Portuguese Math Olympiad | 2019
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- Опубліковано 1 жов 2024
- We look at a nice geometry problem from the 2019 Portuguese Math Olympiad.
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I took an analytical approach:
I assigned both lines to a function:
f(x) = 1/2 x
g(x) = -2x + 10
Thus the Area is equal to:
A = 1/2 * 5 * y(f=g)
A = 1/2 * 5 * 2 = 5
Multiply that with 8 and subtract that from 100 you get 60.
And that's a good place to stop.
I followed the same approach. Coordinates of interception point C are (4;2).
Same for me. But in my mind, the geometric approaches, like Michael's one, are always more elegant.
@@egillandersson1780 this is how I did this too, but I'm an engineer so elegance to me is "as little duct tape as possible"
I followed the same
i labeled the area of triangle ABC x, and noticed that triangle ABC is similar to triangle PAC, and that the ratio between their sides is 1:2. therefore, the area of triangle PAC is 4x, which means that the area of triangle PAB is 5x. but the area of triangle PAB is obviously 5*10/2=25, so x=5. badaboom badabing, the shaded area is 60 units.
Finally something I can solve.
ok..
Yeah like, it’s there right in front of you, dummy.
Lol same, I've been watching this channel the whole day, I could solve only 2 problems myself
Wow , I was on This Olympiad!!!This Will probably be the only time ill be able to Say This .Thx for sharing portuguese math problems :)
Fun Fact:This was the easiest exercise of the finals, so yeah , wasnt that hard to solve xD.
Congrats for being in the Olympiad!
A little simpler: you do not need to calculate AC and BC because similar triangles areas are proportional to the square of their proportional constant
By similar triangles, AC = 2 BC, without having to do any further calcuations and then use area formula.
Yes you just need to calculate (AB^2)/(AQ^2+QR^2) which is 1/5 and that's the ratio between the area of AQR and ABC. With the value of the area of AQR the answer is simple.
@@Tiqerboy
I was just about to write the same thing.
Nevertheless, great problem that can really test your mathematical abilities.
Yes! I used the same method.
my solution was:
i guessed all triangles were the same
i guessed the height of the triangle was 2
i saw that there was 8 triangles
5*2/2 = 5
5*8 = 40
100 - 40 = 60
the engineer way won once again
Easy problem. Done in less than a minute. Great job with the videos tho!! I would really enjoy if you could try some imo problems next time. Have a fantastic day!
Yeah i also support
I'll not just use your vids to improve my Math dots, but particularly to learn English.
How the saying goes?
"Two birds, one stone." 💪🏿
Tem br em todo lugar kkkk
ABC and PAC are proportional, sum of their areas is PAB = 1/4 of total area, hypotenuse of ABC is 2 times smaller then the one of PAC => area of ABC is 1/4 od PAC or 1/5 of PAB, that is 1/20 of total area.
BRILLIANT.
y=x/2, y=-2x+5 から交点(2,1)を求め、三角形1つで5*2/2=5、10*10-5*8=60としました。
If I were ever to get in to rock climbing, I'd want Michael Penn as my instructor because he would know when to say "and that's a good place to stop"
10:04 É um bom lugar para parar
Alright I’ve changed my mind from yesterday. I’ll try to pick some questions that doesn’t interfere with the main content. And having comments increase the engagement so let's go...
The sums of five numbers (ℤ) taken in pairs are 0, 2, 4, 4, 6, 8, 9, 11, 13 and 15. What are these five numbers?
-1,1,3,5,10
What do you mean by "whole number"?
@@minh9545 Whole numbers are the ℤ set. I thought that using "integers" word only would lead some people to only consider positive integers
@@minh9545 How are you feeling by the way? Better?
É um bom lugar para parar : great!
This was my quick and dirty method.
Using some rules about similar triangles, you can show that APC is 4*ABC and APB is 5*ABC. APB has area (5 * 10) / 2 = 25, which means ABC has an area of 5.
Thank You. As always, it's "passionnant" And well explained. It's also an offered moment of reflexion.
Hi,
For fun:
2 "and so on and so forth",
2 "let's go ahead and",
5 "let's may be go ahead and",
6 "great", including 1 "ok, great",
1 "and then, the next thing that I want to do",
1 "so the next thing that I want to notice",
1 "now the next thing that I want to look at",
1 "now the next thing that we might want to do"
and the final "and that's a good place to stop".
Find the area of the octagon that you can see in the middle. ( It was one of a tricky, but not so difficult problems in 1989 at a hungarian high school final exam to get in some university There were 7 other problems in that exam...)
