Midpoints of altitudes - a GM problem of Angelescu and more

darij grinberg

I have already posted this in Hyacinthos message #8150, but there was no reaction. Hope that this will be different here. The problem is really beautiful, and I have a rather nice synthetic solution. Here goes the problem:

Let D, E, F be the feet of the three altitudes of a triangle ABC, and let D', E', F' be the midpoints of the segments AD, BE, CF. Let the line E'F' meet the lines AB and CA at the points $B_a$ and $C_a$ , let the line F'D' meet the lines BC and AB at the points $C_b$ and $A_b$ , and let the line D'E' meet the lines CA and BC at the points $A_c$ and $B_c$ .

(a) (original problem by A. Angelescu, Gazeta Matematica 35): If X', Y', Z' are the circumcenters of triangles $AB_aC_a$ , $BC_bA_b$ , $CA_cB_c$ , respectively, then the points D', E', F', X', Y', Z' lie on one circle.

(b) (my observations, which lead to a proof of (a)): Let L be the de Longchamps point of triangle ABC (this is the image of the orthocenter of triangle ABC in the homothety with center at the centroid of triangle ABC and factor -2). Let the lines AL, BL, CL meet the lines BC, CA, AB at the points X, Y, Z, respectively. Then, the points $B_a$ , $C_a$ and D lie on the circle with diameter AX. Similarly for the circles with diameters BY and CZ.

Darij

**Posted:** Sun Jun 19, 2005 1:38 pm

_________________
Now the die is cast, the first step taken, a glimmer of hope lights up our lives
Visions of the past, dreams forsaken forming right under our eyes
We are alive...

yetti

Denote $a = BC, b = CA, c = AB$ the triangle sides and $\alpha = \angle A, \beta = \angle B, \gamma = \angle C$ the triangle angles. WLOG, assume that the angle $\gamma > \alpha, \beta$ is the largest one, the relatively trivial case of an isosceles triangle is disregarded. We will also assume that $CH < HF$ , i.e., the points $C, H, F', F$ follow on the altitude $CF$ in this order. By considering an isosceles triangle with $CH = HF$ ,

$CH = \frac{a \sin \alpha}{2} = \frac{a \cos \alpha}{\tan \alpha} = HF$

$\sin^2{\alpha} = 2 \cos^2{\alpha},\ \ \cos^2{\alpha = \frac 1 3,\ \ \cos \gamma = 1 - 2 \cos^2{\alpha} = \frac 1 3,\ \ \gamma...$

we see that if $CH < HF$ , $\cos \gamma < \frac 1 3$ and $\gamma > 70.53^o$ is the largest triangle angle. The case when $CH > HF$ , while $\gamma$ remains the largest triangle angle, is treated similarly.

Step 1: Let $(O_B), (O_E)$ be the circumcircles of the triangles $\triangle BE'F, \triangle EE'F$ and let $A', B'$ be their intersections with the sides $CA, BC$ other than the points $E, B$ , respectively. Using properties of the orthic triangle $\triangle DEF$ , the triangles $\triangle EFA' \sim \triangle BFB'$ are similar, because the angles $\angle A'EF = \angle B'BF = \beta$ are equal and the angles

$\angle BB'F = \angle BE'F = 180^o - \angle FE'E = \angle EA'F$

are also equal. The similarity coeficient of these 2 triangles is

$\frac{EF}{BF} = \frac{a \cos \alpha}{a \cos \beta} = \frac{\cos \alpha}{\cos \beta}$

The angles $\angle EFA' = \angle EE'A'$ spanning the same arc $FA'$ of the circle $(O_E)$ are equal and the angles $\angle BFB' = \angle BE'B'$ spanning the same arc $BB'$ of the circle $(O_B)$ are also equal. Due to similarity of the triangles $\triangle EFA' \sim \triangle BFB'$ , these angles are equal to each other, hence, the angles $\angle EE'A' = \angle BE'B'$ are vertical, which means that the points $A'E'B'$ are collinear. The triangles $\triangle AFA' \sim \triangle DFB'$ are also similar, because the angles $\angle FAA' = \angle FDB' = \alpha$ are equal and the angles $\angle AA'F = 180^o - \angle EA'F = 180^o - \angle BB'F = \angle DB'F$ are also equal. The similarity coefficient of these 2 triangles is also

$\frac{AF}{DF} = \frac{b \cos \alpha}{b \cos \beta} = \frac{\cos \alpha}{\cos \beta}$

Denote $\omega = \angle BE'F$ . Using the sine theorem for the triangle $\triangle BFE'$ ,

$\frac{BF}{FE'} = \frac{\sin \omega}{\sin{(90^o - \alpha)}} = \frac{\sin \omega}{\cos \alpha}$

The segment $FE'$ is the median of the triangle $\triangle BEF$ from the vertex $F$ . Hence

$FE'^2 = \frac 1 4(2EF^2 + 2BF^2 - BE^2) =$

$= \frac{a^2}{4}(2\cos^2{\alpha} + 2 \cos^2{\beta} - \sin^2{\gamma}) = \frac{a^2}{4}(\cos^2{\alpha} + \cos^2{\beta} - 2 \cos \...$

