Friday, March 30, 2007

Challenging Geometry: Circles Inscribed in Quadrilaterals, Right Triangles

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Update2: See the awesome article in MathWorld on tangential quadrilaterals for more info re problem #2.
Update: See Comments section for some answers, solutions.

Part (a) of each of the following are somewhat difficult questions that can be found in some geometry textbooks. These are numerical exercises and good practice for the more difficult SAT-types of questions or for math contests. The last part of each question is an extension or generalization of the problem. Texts do not often ask students to delve beneath the surface and look for general relationships.

1. A circle of radius 4 is inscribed in a right triangle with hypotenuse 20.
(a) Find the perimeter of the triangle without using the Pythagorean Theorem. Justify your reasoning.
(b) Using the Pythagorean Theorem, show that the triangle is similar to a 3-4-5 triangle.
Note: Many students tend to guess multiples of 3-4-5 when doing these. Sometimes they get lucky but they need to prove it!
(c) PROVE in general that the perimeter of a right triangle is twice the sum of its hypotenuse and the radius of its inscribed circle. Again, no Pythagorean Thm allowed.
Note: There are well-established formulas for the inradius of a triangle. Our objective here is to look at one special case.

2. A circle is inscribed in a quadrilateral which has a pair of opposite sides equal to 12 and 18. Neither pair of opposite sides of the quadrilateral is parallel.
(a) Find the perimeter of the quadrilateral. Justify your reasoning.
(b) PROVE in general that the perimeter of a quadrilateral in which a circle is inscribed equals twice the sum of either pair of opposite sides.
Note:: Not all quadrilaterals have an inscribed circle, so this is a strong condition.

Note: As always, these results need independent verification. I welcome your comments and edits!


Dave Marain said...

Comments,Partial Answers, Solutions:
1. (a) 48
(b) Sides are 12-16-20
(c) Start of proof: Draw radii to the 2 legs of the right triangle. The center of the circle, the two points of tangency and the vertex of the the right angle in the triangle form a square (what theorems are being used?).If the legs are A and B and the radius is R, then the side of length A can be separated into segments of lengths R and A-R, similarly, the other leg into parts R and B-R.Use the theorem on the congruence of tangent segments from an external point to represent the two parts of the hypotenuse A-R and B-R. The conclusion falls out from here.
Note: This kind of problem is sometimes known as a 'Walk-around' problem.

2. (a) 60
More to come...

Unknown said...

Cool problem, Dave.

For (1), part (c), draw all angle bisectors, and the perpendiculars from the intersection point to the sides. Now, consider the pair of triangles that has a vertex of the original triangle in commmon. These two triangles are congruent. Each triangle that has the right vertex in common is a 45-45-90 triangle, with the legs equal to the radius of the incircle. Adding up the different sides, one can arrive at the result.

The similar procedure seems to apply to (2) part (b) also.

Anonymous said...


I have figured out problem 1 somewhat but need help. I have drawn a right angled triangle with hypotenuse 20 and a circle inscribed in the center with radius 4. The legs allow for some ar to be 4 and ap to be 4. In order to get ab it would be x-4 and to get ac it would be y-4.
c+b=28 I know that 28 equals the lengths of the other two sides 12, 16. Not sure how to get the 16. I did plug the 12 and 16 in to figure out that x would be 12 and y would be 8. after all 12+16+20=48. How do I get the 12 and the 16?

Dave Marain said...

Hi Anon!
I'm curious what drew you back to this problem that hasn't been discussed in many months. Your notation confused me a bit, but I basically followed your logic.

First of all, part(a) does not require you to know the actual lengths of the legs. Once you obtained c+b = 28, then it follows that the perimeter is 28+20 = 48.
I'm guessing you're interested in the calculations leading to 12 and 16. Well, we have 2 conditions on b and c:

b+c=28 and

This linear-quadratic system can be solved in several ways but the traditional approach is to solve for, say, b, in the linear equation: b = 28-c
Then substitute this into the 2nd equation:
(28-c)^2 + c^2 = 20, simplify and solve the quadratic by your favorite method. The result will be 12 & 16.
I'm not quite sure if this is what you were asking for. Let me know...

Anonymous said...

Yes Dave that is what I was asking. I wanted to know how to get the 12 and 16. I am using the formula you put in and am still having a bit of trouble. Thanks Anon.

