# easy geometry!

• Nov 19th 2010, 05:05 PM
earthboy
easy geometry!
• Nov 19th 2010, 05:25 PM
earthboy
Well, my teacher gave a solution , but as my trig is horrible (Headbang), ididnt understand it.....
now we know $\displaystyle angles PAB , PBC, QBC, QCB, RAC, ACR = 30$
Now $\displaystyle AP \sin 30 + CQ \sin 30 = \frac{1}{2} AC = AR \sin 30$ , which implies $\displaystyle AP+CQ = AR$ OR, AP+CQ-AR=0
It may be very trivial but i dont understand why $\displaystyle AP \sin 30 + CQ \sin 30 = \frac{1}{2} AC = AR \sin 30$
• Nov 19th 2010, 05:28 PM
Jhevon
Note that $\displaystyle \displaystyle \triangle APB$ and $\displaystyle \displaystyle \triangle BQC$ are both isosceles triangles with the same side lengths. Similar triangles with exactly the same magnitude. (do you see why?)

B bisects AC (why?), and so, the bases of triangles $\displaystyle \displaystyle APB$ and $\displaystyle \displaystyle BQC$ are each half the length of $\displaystyle \displaystyle \triangle ARC$, which is also a similar triangle. Since these are similar, corresponding sides are proportional, and knowing $\displaystyle \displaystyle \triangle ARC$ is double the magnitude of the two other triangles, we know the exact proportion. The upshot of all this is:

$\displaystyle \displaystyle AP + CQ - AR = \frac 12AR + \frac 12 AR - AR = 0$

since AP and CQ are corresponding sides to AR with half its magnitude.

Now I spouted a lot of unproven claims here. It's up to you to fill in the blanks
• Nov 19th 2010, 05:32 PM
Jhevon
Quote:

Originally Posted by earthboy
Well, my teacher gave a solution , but as my trig is horrible (Headbang), ididnt understand it.....
now we know $\displaystyle \angle PAB , PBC, QBC, QCB, RAC, ACR = 30$
Now $\displaystyle AP \sin 30 + CQ \sin 30 = \frac{1}{2} AC = AR \sin 30$ , which implies $\displaystyle AP+CQ = AR$ OR, AP+CQ-AR=0
It may be very trivial but i dont understand why $\displaystyle AP \sin 30 + CQ \sin 30 = \frac{1}{2} AC = AR \sin 30$

Fine, so we want to use trig. Very well. We know the magnitude of all the angles because we see that we had similar triangles here, they are all isosceles, and hence the other angles were 30 degrees each.

construct two perpendicular lines to AC, one passing through P, the other through Q. That will give you the sine relationships your professor mentioned (Remember SOHCAHTOA!) The fact that the sum is 1/2 AC goes back to what I was saying about similar triangles with the factor of 1/2
• Nov 20th 2010, 12:48 PM
bjhopper
Hi earthboy,
Is B supposed to be the midpoint of AC?

bjh
• Nov 20th 2010, 10:30 PM
earthboy
Quote:

Originally Posted by bjhopper
Is B supposed to be the midpoint of AC

the question says, B is any point on the line AC.
but then
Quote:

Originally Posted by Jhevon
B bisects AC (why?)

omg!its becoming more confusing for me (Whew)(Whew) is the solution even correct ?
sorry if i am being silly :-o.
• Nov 21st 2010, 11:28 AM
bjhopper
Hi earthboy,
With B as any point on a line segment AC the statement AP +CQ - AR =0 can be proved by easy geometry (no trig required ).Other info as you described.

bjh
• Nov 21st 2010, 05:07 PM
Jhevon
Quote:

Originally Posted by earthboy
the question says, B is any point on the line AC.
but then omg!its becoming more confusing for me (Whew)(Whew) is the solution even correct ?
sorry if i am being silly :-o.

the problem says B is a point on AC, not "any" point. it ends up being the midpoint with the way the rest of the problem is set up. the secret that tells us this lies in the angles APB and BQC....
• Nov 22nd 2010, 05:41 AM
bjhopper
easy geometry problem
Hello jhevon,
A point on a line can be placed anywhere if there is no spec given.To draw a diagram for a proof lets make AB= 4 cm BC = 6 cm AC =10cm. Points P, Q, R lie on the perpendicular bisectors of AB,BC, AC respectively. Three 30-30-120 isosceles triangles are erected as described. Extend AP and QC to meet the perpendicular bisector of AC at S above AC

ARCS and PBQS are parallelograms AP+QC = AR.The whole is the sum of the parts
• Nov 22nd 2010, 06:56 AM
earthboy
yes ...
sorry if i am being a moron (Thinking) ,but ....
Quote:

Originally Posted by bjhopper
Extend AP and QC to meet the perpendicular bisector of AC at S above AC

suppose AP and QC when extended meet as S. Can we be sure that the perpendicular bisector of AC always passes through the point S ???

another doubt: is the original trig solution ok?

HELP!
• Nov 22nd 2010, 07:27 AM
bjhopper
Hi earthboy,
Did you draw the diagram ? Triangle ACS is an isosceles triangle congruent to ARC.AP and CQ extended meet at S on the perpendicular bisector.I cannot believe you received the trig relations from a math teacher.

bjh