General – High School Geometry

This is a paragraph.

A sample problem from social media I

Examine the following problem from a Facebook group:

Solve the above problem. (See below for enlarged image.)

Sample solution here .

Takeaway

Let $ABC$ be a non-right triangle with side-lengths $a,b,c$ , and let $T_b$ be as defined in today’s post. Then the following statements are equivalent:

$OT_b=\frac{2R^2}{a}$
$(b^2-a^2)^2=(ac)^2+(cb)^2$ .

Task

Let be a triangle with vertices at , , and . Let be as defined in today’s post.
1. Find the coordinates of $T_b$
2. Verify that $AT_b=2R=CT_b$ , where $R$ is the circumradius.

An example of harmonic division II

Let $ABC$

be a right triangle in which $\angle C=90^{\circ}$

, and let $F_c$

be the foot of the altitude from $C$

. The segment $CF_c$

contains two special points: the symmedian point ( $S$

), and the Kosnita point ( $K$

). We show an example where $S$

and $F_c$

divide $CK$

harmonically.

Given $\triangle ABC$

with vertices located at $A(0,6)$

, $B(8,0)$

, and $C(0,0)$

, find the symmedian point.

Observe that the given triangle is right-angled at $C$ . Further, in a right triangle, the symmedian point is the midpoint of the right-angled vertex and the foot of the altitude from the right-angled vertex. In a previous post, we found the foot of the altitude from $C$ to be $F_c\left(\frac{72}{25},\frac{96}{25}\right)$ . Thus, the symmedian point in this case is $S\left(\frac{36}{25},\frac{48}{25}\right)$ .

Given $\triangle ABC$

with vertices located at $A(0,6)$

, $B(8,0)$

, and $C(0,0)$

, verify that $S$

and $F_c$

divide $CK$

harmonically, where $K$

is the Kosnita point, $S$

is the symmedian point, and $F_c$

is the foot of the altitude from $C$

.

We have $C(0,0)$ , $S\left(\frac{36}{25},\frac{48}{25}\right)$ , $K\left(\frac{48}{25},\frac{64}{25}\right)$ , and $F_c\left(\frac{72}{25},\frac{96}{25}\right)$ .

Explicit calculation gives

$CS=\frac{12}{5}, SK=\frac{4}{5}, CF_c=\frac{24}{5},KF_c=\frac{8}{5}$

Thus $S$ and $F_c$ divide $CK$ harmonically, since:

$\frac{CS}{SK}=\frac{3}{1}=\frac{CF_c}{KF_c}.$

Takeaway

In any right triangle, the following statements are equivalent:

the right triangle is isosceles
the Kosnita point coincides with the centroid.

Task

(Equivalent statements) Let be a non-right triangle. PROVE that the three statements below are equivalent:
1. $2\cos A\cos B+\cos C=0$
2. $2\sin A\sin B-\cos C=0$
3. $\tan A+\tan B+2\tan C=0$
(Equal steps) Suppose that triangle $ABC$ satisfies any of the three equivalent statements above. In such a triangle, let $S_c$ be the foot of the symmedian from $C$ , and let $F_a,F_b$ be the feet of the altitudes from $A,B$ in that order. PROVE that $F_aS_c=F_bS_c=\frac{abc}{a^2+b^2}$ .