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Given acute triangle $\triangle ABC$ satisfying $|\overline{AB}| \ne |\overline{AC}|$. Let $D,E$, respectively, be the midpoints of $\overline{AB}, \overline {AC}$. Let $Q, P$ be the intersections of $(\triangle ADE)$ and $(\triangle BCD)$, $(\triangle ADE)$ and $(\triangle BCE)$, respectively. Prove that $|\overline{AP}| = |\overline{AQ}|$.

I have already tried using radical axis but still cannot figure out the solution. Please help me with this. Thanks.

enter image description here

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  • $\begingroup$ Are you familiar with barycentric coordinates? $\endgroup$
    – Anubhab
    Jan 18, 2019 at 18:33

3 Answers 3

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Draw a line through the midpoint between BD perpendicular to BD. Draw a line through the midpoint of CE perpendicular to CE. Let the point where these meet be labeled K. Draw the line AK. From here it should be rather straightforward that H, the center of the circle circumscribing ADE, lies on AK and that AK bisects the angle GAF. Since the triangle AGF is also circumscribed, AG = AF.

Solution

EDIT: Further explanations in the figure below. Draw two lines, not parallel to each other, of which AC and AB are segments, with A being their point of intersection. Draw any two parallel lines crossing the other two, neither of which pass through A. In the figure they intersect at C, B, D, and E. Draw perpendicular lines through the midpoints of the segments CD, AD, BE, AE. Let the points where these meet be G and F. Now G, F, and A are colinear.

Incidentally, the same steps are perfomed when circumscribing the circles above. Hence, A, H, and K are colinear in the figure above.

Bisecctors

EDIT: Now, draw the circle with center K and radius KF, as in the figure below. Note the intersections F and G. From this it should be obvious that KA bisects the angle GAF.

Full Solution

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  • $\begingroup$ If you want to fiddle around with it: ggbm.at/gkt9x2yc $\endgroup$ Jan 17, 2019 at 19:09
  • $\begingroup$ Thank you very much for the strategy. Can you explain a little bit clearer why A, K, H are collinear and AK bisects the angle GAF? $\endgroup$
    – Martin Tr
    Jan 17, 2019 at 22:01
  • $\begingroup$ I've added a first step. This "lemma" can be fiddled with here: geogebra.org/geometry/xdfts74y $\endgroup$ Jan 18, 2019 at 7:46
  • $\begingroup$ I've edited the answer again to add a second step. $\endgroup$ Jan 18, 2019 at 16:36
  • $\begingroup$ Thank you so much. I’ve totally understood!!! $\endgroup$
    – Martin Tr
    Jan 22, 2019 at 14:22
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The most clean solution (from my point of view, ofc) is using the inversion:

enter image description here

  1. Draw a sufficiently large circle with the center in $A$ and invert points $B,C,D,E,P,Q \to B',C',D',E',P',Q'$
  2. Blue circle $ADE\to$ line $D'E'$, circles $BCD$ and $BCE$ will transform to corresponding circles (red and green).
  3. Segment $B'C'$ will be a midline in $\triangle A'D'E'$.
  4. Let's take $MN$ a perpendicular bisector of $B'C'$. Since centers of green circles lie on $MN$, points $D'$ and $Q'$ are symmetric around $MN$. This is also true for points $P'$ and $E'$.
  5. Everything else is just algebra:

$$HQ' = HN+NQ'=HN+ND'=HN+(HN+HD')=2\frac{HE'-HD'}4+HD'=\\=\frac{HE'+HD'}2=\frac{E'D'}2=B'C'$$

  1. The same is true for $HP'=B'C'$. Thus in $\triangle AP'Q'$ altitude $AH$ is also a median, so $AP'=AQ' \Rightarrow AP=AQ$
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We shall use Barycentric Coordinates with reference triangle $ABC$, that is $P=(x,y,z) \stackrel{def}{\iff} \vec P=x\vec A+y\vec B +z \vec C$ and $x+y+z=1$. Also, $(x:y:z)=(\frac{x}{x+y+z},\frac{y}{x+y+z},\frac{z}{x+y+z})$. Let $a=BC,\ b=CA,\ c=AB.$

The general equation of a circle(G.E.C.) is $-a^2yz-b^2zx-c^2xy+(x+y+z)(ux+vy+wz)=0.$

Let $T=PB\cap QC$. $AP=AQ\iff\angle AEP=\angle ADQ \iff \angle PBC= \angle QCB \iff K$ lies on perpendicular bisector of $BC$.

$A=(1:0:0),\ D=(1:1:0),\ E=(1:0:1)$. Plugging into G.E.C., we obtain $(ADE): -a^2yz-b^2zx-c^2xy+\frac{(x+y+z)}{2}(c^2y+b^2z)=0$. Also, $B=(0:1:0)$ and $C=(0:0:1)$. Therefore, $(BEC): -a^2yz-b^2zx-c^2xy+\frac{(x+y+z)}{2}(b^2x)=0$.

Without loss of generality, let $P=(1:y_1:z_1)$. Plugging $P$ into $(ADE)$ and $(BCE)$ and solving for $z_1$, $z_1=\frac{b^2-c^2}{2a^2}$. By symmetry, if $Q=(1:y_2:z_2)$, $y_2=\frac{c^2-b^2}{2a^2}$.

Therefore, $T=PB\cap QC=(1:\frac{c^2-b^2}{2a^2}:\frac{b^2-c^2}{2a^2})=(2a^2:c^2-b^2:b^2-c^2)$. Clearly, $T$ satisfies the equation of perpendicular bisector of $BC$ : $a^2(y-z) +x(b^2-c^2)=0$

$\blacksquare$

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