A, D, N are collinear iff BAC = 45º

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A, D, N are collinear iff BAC = 45º

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Source: Problem 1, Brazilian MO 2015

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82 posts

#1 h Oct 20, 2015, 9:00 PM • 3 Y

Y by Davi-8191, jhu08, Adventure10

Let $\triangle ABC$ be an acute-scalene triangle, and let $N$ be the center of the circle wich pass trough the feet of altitudes. Let $D$ be the intersection of tangents to the circumcircle of $\triangle ABC$ at $B$ and $C$ . Prove that $A$ , $D$ and $N$ are collinear iff $\measuredangle BAC = 45º$ .

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617 posts

#2 h Oct 20, 2015, 10:21 PM • 3 Y

Y by jhu08, Adventure10, Mango247

My solution:
The circle which passes through the feet of altitudes is none point circle. Now let $O$ be circumcerntre of $ABC$ and $H$ it's orthocentre. We have $NH=NO$ and $A,D,N$ are collinear. We have that $OD=AH$ which implies $AO=HD$ and now from triangle $HAD$ implies $\angle BAC=45$ . Done.

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241 posts

#3 h Oct 21, 2015, 3:18 AM • 3 Y

Y by lucaslcmlmo, jhu08, Adventure10

My solution:
Let $H$ and $O$ be circumcenter and orthocenter of $\triangle ABC$ $\Longrightarrow$ $NH=NO$
Let $M$ be a midpoint of $BC$ and $\angle BOC=90^\circ$ $\Longrightarrow$ $OM=MD$
We get $OD=AH...(1)$ since $2{OM}=AH$ and $AH\parallel OD...(2)$
By $...(1)$ and $...(2)$ we get $A,N,D$ are collinear...

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pi37

#4 h Oct 21, 2015, 4:38 AM • 4 Y

Y by jaobundao, jhu08, Adventure10, Mango247

Let $O_A$ be the center of $(BOC)$ . Note that $AO_A$ is isogonal to $AN$ (since inverting about $A$ maps $O$ to the reflection of $A$ over $B'C'$ , easily yielding the isogonality). So $AN$ is a symmedian iff $AO_A$ is a median, iff the center of $(BOC)$ is the midpoint of $BC$ , iff $\angle BOC=90$ , iff $\angle BAC=45$ .

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#5 h Oct 21, 2015, 9:34 PM • 5 Y

Y by rkm0959, jaobundao, jhu08, Adventure10, Mango247

Since $D$ is the unique intersection of $AN$ with the perpendicular bisector of $\overline{BC}$ , it is well-known that $D$ is the reflection of the circumcenter $O$ in $BC.$ Therefore, $\overline{OD}$ is a diameter of the circle passing through $O, D, B, C$ , and $BC$ is the perpendicular bisector of $\overline{OD}.$ It follows that $OBDC$ is a square, so $90^{\circ} = \angle BOC = 2\angle A.$ $\square$

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977 posts

#6 h Oct 22, 2015, 1:04 AM • 3 Y

Y by jhu08, Adventure10, Mango247

This is utterly trivial with complex bashing.

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105 posts

#7 h Oct 26, 2015, 1:11 PM • 3 Y

Y by jhu08, Adventure10, Mango247

I will prove the only if part since the if part is fairly easy.In the proof $H,G,O$ have their usual meaning and $M$ is the midpoint of $BC$ .
It is known by the nine-point circle that $N$ is the midpoint of $OH$ .It is also known by the Euler line that $G\in OH$ and since $AM$ is a median, $G'\equiv OH\bigcap{}AM\equiv G$ .Now since $A,D,N$ are collinear,AN is the symmedian of triangle $ABC$ and hence $\angle BAD=\angle CAM\Leftrightarrow\angle BAD-\angle BAH=\angle CAM-\angle CAO\Leftrightarrow \angle DAH=\angle MAO$ (since $O$ and $H$ are isogonal).So $AG$ is the symmedian of triangle $AHO$ and thus we obtain $\frac{AH^2}{AO^2}=\frac{GH}{GO}=2\Rightarrow \frac{AH}{AO}=\sqrt{2}\Rightarrow AH=AO\sqrt{2}$ (well-known property of Euler line) which implies (easy,use $AH=2OM$ ) that $\angle OBM=\angle OCM=45º$ and hence $\angle BOC=90\Rightarrow \measuredangle BAC =\frac{\angle BOC}{2}=45º$ and we are done.