You could also put a coordinate system ontop, with the origin bottom left. I then looked at the triangle one step clockwise from yours and calculated the height as the x-value of the intersection between the lines y=2x and y=-1/2 x+10. Quickly giving the heigt of 2, and then area=1/2*2*5=5, avoiding squareroots and the necessity for similar triangles :-) but I definately like your pure geometric calculation as Well 🥳
Estava a observar o problema, e depois percebi...
Eu já o tinha visto antes
In general for a sidelength 'm' the area of the star figure will be A = 3/5*m^2... which is a neat fractional amount, I guess.
There's a simple solution. We extend the top side of the square and the line AR. They cross at point F and then we have two similar triangles: BCA and PCF. There bases are 5 and 20 and their heights equal h and 10-h. Then you make a simple proportion (h/5)=(10-h)/20. Then h=2 and the area of BCA equals 5. Done.
no gougu here, so great!! 😘
😀
Good Lord. One of your problems that I could actually solve! :-) Thank you for the videos.
07:41 Quicker way to calculate AC:
In triangle AQR we have AQ = 2 * QR.
Because triangle ABC is similar to the triangle AQR,
in triangle ABC the corresponding two sides are also in this relation.
So, in triangle ABC we have:
AC = 2 * BC
AC = 2 * sqrt(5)
Thanks for the tips. I was able to get the area of the tiny triangle by finding where the two lines intersect, and from there do the base*height
Somewhere in world
Teacher: find the shaded area
Student: there it is
Teacher: and calculate its size
Student: Oh..
Hello 60% Portugal star! 🌟
Hello √5/5.
But I wonder if this star was in a hexagon, would it take the same percentage?
I think its area then will be 90√3
What do you think?
I made two kites in each corner. One within the other in each corner and subtracted to differences to find the area of the eight triangles.
Can someone explain to me this
If ABC is a right triangle
Then by Pythagoras theorem: square root( 5^2 - square root(5)^2) should give us 5 which is the hypotenuse, however it doesn’t give us 5
I tried solving it by cos(60)=BC/5, which in turn gives us BC=5/2
Then sin(60)=AC/5, which gives us 5(square root(3))/2
And by Pythagoras Theorem, the hypotenuse IS 5
Anyway my final answer wasn’t that far off (56,70) but Im still confused
Anyone?
We can Calculate AC by using in right triangle ABC (1/AC*2)=(1/AB*2)+(1/AP*2) I mean by AB*2 AB square then using pythagore in triangle ABC & Calculate thé aria of ABC
I think it's a little simpler (but essentially the same approach) to first consider a 1,2,sqrt(5) right-angled-triangle in abstract, and show cos and sin are 2*sqrt(5)/5 and sqrt(5)/5 respectively. The top of those absence triangles are clearly a right-angle because the lines making up that corner can be rotated by a right-angle and then displaced into each other by the symmetry of the setup (equivalent to the start of your similar triangles argument). But then, as you have the hypotenuse length of 5 you can use base*height/2 to give an area of 2*sqrt(5)/5*5*sqrt(5)/4*5/2 which is 5. And from then on you're done and get 55.
A much cleaner resolution:
Area[ABR]=5x10/2=25 r=5/sqr(100+25)=1/sqr(5) Area[ABC]=(1/sqr(5) ^2x25=5 A[star]=100-8x5=60
Sir Respect from India,
Sir I am in 12 grade. Can u suggest me some tips to become a good mathematician. I want to teach in universities too like u. U are my inspiration sir.
Mee too from india also in class 12,and are you preparing for jee
@@gurusamy2911yes brother but I want to peruse mathematics in my undergraduate
@@rounaksinha5309 oh ok brother then you prepare for isi or cmi
@@gurusamy2911 thanks brother
Michael penn sir is our inspiration. Hats off to him.
For a general parallelogram with base b and height h, with h=kb, the area of the star is: A = bh(4k-1)/(4k+1) and it is not necessary alpha=beta. In this particular case the parallelogram is a square, so b=h=10 and k=1, then A=bh(3/5)=100(3/5); A=60
Let point X be the reflection of A across R. CXP and CAB are directly similar with common ratio 4. So the distance from C to AB is 10*1/(1+4)=2, so the area of ABC is 5*2/2=5 so 100-8*5=60, much cleaner.
Very interesting geometrical figure.
The "star" covers 3/5 of the total area.
The tilted squares cover 1/5 each.
And the 2 squares defined by the central octogon cover 1/8 and 1/9 respectively.