$\sin^2{\omega} = \frac{BF^2}{FE'^2}\ \cos^2{\alpha} = \frac{4 \cos^2{\alpha} \cos^2{\beta}}{\cos^2{\alpha} + \cos^2{\beta} - ...$

$\cot^2{\omega} = \frac{1 - \sin^2{\omega}}{\sin^2{\omega}} = \frac{\cos^2{\alpha} + \cos^2{\beta} - 4 \cos^2{\alpha} \cos^2{\...$

$= \frac{\cos^2{\alpha} (1 - \cos^2{\beta}) + \cos^2{\beta} (1 - \cos^2{\alpha}) - 2 \sin \alpha \cos \alpha \sin \beta \cos \...$

$= \frac 1 4 (\tan^2{\alpha} + \tan^2{\beta} - 2 \tan \alpha \tan \beta) = \frac 1 4 (\tan \alpha - \tan \beta)^2$

Let $C'$ be the intersection of the altitude $CF$ with the circle $(O_B)$ other than the point $F$ . Since the angle $\angle BFC' = 90^o$ is right, $BC' = 2O_BB$ is a diameter of the circle $(O_B)$ and

$FC' = \sqrt{BC'^2 - BF^2} = \sqrt{4O_BB^2 - BF^2}$

Radius of the circle $(O_B)$ is equal to $O_BB = \frac{BF}{2 \sin \omega}$ and the power of the vertex $C$ to this circle is

$CB \cdot CB' = CF \cdot CC',\ \ \ CB (CB - BB') = CF (CF + FC')$

$BB' = CB - \frac{CF (CF + \sqrt{4O_BB^2 - BF^2})}{CB} =$

$= a - a (\sin^2 \beta + \sin \beta \cos \beta \cot \omega}) = a (\cos^2{\beta} - \sin \beta \cos \beta \cot \omega)$

Let the line $A'B'$ cut the altitude $AD$ at a point $D''$ . Using Menelaus' theorem for the triangle $\triangle ACD$ cut by the line $A'B'$ ,

$\frac{A'A}{A'C} \cdot \frac{B'C}{B'D} \cdot \frac{D''D}{D''A} = 1$

$\frac{D''D}{D''A} = \frac{A'C}{A'A} \cdot \frac{B'D}{B'C} = \frac{\cos \beta}{\cos \alpha} \cdot \frac{A'C}{BD'} \cdot \frac{...$

$= \frac{\cos \beta}{\cos \alpha} \cdot \frac{BB' \frac{\cos \alpha}{\cos \beta} + a \cos \gamma}{a - BB'} = \frac{BB' \cos \a...$

$= \frac{a (\cos^2{\beta} - \sin \beta \cos \beta \cot \omega) \cos \alpha + a \cos \beta \cos \gamma}{[a - a (\cos^2{\beta} -...$

$= \frac{(\cos^2{\beta} - \sin \beta \cos \beta \cot \omega) \cos \alpha + \cos \beta \cos \gamma}{(\sin^2{\beta} + \sin \beta...$

$= \frac{\tan \alpha - \cot \omega}{\tan \beta + \cot \omega} = \frac{\tan \alpha - \frac 1 2 (\tan \alpha - \tan \beta)}{\tan...$

Thus we proved that $D''A = D''D$ , i.e., $D'' \equiv D'$ is the midpoint of the altitude $AD$ lying on the line $A'B'$ . Since the midpoint $E'$ of the altitude $BE$ also lies on the line $A'B'$ , the points $A' \equiv A_c$ and $B' \equiv B_c$ are identical. As a result, the quadrilateral $CA_cFB_c$ is cyclic, because

$\angle A_cFB_c = \angle A_cFE + \angle EFD + \angle DFB_c =$

$= (180^o - 2 \gamma) + \angle A_cFE + \angle AFA_c = 180^o - 2 \gamma + \gamma = 180^o - \gamma$

This means that the circumcircle $(Z')$ of the triangle $\triangle CA_cB_c$ passes through the foot of the altitude $CF$ . Consequently, this circle is centered on the triangle midline $A_1B_1 \parallel AB$ , where $A_1, B_1$ are the midpoints of the sides $BC, CA$ . Similarly, quadrilaterals $AB_aDC_a, BA_bEC_b$ are also cyclic, the circles $(X'), (Y')$ pass through the altitude feet $D, E$ and they are centered on the triangle midlines $B_1C_1 \parallel BC, C_1A_1 \parallel CA$ , where $C_1$ is the midpoint of the remaining triangle side $AB$ . The power of the orthocenter $H$ to the circlec $(X'), (Y'), (Z')$ is $HA \cdot HD = HB \cdot HE = HC \cdot HF$ , i.e., the orthocenter is the radical center of these 3 circles.

To be continued.