This post has been edited 2 times. Last edited by john111111, Oct 27, 2015, 7:52 AM

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1721 posts

#8 h Feb 24, 2016, 8:56 AM • 2 Y

Y by jhu08, Adventure10

Below is a horrible solution. ~~I should've done synthetic~~
A well-known characterization of the $A$ -symmedian is $\frac{d(X,AB)}{d(X,AC)} = \frac{AB}{AC}$ . Also, $AD$ is the $A$ -symmedian.
Note that $N$ is the midpoint of $OH$ . $N$ lies on the $A$ -symmedian $AD$ if and only if $\frac{d(N,AB)}{d(N,AC)}=\frac{d(O,AB)+d(H,AB)}{d(O,AC)+d(H,AC)} = \frac{AB}{AC}$
Using simple trig formulas, and the well-known $AH=2OM$ we find $\frac{d(O,AB)+d(H,AB)}{d(O,AC)+d(H,AC)} = \frac{a \sin B - R \cos C}{a \sin C - R \cos B} = \frac{ab-2R^2 \cos C}{ac-2R^2 \cos B} = \frac{ab-2 \cdot \frac{a^2b^2c^2}{2(a^2b^2+b^2c^2+c^2a^2)-(a^4+b^4+c^4)} \cdot \frac{a^2+b^2-c^2}{2ab}}{ac-2 \cdot \frac{a^2b^2c^2}{2(a^2b^2+b^2c^2+c^2a^2)-(a^4+b^4+c^4)} \cdot \frac{a^2+c^2-b^2}{2ac}}$ Simplifying the R.H.S, we have $\frac{2a^2b^3+a^2bc^2+b^3c^2-a^4b-b^5}{2a^2c^3+a^2b^2c+b^2c^3-a^4c-c^5}$ Now we see when this is equal to $\frac{c}{b}$ . We cross multiply, delete same terms, then find $(b-c)(b+c)(2a^2b^2+2a^2c^2-a^4-b^4-c^4)=0$ Solving for $a^2$ , we find $a^2=b^2+c^2 \pm \sqrt{2} bc$ , and $\triangle ABC$ is acute, giving $\angle A = 45$ as desired. $\blacksquare$

This post has been edited 2 times. Last edited by rkm0959, Feb 24, 2016, 8:57 AM

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#9 h Feb 24, 2016, 9:10 AM • 3 Y

Y by jhu08, Adventure10, Mango247

Or you could use barycentric coordinates:
The barycentric co ordinates of nine point centre are $(a^2(b^+c^2)-(b^2-c^2)^2:\text{etc.}$ . So $AN$ divides $BC$ in the ratio $\frac{b^2(c^2+a^2)-(c^2-a^2)^2}{c^2(a^2+b^2)-(a^2-b^2)^2}$ We want this ratio to be $\frac{b^2}{c^2}$ , so we have $\frac{b^2(c^2+a^2)-(c^2-a^2)^2}{c^2(a^2+b^2)-(a^2-b^2)^2}=\frac{b^2}{c^2}$ .
Simplifying (

) , we get the same equation as above.

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1979 posts

#11 h Sep 9, 2016, 6:45 PM • 3 Y

Y by jhu08, Adventure10, Mango247

We simply note that the point symmetric to the circumcenter $O$ in line $BC$ lies on the line $AN$ . This may be seen by using complex numbers or any other methods. Now, we get $\angle BOC=180-\angle BOC$ and so $2\angle A=\angle BOC=90^{\circ}$ and thus, $\angle A=45^{\circ}$ . It is clear that all implications are reversible.

