What Is The Area? Challenge From Croatia
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- čas přidán 19. 06. 2024
- Thanks to Matej for suggesting this problem! This is from a math competition for 14 year olds in Croatia. While the problem is challenging, "you should be able to solve it" using elementary geometry and algebra. Can you figure it out?
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Spoken typo: at 2:28 you say b1 plus b2 when you mean b1 over b2.
Thanks, you get the award for being the most attentive! As a general response to other comments, some people have been talking about the "ladder theorem" and "mass point geometry." If you guys can provide links to explain that would help the rest of us who don't know these concepts, thanks.
@@MindYourDecisions czcams.com/video/rTMMZQCK2ZE/video.html
Just Curious
You only said to draw a line segment
A line segment from a point to side of a triangle, doesn't necessary mean it becomes the height of a triangle inside
Can u please explain?
@@bamgm14 You are correct that the line segment is not necessarily the height, but that doesn't matter. If you draw a line segment from one vertex to the opposite base, it will always form two triangles that share their height, and also share that height with the original triangle. You don't need to know the height to know that's true.
@@Quintaner I didn't know that.... Thanks for the clarification. I was just using my basic knowledge of maths
I just used the phrase "We can math the world a better place" and my parents stopped calling me son.
There was a time your parents put you down after picking you up... That's right. Your parents put you down for the last time and said "enough".You're busting camels backs brah.
Why?
JBTechCon Why? Did they disown you for being so pretentious?
*Do they call you sum or sin?*
@@MKD1101 Sign
😂
these videos make me feel like the biggest dimwit on the face of the earth
blazejecar this is painfully relatable
However we still kept on watching
@@ambientamben5059 true
sem
Same here
To save you 2 minutes, at #4.21 You don’t have to solve for x and y individually. you can get x+y directly by multiplying the 1st eq with 6 and the second equation with 7. you will only have x+y as the only variable. Then you get x+y = 78/10 = 7.8
My college algebra professor, Dr. Nelson, always told us to draw things in such a way that you wouldn't assume things that weren't true. If a line wasn't perpendicular to another, it shouldn't be drawn as close to perpendicular. If a line segment isn't bisected, it shouldn't be drawn to look like it is.
I saw the diagram in the thumbnail and thought, "oh, that's easy. Split the quadrilateral in half, and you end up with triangles of area 2 and 3, so it's 5."
But the line segments don't necessarily bisect the sides of the triangle, despite the diagram appearing that way. I actually had to rewind and listen to the question definition again.
Don’t assume things then, there is a reason we (not drawn to scale)
Fool
That is exactly my philosophy. Whenever I needed to draw a trapezium for example, I would definitely make it non-isosceles!
Even though your professor was right in that it's generally bad practice, it's your own fault for assuming and not the problem maker's fault for drawing it so that you can assume. There were no marked right angles, no markings indicating that someting is equidistant or that it's an isosceles triangle. Problems can intentionally fool you into assuming things, because it's also important to teach you to not trust any pictures that do not claim to be drawn in correct scale. Only properties that are explicitly stated should be taken into consideration.
You could also have easily noticed a problem in your solution. If both lines bisected the triangle, the quadrilateral would have had to be 3 and 5 at the same time. The areas 2 and 3 would have had to be equal for this to be possible with bisecting lines, in this case 2 and 2.
said right. completely
All right, so I am from Croatia and I have the same name as the person who suggested this problem. Hearing him say my name like that caused me to actually fall out of my chair laughing! I simply love to hear people from english speaking countries try to pronounce Croatian names, even more so my own! Presh, you made my day!
You mean Serbian names.
@@LeonEdwardsGoat ;)
@@LeonEdwardsGoat lmfao I just realized that I replied with winking smile to someone named Winks
Presh Talwakar is Indian :)
what a beautiful solution. Thank you!
A simpler way:
x:(y+2) = 3:4
y:(x+3) = 2:4
Exactly. I was just going to write the same comment.
The answer will be different from the video. Wont it?