Yetti

_________________
Carthage must be destroyed.

yetti

Step 2: Earlier, we found that

$BB_c = a (\cos^2{\beta} - \sin \beta \cos \beta \cot \omega) = a \left(\cos^2{\beta} - \sin \beta \cos \beta\ \frac{\tan \alp...$

By cyclic exchange, we obtain

$CC_a = b \left(\cos^2{\gamma - \sin \gamma \cos \gamma\ \frac{\tan \beta - \tan \gamma}{2}\right)$

Comparing the segments $CC_a, AA_c$ , we have

$DB_c = BD - BB_c = c \cos \beta - a (\cos^2{\beta} - \sin \beta \cos \beta \cot \omega) =$

$= a \left(\frac{\sin \gamma}{\sin \alpha} \cos \beta - \cos^2{\beta} + \sin \beta \cos \beta \cot \omega \right) =$

$= a (\cos^2{\beta} + \cot \alpha \sin \beta \cos \beta - \cos^2{\beta} + \sin \beta \cos \beta \cot \omega) =$

$= a \sin \beta \cos \beta \left(\cot \alpha + \frac{\tan \alpha - \tan \beta}{2}\right)$

Removing the angle $\alpha$ from the expression for $AA_c$ ,

$AA_c = \frac{\cos \alpha}{\cos \beta}\ DB_c = b \sin \alpha \cos \alpha \left(\cot \alpha + \frac{\tan \alpha - \tan \beta}{2...$

$= b \left(\cos^2{\alpha} + \frac{\sin^2{\alpha}}{2} - \sin \alpha \cos \alpha\ \frac{\tan \beta}{2}\right) =$

$= \frac b 2 [1 + (\cos \beta \cos \gamma - \sin \beta \sin \gamma)^2\ +$

$+ (\sin \beta \cos \gamma + \cos \beta \sin \gamma)(\cos \beta \cos \gamma - \sin \beta \sin \gamma) \tan \beta}] =$

$= \frac b 2 [1 + \cos^2{\beta} \cos^2{\gamma} - 2 \sin \beta \sin \gamma \cos \beta \cos \gamma + \sin^2{\beta}\sin^2{\gamma}...$

$+ (\sin \beta \cos \beta \cos^2{\gamma} - \sin^2{\beta} \sin \gamma \cos \gamma + \cos^2{\beta} \sin \gamma \cos \gamma - \si...$

$= \frac b 2 (1 + \cos^2{\beta} \cos^2{\gamma} - 2 \sin \beta \sin \gamma \cos \beta \cos \gamma + \sin^2{\beta}\sin^2{\gamma}...$

$+ \sin^2{\beta} \cos^2{\gamma} - \sin^2{\beta} \tan \beta \sin \gamma \cos \gamma + \sin \beta \cos \beta \sin \gamma \cos \g...$

$= \frac b 2 (1 + \cos^2{\gamma} - \sin \beta \sin \gamma \cos \beta \cos \gamma - \sin^2{\beta} \tan \beta \sin \gamma \cos \...$

$= \frac b 2 [2 \cos^2{\gamma} + \sin \gamma \cos \gamma (\tan \gamma - \sin \beta \cos \beta - \sin^2{\beta} \tan \beta)] =$

$= b \left(\cos^2{\gamma} - \sin \gamma \cos \gamma \frac{\tan \beta - \tan \gamma}{2}\right) = CC_a$

Thus we proved that the segments $AA_c = CC_a$ are not only equal, but either both inside or both outside of the triangle side $CA$ (this follows from the signs of the tangents). Similarly, we could show that the segments $BB_c = CC_b, AA_b = BB_a$ are equal and either both inside or both outside of the triangle sides $BC, AB$ . As a result, the powers of the midpoint $C_1$ of the side $AB$ to the circles $(X'), (Y')$ are equal. For example, if the points $A_b, B_a$ are both outside of the side $AB$ ,

$C_1A \cdot C_1B_a = C_1A (C_1B + BB_a) = C_1B (C_1A + AA_b) = C_1B \cdot C_1A_b$

and consequently, the radical axis of the circles $(X'), (Y')$ passes through the midpoint $C_1$ . Similarly, the radical axes of the circle pairs $(Y'), (Z')$ and $(Z'), (X')$ pass through the midpoints $A_1, B_1$ of the sides $BC, CA$ .

To be continued.

Yetti

_________________
Carthage must be destroyed.

yetti

Step 3: The angles $\angle B_cA_cF \equiv \angle E'A_cF = \angle E'EF \equiv \angle BEF$ and the angles $\angle A_cB_cF \equiv \angle E'B_cF = \angle E'BF \equiv \angle EBF$ spanning the same arcs $E'F$ of the circles $(O_B), (O_E)$ are equal. Hence, the triangles $\triangle ABH \sim \triangle EFB \sim \triangle AFD \sim \triangle A_cFB_c$ are all similar, because

$\angle FAD \equiv \angle BAH = 90^o - \beta,\ \ \angle B_cA_cF = \angle BEF = 90^o - \angle FEA = 90^o - \beta$

$\angle FBE \equiv \angle ABH = 90^o - \alpha,\ \ \angle A_cB_cF = \angle EBF \equiv \angle EBA = 90^o - \alpha$

