Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 801.

## Saturday, September 1, 2012

### Problem 801: Equilateral Triangles, Cevian, Midpoint, Distance, Measurement, Metric Relations

Labels:
equilateral,
measurement,
metric relations,
midpoint,
triangle

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Let AB=a,CD=b and BD=y

ReplyDelete1- Δ BCE

a²+b²+ab=100

2- Δ BCD

a²+b²-ab=y²

2(a²+b²)=100+y²

3- ΔBDC => theorem of apollonius

2(a²+b²)=4x²+y²

4x²+y²=100+y²

4x²=100

X=5

Erina-NJ

ED meets AB at H. CH=BD and CF=BG, where F and G are mindpoints of BD and CH respectivly. It is obvious that CF=BG=5.

ReplyDeleteYou can see this here: http://mtz256.files.wordpress.com/2012/09/801.jpg

Let AB=BC=Ac=a and CD=CE=DE=b

ReplyDeleteApply cosine formula in triangle BCE we have BE^2=a^2+b^2+a.b= 10^2

Apply cosine formula in triangle BCD we have BD^2=a^2+b^2-a.b

In triangle BCD apply formula for calculate median we have 2.CF^2=BC^2+CD^2-BD^2/2

Or 2.x^2=1/2(a^2+b^2+a.b)= ½.BE^2=50

So x=5

1) continue CF until CF=FP => CF = CP/2

ReplyDeleteBCDP is a paralellogram

2) E, D, P are collinear. BCEP is an isosceles trapezoid.

PC=BE=10 => CF=5

Erina-NJ

Problem 801 solution: This solution was submitted by Michael Tsourakakis from Greece

ReplyDeleteThanks Mixalis.

Sea M el punto medio del lado BC. Entonces el segmento FM es paralelo a DC y mide DC/2, y el triángulo CFM es semejante al triángulo BEC, por el criterio LAL, CM=BC/2, FM=DC/2=EC/2, ángulo FMC = ángulo ECB = 120º, así pues x=CF=BE/2=5.

ReplyDeleteMigue.

Complete the parallelogram BCDG.

ReplyDeleteTriangles BCG, BCE are congruent.

2x = GC = BE = 10.

x = 5

Construind simetricul punctului D fata de C si notandu-l cu P obtinem ca

ReplyDeleteΔ BCE=Δ BCP (LUL) BE=BP si CF linie mijlocie in

Δ BDP <=> x=CF=BE/2=5.

BP is parallel to FC such that P is on AC. DC=CP but also DC=EC so CP=CE. <BCE=60+60=120, but also <BCP=180=60=120 so <BCE=BCP. These equalities show that BCE and BCP are congruent, making BC=10=BP=2x so x=5.

ReplyDeleteSimple pure geometry solution

ReplyDeleteDraw FG // to DC where G is on BC

Consider Tr.s BCE & FGC

< C = < G = 120

BC = 2 GC

CE = 2 FG ( last 2 results from applying mid point theorem to Tr. BCD)

The 2 triangles are therefore obviously similar and the sides of Tr. FGC are half that of Tr. BCE

Hence x = 10/2 = 5

Sumith Peiris

Moratuwa

Sri Lanka