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Source: USA Team Selection Test 2002, Day 2, Problem 5

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hxtung

#1 h Jun 9, 2004, 8:25 AM • 4 Y

Y by Amir Hossein, Adventure10, Mango247, and 1 other user

Consider the family of nonisosceles triangles $ABC$ satisfying the property $AC^2 + BC^2 = 2 AB^2$ . Points $M$ and $D$ lie on side $AB$ such that $AM = BM$ and $\angle ACD = \angle BCD$ . Point $E$ is in the plane such that $D$ is the incenter of triangle $CEM$ . Prove that exactly one of the ratios
$\frac{CE}{EM}, \quad \frac{EM}{MC}, \quad \frac{MC}{CE}$
is constant.

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#2 h Jun 18, 2004, 10:35 AM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

Here's a sketch of the proof:

Try to prove (by using the relation and the median bisector) that $\frac {GA}{GB}=\frac {CA}{CB}$ , where $G$ is the centroid. This means that $G$ is on the Apollonius circle of $ABC$ corresponding to $C$ . It's prety obvious that if we take a point $X$ on the apollonius circle corresponding to $C$ , then $X'$ , the reflection of $X$ in $AB$ will be situated on the cevian which is isogonal to $CX$ . We do this for $X=G$ and find that the reflection of $G$ in $AB$ is on the symmedian from $C$ . It is now clear that $E$ is exactly this reflection of $G$ in $AB$ . This means that $ME=MG=\frac {MC}3\Rightarrow \frac {EM}{MC}=\frac 13$ , and this shows that $\frac {EM}{MC}$ is constant.

I didn't spend time thinking about how to show that the others aren't constant, but that can't be too hard (if another of those ratios were constant it would mean that all triangles $EMC$ are similar). Come to think of it, it would mean that the angle $\angle CMD=\angle CMA$ is constant, and this is obviously false.

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#3 h Nov 18, 2005, 12:50 PM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

Further properties of such triangles are discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=28877 . I am planning to find and post further proofs of these when I will find time (however, probably not too soon).

Darij

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probability1.01

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#4 h Dec 29, 2006, 11:02 PM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

Hrmmm ok well here was a pretty straightforward proof:

Reflect C across AB to C'. Let P be the point such that PA and PB are tangent to the circumcircle of ABC. From the definitions and the fact that CP is a symmedian, it follows that the intersection of CP and C'M is E. In order to prove that $\frac{EM}{MC}$ is constant, it suffices to show that $\frac{C'E}{EM}= \frac{CC'}{MP}$ is constant.

From the condition, after subtracting the law of cosines equation, we get

$2 \cdot AC \cdot BC \cos C = AB^{2}$
$2 = \frac{c^{2}}{ab \cos C}$ .

Now the ratio $\frac{CC'}{MP}$ is equal to

$\frac{[CAC'B]}{[ABP]}= \frac{ab \sin C}{\frac{c^{2}\tan C}{4}}$
$= \frac{4ab \cos C}{c^{2}}= 2$

so we win. The other part of the problem, as grobber noted, is necessarily routine.

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#5 h Jun 24, 2013, 11:16 PM • 7 Y

Y by applepi2000, ahaanomegas, dantx5, Binomial-theorem, Amir Hossein, Adventure10, and 1 other user

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We use barycentric coordinates. Let $A = (1,0,0), B=(0,1,0), C = (0, 0, 1)$ , and let $BC = a, CA = b, AB = c$ .

We are given that $b^2+a^2=2c^2$ . Since $M$ is the midpoint of $AB$ , we have $M = (1:1:0)$ . $D$ is the $C$ -trace of the incenter $I$ of $\triangle ABC$ , so $D = (a:b:0)$ .

Since $\angle DCM = \angle ECD$ , the symmedian point $K = (a^2:b^2:c^2)$ , the isogonal conjugate of the centroid $G$ , lies on $EC$ . Hence line $EC$ has equation $b^2x-a^2y = 0$ .