The segments $FE', HC_1$ are the corresponding medians of the similar triangles $\triangle EFB \sim \triangle ABH$ , hence, the angles $\angle FC_1H \equiv BC_1H = \angle FE'E = 180^o - \angle FE'B = \omega$ are equal. The angles $\angle FE'B = \angle FB_cB = 180^o - \omega$ spanning the the same arc $BF$ of the circle $(O_B)$ are also equal. Let the circle $(Z')$ intersect the line $AB$ at a point $Z$ different from $F$ . Since the angle $\angle CFZ \equiv CFB = 90^o$ is right, $CZ$ is a diameter of the circle $(Z')$ . The angles $\angle FZC = \angle FB_cC = 180^o - \angle FB_cB = \omega$ are also equal. Thus the right angle triangles $\triangle CFZ \sim \triangle HFC_1$ are centrally similar with the similarity center $F$ and consequently, the lines $HC_1 \parallel CZ$ are parallel. As a result, the circumcenter $Z'$ of the triangle $\triangle CA_cB_c$ is identical with the intersection of the midline $A_1B_1$ with the line $CZ \parallel HC_1$ parallel to the radical axis of the circles $(X'), (Y')$ . Since the cevians $A_1H, B_1H, C_1H$ of the medial triangle $\triangle A_1B_1C_1$ , the pairwise radical axes of the circles $(X'), (Y'), (Z')$ , meet at the orthocenter $H$ of the original triangle $\triangle ABC$ , by central similarity of these 2 triangles with the similarity center at the triangle centroid $G$ and similarity coefficient -2, the cevians $AX, BY, CZ$ of the triangle $\triangle ABC$ parallel to $A_1H, B_1H, C_1H$ meet at a point $L$ on the line $GH$ (Euler line) opposite to the orthocenter $H$ , such that $\frac{GH}{LH}\right = 2$ (Longchamps point, never heard of it) or $\frac{OH}{OL} = 1$ , i.e., the reflection of the orthocenter $H$ in the circumcenter $O$ . Since the angles $\angle ADX = \angle BEY = \angle CFZ = 90^o$ spanned by the appropriate arcs $AX, BY, CZ$ of the circumcircles $(X'), (Y'), (Z')$ are right, the circumcenters $X', Y', Z'$ are the midpoints of the cevians $AX, BY, CZ$ .

The right angle triangle $\triangle CFZ$ is half the size of the right angle triangle $\triangle CF'Z'$ , hence, the triangles $\triangle CFZ \sim \triangle HFC_1$ are also similar with the similarity coefficient

$\frac{CF'}{HF} = \frac{CF}{2HF} = \frac{a \sin \beta}{2 a \cos \beta \cot \alpha} = \frac{\tan \alpha \tan \beta}{2}$

$Z'F' = \frac{\tan \alpha \tan \beta}{2}\ C_1F = \frac{\tan \alpha \tan \beta}{2} \left(a \cos \beta - \frac c 2\right) = \fra...$

$= \frac{a \tan \alpha \tan \beta}{4} \left(2 \cos \beta - \cos \beta - \frac{\sin \beta}{\tan \alpha}\right) = \frac{a \sin \...$

$A_1Z' = A_1F' - Z'F' = \frac{a \cos \beta}{2} - \frac{a \sin \beta}{4}(\tan \alpha - \tan \beta) =$

$= \frac{a}{4}\left[\cos \beta (1 - \tan \alpha \tan \beta) + \cos \beta (1 + \tan^2{\beta}) \right] = \frac{a}{4 \cos \beta} ...$

$A_1F' \cdot A_1Z' = \frac{a \cos \beta}{2} \cdot \frac{a}{4 \cos \beta} \left[\cos^2{\beta} (1 - \tan \alpha \tan \beta) + 1\...$

$= \frac{a^2}{8} \left[\cos^2{\beta} (1 - \tan \alpha \tan \beta) + 1\right] = \frac{a^2}{8}\left(1 - \frac{\cos \beta \cos \g...$

Let $Y$ be the intersection of the circle $(Y')$ with the line $CA$ other that the points $E$ . For similar reasons as above, the right angle triangles $\triangle BE'Y' \sim \triangle BEY \sim \triangle HEB_1$ are similar, the similarity coefficient of $\triangle BE'Y' \sim \triangle HEB_1$ equal to

$\frac{BE'}{HE} = \frac{BE}{2HF} = \frac{a \sin \gamma}{2 a \cos \gamma \cot \alpha} = \frac{\tan \gamma \tan \alpha}{2}$

$Y'E' = \frac{\tan \gamma \tan \alpha}{2}\ B_1E = \frac{\tan \gamma \tan \alpha}{2} \left(\frac b 2 - a \cos \gamma\right) = \...$

$= \frac{a \tan \gamma \tan \alpha}{4} \left(\frac{\sin \gamma}{\tan \alpha} + \cos \gamma - 2\cos \gamma\right) = \frac{a \si...$

$A_1Y' = AE' + Y'E' = \frac{a \cos \gamma}{2} + \frac{a \sin \gamma}{4}(\tan \gamma - \tan \alpha) =$

$= \frac{a}{4}\left[\cos \gamma (1 + \tan^2{\gamma) + \cos \beta (1 - \tan \gamma \tan \alpha) \right] = \frac{a}{4 \cos \gamm...$

$A_1E' \cdot A_1Y' = \frac{a \cos \gamma}{2} \cdot \frac{a}{4 \cos \gamma} \left[\cos^2{\gamma} (1 - \tan \gamma \tan \alpha) ...$

$= \frac{a^2}{8} \left[\cos^2{\gamma} (1 - \tan \gamma \tan \alpha) + 1\right] = \frac{a^2}{8}\left(1 - \frac{\cos \beta \cos ...$

