Geometry Problem that I don't know where to begin.

evene · 2015-10-22 12:50:15

Let ABCD be a square of side length 1. Let P be a point on side CD such that angle DAP = 20 degrees. Let Q be a point on side BC such that angle BAQ = 25 degrees. Find the perimeter of triangle CPQ.

bobbym · 2015-10-22 15:17:55

Hi;

I am getting 2.

evene · 2015-10-23 00:26:47

... what? How? Can you show me your thinking please?

Bob · 2015-10-23 00:40:08

hi evene,

Draw a circle, centre P and radius PD.

As angle PDA is 90, AD is a tangent to this circle. Let AR be the second tangent, so that triangles PDA and PRA are congruent.

Hence angle PAR = 20 => QRA = 25 and RA = 1

Hence triangles QRA and QBA are congruent => QB = QR.

The perimeter value of 2 follows from this. (CQ + QR + RP + PC)

I think you probably should show that PRQ is a straight line, but that's easy to do by considering angles RPC and RQC.

Bob

evene · 2015-10-23 10:20:09

oohh! I never thought of it this way! You guys are brilliant!

Bob · 2015-10-23 19:46:47

hi evene,

Thanks for the comment.

I'll take you through my approach ... it'll take away some of the 'magic' and give you some hints for future problems.

A good diagram always helps. I have a program called 'Geometer's Sketchpad' , and I made the above diagram using it. It's a vector geometry program which means it preserves geometric properties accurately even if you vary the positioning of points for example. It's a paid for program, but there is a free program called Geogebra which does a similar job. Each program has advantages, but you can certainly do this problem using either.

I could see that you could get an answer using trigonometry. BQ = 1.tan(25) so CQ = 1 - tan(25). You can get CP similarly, and then PQ using Pythagoras theorem.

But such a method will involve decimals and I was struck by bobbym's answer of exactly 2. That suggested to me that an approach using basic geometry would also be possible. I noticed that 25 + 20 = 45, exactly half of angle A.

On my diagram I checked the perimeter and it came out close to 2 (due to accuracy issues maybe). I wondered what would happen if I varied the angles. I found that the perimeter didn't stay as a 'nice' value with different angles, so this suggested to me that I should be exploiting the 25 + 20 property somehow.

I also realised that there would have to be a point, R, on PQ so that QB = QR and PD = PR. So finding R became my target.

That's why I drew the circles. It looks like R is the point where the circles touch, making AR a tangent to both. If that appearance is a true fact then the perimeter result follows straight away so I had my geometrical method.

But there is a danger with an accurate diagram. If something looks true it is easy to fall into the trap of assuming it is true. You can easily end up with a false proof because you've assumed what you are supposed to be proving.

My method finds a point R with CQ + QR + RP + PC = 2, but it is still necessary to show that PRQ is a single straight line, otherwise I've found the perimeter of a quadrilateral! As QR is perpendicular to AR which is perpendicular to RP, this is OK but it is a necessary part of the proof. Or you can work out angle RPC ( = 40) and RQC ( = 50) and hence show that CQRP is a triangle (40 + 50 + 90)

Hope this helps,

Bob

evene · 2015-10-24 13:00:29

About trigonometry. I haven't learned it yet, so try avoiding it with me until I say I've learned it, or people will be looking at me like I need to get outside more often. And thanks, Geogebra, I'll need to see if I can have download it...

Bob · 2015-10-24 20:23:57

http://www.geogebra.org/cms/

Bob

evene · 2015-10-25 01:29:20

How does geogebra work?

phanthanhtom · 2015-10-25 04:20:23

That perimeter will always be 2 given that angle PAQ = 45.

Let E be a point on the ray opposite of ray DC such that DE = BQ. Then triangle ABQ = triangle ADE so AQ=AE and also angle QAP equals angle PAE equals 45. Therefore triangle AQP equals triangle AEP (s.a.s.). So QP = EP = ED + DP = QB + DP. Therefore the perimeter of triangle CPQ would be CP + PQ + QC = CP + PD + BQ + QC = CD + CB = 2.

phanthanhtom · 2015-10-25 04:21:28

My solution doesn't use 20 or 25 degrees, but only uses angle PAQ = 45

Bob · 2015-10-25 04:24:21

hi phanthanhtom

I wondered if that was the case. Thanks.

Bob

Math Is Fun Forum

#1 2015-10-22 12:50:15

Geometry Problem that I don't know where to begin.

#2 2015-10-22 15:17:55

Re: Geometry Problem that I don't know where to begin.

#3 2015-10-23 00:26:47

Re: Geometry Problem that I don't know where to begin.

#4 2015-10-23 00:40:08

Re: Geometry Problem that I don't know where to begin.

#5 2015-10-23 10:20:09

Re: Geometry Problem that I don't know where to begin.

#6 2015-10-23 19:46:47

Re: Geometry Problem that I don't know where to begin.

#7 2015-10-24 13:00:29

Re: Geometry Problem that I don't know where to begin.

#8 2015-10-24 20:23:57

Re: Geometry Problem that I don't know where to begin.

#9 2015-10-25 01:29:20

Re: Geometry Problem that I don't know where to begin.

#10 2015-10-25 04:20:23

Re: Geometry Problem that I don't know where to begin.

#11 2015-10-25 04:21:28

Re: Geometry Problem that I don't know where to begin.

#12 2015-10-25 04:24:21

Re: Geometry Problem that I don't know where to begin.

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