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

evene
Member
Registered: 2015-10-18
Posts: 272

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

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.

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

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

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

Hi;

I am getting 2.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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

evene
Member
Registered: 2015-10-18
Posts: 272

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

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

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

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

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

hi evene,

eyn0qKC.gif

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


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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

evene
Member
Registered: 2015-10-18
Posts: 272

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

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

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

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

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

hi evene,

Thanks for the comment.  smile

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


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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

evene
Member
Registered: 2015-10-18
Posts: 272

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

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...

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

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

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


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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

evene
Member
Registered: 2015-10-18
Posts: 272

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

How does geogebra work?

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

phanthanhtom
Member
Registered: 2012-06-22
Posts: 290

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

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.

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

phanthanhtom
Member
Registered: 2012-06-22
Posts: 290

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

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

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

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

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

hi phanthanhtom

I wondered if that was the case.  Thanks.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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