Thus we proved that $A_1E' \cdot A_1Y' = A_1F' \cdot A_1Z'$ , i.e., that the quadrilateral $E'F'Y'Z'$ is cyclic. In an entirely similar way, we can show that $B_1F' \cdot B_1Z' = B_1D' \cdot B_1X'$ and $C_1D' \cdot C_1X' = C_1E' \cdot C_1Y'$ , i.e., that the quadrilaterals $F'D'Z'X', D'E'X'Y'$ are also cyclic. This still leaves the possibility that the circumcircles of these 3 cyclic quadrilaterals are different form each other and from the circumcircle of the triangle $\triangle D'E'F'$ . Assume that the circumcircles $(P_A), (P_B)$ of the cyclic quadrilaterals $E'F'Y'Z', F'D'Z'X'$ intersecting at the points $F', Z'$ are different from each other. Since $C_1D' \cdot C_1X' = C_1E' \cdot C_1Y'$ , the point $C_1$ has the same power to the circles $(P_A), (P_B)$ , which means that it lies on their radical axis $F'Z' \equiv A_1B_1$ , i.e., that the midpoints $A_1, B_1, C_1$ of the triangles sides $BC, CA, AB$ are collinear, which is clearly impossible. As a result, the circumcircles of the cyclic quadrilaterals $E'F'Y'Z', F'D'Z'X', D'E'X'Y'$ are identical. This concludes the proof of the problem proposition.

Perhaps we should mention the case, when the angle $\gamma = 90^0$ is right. The triangle $\triangle D'E'F'$ then degenerates into a line segment, its circumcircle into the midline $A_1B_1$ of the right angle triangle $\triangle ABC$ , the circles $(X'), (Y')$ degenerate into the lines $CA, BC$ and the circle $(Z')$ becomes identical with the triangle 9-point circle. Hence, the circumcenters $X', Y'$ move to infinity and the circumcenter $Z'$ , the midpoint of the segment $CC_1 \equiv OH$ lies on the midline $A_1B_1$ , the degenerate circumcircle of the degenerate triangle $\triangle D'E'F'$ .

To be continued.

Yetti

_________________
Carthage must be destroyed.

yetti

Bonus: Earlier, we proved that the circumcenters $X', Y', Z'$ lie of the (extended) triangle midlines $B_1C_1, C_1A_1, A_1B_1$ . The midpoints $D', E', F'$ of the altitudes $AD, BE, CF$ also lie on these midlines. The cevians $A_1X', B_1Y', C_1Z'$ of the medial triangle $\triangle A_1B_1C_1$ cut its (extended) sides at the points $X', Y', Z'$ in the ratios $\frac{X'B_1}{X'C_1},\ \frac{Y'C_1}{Y'A_1},\ \frac{Z'A_1}{Z'B_1}$ .

$Z'B_1 = A_1B_1 - Z'A_1 = \frac c 2 - \frac{a}{4 \cos \beta} \left[\cos^2{\beta} (1 - \tan \alpha \tan \beta) + 1\right] = \fr...$

$= \frac a 4 \left(2 \cos \beta + 2 \sin \beta \cot \alpha - \cos \beta + \tan \alpha \sin \beta - \frac{1}{\cos \beta}\right)...$

$= \frac{a \sin \beta}{4} (2 \cot \alpha + \tan \alpha - \tan \beta) = \frac{a \sin \beta}{4 \sin \alpha \cos \alpha} \left[\c...$

$\frac{Z'A_1}{Z'B_1} = \frac{\sin \alpha \cos \alpha}{\sin \beta \cos \beta } \cdot \frac{\cos^2{\beta} (1 - \tan \alpha \tan ...$

By cyclic exchange (and paying attention to the signs of the tangents),

$\frac{X'B_1}{X'C_1} \cdot \frac{Y'C_1}{Y'A_1} \cdot \frac{Z'A_1}{Z'B_1} =$

$= \frac{\cos^2{\gamma} (1 - \tan \beta \tan \gamma) + 1}{\cos^2{\beta} (1 - \tan \beta \tan \gamma) + 1} \cdot \frac{\cos^2{\...$ . $1$

because all terms cancel out, for example,

$(?)\ \ \cos^2{\gamma} (1 - \tan \beta \tan \gamma) + 1 = \cos^2{\alpha} (1 - \tan \alpha \tan \beta) + 1$

$(?)\ \ \cos^2{\gamma} - \cos^2{\alpha} = (\sin \gamma \cos \gamma - \sin \alpha \cos \alpha) \tan \beta = \frac{(\sin \gamma ...$

$(?)\ \ \cos^2{\gamma} - \cos^2{\alpha} = \frac{(\sin^2{\gamma} - \sin^2{\alpha}) \cos \alpha \cos \gamma + (\cos^2{\gamma} - ...$

$(?)\ \ \cos^2{\gamma} - \cos^2{\alpha} = \frac{(\cos^2{\gamma} - \cos^2{\alpha}) (\sin \alpha \sin \gamma - \cos \alpha \cos ...$

and the last equation is an obvious identity. By Ceva's theorem, the cevians $A_1X', B_1Y', C_1Z'$ of the medial triangle $\triangle A_1B_1C_1$ concur at a point $Q$ , so-called Yetti's point of the triangle $\triangle ABC$ (hey, everybody's got a triangle point and there's plenty more available Rotfl

).