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308 posts

#12 h Dec 25, 2018, 7:59 PM • 2 Y

Y by jhu08, Adventure10

Recall that $N$ is the center of the nine-point circle and $N$ is the midpoint of $OH$ , where $O$ is the circumcenter of $\triangle ABC$ and $H$ the the orthocenter of $\triangle ABC$ . Let $R$ be the length of the circumradius of $\triangle ABC$ and let $M$ be the midpoint of $BC$ .

Claim 1: $AH\parallel OD$
Proof: Let $X$ be the perpendicular from $D$ to line $BC$ . By equal tangents, we have $BD=DC$ . By HL similarity, we have $\triangle BXD\sim\triangle CXD$ so $BX=XC$ . Hence, $D$ lies on the perpendicular bisector of $BC$ . Similarly, we can show that $O$ lies on the perpendicular bisector of $BC$ . Thus, $OD\perp BC$ . By definition, $AH\perp BC$ so we have $AH\parallel OD$ . $\Box$

Claim 2: $AH=2R\cos A$
Proof: Let $E, F$ be the feet of the altitudes from $B,C$ to $AC, AB$ respectively. Note that $AEHF$ and $BFEC$ is cyclic. Because $\angle AEF=\angle B$ , we have $\triangle AEF\sim \triangle ABC$ . Thus, the ratio of the lengths of the corresponding sides is then congruent to the ratio of the lengths of the corresponding diameters, or $\frac{AE}{AB}=\frac{AH}{2R}\implies AH=2R\cos A.\Box$

Assume that $\angle BAC=45^{\circ}$ . Now, $\angle BOC=2\angle BAC=90^{\circ}.$ Since $BOCD$ is cyclic, this makes $\angle BDC=90^{\circ}$ , $BOCD$ is a square with side length $R$ . Thus, the diagonal of this square, $OD$ , has length $R\sqrt{2}$ . By Claim 2, $AH=R\sqrt{2}.$ Because $AH=DO$ and $AH\parallel DO$ (Claim 1), we have $AHDO$ is a parallelogram so its diagonals bisect each other. SInce $NH=NO$ , the diagonal $AD$ must contain $N$ so $A, D, N$ are collinear.

Now, assume that $A, D, N$ are collinear.

Claim 3: If $A, D, N$ are collinear, $AHDO$ is a parallelogram.
Proof: By Claim 1, $AH\parallel DO$ . Since $A, D, N$ are collinear, we have $\angle AHN=\angle NOD$ and $\angle HNA=\angle OND$ . Since $NH=NO$ , by ASA congruence, we have $\triangle AHN\cong \triangle DON$ . Thus, $AH=DO$ and $AH\parallel DO$ gives that $AHDO$ is a parallelogram . $\Box$

Hence, by Claim 3, $OD=HA=2R\cos A$ . Because $\triangle DBO\cong \triangle DCO$ , we have $\angle DOC=\frac{1}{2}\angle BOC=\angle A.$ Thus, $OM=R\cos A$ . Notice that $\angle DCB\angle A$ so $DM=CD\sin A.$ Note that $OD=OM+MD=R\cos A+CD\sin A=2R\cos A\implies CD\sin A=R\cos A.$ By the Pythagorean Theorem on $\triangle OCD$ , we have $CD=\sqrt{4R^2\cos^2A-R^2}=R\sqrt{(4\cos^2A-1)}$ and plugging this in the previous equation yields $\sqrt{(4\cos^2A-1)}\sin A=\cos A.$ Squaring and expanding gets $4\cos ^2A\sin^2 A=\sin^2A+\cos^2A=1$ . This simplifies into $(\sin 2A)^2=1$ so $\angle A=45^{\circ}$ . $\blacksquare$