@@Bkthomte simplifying the two equation, we have
4x=3y+6 -----(1)
x=2y-3 -------(2)
The answer should be the same.
Got it.. Thanks
Please tell me the name of this method.
The person who drew the triangle put the top area smaller than the triangle of area 4. Amazing
For extra credit: How could this problem be drawn to scale, when you're starting out with just areas?
@@pbierre Use ladder theorem: 1/(3+4) +1/(2+4) = 1/4 + 1/Area(Large triangle)
I immediately noticed when looking at the problem that the height of the triangle with area 4 must be way under half the height of the overall triangle, because the overall triangle's area is always significantly larger than 4+3+2, which is already more than 2x4.
It is slightly easier to divide the unknown area a different way. Connect the other 2 vertices of the quadrilateral to form 2 triangles. You can easily show that the bottom triangle (of the 2 just formed) has an area of 1.5. Then solve for the area of the other triangle much like Presh did in this video.
man !!
u'r awsome
the best channel ever !
finaly something intresting on youtube !!
What a brilliant problem....I would have never thought of that ratio of area - well done!
Elegant and beautiful!
Now I feel the need to refresh my geometry skills. :D
Thanks for a great puzzle!
I used a similar technique, drawing a line segment as you did. But I used the internal lines for the bases of triangles rather than the outer edges of the large triangle, so I ended up with different pairs of triangles and different equations, but of course, I got the same answer in the end.
How did you come up with the two line segments as perpendicular to the sides? The problem only stated "a triangle". It did not say that the line segments are "perpendicular" as well.
They are not be assumed as perpendicular. He took the ratio of their areas and base because they have the same height:
A1 = 0.5b1(h)
A2 = 0.5b2(h)
Hence:
h = A1/0.5b1 = A2/0.5b2
= A1/A2 = b1/b2
@@nellvincervantes3223 yes but the height always has to be perpendicular to the base.
theorem used here is when two traingles are between the same parallels. u dont have to
imagine the middle line to be height . you can also construct your perpendicular line.🙂
Carl Edward Mapanao he probably forget to mention that the segments are the heights , because I don’t know if there is any other way to prove that is indeed the heights
It didnt have to be perpendicular. You can see in the link bellow tha triangles ABC and ACD have the same height, even tought AČBAČD90⁰. They share the side AC, but the height isn't AC. In this case, the height is AB, since the angle ABC=90⁰.
math.stackexchange.com/questions/746084/two-triangles-only-one-different-side-same-area
Nice one! And well presented & explained. Thank you.
Fred
Hi, congratulations on your videos. One question, what program do you use to make them? I want to prepare my most interesting classes. I am a teacher of Mathematics, thank you
We can use barycentric coordinates (specifically, areal coordinates). Let the coordinates of the intersection of the lines be (a,b,c) (where (1,0,0) is the top corner of the triangle, (0,1,0) is the bottom left corner, and (0,0,1) is the bottom right corner). The fact that the bottom right two regions are split 4:2 means that the cevian from the bottom left corner is also split 4:2, so b = 1/3. Similarly, the cevian from the bottom right corner is split 4:3, so c = 3/7, and hence a = 1-1/3-3/7=5/21. So the coordinates of this middle point are (5/21, 7/21, 9/21). The cevian from the bottom right corner intersects the upper left side of the triangle at a point whose third coordinate is 0 and whose first two coordinates are in the same ratio as those of the central intersection, so it intersects at (5/12, 7/12, 0). This means the ratio of lengths of the two pieces of the upper left side are 7:5, with the 7 closer to the top corner. Thus, the areas of the two larger triangles (upper right and bottom left) are also in a 7:5 ratio. The area of the bottom left triangle is 7, so the area of the upper right triangle is 49/5. Take away 2 and we get the area of the quadrilateral: 39/5, or 7.8.
