# Area and circles

I need help with this hard geometry problem.

Given a $\triangle ABC$. M is the midpoint of AB. $k_1$ and $k_2$ are the circles with diameters AC and BC respectively. $A_1$ and $A_2$ are the midpoints of the two arcs AC of $k_1$. $B_1$ and $B_2$ are the midpoints of the two arcs $BC$ of $k_2$. Prove that ${\triangle A_1A_2M}$ and ${\triangle B_1B_2M}$ have equal area.

-
Area of the triangle –  Adam Jun 1 '12 at 9:53

## 3 Answers

A proof using purely Euclidean geometry:

Define $A_3$ and $B_3$ to be the midpoints of $AC$ and $BC$ respectively (they are also the intersection of $AC$ with $A_1A_2$, and $BC$ with $B_1B_2$). Clearly $A_1,A_2,A_3$ are collinear. Similarly for the $B$'s. And $A_3$ is the midpoint of $A_1A_2$. So the area of triangle $A_1A_2M$ is the same as twice the area of the triangle $A_1A_3M$. So it suffices to show that the magenta triangle and the blue triangle in the above picture have the same area.

Assume that $A_1$ is on the same side of $AC$ as $B$, similarly $B_1$ on the same side of $BC$ as $A$.

The angle $AA_3A_1$ is a right angle. Similarly $BB_3B_1$.

The triangles $AA_3M$ and $MB_3B$ are congruent. (They are both similar to $ACB$ with lengths halved.)

Hence the angles $MA_3A_1$ and $MB_3B_1$ are equal.

Now, using that $A_3M$ has the same length as $B_3B$ and hence as the same length as $B_3B_1$, and similarly $B_3M$ has the same length as $A_3A$ and so as $A_3A_1$, we have that the triangles $A_3MA_1$ and $B_3B_1M$ are congruent (by the Side-Angle-Side theorem) and hence have the same area.

-
In case anyone cares, the image above is produced using kseg, a program I highly recommend. –  Willie Wong Jun 1 '12 at 10:54
Looks good. Nice diagram, too! (+1) –  robjohn Jun 1 '12 at 19:53

Let $D$ be the circumcenter of $\triangle ABC$; note that $D$ lies on the perpendicular bisector of each side of $\triangle ABC$. Drop the perpendicular from $C$ onto $\overline{AB}$ at $E$.

Since $\overline{CE}$ and $\overline{DM}$ are perpendicular to $\overline{AB}$, they are parallel.

$\hspace{4mm}$

In the diagram on the left:

Drop the perpendicular from $M$ onto $\overline{A_1A_2}$ at $F$.

Since $\overline{CA}$ and $\overline{FM}$ are perpendicular to $\overline{A_1A_2}$, they are parallel. Therefore, the right $\triangle MFD$ is similar to the right $\triangle CEA$. Thus, $$\frac{\overline{FM}}{\overline{DM}}=\frac{\overline{CE}}{\overline{CA}}\tag{1}$$ Since $\overline{A_1A_2}=\overline{CA}$, we get that the area of $\triangle A_1A_2M$ is $$\tfrac12\overline{A_1A_2}\;\overline{FM}=\tfrac12\overline{CA}\;\overline{FM}=\tfrac12\overline{DM}\;\overline{CE}\tag{2}$$ In the diagram on the right:

Drop the perpendicular from $M$ onto $\overline{B_1B_2}$ at $G$.

Since $\overline{CB}$ and $\overline{GM}$ are perpendicular to $\overline{B_1B_2}$, they are parallel. Therefore, right $\triangle MGD$ is similar to right $\triangle CEB$. Thus, $$\frac{\overline{GM}}{\overline{DM}}=\frac{\overline{CE}}{\overline{CB}}\tag{3}$$ Since $\overline{B_1B_2}=\overline{CB}$, we get that the area of $\triangle B_1B_2M$ is $$\tfrac12\overline{B_1B_2}\;\overline{GM}=\tfrac12\overline{CB}\;\overline{GM}=\tfrac12\overline{DM}\;\overline{CE}\tag{4}$$

Thus, the areas of $\triangle A_1A_2M$ and $\triangle B_1B_2M$ are both $\tfrac12\overline{DM}\;\overline{CE}$.

-

Drop a perpendicular from $M$ to $BC$ to get point $F$. With $M_A$ the midpoint of $BC$, and $BC \perp B_1 B_2$, we have that the altitude of $\triangle B_1 B_2 M$ w.r.t. side $B_1 B_2$ is congruent to $FM_A$.

Now, \begin{align} |FM_A| = |BM_A| - |BF| &= \frac{1}{2}|BC| - |BF| \\ &=\frac{1}{2}|BC|-|BM|\cos B \\ &=\frac{1}{2}\left(|BC|-|BA|\cos B \right) \\ &= \frac{1}{2}|CA| \cos C \end{align}

so that, since $B_1 B_2 \cong BC$,

$$\text{area } \triangle B_1 B_2 M = \frac{1}{4} |B_1 B_2||CA|\cos C=\frac{1}{4}|BC||CA|\cos C$$

The final expression is symmetric in $A$ and $B$, and thus must also apply to $\text{area }\triangle A_1 A_2 M$.

-
I tried this method first and arrived at the formula for the areas of the triangles in question being $$\tfrac18(|AC|^2+|BC|^2-|AB|^2)$$ –  robjohn Jun 2 '12 at 0:57
@robjohn: Same thing. –  Blue Jun 2 '12 at 1:01
Yes, using the Law of Cosines to get $\cos(C)$ from the side lengths. –  robjohn Jun 2 '12 at 1:04