Segment $EM$ lies on the reflection of line $CM$ over $AB$ . The projection $P$ of $C$ onto $AB$ is the $C$ -trace of the orthocenter $H$ of $\triangle ABC$ . Thus $P = (S_B:S_A:0)$ , where $S_A = \frac{b^2+c^2-a^2}{2}$ . The reflection of $C = (0:0:S_B+S_A)$ over $AB$ has coordinates $C' = (2S_B:2S_A:-S_B-S_A)$ , so the equation of $EM$ is
$(S_B+S_A)x - (S_B+S_A)y + 2(S_B-S_A)z = 0$
But $S_B+S_A = c^2$ and $S_B-S_A = a^2-b^2$ , so the equation is equivalently $c^2x-c^2y+2(a^2-b^2)z = 0$ . Intersecting lines $EC$ and $EM$ yields $E = (2a^2:2b^2:-c^2)$ .

Applying the distance formula to points $C, E, M$ yields:
$\begin{align*} (2a^2+2b^2-c^2)^2\cdot EC^2 &= -a^2(2b^2)(-2a^2-2b^2)-b^2(-2a^2-2b^2)(2a^2)-c^2(2a^2)(2b^2) \\&= 8a^4b^2+8a^2b^4-4a^2b^2c^2\\&=6a^2b^2(a^2+b^2) = 12a^2b^2c^2\\ (2a^2+2b^2-c^2)^2 \cdot CM^2 &= (2a^2+2b^2-c^2)^2\left(\frac{1}{2}a^2+\frac{1}{2}b^2-\frac{1}{4}c^2\right) \\&= (3c^2)^2\frac{3}{4}c^2 = \frac{27}{4}c^6\\ (2a^2+2b^2-c^2)^2 \cdot ME^2 &= \frac{1}{4}(-2a^2(-2a^2+2b^2+c^2)(-c^2)-2b^2(-c^2)(2a^2-2b^2+c^2)\\&-c^2(2a^2-2b^2+c^2)(-2a^2+2b^2+c^2)) \\&= \frac{1}{4}(2a^2c^4+2b^2c^4-c^6) = \frac{1}{4}c^4(2a^2+2b^2-c^2) = \frac{3}{4}c^6 \end{align*}$
Hence
$\begin{align*} \left(\frac{EM}{MC}\right)^2 = \frac{\frac{3}{4}c^6}{\frac{27}{4}c^6} = \frac{1}{9} \end{align*}$
so $\frac{EM}{MC} = \frac{1}{3}$ is constant. In addition,
$\left(\frac{MC}{CE}\right)^2 = \frac{\frac{27}{4}c^6}{12a^2b^2c^2} \Longleftrightarrow \frac{MC}{CE} = \frac{3c^2}{4ab}$ $\left(\frac{CE}{EM}\right)^2 = \frac{12a^2b^2c^2}{\frac{3}{4}c^6} \Longleftrightarrow \frac{CE}{EM} = \frac{4ab}{c^2}$
and so neither $\frac{MC}{CE}$ nor $\frac{CE}{EM}$ are constant; thus we are done.

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#6 h Jun 25, 2013, 3:39 PM • 6 Y

Y by Binomial-theorem, Amir Hossein, NewAlbionAcademy, pi37, Adventure10, and 1 other user

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Let $G$ be the centroid of $\triangle ABC$ (this was really enticed by the midpoint $M$ ) and let the midpoints of $BC, AC$ be $K, N$ (so $\triangle KNM$ is the cevian triangle of $G$ ). Also, let $w$ be the circumcircle. We have by the median relations that:

$BN^2 = \dfrac{BC^2 + AB^2}{2} - \dfrac{AC^2}{4} = \dfrac{3AB^2}{2} - \dfrac{3AC^2}{4} = \dfrac{3BC^2}{4}$

where the last inequality followed from $BC^2 + AC^2 = 2AB^2 \implies \dfrac{3BC^2}{4} + \dfrac{3AC^2}{4} = \dfrac{3AB^2}{2}$

Thus, $BN^2 = \dfrac{3AB^2}{4}$ and similarly $AK^2 = \dfrac{3AC^2}{4}$ . So,

$\left(\dfrac{AK}{BN}\right)^2 = \left(\dfrac{GA}{GB}\right)^2 = \left(\dfrac{AC}{BC}\right)^2 \implies \dfrac{GA}{GB} = \dfrac{AC}{BC}$

Now, let the tangents at $B, C$ meet at $T$ . It is well-known that $CT$ is a symmedian. Let $CT \cap w = E'$ . We know that $CBE'A$ is a harmonic quadrilateral, so

$\dfrac{E'B}{E'A} = \dfrac{CB}{CA} = \dfrac{GB}{GA}$

So, $E'$ is the reflection of $G$ over $AB$ (by similar triangles $\triangle GMB \sim \triangle CMB$ ). Furthermore, since $CD$ bisects $\angle MCE'$ and $MA$ bisects $\angle E'MC \implies D$ is the incentre of $\triangle CME'$ . Therefore, $E' = E$ .