Hey I did the same thing except I called it MASS POINTS
We got in trouble at MOSP when we used mass points without phrasing it in terms of barycentric coordinates, so my write-up here was phrased as such. But yeah, I worked it in my head with mass points. (As an aside to anyone else reading, mass points lets you do this problem in your head. Yeah, it's that useful.)
Damn I've heard of the uses of barycentrics over and over in JMO, USAMO, MOP, IMO etc. yet I never bothered to learn this. This would definitely be a useful tool to put in my toolbox - I'll read Evan Chen.
Interesting (and good) approach.
I have never heard of barycentric coordinates before. Thanks to you, I am watching some youtube videos mainly to understand the basics and have an appreciation for the approach. In any case, in your solution, once a=5/21 is calculated, I think we can get the total area as 4*(5/21)=16.8
Then, subtract (4+3+2) to get the remaining area as 7.8
Thanks for mind your decision channel, becuase I wanted answer of its same question, now I got answer from this video with correct explanation
General rule (you can find it by doing exactly the same method as shown in the video) : If you have the left triangle's area = c, the right triangle's area = e, and the bottom triangle's area = d, then the top quadrilateral's area is (ec^2+ce^2+2ced)/(d^2-ce)
Great vid presh! 👍
You have solved much longer than it needs to be. Simply draw a line between two middle points. Calculate the area 2/4 3/1,5... call upper triangle s area “x”. Than 3,5/x equal to 6/4,5+x... x equal 6.3 plus 1.5 answer is 7.8... much fast and shorter!
Between two middle points? Are we talking about the points where b1 and b2 meet (and b3 with b4)? Where does this 1.5 come from?
Love these videos. Thanks.
Are the line segments perpendicular to the side that they meet?
It seems to me that the solution presumes this although this is not stated in the problem.
I am curious if it is possible for other solutions to materialize if this requirement is relaxed. I suspect not but I cannot prove it.
P.S. I am glad I did not learn maths in Croatia :-)
The great part is the line segments don't need to be perpendicular. The height is the vertical distance from the base to the opposite vertex--or the length of the corresponding altitudes. From the emails I get, it does seem many countries have harder/stronger geometry instruction compared to America.
@@MindYourDecisions
Yes in India there are 2 exams called rmo and inmo.you will be given 3 hours for just solving 6 questions and they are really very very tough!
@@MindYourDecisions doesn't vertical distance mean perpendicular to base dropped from vertex
@@MindYourDecisions
The statement that "other countries have harder geometry than America" maybe wrong. Because the problems are sent here may not be from normal curriculum but from advanced curriculum. I myself have learned this technique from 5th grade but from advanced class, in Viet Nam. So in America, in advanced classes, how hard is the mathematics?
@@AnTran-ie7vq we are talking just comparing how hard are questions in the same age group between countrues.Americans have easier questions
it can also be solved by using ceva's and menalau's theorem
I immediately thought it was 3, because if you rotate the triangle, it looks like a triangle cut in half, so
4 + 2 = 3 + x
and
x = 3
lol
I also answered that
Same here!
Same xD
We're a more realistic people lol
same
I used different set of triangles. "x" and "y + 2" triangles that gives x/(y+2)=3/4, and "3+x" and "y" triangles giving (3+x)/y=4/2.
I like very much you channel! Since I have discovered it I usually solve one geometry problem every week. Just to entertain and exercise my chemistry mind.
very interesting. and very simple to understand. great job go ahead.
Croatia should've won
@physics Insider probably referring to the football championship, they got to the finals.
Sir platinium Ironically, Croatia will play France, the World Cup final opponent, in the final of the Davis Cup.
@@toddbiesel4288 I have never heard of such a cup
Sir platinium It is a year-long men's tennis tournament.
@@toddbiesel4288 oh thank you
That area sure don't look almost twice as big as the lower one..!
Because the proportions are actually wrong. The picture does not represent what was calculated.
@@krzysztofmatuszek that's true. I can't however really picture the shape of the triangle that's described in the problem... :)
@@sk4lman Imagine B1 being far longer than b2 and b3 far longer than b4, that's your answer
@@krzysztofmatuszek Ah, that was in a way too simple an answer.
