Japan Math Olympiad | A Nice Geometry Problem | 2 Methods
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- Опубліковано 12 чер 2024
- Japan Math Olympiad | A Nice Geometry Problem | 2 Methods
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My method was similar to first one, but using triangles ABD and ABC instead.
Once I've determined ∠ADB = 60° and ∠BAD = 80°, then I used law of sines:
In △ABD
AB/sin 60 = x/sin 40 → AB = x sin 60 / sin 40 = x√3/(2 sin 40)
In △ABC
AB/sin 50 = 6/sin 80 → AB = 6 sin 50 / sin 80 = 6 cos 40/(2 sin 40 cos 40) = 3/sin 40
Combining results, we get:
x√3/(2 sin 40) = 3/sin 40
x = 2·3/√3
x = 2√3
x=3.464 Easy problem
For the triangle below with the degrees 10, 50, 120 and the side 6, let the unknow side between 10 and 120 degreen=n
then using the law of sines,
6/sine 120 = n/sine 50
6* sine 50/sine 120 (degrees) = n
5.3073115
For the triangle on top with the degrees 40, 60, 80 and using the law of sines again
x/sine 40 = 5.307315/sine 80
x = 5.307315 * sine 40 degrees/ sine 80 degrees
x= 3.464 Answer
Solution #1 in the video was unnecessarily long.
At 5:00 you had xsin80/sin40 = 6sin50/sin120. Cross multiply to give x = 6sin40sin50/sin120sin80. Since sin50 = cos 40, sin120=sin60=rt3/2, this gives x = 6sin80/rt3sin80 = 2rt3. Quite a bit quicker.
AB =BC
AD Perpendicular to BC
ANGLE 50 deg
AC =X
DC =3
Drop a perpendicular bisector AP on BC, intersecting BD at O. Drop another perpendicular AE on BD.
∠ABO= ∠BAO =40°, hence ∆AOB is Isosceles with AO= BO.
∠OBP =∠OAE =10°, hence ∆OPB is Congruent to ∆OEA with AE=BP=3
In the (30,60,90)∆AED,
X*Sin60= AE = 3
X =2√3
*Ιt's a difficult exercise..... after I met Scylla and Charybdis like a contemporary Odyssey* 😊
I came to the following solution :
Let AP⊥BC and AQ ⊥BD
ABPQ is quadrilateral inscribed in a circle because side ΑΒ appears at right angles.
So
From A draw a segment AE perpendicular to BD (E on BD) so triangle ADE is 30-60-90, also notice that AE is half of BC because angle BAE is 50°
Drop a perpendicular from A to BC and label the intersection as point P. Drop a perpendicular from A to BD and label the intersection as point R. From sum of interior angles of a triangle equals 180°,
Very nice demonstration on second method. Congrat.
Q is the circumcenter of triangle BDC and could have been defined that way from the start! Thank you for this nice exercise !
drop a perpendicular from A to BD, AE, we will get a triangle BEA , which is equal to the triangle BPA. AE = BP. and so on.
Try this method-much easier due to less of additional constructions.
Thank you for drilling me into being able to solve problems like this in 3 minutes. I used to take up to 30 minutes just a month ago.
Angle ABC = Angle ACB = 50 => triangle ABC is isosc.
AB=3/cos 50=3/sin 40
Angle ABD = 60
Law of Sine gives x / sin 40 = AB / sin 60
x = sin 40 * (3 /sin 40) / sin 60 = 3 / (sqrt 3 / 2) = 2 sqrt 3
It should take one minute if you take the right approach. Nothing fancy is involved at all.
If you drop a perpendicular from A to BC and call the intersection P, A to BD and call the intersection M,
triangle ABP is congruent to triangle BAM => AM=BP=3 => x = 2 sqrt 3
This also takes one minute.
This is 2 method example of something that is easier than it looks. I suspect that the second method is really using a cyclic quadrilateral, am I right? And that pretty results in corresponding angles being congruent as well as sides being congruent. I hope that this is a sufficient summary.
X=2√3
asnwer=8cm isit
asnwer=2/3 is it