We get $\dfrac{EM}{MC} = \dfrac{MG}{MC}= \dfrac{1}{3}$ , so it is constant.

If any of the others are constant, then it would follow that all triangles $\triangle CME$ are similar, meaning $\angle EMA = \angle EBC$ (property of harmonic quad) is constant.
Suppose we fix $BC$ . Then by this, the locus of all points $E$ lies on $w$ . This obviously wrong by definition of $E$ , because this would imply that the circumcircle is fixed. Alternatively, I am sure it is not hard to find a sine/bash method (note that $CM = \dfrac{\sqrt{3}AB}{2}$ ).

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#7 h Aug 24, 2013, 9:24 PM • 12 Y

Y by dantx5, NewAlbionAcademy, djmathman, r31415, El_Ectric, pi37, superpi83, aZpElr68Cb51U51qy9OM, Binomial-theorem, Adventure10, Mango247, and 1 other user

We do not use barycentric coordinates.

Consider the point $X$ on the C-symmedian such that it lies on the circle of $ABC$ . We know that $ACBX$ is a harmonic quadrilateral, thus $XC$ is the symmedian of triangle $AXB$ (and $ACB$ ). Furthermore, $\frac{AX}{BX}=\frac{AC}{BC}=\frac{AD}{DB}$ , thus $\angle CXD=\angle DXM$ (as angle bsiector with symmedian) and we know $\angle XCD=\angle MCD$ . Therefore $E=X$ and notice by the ratio-lemma that

$\frac{EM}{MC}=\frac{\sin\angle EAB}{\sin\angle BAC}\cdot\frac{AX}{AC}=\frac{\sin\angle ACM}{BAC}\cdot \frac{\sin\angle MCB}{\sin\angle ABC}=\frac{AM^2}{CM^2}$

We now apply Stewart's theorem, and $AM(BC^2+AC^2)=2AM(CM^2+AM^2)$ , or $AM(8AM^2)=2AM(CM^2+AM^2)\rightarrow CM^2=3AM^2$ and we have a fixed ratio for $\frac{EM}{CM}$ .

If the other ratios were fixed, then all such $CEM$ would be similar and this is easy to see is not true.

This post has been edited 1 time. Last edited by sjaelee, Aug 27, 2013, 7:18 PM

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NewAlbionAcademy

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#8 h Aug 27, 2013, 7:06 PM • 12 Y

Y by dantx5, sjaelee, djmathman, r31415, El_Ectric, pi37, superpi83, aZpElr68Cb51U51qy9OM, Binomial-theorem, Adventure10, Mango247, and 1 other user

We do not use barycentric coordinates.

Since $CE$ is a symmedian, we intersect it with the tangents to the circumcircle at $A$ and $B$ and call this point $X$ . Let $CE \cap AB=F$ . Since $\angle CMF=\angle EMF$ and $\angle XMA=\dfrac{\pi}{2}$ , we get that (CFEX) is harmonic, implying that $E$ is on the circumcircle.

Then we have $\dfrac{EM}{MC}=\dfrac{EF}{FC}=\dfrac{FB}{AF}\cdot\dfrac{FE^2}{FB^2}=\dfrac{a^2}{b^2}\cdot\dfrac{\sin^2 \angle MCB}{\sin^2 \angle CAB}$ . Applying Stewart's Theorem, we get that $CM=\dfrac{\sqrt{3}c}{2}$ , so we get $\sin \angle MCB=\dfrac{\sin\angle ABC}{\sqrt{3}}$ . Thus we have $\dfrac{EM}{MC}=\dfrac{1}{3}$ .

On the other hand, a quick computation using the fact that $\triangle BCM \sim \triangle ECA$ gives that $\dfrac{MC}{CE}=(\dfrac{3}{8})(\dfrac{a^2+b^2}{a^2b^2})$ , which is clearly not constant. Multiply this by $\dfrac{EM}{MC}=\dfrac{1}{3}$ to see that $\dfrac{CE}{EM}$ is not constant as well. Thus, we are done.

QED

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