Anonymous said...

Oh to answer your question I just found your challengin site. This is cool. Thanks Dave.

Anonymous said...

Ok Dave what about this?

so, b+c=28

Dave Marain said...

Welcome, Anon!
Other than a couple of typos, your work looks fine. As you did, I reached the quadratic equation c^2 - 28c + 192 = 0, then solved it by factoring: (c-16)(c-12) = 0. This leads to 2 solutions:
c = 16, b = 12
c = 12, b = 16
These of course are equivalent to each other and so we have our 12-16-20 right triangle. Note that this happens to be a multiple of the ever-popular 3-4-5.

I'm glad you enjoyed this. I hope you will spread the word to your friends and keep visiting. Happy Thanksgiving!

Anonymous said...


I was wondering about problem 2. I was wondering how a person would explain how you got the perimeter of 60.

The following was kind of my line of thinking: I have the shape as being a Kite and the perimeter being 60. If you draw a diagonal connecting the two center vertices this would divide the shape into two apparently congruent Isosceles triangles, with two sides being 18 in length and the base being 12. Also, I noticed that if I draw a diagonal connecting the other two vertices the origianl diagonal seems to be perpendicularly bisecting the the second diagonal. Also after drawing the second diagonal the original kite shape seems to be cut into 4 congruent right angle triangles. I'm just having trouble trying to connect all this info with proving what the measurments and perimeter are. Could I use the ASA method to prove the perimeter? Any help would be much appreciated.

Dave Marain said...

These are commonly known as 'walk-around' problems in geometry. There are many ways to get started and different strategies using a number of possible variables. All of these depend on one key theorem about circles,
The Two Tangent Theorem:
The two tangent segments from an external point are congruent, a theorem which is proved by HL (and the fact that radii drawn to the point of contact are perpendicular to the tangent).

I'll try to outline a possible approach, but, without a diagram, it may be hard to follow:

Divide the side of length 12 into 2 parts of lengths x and 12-x. Similarly, divide the opposite side of length 18 into y and 18-y. Now use the Two Tangent Theorem several times, in effect, walking around the sides of the quadrilateral in both directions. The 'other' pair of sides will then be divided into parts (the following depends on where you place the x and y but the final result will be the same):

One side will be divided into parts of lengths 12-x and y.
The other side into parts x and 18-y.
The sum of all 4 of these parts will still be 30, since the x's and y's cancel!

Thus, each pair of opposite sides will have a sum of 30 and the perimeter will be 60. This may be very difficult to follow without a picture but I hope it makes sense anyway.

When first learning this (or teaching this) it often helps to choose random values for the parts. Thus, say the point of tangency divides the side of length 12 into parts of lengths 8 and 4, while the '18' side is divided into parts 11 and 7. Place these values on the quadrilateral and 'walk around' the figure to convince yourself the each pair of opposite sides will have a sum of 30. Change the parts to other values and you'll see the beauty of this. Of course, algebra provides a 'proof!' Your approach of using a special case (kite) was a good way to get started. I often encourage students to begin with particular cases before looking for a general method.

I hope this helps.

Anonymous said...


Problem 2.

COuld you break down the Algebra part of the problem a little more please?

I have the shape divided into 8 parts altogether steaming from the points of tangency: y, 18-y,y and 18-y. the other side is 12-x, x, x and 12-x.

Any other help would be very much appreciated.

Anonymous said...

Disregard that last post. I now see what your saying with each pair of opposite sides equal 30.

I still don't feel like I understand why? Why on the side with 18-y do we use x for the unknown length and not y?

Anonymous said...

Frustrated...need more help with problem 2. Where you at Dave?

Dave Marain said...

Top side divided into x and 12-x.

Bottom side: 18-y and y
This is arbitrary, i.e., we could reverse it and the final result would be the same (see work below)

Left side: 18-y and x
The 18-y comes from the equal tangent segment on bottom; the 'x' comes from the equal tangent segment on top

Right side: 12-x and y

Sum of left and right: 30

If we choose to reverse the bottom and label the parts y and 18-y from left to right:

Left side: x and y
Right side: 12-x and 18-y
Sum of left and right: 30

No matter what combination we use, the final result will be the same.

I hope we've 'walked around' this problem sufficiently!

Anonymous said...

Thanks Dave. That helped a lot.