I think there is an immediate solution, just considering we need the AREA of ABC and not its sides: AQR and ABC are equivalent, thus their areas are proportional to the square of any of their sides, let's take the hypotenuses. These are: 125 and 25, thus 5 to 1. Since the AQR area is 10*5/2=25, the ABC area is 25/5=5. And that's a good place to stop!
Finding the intersection of
y = -2x + 10
and y = x/2
Wet point y(C) = 2
Thus the area of a triangle is 5*2/2 = 5
And the area of the shaded shape is 100 - 8*5 = 60
Their is a simpler way to find the area of the triangle ABC
Once you have found all the angles you can conclude that is similar to the triangle AQR
This mean ratio ac/bc =2
The triangle is als a right trial so combing this we get
(2x)^2 +x^2 =5 ^2
We could solve the for x but we als know that the area of the triangle is 1/2*2x*x=x^2
So we get tge answer of 5 imidiatly
even simpler: the areas of ABC and AQR are in the same ratio as the squares on the hypothenusas, namely 25 to 100+25=125. That is 1:5, so the small triangle's area is (100/4)/5=5.
I was gearing up to find the areas of all the different shaped before I realised I only needed that one triangle. A fun one!
Here I was thinking you could just throw the shoelace algorithm at it......it did not work
Since each purple triangle turned out to be 5% of the square, I made a tiling of the square with 20 of these
triangles: imgur.com/gallery/QEe3pZb
Lol this is so easy. Fourth grader work.Anyway i did use double integration.
In this example, it is interesting geometric interpretation. The large square on page 10 can be divided into parallelogram side 5 and a height 10 and two identical right triangles having sides of 5 to 10.
The content of unknown formation can be computed as 10 * 10-8 * of a triangle with sides a and b.
It is seen that the content of a small square inside a large square is the same as the content of four triangles with sides a and b. This can be achieved by moving them.
Compared also formations in two triangles and a parallelogram.
It follows that b = 2 * a. From the Pythagorean theorem we calculate b, a.
👏 A different, but really valid aproach.
And you can prove that the middle smaller square's area is exactly ¹/₅ of the bigger square's area.
➾ 100/5 = 20, but there are 4 triangles in it, so
20/4 = 5;
8·5 = 40;
100 − 40 = 60 ✅ Solved❕😁
@@ciberiada01 This is still just a consequence of the ratio b / a = 2/1 in the small right triangle. On the picture you can see 2 triangles, they are similar, first of them has sides a, b,5. Second b,sqrt(2ab)+b,10. (aaa).You see immediately that b=2a without calculation.
after 5:50...
find AR by Pythagorean.. easy peasy
area ratio = square of (any side ratio)
I arrived at the area of the purple shaded triangle by co-ordinate geometry. Set the bottom left corner of the square to be the origin and find the point of intersection between the two lines. Then use the given 5 as the base and the y-value of the point of intersection as the height of the triangle calculate the area.
Finally, Portugal!
Yup :)
I was waiting for this day for a while!
gajo ta empolgado
PORTUGAL CARAAAAALHO!
On a related note, if I were in the competition, I would have gone for the cartesian plane approach.
Omg I was there! This was the second time I had participated in the olympiads and I was able to solve it! It was also one of the easiest exercises on there
Mr Michael Penn, I wonder if on your grove there will be your famous words:
"And that's a good place to stop"
That's going to be many years from now. He takes fitness seriously, you can tell.
They way you end your videos does not work with the non-removable overlays in the ends. You need to rethink you outro or the editing of the end of the video.
I used coordinate geometry to solve this problem
For the triangle
(x^2+(2x)^2)=25
A=x^2
Wowser
This is the sort of English I want to hear!
I literally could get every single word out from his mouth.
Amazing, amazing, amazing!
Greetings from portugal
Greetings from Germany
anyone else get 68^2 i might of made a cal error
Thank you from Morocco. شكرا
First line: y=x/2, second: y=10-2x, 10-2x=x/2, 5x/2=10, x=4, y(4)=2, S=2*5/2=5, allS=8*5=40, needS=10*10-40=60
Segment AR is y=0.5x and PB is y=-2x+10
So the 2 segment are crossing in (4;2) and the area of the triangle is 5*2/2=5
5*8=40 and total area is 100-40=60
The shaded area, there is the point reflection. We can see small square into the picture.See you on points (0,0),(0,10) and (0,5),(0,10).So right triangles 10,b,sqrt(100-b^2) and 5,a, sgrt[25-a^2] are similar (aaa). So b=2a.QUICK SOLUTION
I solved by 2 graphs.