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Ilove_mathematics

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Ilove_mathematics

#14 h Dec 26, 2018, 12:41 AM • 4 Y

Y by Mathcat1234, jhu08, Adventure10, Mango247

Let, $H$ be the orthocenter of $\triangle ABC$ respectively. Note that $DB = DC \implies OD$ is the perp. bissector of $BC \implies OD \perp BC$ and
$AH \perp BC \implies AH||OD$ . As $N$ is the nine-point circle center of $\triangle ABC$ , then $N$ is the medium point of $OH$ . Hence:
$A, D, N$ are collinear $\iff \angle ANH = \angle DNO$ , as $HN = NO$ and $\angle NOD = \angle NHA \implies A, D, N$ are collinear $\iff \triangle DNO \equiv \triangle ANH \iff OD = AH$ . Now look that $\angle OCD = 90º$ and $\angle DOC = \frac{\angle BOC}{2} = \angle A$ . So, by law of sines:
$\rightarrow \triangle OCD \implies cos(A) = \frac{OC}{OD} = \frac{R}{OD} \implies OD = \frac{R}{cos(A)}.$
$\rightarrow \triangle AHB \implies \angle ABH = 90º-A$ and $\angle AHB = 180º-C,$ thus: $\frac{sin(C)}{AB} = \frac{cos(A)}{AH}.$
$\rightarrow \triangle ABC \implies \frac{sin(C)}{AB} = \frac{1}{2R} \implies \frac{1}{2R} = \frac{cos(A)}{AH} \implies AH = 2.R.cos(A).$

Hence, $A, D, N$ are collinear $\iff AH = OD \iff \frac{R}{cos(A)} = 2.R.cos(A) \iff \frac {1}{2} = cos^2(A) \iff cos(A) = \frac{\sqrt{2}}{2} \iff \angle A = 45º.$ Because $A \in (0, \pi). \blacksquare$

This post has been edited 1 time. Last edited by Ilove_mathematics, Dec 26, 2018, 1:09 PM

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trying_to_solve_br

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trying_to_solve_br

#17 h Sep 16, 2021, 8:49 PM • 1 Y

Y by jhu08

Let $H,O$ be the orthocenter, circumcenter of $ABC$ and $M$ the midpoint of $BC$ .
Main fact: $AH=2.OM$
Note $OD || AH$ and $AD$ bissects $OH$ . Thus $AHDO$ is a parallelogram. Thus $OD=AH$ and $OD=OM+DM=2.OM \implies OM=DM$ and $OCD$ is rectangle because of the tangency. Thus $\angle A=45$ .

Now, as $\angle A=45$ , $OM=MD$ and then $OD=AH$ and as $OD || AH$ we have $AHDO$ parallelogram and thus $AD$ bissects $OH$ .

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#18 h Dec 22, 2021, 12:59 AM

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Complex bash :
Let the $(ABC)$ being the unit circle, with $A=a, B=b,C=c$ and let the real axis as parallel to AB. So, we have that $a-b$ = $\overline{a} - \overline{b}$ , because $a - b$ is a real number.We have $N$ that is the center of the nine-point circle, so It's given by $\frac{a+b+c}{2}$ and D its the meeting point of the tangents to B and C, so its $\frac{2bc}{b+c}$ . By the collinearity formula, $A, D,N$ are collinear if and only if $\frac{a - \frac{a+b+c}{2}}{a - \frac{2bc}{b+c}}$ = $\frac{ab + ac -b^2 -c^2 - 2bc}{2ab + 2ac - 4bc}$ and It's conjugate is $\frac{abc^2 + ab^ 2 c - a^2c^2 -a^2b^2 -2a^2bc}{2abc^2 + 2ab^2 - 4a^2bc}$ that are not equal. Using the fact that $\angle BAC = 45$ we get the first thing as $\frac{-1+1-b^2 -c^2 -2bc}{-2+2 - 4bc}$ = $\frac{b^2 + 2bc + c^2}{4bc}$ that has conjugate $\frac{\frac{1}{b^2} + \frac{2}{bc} + \frac{1}{c^2}}{\frac{4}{bc}}$ = $\frac{c^2 + 2bc + b^2}{4bc}$ . And now its equal so its an if and only if argument that, if $\angle BAC =45$ , A , D , N are collinear.

This post has been edited 1 time. Last edited by Humberto_Filho, Dec 22, 2021, 12:59 AM
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