Proposition: Prove that the common circumcenter $P$ of the triangles $\triangle D'E'F', \triangle X'Y'Z'$ lies on the line $QH$ connecting Yetti's point $Q$ with the triangle orthocenter $H$ .

Yetti

Note: All trigonometric formulas in the entire proof check with the Sketchpad.

**Posted:** Mon Jun 27, 2005 3:40 am

_________________
Carthage must be destroyed.

pestich · Poincare Conjecture Offline Joined: 03 Mar 2004 Posts: 181

http://forumgeom.fau.edu/FG2004volume4/FG200405index.html

The above article by JPE may bring some clearity into the subject roughed-up by previous
replies by our own Yetti. A complete quad where the 4th side is a line thru midpoints of
altitudes in a triangle made by the other 3 lines should yield some interesting properties.

Maj. Pestich

**Posted:** Mon Jun 27, 2005 8:19 pm

pestich · Poincare Conjecture Offline Joined: 03 Mar 2004 Posts: 181

I still hope that in some geo problem the Fox-Talbot theorem will be used.
Is it possible this is the one?

Maj. Pestich

**Posted:** Mon Jun 27, 2005 8:39 pm

yetti

**Posted:** Mon Jun 27, 2005 8:55 pm

_________________
Carthage must be destroyed.

pestich · Poincare Conjecture Offline Joined: 03 Mar 2004 Posts: 181

My dear Yetti,

You do not have to rough me up. I was the first one to agree that it was a wonderful solution indeed.
Put your customary doubts about your own (I had to use this word again, sorry) solutions aside: this
time it is a bona fide one. I'd say it is even more than a solution to a tough problem (those people at Hya-group
wouldda never imagined the circumcenters on midlines), it is a novel to read, with a hero, villans, chapters,
paragraphs and such. But you are a bit lazy, and if I do not tell you, nobody will. Where are the table of contents,
index, and referrences?

Sincerely,

Maj. Pestich

**Posted:** Tue Jun 28, 2005 3:04 pm

yetti

**Posted:** Wed Jun 29, 2005 8:27 pm

_________________
Carthage must be destroyed.

lomos_lupin

jesus christ , jesus christ yetti
i havent seen a stonger one in calculating than you in my life even never heard of
and this skill of you is so valuable but please dont take these in a bad manner but
do you think that all of these are necessary Question

in some cases some uses of geometrical ideas would really make the calculations shorter, but in all the ways no one can disagree your abilities O mighty!

**Posted:** Sat Jul 02, 2005 7:50 pm

_________________
Mind likes to fly to mysterious realms.

yetti

No, I do not think that all these calculations are necessary. The fact that everything came out in closed forms (i.e., I did not have to resort anywhere to numerical calculations) indicates that the problem has underlying symmetries, which can be used in the solution (reflections in lines?). After I identified the basic steps of the proof, I was not able to see those symmetries, hence, I used trigonometry. But trigonometry itself is a (powerful) geometric tool. If the sine and cosine functions are defined as the ratios of the opposite and adjacent legs of a right angle triangle to the hypotenuse, geometric proofs of any trigonometric formulas do exist. Trigonometry works even if nothing else does, i.e., if the problem does not have underlying symmetries and you are willing to use numerical methods to get the result. Moreover, after the solution is found by any means, for example by trigonometry, it might be easier for someone else to see the underlying symmetries. In this particular case, it is a mute proposition, because the problem was not posted in the Unsolved Problems section and Darij said that he already has a synthetic solution.

I will give you an example from my line of work. Suppose I have to transport a particle beam in an accelerator over huge distances in a beamline of a rather small diameter without losing too much by hitting the beamline walls. The beamline has to be evacuated and any increase in the beamline diameter calls for larger magnets, the principal item in the cost of building an accelerator. So, the beamline diameter really has to be rather small. The phase space volume of the beam is the "volume" in the space with one coordinate being the beam displacement from the beamline axis and the other coordinate the particle radial velocity toward or away from the beamline axis, proportional to the particle divergence, the angle between the particle path and the beamline axis. This phase space volume is called the beam emittance and it looks like an ellipse tilted to these coordinate axes. From Liouville's theorem of analytical mechanics, it follows that this space volume remains constant, i.e., it cannot be decreased by any focusing element in the beamline. The ellipse can be only rotated or "squashed" without decreasing its area. Consequently, to get a good beam, which can be successfully transported over those huge distances, the beam emittance has to be small to begin with, as the beam emerges from the ion source. Also, the beam emittance is an important beam parameter to measure. The measurement methods depend on whether the beam can be "consumed" for the measurement (for a while), or whether it has to be done on the fly and usually, they are very ingenious. The beam focusing elements are almost exclusively magnetic quadrupole lenses, because at relatively low particle energies, they overcome the focusing capabilities of other elements, such as electrostatic lenses, etc. The action of a magnetic quadrupole lens on the beam is described by a matrix operator with elements that are either normal or hyperbolic trigonometry functions. You can have tens to hundreds of these lenses, which rotate and squash the emittance ellipse in various ways. One of the non-destructive methods of measuring the beam emittance consists of slowly moving a thin wire across the beam and measuring the (very small) beam current. In principle, this yields the position of the vertical tangent to the emittance ellipse at this spot. Several thin wires at several different places in between the quadrupole lenses can be moved simultaneously, thus obtaining several vertical tangents to several different ellipses, which are all images of some original ellipse transformed by the quadrupole lenses. Using the inverse operators for the quadrupole lenses, these tangents can be transformed back to one spot and in this transformation, they become rotated and no longer vertical. To get the beam emittance, an ellipse has to be inscribed into these tangents and its area calculated. For this, at least 3 different tangents are necessary (because the ellipse center is known), but since all measurements carry some errors, more than 3 tangents are measured and the best possible ellipse is inscribed into these lines by the least squares method. This is an interesting application of geometry and I doubt that it could be done without both the normal and hyperbolic trigonometry.

