2016 SMO Open Geometry

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vlwk

#1 h Jul 5, 2016, 1:36 AM • 2 Y

Y by Adventure10, Mango247

Let $D$ be a point in the interior of $\triangle{ABC}$ such that $AB=ab$ , $AC=ac$ , $BC=bc$ , $AD=ad$ , $BD=bd$ , $CD=cd$ . Show that $\angle{ABD}+\angle{ACD}=60^{\circ}$ .

Source: 2016 Singapore Mathematical Olympiad (Open) Round 2, Problem 1

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12 posts

vlwk

#2 h Jul 5, 2016, 1:39 AM • 1 Y

Y by Adventure10

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4071 posts

#3 h Jul 5, 2016, 1:51 AM • 2 Y

Y by vlwk, Adventure10

Beautiful problem

vlwk wrote:

Let $E$ be a point outside $\triangle{ABC}$ such that $\triangle{ABD} \sim \triangle{ACE}$ .

We get $CE=BD\cdot \frac{AC}{AB}=cd,AE=AD\cdot \frac{AC}{AB}=\frac{acd}{b}$
We get $\angle{BAC}=\angle{DAE}$ and $\frac{AB}{AC}=\frac{b}{c}=\frac{AD}{AE}$
So $\triangle{ADE} \sim \triangle{ABC}$ give us $DE=cd$
So $\triangle{CDE}$ is equilateral triangle
So $\angle{ABD}+\angle{ACD}=\angle{DCE}=60^{\circ}$

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1097 posts

#4 h Aug 13, 2020, 1:27 AM

Y by

related:
https://artofproblemsolving.com/community/c6h112625

this problem should have been posted in HSO forum instead cos it is an olympiad problem.

This post has been edited 2 times. Last edited by dominicleejun, Aug 13, 2020, 2:39 AM

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3 posts

#5 h Jun 10, 2022, 3:07 AM

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Here's my solution using mainly algebra instead of geometry.

Using the cosine rule on point C, we obtain (ab)^2 = (ac)^2 + (bc)^2 - 2(ac)(bc)cos(C)
Divide by c^2 on both sides to get (ab/c)^2=a^2+b^2-2ab(cos(C)) - which equals c^2 for a given triangle A'B'C' with
this means that ab/c=c --> ab=c^2
this can be reapplied to points A and B to obtain the symmetric algebraic equations
ab=c^2
bc=a^2
ac=b^2

We know that a^2+b^2+c^2 >= ab+ac+bc, equality occurring when a=b=c. We know that it is equal due to the trio of equations above, and a=b=c.
This means that ABC with side lengths ab, bc, ac, are all the same and ABC is an equilateral triangle.

From there we can just use perpendicular bisectors on AB, BC and AC to show that they intersect as well, to show that the intersection of points is Point D, which is both the incenter and circumcenter of triangle ABC. This means that the perpendicular bisector passing through its respective angle bisects it as well, and both angles ABD and ACD are 30 degrees. This means that angles (ABD+ACD)= 60 degrees.

...yes I don't know how to use LaTeX, how could you tell

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86 posts

#6 h Apr 25, 2025, 11:24 AM

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Using the cosine rule at point $C$ , we obtain:
$(ab)^2 = (ac)^2 + (bc)^2 - 2(ac)(bc)\cos C$
Dividing both sides by $c^2$ , we get:
$\left(\frac{ab}{c}\right)^2 = a^2 + b^2 - 2ab\cos C$
Now consider a triangle $A'B'C'$ with side lengths $a, b, c$ and angle $C$ , so that:
$a^2 + b^2 - 2ab\cos C = c^2$
Thus,
$\left(\frac{ab}{c}\right)^2 = c^2 \Rightarrow \frac{ab}{c} = c \Rightarrow ab = c^2$
This process can be repeated symmetrically at the other vertices to obtain:
$ab = c^2 \\ bc = a^2 \\ ac = b^2$
Now consider the inequality:
$a^2 + b^2 + c^2 \geq ab + bc + ac$
Equality occurs when $a = b = c$ , and we know from the system above that equality holds, so:
$a = b = c$
Therefore, triangle $ABC$ , with side lengths $ab, bc, ac$ , must be equilateral.

Now, using perpendicular bisectors on $AB, BC,$ and $AC$ , they intersect at a single point, which we call point $D$ . Since triangle $ABC$ is equilateral, $D$ is both the incenter and the circumcenter.

Hence, each perpendicular bisector also bisects the corresponding angle, so:
$\angle ABD = \angle ACD = 30^\circ$
Thus,
$\angle ABD + \angle ACD = 60^\circ$

This post has been edited 1 time. Last edited by mqoi_KOLA, Apr 25, 2025, 9:39 PM

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