Triangle property [A = 30° yields AT = 2 * BC]

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Triangle property [A = 30° yields AT = 2 * BC]

G H J

Reveal topic tags

geometric transformation

L

G H BBookmark kLocked kLocked NReply

Source: INMO 1995 Problem 1

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1592 posts

Rushil

#1 h Oct 6, 2005, 12:09 PM • 2 Y

Y by Adventure10, Mango247

In an acute angled triangle $ABC$ , $\angle A = 30^{\circ}$ , $H$ is the orthocenter, and $M$ is the midpoint of $BC$ . On the line $HM$ , take a point $T$ such that $HM = MT$ . Show that $AT = 2 BC$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

6555 posts

#2 h Oct 6, 2005, 1:38 PM • 6 Y

Y by Dytto, Ramanujanjr_, SahilAgrawal753, Adventure10, Mango247, and 1 other user

Rushil wrote:

In an acute angled triangle $ABC$ , $\angle A = 30 ^{\circ}$ , $H$ is the orthocenter, and $M$ is the midpoint of $BC$ . On the line $HM$ , take a point $T$ such that $HM = MT$ . Show that $AT = 2 BC$ .

Since HM = MT, the point T is the reflection of the point H in the point M. On the other hand, since M is the midpoint of the segment BC, the point C is the reflection of the point B in the point M. Since reflection in a point maps any line to a parallel line, we thus have CT || BH. But since H is the orthocenter of triangle ABC, we have $BH\perp CA$ , and thus $CT\perp CA$ . Thus, < ACT = 90º. Similarly, < ABT = 90º. Hence, the points C and B lie on the circle with diameter AT. In other words, the circle with diameter AT passes through the points A, B, C; thus, this circle is the circumcircle of triangle ABC. Hence, the segment AT is a diameter of the circumcircle of triangle ABC, so that we have AT = 2R, where R is the radius of the circumcircle of triangle ABC. But on the other hand, by the extended law of sines, $BC=a=2R\sin A=2R\sin 30^{\circ}=2R\cdot\frac12=R$ , so that $AT=2R=2\cdot BC$ . That's all.

darij

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

madhusudan kale

23 posts

madhusudan kale

#3 h Feb 3, 2011, 3:35 PM • 2 Y

Y by Adventure10, Mango247

good solution!

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

276 posts

#4 h Mar 3, 2011, 1:35 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

intresting problem thanks

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

316 posts

#5 h Mar 12, 2011, 4:50 PM • 4 Y

Y by darthsid, Ramanujanjr_, Adventure10, Mango247

Just for the end of the solution : instead of use of the law of sines, we have $\angle BOC=2\angle BAC$ (angle inscribed !) and then $\angle BOC=60^{\circ}$ . Then triangle $BOC$ is equilateral, so $BC=BO=\frac{AT}{2}$
( $O$ is the center of the cercle circumscribed to $ABC$ .)

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

904 posts

#6 h Apr 28, 2011, 5:36 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Rushil wrote:

In an acute angled triangle $ABC$ , $\angle A = 30^{\circ}$ , $H$ is the orthocenter, and $M$ is the midpoint of $BC$ . On the line $HM$ , take a point $T$ such that $HM = MT$ . Show that $AT = 2 BC$ .

Well, the homothety $H(H,2)$ take the nine point circle to the circumcircle, since the midpoints of $AH,BH,CH$ lie on the nine point circle. Also, it is well known that if $P$ is the midpoint of $AH$ , then $PM$ is a diameter of the nine point circle. Therefore, $M$ goes to the point diametrically opposite to $A$ . Thus, $AT = 2R$ and since $BC = R$ by easy law of sines, we're done.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

131 posts

Shravu

#7 h Jun 8, 2011, 7:04 AM • 1 Y

Y by Adventure10

My solution:
ATC is right triangle;
TC=BH...(TBCH is a parallelogram)
BH=2OE (O is circumcenter;E is midpoint of AC)
OAE is a right traingle
OE^2=R^2-1/4*b^2
AT^2=AC^2+CT^2
=b^2+4(OE^2)
=b^2+4(R^2-b^2/4)
=4R^2
AT=2R