So, three cheers for trigonometry !!! ThumbUp

Regards, Yetti

**Posted:** Sun Jul 03, 2005 6:37 am

_________________
Carthage must be destroyed.

Virgil Nicula

Smeenk show (in Crux Mathematicorum, the problem nr. 2379/98) that the lines MD', NE', PF' are conccurent, where M, N, P are the middlepoints of the sides BC, CA, AB respectively. This fact may be used ?

**Posted:** Tue Jul 05, 2005 4:33 pm

treegoner

After a few hours, I finally got a purely synthetic solution to Darij 's observation. It seems that I am getting very close to proving the 3 circumcenters lying on the circumcirlce of $D'E'F'$ . I will post the solution in a next few days. I am very busy right now.

**Posted:** Fri Jul 08, 2005 2:20 pm

_________________
Life doesn't always follow what I have planned or expected. No matter what way my life leads me, I am still thankful.

treegoner

Finally, everything is clear. And here is the solution (guess this is the longest I have ever posted). I won't write just only the solutions of this problem but also others nice theorems relating to this wonderful sketch.
So, let me first redenote some points so that the problem accords with my figure. (I am sure you don't mind Smile

).
Let $ABC$ be a triangle whose circumcenter and orthocenter are $O$ and $H$ . Let L be the de Longchamp point, i.e the point on $HO$ such that $\frac{OL}{OH} = -1$
Let $A'B'C'$ be the orthic triangle and $MNP$ be the medial triangle. Call $D, E, F$ the midpoints of $AA', BB', CC'$ . Suppose the line $EF$ intersects the line $AB$ and $AC$ at $A_b$ and $A_c$ respectively. Analogously, we define the point $B_a$ , $B_c$ , $C_a$ , $C_b$ .
Let $O_a, O_b, O_c$ be the circumcenters of $AA_bA_c, BB_aB_c, CC_aC_b$ respectively.
All right ! Let 's do some small observation of the de Longchamp point for a moment. Well, by calling $T$ the antipode of $A$ in the circle $(O)$ ., we have $AHTL$ is a parallelogram since the diagonals bisect each other. On the other hand, we have $BHCT$ is a parallelogram as its opposite sides are parallel. Hence $H, M, T$ are collinear and therefore $HM // AL$ . Suppose we call $O_a'$ and $A_1$ the intersections of $AL$ with $NP$ and $BC$ , then surely we have $O_a'$ is the midpoint of $AA_1$ . Thus if we call $U$ the midpoint of $HM$ , since $HM // AA_1$ ,we obtain that $A', U, O_a'$ are collinear. (Later, we shall see that this point $O_a'$ is the point $O_a$ ). So let 's official state it as a theorem :
Theorem 1 $A, U, O_a'$ are collinear.
Let 's next consider the point $U$ . Since $\measuredangle{MEH} = \measuredangle{MFH} = \measuredangle{MA'H} = 90^0$ , we obtain five points $H, E, A', M, E$ lie on the circle $(U)$ with diameter $HM$ . Therefore $\measuredangle{A'EF} = \measuredangle{A'HF} = \measuredangle{A'BA_b}$ . This implies that $BA'EA_b$ is cylic. By similar argument, we obtain that $CA'FA_c$ is cyclic and we have a theorem
Theorem 2 $BA'EA_b$ is inscribed in $(b)$ and $CA'FA_c$ is inscribed in $(c)$
Let 's derive some more properties from this cool theorem. From theorem 2, we have $\measuredangle{A'A_bA_c} = \measuredangle{A'A_bE} = \measuredangle{A'BE} = \measuredangle{A'AA_c}$ . Hence $AA_bA'A_c$ is cyclic, i.e the circumcircle $(O_a)$ of $AA_bA_c$ passes through $A'$ . This means that $O_a$ lies on $NP$ . Now let 's prove that this point lies on $AL$ by showing that $AO_a // HM$ . Let $A_1$ be the second intersection different from $A'$ of the circle $(O_a)$ and $BC$ . Then $AA_1$ is the diameter of $(O_a)$ . Hence $\measuredangle{A_1A_cC} = 90^0$ . Now according to theorem 1 and 2, we have $\measuredangle{A'AA_1} = \measuredangle{A'A_cA_1} = \measuredangle{A'A_c_C} - 90^0 = \measuredangle{A'FC} - 90^0 = \measureda...$ . This yields that $HM // AA_1$ , and therefore $AO_a$ passes through the de Longchamp point $L$ . And Darij 's observation is already proved. Let 's state it
Theorem 3 The circumcircle $(O_a)$ of $AA_bA_c$ passes through $A'$ and $A, O_a, L$ are collinear.
In my next posts, we will give some more observations on the relations of this circle with the circle $(U)$ , $(b)$ and $(c)$ .

