MathDB

5

Part of 2007 May Olympiad

Problems(2)

Angles and segments

Source: May Olympiad(Olimpiada de Mayo)2007

2/1/2018
In the triangle ABCABC we have ∠A=2∠C\angle A = 2\angle C and 2∠B=∠A+∠C2\angle B = \angle A + \angle C. The angle bisector of ∠C\angle C intersects the segment ABAB in EE, let FF be the midpoint of AEAE, let ADAD be the altitude of the triangle ABCABC. The perpendicular bisector of DFDF intersects ACAC in MM. Prove that AM=CMAM = CM.
geometry
paper folding a pentagon (May Olympiad 2007 L1)

Source:

5/11/2019
You have a paper pentagon, ABCDEABCDE, such that AB=BC=3AB = BC = 3 cm, CD=DE=5CD = DE= 5 cm, EA=4EA = 4 cm, ∠ABC=100o\angle ABC = 100^o ,∠CDE=80o \angle CDE = 80^o. You have to divide the pentagon into four triangles, by three straight cuts, so that with the four triangles assemble a rectangle, without gaps or overlays. (The triangles can be rotated and / or turned around.)
geometrypaperrectanglepentagon