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

102 posts

#8 h Jan 2, 2014, 3:05 PM • 4 Y

Y by darthsid, Ramanujanjr_, Adventure10, Mango247

Let the circumcentre of $\Delta ABC$ be at the origin.If $z_1,z_2,z_3$ be the complex numbers representing $A,B,C$ respectively.Then, $H=z_1+z_2+z_3$ and $M=\frac{z_2+z_3}{2}$ .If $x$ represents the point $T$ ,then $\frac{x+z_1+z_2+z_3}{2}=\frac{z_2+z_3}{2}$
$\Rightarrow x=-z_1$ .Therefore $AT=2R$
Since $BC=2RSinA=R$ .
$AT=2BC$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

607 posts

#9 h Jan 9, 2014, 9:13 AM • 2 Y

Y by Adventure10, Mango247

$BTCH$ is a parallelogram so $\angle TBM=\angle MCH$ ,but $\angle MCH =\angle ABH$ ,so $BT$ is a tangent of $(ABD)$ { $D=AH \cap BC$ },so $AB$ is perpendicular to $BT$ ,so $AT$ is the diameter of $ABTC$ , ( $ABTC$ cyclic because $\angle BTC=\angle BHC$ ),also $BOC$ is regular( $O$ is the circumcenter),so $BO=BC$ ,hence done.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

74 posts

#10 h Dec 12, 2016, 3:25 PM • 1 Y

Y by Adventure10

BM=MC and HM=MT.
so,BHCT is a parallelogram.
let X be a point on AC passes through the point H.
angle BXC=90 degree.
so, angle TCA is 90 degree.
And Y is a point on AB also passes through point H.
angle TYB=90 degree.
so,angle TBA=90 degree
Now we know that ABTC is cyclic.
AT is the diameter of that circle.
Let O be the circum center of that circle.so, angle BOC=2*angle BAC=60 degree.
OB=OC. and angle BOC=60 degree.
Then angle OBC=angle OCB=60 degree.
Now,OBC is a equilateral triangle.
AT=2*AO
AO=BO=CO.,
BC=OB=OC.
so,AT=2*BC.(proved)

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

LoveMaths26102003

84 posts

LoveMaths26102003

#11 h Dec 22, 2017, 10:01 AM • 1 Y

Y by Adventure10

Take a point F to be the midpoint of AH.Then AT = 2FM. So the problem reduces to show FM = BC.Now consider the Circumcentre O of the Triangle.
Then, 2OM = AH => OM = FH. Also angle chasing shows angle AMH = 90. Therefore, FM = FH = OM = BC (BM = MC = OM/2).So we are done.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

101 posts

#12 h Mar 8, 2020, 6:04 AM

Y by

It follows directly from EGMO lemma

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

66 posts

#13 h Mar 15, 2024, 2:42 PM

Y by

Given, $HM=MT,\quad BM=MC$ .
Thus, if we join $HC,CT,TB,BH,$ the diagonals of quadrilateral $CTBH$ bisect each other $\implies HCTB$ is a parallelogram.
In $\Delta ABE, \angle AEB= 90^{\circ}$ (as $BE$ is altitude). By angle sum property, $\angle ABE= 90^{\circ}-A$ .
Similarly in $\Delta CEB$ , $\angle CBE=90^{\circ}-C$ . In $\Delta FCB, \angle FCB=90^{\circ}-B$ .

$BH \parallel CT$ (parallelogram), $HC\parallel BT$
$\implies \angle HCB=\angle CBT=90^{\circ}-B$ (alternate angles)
$\implies \angle HBC=\angle TCB=90^{\circ}-C$ (alternate angles)
Therefore, $\angle ACT=270^{\circ}-(A+B+C)=90^{\circ}$
$\angle ABT= 90^{\circ} \implies ABTC$ is cyclic.
As $\angle ACT=90^{\circ}$ , diameter= $AT=2R$ . $T$ lies on the circumcircle of $\Delta ABC$ .
$\angle A=30^{\circ}$
By sine rule, $\frac{a}{\sin A}=2R \implies \frac{BC}{\sin 30^{\circ}}=AT$

$\implies BC=AT/2 \implies AT=2BC$ . Hence, proved!

Attachments:

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

993 posts

#14 h Mar 24, 2024, 12:33 PM • 1 Y

Y by Math_DM

Observe that from the Orthocentre Reflection Lemma in EGMO Ch 1, $AT$ is nothing but the circumdiameter of $\Delta ABC$ . Now letting $R$ be the circumradius, and $a$ to be the length $BC$ , we get $2BC = 2a = 2 \cdot (2R \sin A) = 2 \cdot \left(2R \sin \frac{\pi}{6} \right) = 2R = AT,$ so we are done. $\square$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a