Download ARML Techniques: Working with Obscure Congruent Angles By

ARML Techniques: Working with Obscure Congruent Angles
By: Alex Anderson
Here is one specific strategy that I use in geometry problems involving congruent angles. It is:
Draw a parallel line to a ray of the bisected angle, and extend the angle bisector.
Example : (Chicago 2006 ARML Homework #2, Problem #2)
The length
of 2 sides of a triangle are 6 and 9. The length of the angle bisector of the included
√
angle is 24. Compute the length of the third side.
Comment :
Usually when people see the congruent angles and the angle bisector theorem does not help them
directly, they resort to trig or other techniques that produce a lot of algebra. However, there is
almost always a geometric solution to these types of problems.
A
9
6
root(24)
B
D
C
H
E
Solution :
√
4ABC has D on BC such that AD bisects ∠BAC. Let AB = 6, AC = 9, and AD = 2 6.
Draw a line parallel to AB through C. Extend AD to hit this line at E.
Using the parallel lines and the angle bisector: ∠CAE = ∠CEA. Then AC = CE = 9.
Clearly, we have 4ABD ∼ 4ECD by AA similarity. Then:
BD
CD
=
AB
EC
=
AD
DE
=
6
9
=
√
2 6
DE .
√
This implies DE = 3 6.
Let the foot
√ of the perpendicular from C to AE be H. 4ACE is isosceles, so H bisects AE.
Then DH = 26 . By the pythagorean theorem,
CH 2 = CE 2 − EH 2 = DC 2 − DH 2 .
√
3
2
81 − 75
2 = DC − 2 Then: DC = 3 5. From above,
BD
CD
=
2
3
√
√
so BD = 2 5 and BC = 5 5 .
Now obviously: ‘So what?’ There were many ways of solving this problem including Heron’s
forumla, Stewart’s theorem, and the Law of Cosines, but this offers a purely geometric solution. It
seems that there is no reason to go through all this thinking, but consider: this technique lessens
algebra, saving time. Also in more some more difficult problems, you will not be able to brute
force like you could in this problem. (Be sure you see how the parallel line helped us to use the
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congruent angles.) Here are some problems that have elegant solutions using this technique. The
first problem is a pretty direct application of this technique. The second problem requires you
to think a little more, but it still is pretty straightforward. The last problem is somewhat more
difficult, but this technique prompts an elegant solution to it.
1. (ICTM State 02 Division AA #19)
In triangle ABC, H is on line AB beyond A and ∠ACB > ∠ABC. The angle bisector of ∠HAC
intersects line BC at D. AC = 12, AB = 16, and BD = 46.4. Give BC as an exact decimal.
2. (#6 on the Mock AIME on AoPS by beta)
Let ABC be a triangle such that AB = 13, AC = 14 and BC = 15. Extend √AC through A to point
D. Let E be the intersection of the bisectors of ∠DAB and ∠ACB. EB = a c b where a, b, c are positive integers, a, c are relatively prime, and b is not divisible by the square of a prime. Find a+b+c.
3. (USA AIME II 2005 #14)
In triangle ABC, AB = 13, BC = 15, and CA = 14. Point D is on segment BC such that CD = 6.
Point E is on BC such that ∠BAE = ∠CAD. Given BE = pq where p and q are relatively prime
positive integers, find q.
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