Have a go at this question ... it was from AIMO 2007 Intermediate Paper.

A point P is marked inside a regular hexagon ABCDEF so that angle BAP = angle DCP = 50 degrees. Find angle ABP.

In my opinion, its harder than it looks.

Printable View

- Aug 6th 2008, 05:14 AMBG5965Hexagon Geometry
Have a go at this question ... it was from AIMO 2007 Intermediate Paper.

A point P is marked inside a regular hexagon ABCDEF so that angle BAP = angle DCP = 50 degrees. Find angle ABP.

In my opinion, its harder than it looks. - Aug 6th 2008, 04:05 PMSerena's GirlLong solution...
I'm sure there is a shorter solution to this problem. However, this long solution is the only solution I can come up with.

http://farm4.static.flickr.com/3205/...1b0b31.jpg?v=0

I isolated the quadrilateral ABCP. From the problem, BAP is 50 degrees. Since DCP is also 50 degrees, and since the interior angles of a regular hexagon are 120 degrees each, then BCP is 70 degrees. ABC is 120 degrees, and since the angles of a quadrilateral sum up to 360 degrees, then APC is also 120 degrees.

Triangle ABC is an isosceles triangle. AB = BC. Hence, BAC = BCA = 30 degrees.

That makes PAC = 20 degrees and PCA = 40 degrees.

Let's say AB = BC = 2. Using trigonometry (special triangles) on triangle ABC, AC = 2 * sqrt(3).

Using sine law on triangle APC, AP = 4 * (sin 40)

Using cosine law on triangle BAP, we find BP = 2. Hence, APB = 50 degrees and**ABP = 80 degrees**

AB = BP = BC, but I don't know how to make that conclusion without solving it the long way.