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I have this problem, which is about this right triangle below. It says that $|AB|$ and $|BD|$ (which is the diameter of the circle) are equal and that the circle is touching the side $|AC|$. Now I have to determine the fraction $\dfrac{|AB|}{|BC|}$

enter image description here

I have tried a couple of things and I got the result $\frac{1}{\sqrt{15}}$. Is that correct? If not I will be pleased if somebody would give me some hints.

And an important detail: I may not use trigonometry!

JohnD
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Nick Podowalski
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    It helps to learn how to check answers to see if they make sense. In this figure it is easy to see that $|BC|$ must be less than two times $|AB|,$ for otherwise $AC$ could not be tangent to the semicircle. The answer must then be greater than $\frac12.$ Since $\frac1{\sqrt{15}}<\frac12$ you can see right away you have to try for a different solution. – David K Dec 21 '14 at 16:23

2 Answers2

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Let $E$ be the point where the circle touches the hypotenuse. Then we have $|AE| = |AB|$. We also have $|CE|^2 = |CD|\cdot |BC| = |BC|(|BC| - |AB|)$ (this quantity is called the power of the point $C$ with respect to that circle). With all these sizes accounted for, the Pythagorean theorem gives $$ |AB|^2 + |BC|^2 = |AC|^2\\ |AB|^2 + |BC|^2 = \left(|AB| + |CE|\right)^2\\ |AB|^2 + |BC|^2 = |AB|^2 + 2|AB||CE| + |BC|^2 - |BC||AB|\\ |BC||AB| = 2|AB||CE|\\ |BC|= 2|CE| \\ |BC|^2 =4|CE|^2 = 4|BC|(|BC| - |AB|)\\ |BC| = 4(|BC| - |AB|)\\ \frac{|AB|}{|BC|} = \frac{3}{4} $$

Arthur
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Here's a solution which doesn't use the Pythogorean theorem:

Let $E$ be the point, as in Arthur's solution, where the semicircle touches the hypothenuse, and $O$ the center of the semicircle. Then note that the triangles $\Delta ECO$ and $\Delta BAC$ are similar, with the rate $2$, i.e. if $|OC| = x$, then $|AC| = 2x$ (note that they have the same angles and the angle $\angle BCA$ sees an edge of length $r$ in one triangle, and an edge of length $2r$ in the second).

Similarly, if we let $|BO| = r$ (i.e. $|AB| = 2r$) then we have $|EC| = (r + x)/2$, and $|AE| = 2x - (r + x)/2 = (3/2) x - r/2$.

Using the congruence of the triangles $\Delta OBA$ and $\Delta OEA$ we obtain

$$2r = (3/2) x - r/2$$

And conclude that $x = (5/3) r$. Using $|AB| = 2r$ and $|BC| = r + x$ we obtain

$$\frac{|AB|}{|BC|} = \frac{3}{4}$$

xxxx
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