I just took the AMC 8. I scored a 20/25


4 of the incorrect answers were just silly mistakes and miscalculation


One of the questions, the last problem, I was completely stumped.


Note: I haven't learned geometry yet, I just have scattered knowledge across all math subjects.


Do not just tell me to look at the solution on art of problem solving.




Question: A semicircle is inscribed in an isosceles triangle with base 15 and height 16 so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?



A:  \(4\sqrt{3}\)


B: \(\frac{120}{17}\)


C: 10


D: \(\frac{17\sqrt{2}}{2}\)


E: \(\frac{17\sqrt{3}}{2}\)

 May 27, 2019
edited by Guest  May 27, 2019
edited by CalculatorUser  May 27, 2019


Label the drawing along with me.


Label the bottom left vertex of the triangle A

Label the top vertex of the triangle B

Label the bottom right vertex of the triangle C


Label the point of tangency on the right D

Bisect the base of the triangle and label the point E - this is also the center of the circle


A line from the center of a circle to the point of tangency of a line is a right angle

So the little triangle CDE is a right triangle


The large triangle ABC is almost an equilateral triangle.  If it were equilateral, angle DCE would be 60o and DEC 30o 

The sides are slightly longer than the base, so DCE is slightly larger than 60o but for now let's call it 60o - we'll fix it later.


Calling DCE 60o makes DEC 30o (remember, it's just for now) 


Recalling our trig, we know that the cosine of 30o is the (square root of 3) over 2


The cosine of our small triangle, CDE, is DE over CE

The cosine of our small triangle, CDE, is (sqrt 3) over 2


Set them equal                            DE / CE = (sqrt 3) / 2

Multiply both sides by CE            DE = CE * (sqrt 3) / 2

But CE is half of 15 so                 DE = 7.5 * (sqrt 3) / 2

                                                    DE = 3.75 * (sqrt 3)


But remember angle CED is a little smaller than 30o so that makes the cosine a little larger


Looking at the multiple choices, 4  * (sqrt 3) is a little larger than 3.75 * (sqrt 3)


None of the other choices is even close, therefore A has to be the right answer .... 4 * (sqrt 3) 


 May 27, 2019

This is what I found....


Let's label the figure like this:


A line from the center of a circle to a point of tangency is perpendicular to its tangent line, so  CD  is perpendicular to  AB .

And so  △CDA  is a right triangle  and  △CDB  is a right triangle.


By the Pythagorean theorem:












Now we have 3 equations and 3 unknowns.


Now I am really confused....because when I used WolframAlpha to solve the system of equations, it found that




See link here!!


But that is not any of the options, but this result IS slightly larger than 3.75√3 , so it does not conflict with Guest's logic.

Maybe the way Guest did it is the intended way.......???

 May 27, 2019

I also get \(r=\frac{240\sqrt{1249}}{1249}\)


I did it by writing equations for the left side of the triangle, and the perpendicular to this that passes through the midpoint of the base.  Equating these gave me the coordinates of the intersection.  The length between this intersection and the midpoint of the base is the radius of the semicircle.

Alan  May 27, 2019
edited by Alan  May 27, 2019

Hectictar and Alan are right...


I accidentally switched the base and height.


However, their method makes sense and I will make sure to use that method next time I see a problem like this.

 May 27, 2019

Using hectictar's diagram..and switching the base and height per CU...we have that


triangle   CBA  is similar to triangle DBC




AC / AB  =  DC / BC  subbing we have that


15/ sqrt ( 15^2 + 8^2 ) =  r  /  8


8 * 15  / sqrt (289)  = r


8*15 / 17  =  r  =   120 / 17


cool cool cool

 May 27, 2019

I might as well post my solution.


We can use a very handy formula, which is very important to remember, \(A=rs\), where \(A\) is the area of the figure, \(r\) is the inradius when it is a complete circle, and \(s\), the semi-perimeter of the figure, or you can call it as, "half the perimeter."


Now, this is where the tricky part comes in. we can reflect the second part of the figure, like this:


After drawing this figure, we can find the area of one triangle, the multiply it by two, to find the area of the rhombus.  Drawing the altitude of sixteen to the base, we find that the area is \(\frac{1}{2}*16*15=120\). Multiply it by two since there are two triangles, to get \(2*120=240.\)And, using the Pythagorean Theorem, that gives us \(\sqrt{8^2+15^2}=\sqrt{17^2}=17.\) Multiply it by four to find the whole(full) perimeter, and divide it by two to find the semi-perimeter, to get \(17*2=34.\) Now, we found \(A\) and \(s\), and we divide it to get \(r\), to get \(\frac{240}{34}=\boxed{\frac{120}{17}}.\)

 May 28, 2019

Very crafty, tertre!!!


I like that approach, too  !!!



cool cool cool

CPhill  May 28, 2019

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