Geometry Proof - Two Intersecting Circles Problem

Question

This problem has been bugging me for the last week - anyone find the solution?!

Given:

The points A,B and C are on circumference of the larger circle which has tangent $DE$ at $C$. $O$ is the circumcenter. The smaller circle passes through the center of the larger circle. Angle $BCE = 48^0$ between tangent $CE$ and chord $CB$.

Answer 1

If fixing positions of $A,B,C$ is required, note that angle $A$ in the alternate segment equals $48^0$ as angle between chord/tangent and so vertex $A$ can be placed anywhere on the large arc segment $CB.$ I have drawn several circles which are arbitrarily placed. Why should an arbitrary circle cut a given tangent line specifically along its diameter? It need not even cut at all.

To develop thought clarity one has to put all that is known on one side and all that is to be solved for, on the other. Then connect them up.

Answer 2

Any circle whose center lies along the line $DE$ (so that the line $DE$ acts as a diameter) and which passes through the center $O$ of the "larger" circle necessarily has radius at least as large as the "larger" circle. So if the circle in question is indeed smaller than the circle centered at $O$ with the line $DE$ tangent at $C$, its center cannot lie along the line $DE$, and hence $DE$ cannot be a diameter of it.

Note, this has nothing at all to do with the triangle $\triangle ABC$ or the measure of the angle $\angle BCE$. All that information is a red herring.

Remark: I just noticed, this answer essentially duplicates a remark made by Alain Remillard below the OP.

Answer 3

If the diameter of the circle-to-be-constructed lies on DE, then we have the following two cases:-

(1) C is its center with CO as its radius --- Then, the green circle is formed with its size the same as the red one but NOT larger.

(2) $C’$ (an arbitrary point on DE other than C) is its center with $C’O$ as its radius --- Then, the purple circle is formed with its size larger than the green one because $C’O$ is the hypotenuse of $\triangle COC’$.

The conclusion is:-

“any circle drawn through O with diameter lying on DE must be smaller than (and at most equal to) circle $ABC$”.

Answer 4

For us to define a circle such that one is smaller than the other will require the distance and betweenness axioms. If you don't have those available to you than I suggest doing something different. I will continue with those axioms assumed.

Let $\bigodot (O,r)$ and $\bigodot (P,s)$ such that $\bigodot (O,r)$ is the circumcircle of $\triangle ABC$ with radius $r$, and $\bigodot (P,s)$ is the smaller circle who's radius is $s$, and center is $P$ and intersects at $O$. Note that this implies $r>s$.

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Suppose $DE$ is the diameter of $\bigodot (P,s)$. This implies that there is a $P$ on $DE$ such that $P$ bisects $DE$. Construct line $PO$. As $r > s$ there exists a point $P^\prime$ such that $P^\prime O\cong CO = r$. Construct line $P^\prime C$. Note $P^\prime O\cong CO$ implies $\triangle COP^\prime$ is isosceles. By pons asinorum $\angle CP^\prime O \cong \angle P^\prime CO$. Since $$\angle P^\prime CO = \angle P^\prime CP+\angle PCO=\angle P^\prime CP+ 90^o,$$ this implies that, $$\begin {array}{ccc} \angle CP^\prime O+ \angle P^\prime CO+\angle COP^\prime &=& 2\angle P^\prime CO+\angle COP^\prime \\ &=& 2(90^o)+2\angle P^\prime CP+\angle COP^\prime \\ &>& 180^o \end{array}$$ a contradiction.