**Posted:** Sat Jul 09, 2005 1:31 pm

_________________
Life doesn't always follow what I have planned or expected. No matter what way my life leads me, I am still thankful.

treegoner

OK. Let 's continue with our work.
We have $\measuredangle{A_bA'F} = \measuredangle{A_bA'E} + \measuredangle{EA'F} = \measuredangle{A_bBE} + \measuredangle{EHF} = \measu...$ . Hence $A_bA'FC'$ is cylic. Similarly, we have $A'EB'A_c$ is cylic.
Theorem 4 $A_bA'FC'$ and $A'EB'A_c$ are cylic.
By applying the Menelaus 's theorem to triangle $ABB'$ and three collinear points $A_b, E, A_c$ , we have $\frac{AA_b}{AA_c} = \frac{A_bB}{B'A_c}$ . On the other hand, we have $\frac{A_bB}{A'A_b} = \frac{sin{ \measuredangle{BA'A_b}}}{sinB} = \frac{sin{ \measuredangle{BEA_b}}}{sinB} = \frac{sin{ \measu...$
Reason
Hence $\frac{A'A_b}{A'A_c} = \frac{A_bB}{B'A_c}$ . Thus $\frac{AA_b}{AA_c} = \frac{A'A_b}{A'A_c}$ . This implies that $AA_bA'A_c$ is a harmonic quadrilateral. We will see its application later to prove the original problem.
Theorem 5 $AA_bA'A_c$ is a harmonic quadrilateral.
We will finish our proof in the next post.

**Posted:** Sat Jul 09, 2005 1:51 pm

_________________
Life doesn't always follow what I have planned or expected. No matter what way my life leads me, I am still thankful.

treegoner

Let 's consider the relationship between the two circles $(b)$ and $(c)$ . We have $\measuredangle{A'O_aA_c} = 2.\measuredangle{A'AA_c} = 2.\measuredangle{A'AC} = \measuredangle{A'NA_c}$ . Thus $A'O_aNA_c$ is cylic. Hence $\measuredangle{O_aA'A_c} = \measuredangle{O_aNA} = \measuredangle{PNA} = \measuredangle{BCA} = \measuredangle{A'CA_c}$ . This yields that $A'O_a$ touches the circle $(c}$ at $A'$ . Similarly we have $A'O_a$ touches the circle $(b}$ at $A'$ . And theorem 6 follows
Theorem 6 $(b)$ and $(c)$ touches each other and the line $A'O_a$ at $A'$ .
All right, we are 90% done.
Let $l$ be the line joining the centers of $(b)$ and $(c)$ . Then $l$ passes through $A'$ and is perpendicular to $O_aA'$ . Hence $l$ touches the circle $(O_a)$ at $A'$
Recall theorem 1 that $A', U, O_a$ are collinear. Thus, the circle $(U)$ touches the circle $(O_a)$ at $A'$ . Therefore theorem 7 follows
Theorem 7 The circle $(O_a)$ and $(U)$ touches each other and the line $l$ at $A'$ .
Again, recall from theorem 5 that $AA_bA'A_c$ is harmonic, we obtain that the tangents at $A$ and at $A'$ (i.e the line $l$ ) are concurrent with the line $A_bA_c$ at $R$ . Since $DO_a$ is the perpendicular bisector of the segment $AA'$ , we obtain that $DO_a$ passes through $R$ . Therefore $RA'^2 = RE.RF = RA_b . RA_c = RD. RO_a$ , we obtain that $D, E, F, O_a$ lie on a circle. By similar argument, it follows that
Theorem 8 $D, E, F, O_a, O_b, O_c$ lie on a circle.
Note that from the equality above, we also obtain that $A_bDO_aA_c$ is cyclic. Since $O_aA_b = O_aA_c$ , we have
Theorem 9 $DA'$ is the bisector of $\measuredangle{A_bDA_c}$ .
All right ! Let me finish this post by introducing some more nice properties relating to this sketch in theorem 10
Theorem 10
(i) $R$ is the excenter of similitude of the two circles $(b)$ and $(c)$ .
(ii) $(b)$ and $(U)$ is orthogonal, $(c)$ and $(U)$ is orthogonal.
(iii) The circumcircle of $ADA_b$ touches $AC$ at $A$ , The circumcircle of $ADA_c$ touches $AB$ at $A$
Hint
Khoa Lu Nguyen.

**Posted:** Sat Jul 09, 2005 2:25 pm

_________________
Life doesn't always follow what I have planned or expected. No matter what way my life leads me, I am still thankful.

jayme · Yang-Mills Theory Offline Joined: 20 Nov 2007 Posts: 681

Dear Mathlinkers,
A new proof of "the Mineur's circle" has been put on my website

http://perso.orange.fr/jl.ayme

see contenu, vol. 3
Sincerely
Jean-Louis

**Posted:** Thu May 08, 2008 11:25 pm