and CASCAS are equilateral triangles. a) Find the area of ​​the quadrilateral PQRSPQRS. b) If all the elements of the figure are deleted except the points P,Q,RP, Q, R and SS, indicate a procedure to reconstruct the square ABCDABCD, using only the ruler and compass.

18th Cabri Clubs 2007, round 1, level 2, 3 problems, Argentinian geo contest

level 2
p4. Let ABCDABCD be an isosceles trapezoid with bases ABAB and CDCD . It is known that ∠ABC=60o\angle ABC = 60^o, AB=6AB = 6 and BC=1BC = 1. Find the length of side CDCD and the area of ​​the trapezoid ABCDABCD. https://cdn.artofproblemsolving.com/attachments/6/d/79137a7d2428005454402b55adf23711c5dd8b.jpg
p5. There is a pentagon ABCDEABCDE such that: ∙\bullet ACEACE is an equilateral triangle. ∙\bullet AB=BC=CD=DEAB = BC = CD = DE ∙\bullet ∠ABC=∠CDE=120o\angle ABC = \angle CDE = 120^o a) Construct the figure. b) If AE=1AE = 1, find the length of BDBD.
p6. Let ABCDABCD be a parallelogram with AB=3AB = 3, BC=5BC = 5 and ∠ABC=60o\angle ABC = 60^o. Construct a parallelogram PQRSPQRS with P,Q,RP, Q, R and SS in AB,BC,CDAB, BC, CD and DADA, respectively, and that fulfills that PQ=2QRPQ = 2QR and ∠PQR=120o\angle PQR = 120^o.