1. Y=0.5X 2. Y= -2X+10
the cross point is (4,2)
So the area of the purpel triangel is 5x2 /2 = 5
We have 8 triangels so they all togehter 40
Finaly 100-40 = 60.
T.I.A.G.P.T.S
That is a good place to stop...
You look like Zuckerberg
I solved it with the Pythagorean theorem.
We have the big triangles with sidelength 5 and 10. With a^2+b^2=c^2
The length c is 5*(5)^0,5 or 5^1,5
As the side hypthenuse of the gray triangles is 5. And it being a similar triangle, regarding the angles.
It is down scaled by the factor square root of 5.
Meaning a and b of it are 2*(5)^0,5 and 5^0,5
Multiplying a*b=10, we have the area of triangle pair or rectangle.
We can have 4 pairs.
Meaning 10^2-40= 60
So the shaded area is 60% of the rectangle.
There's a simpler argument just using symmetry and the ratio of sides & areas (using only given side lengths)
By the symmetry of the problem, area(ABP) is 25
By rotational symmetry (e.g. rotating AR about the centre of the square onto PB), ABC and ACP are both right ∆ with the same internal angles, so they are similar, and have hypotenuse 5 and 10 respectively. This means area(ABC) = 4 * area (ACP). Since area(ABP) = area(ABC) + area(ACP) = 5 * area(ABC) we have that area(ABC) = 5 and the shaded area is 100 - 8 * 5 = 60
A bit complicated in such approach.
Actually we can partition the big triangle into 2 small similar triangles with hypotenuses equal 5 and 10 respectively, i.e. length in 1:2 and so area in 1:4. So we easily deduce small triangle (hypotenuse 5) area is 1/5 of the big triangle = 25/5 = 5, so total shaded area = 100 - 5 x 8 = 60
@MichaelPenn you can solve it only using Pythagoras theorem on APC, APB and ABC to solve the sides of the purple triangle, and then use it a couple of times more by dividing the triangle in two to calculate its height. Only with geometry and only with Pythagoras theorem.
I solved this using the fact the tangent values of the accute angles in the small triangle are 2 and 1/2. Then x(1/2) = h = (5-x)(2) ➡️ x=4 and h = 2. Where h is the altitude of the triangle, so area = 5*2/2 =5. Then total area is 100 - 8(5) = 60
Set up a coordinate system and that’s a good place to stop XD but probably cuz I was trained to be good at geometry back when I was in middle school, solving with pure geometry theorems is more of a fun (and more complex at the same time).
AR is y= x/2 PB is y= -2x+10. C is (4,2). Area triangle ABC is 5*2/2 =5. Area of star is 100-5*8=60. Cheers.
BC is wrong, not sqrt 5
Seems like it would be much more straightforward starting with arctan(5/10) ?
nice hint
To summarize the results, isn’t it just that
area of ABC = area of ARQ * (AB/AR)^2 ?
Nice
I used the fact that ABP is just ARQ are rotated 90 degrees from one another, thus line segments BP and AR are at 90 perpendicular to each other, thus angle ACB is a right angle. And since ABC and ARQ also share the same angle centered on vertex A, along with both having right angles, I used the triple angle rule for similar triangles.
Given a regular n-gon, place two circles on each side such that the center of each circle rests on a side, but the edge of the circle cannot cross any other circle or any side other than the one it is attached to. Next, draw a large exterior circle that is at most tanget to the smaller circles. Find sum of the maximum area of the smaller circles in terms of the area of the larger circle and n.
Finally found one easy enough that I can do in 10 mins w/o pens and paper. c:
i did it using coordinate geometry, i.e. point A is the origin and then find equations of lines going thru AR and PB
Trivial but cute problem. It reminds me of when I was 12.
PORTGUAL CARALHO!
Maybe try some imo problems
It is straight forward to prove that AC is perpendicular to PB so using a property of right triangle we have 1/AC^2=1/AP^2 + 1/AB^2. From this AC is obtained. The area of the purple triangle is then ready to get
7:50 By similarity of triangles triangle ABC and triangle ARQ we have AC = 2*BC = 2*sqrt(5).
Can you explain what sqrt means?
@@JPL454 It's short for the (principal) square root. The (principal) square root of 4 is sqrt(4) = 2 as 2x2 = 4. Does that make sense?
@@DavidCorneth Maybe i need to learn English...
@@JPL454 What's your native language?
@@DavidCorneth Portuguese
Me with 8 grade starting solves the problem immediately corectly