Euclidea 6.9

Euclidea 6.9 Nine Point Circle
Construct a circle that passes through the midpoints of sides of the given acute triangle $$\triangle{\rm ABC} $$.



This problem uses a nice little Lemma that I actually didn't know before. The title of the problem is very suggestive: nine-point circle. Which nine points? It turns out that if we call the midpoints of the three sides as $$ {\rm D, E, F} $$, and the projection points of each of the vertices $$ {\rm A, B, C} $$ to the opposite side as $$ {\rm A', B', C'} $$, then all these six points, $$ {\rm D, E, F, A', B', C'} $$ are on the same circle!

To see why this is true, we prove only one of these points, $$ {\rm C'} $$, which we call $$ {\rm  G} $$ in this figure, lies on the circle passes through $$ {\rm D, E, F} $$.

In the figure, we draw $$ {\rm D, E, F} $$ as the middle points, and easily have $$ \overline{\rm EF} \parallel \overline{\rm AB}$$. We drop a line $$ \overline{\rm CG} \perp \overline{\rm AB}$$, and draw the perpendicular bisector of $$ \overline{\rm EF} $$ as $$ \overline{\rm HJ} $$.

Now since $$ \overline{\rm HJ} \parallel \overline{\rm CG}$$, and $$ \overline{\rm DH} = \overline{\rm CH}$$, we have that $$ \overline{\rm DJ} = \overline{\rm JG}$$, which means $$ \overline{\rm HJ} $$ is also the bisector of $$ \overline{\rm DG} $$. That is, any point on $$ \overline{\rm HJ} $$ has equal distanc to $$ {\rm D} $$, and to $$ {\rm G} $$. The center of the desired circle, the one passes through $$ {\rm D, E, F} $$ actually is on $$ \overline{\rm HJ} $$. It has equal distances to $$ {\rm D, E, F} $$ and thus to $$ {\rm G} $$, which means $$ {\rm G} $$ is indeed on the desired circle. Similarly, the other two projection points of the vertices are also on the same circle.

The trick of this problem is to pick the right 3 out of the 6 points on the same circle, to reduce the budget.

Solution 5L(a)

 * 1) Draw bisector of $$\overline{\rm AB}$$ to find the midpoint $${\rm D}$$;
 * 2) Draw circle $$ \odot{\rm D} $$ with radius $$ \overline{\rm AD} $$, intersecting the other two sides of the triangle at $${\rm E}$$ and $${\rm F}$$. At this point, since $$ \overline{\rm AB} $$ is the diameter of $$ \odot{\rm D} $$, we have $$ \overline{\rm BE} \perp \overline{\rm AC}$$, and $$ \overline{\rm AF} \perp \overline{\rm BC}$$. That is, points $${\rm E}$$ and $${\rm F}$$ are the projection points of $${\rm B}$$ and $${\rm A}$$, respectively. Now we have found 3 out of the 6 points on that desired circle, with only 2L, and all that is left to do is to draw a circle that goes through $${\rm D,E}$$ and $${\rm F}$$.
 * 3) Now this is standard, we draw bisectors of $$\overline{\rm DE}$$ and $$\overline{\rm DF}$$ to intersect at $${\rm G}$$, which is the center of that circle.

Solution 5L(b)

 * 1) Alternatively, we can draw $$ \overline{\rm AF} \perp \overline{\rm BC}$$ and $$ \overline{\rm BE} \perp \overline{\rm AC}$$, this makes $$ {\rm A, E, F, B} $$ on the same circle with $$ \overline{\rm AB} $$ as the diameter;
 * 2) Draw bisector of $$ \overline{\rm EF} $$ to intersect $$ \overline{\rm AB} $$ at $$ {\rm D} $$. $$ {\rm D} $$ is the midpoint of $$ \overline{\rm AB} $$. At this point, we have found $$ {\rm D, E} $$, and $$ {\rm F} $$ with a cost of 3L. However, we have the extra benefit that the bisector of $$ \overline{\rm EF} $$ is already drawn and can be reused;
 * 3) Draw the bisector of $$ \overline{\rm DE} $$ to intersect the previous bisector at $$ {\rm G} $$, which is the center of the desired circle.

Solution 9E

 * 1) Take the same first 2 steps as solution 5L(a) to find points $${\rm D, E, F} $$, this cost 2L 4E;
 * 2) Now the point is that $$ \odot{\rm D} $$ can be reused in finding the bisectors: draw $$ \odot{\rm E} $$ with radius $$ \overline{\rm DE} $$, to intersect $$ \odot{\rm D} $$ at $$ {\rm H} $$ and $$ {\rm J} $$, $$ \odot{\rm F} $$ with radius $$ \overline{\rm DF} $$, to intersect $$ \odot{\rm D} $$ at $$ {\rm K} $$ and $$ {\rm L} $$;
 * 3) Connect $$ \overline{\rm HJ} $$ and $$ \overline{\rm KL} $$ to intersect at $$ {\rm G} $$, which is the center of the desired circle.