I somehow assumed the lines from the vertices to the opposite sides met at the middle of said sides.
That would be a completely different problem, now that I think about it.
I'm going to stand silently over here now.
You are right sk4lman; I think the teacher is wrong !
The real value is 13/3, and I could prove it
You could have used the Ladder Theorem. 1/T+1/4 = 1/(4+2) + 1/(4+3). You find that the Total area (T)= 84/5. To find the missing area, you just subtract the three knowns ?=T-2-4-3= 84/5-9=(84-45)/5=39/5. The Ladder Theorem, Stewart's Theorem, Ceva's Theorem, Menelaus Theorem and Ptolmey's Theorem all come up frequently in Olympiad questions related to geometry involving circles, areas and cyclic quadrilaterals.
Beautiful problem with elegant solution. Thanks......
Great one bro😘
me, an intellectual: It's 4 because it looks twice as big as the small triangle
There are other approaches as well to solve this question but it was appeared for 14 years old.
So those who uses other approaches should use maths concept of upto 8-9 class only.
For me this is the best approach
More generally, if the 3 known areas are a, b and c (3, 4 and 2 above), then the same solution gives the unknown area as ac(a+2b+c)/(b²-ac). This implies that
b² must be > ac;
so the sizes of the 3 areas aren't completely arbitrary.
*I’m from Croatiaaaa plus I loved this puzzle* 😁
Sir , this question can easily be solved by mass point theorem .
Just balance the weights and yes answer is 7.8
thank u for all these videos
Amazing video Presh!
Thanks for the puzzle, Presh!
Consider the two triangles of areas 4 and 2 and treat the line that they share as their baseline. (Rotate them if you need help visualizing it) They have the same top vertex, so they have the same height. Thus each one's area is determined by the length of the baseline that it subtends. So those line segments are in ratio 4:2.
Similarly, the two triangles of areas 4 and 3's bases subtend line segments in a ratio of 4:3 on the other diagonal.
If we parallel-project the diagonal lines exactly onto the "real" baseline, the horizontal one, we must get the same ratios again. This splits the horz baseline into 3 sections, which we'll label in the obvious way as {A,B,C}, and we know that:
length(A) /( length(B) + length(C)) = 3:4
(length(A) + length(B))/length(C) = 4:2
This gives us A = 9/7 C and B = 5/7 C. This let's us solve for the area of the little triangle that is similar to the entire triangle. Its base is B, and:
B/(A+B+C) = 5/21
so its area is 4*5/21 or 20/21. Scaling up, the entire triangle has area 84/5. Subtracting the 3 other areas, the area of the top quadrilateral is:
84/5 - (4+3+2) = 39/5
To help visualize what I did, I made this diagram where the little similar triangle is shown:
[Edit: better version]
instagram.com/p/BqOUBRqAh56/?
Send him an email maybe he'll include it in a video
@@RajShekhar-jy2zi No, it's the same puzzle as the video. Just I diagrammed it with a few helper lines to visualize my solution.
@@Tehom1 oh my bad i thought it was a question , didn't read it
For a less confusing solution, you could focus on the upper half triangles. After drawing the "magic line", using the same base-area ratio logic, I deduced y=(3+x)/2 and x/3=(y+2)/4. Solving these equations also gave me 4,2 for x and 3,6 for y.
Same as me. Few people noticed that.
Exactly. Much simpler.
Good example how to use algebra and geometry together to improve ability to find solutions for such tasks. Thanks a bunch!
Excellent idea, really you've a great mind.
I think I'm confused, at the beginning of the video u divide the full triangle into a half, we know that the bottom half is 4+2=6 WHICH MEANS THE SECOND HAVE GONNA BE 6 TOO SO THE MISSING TRIANGLE IS 3
pramer 10 he never said it was cut in half
*Extended ladder theorem,*
1/A + 1/4 = 1/(3+4) + 1/(2+4)
A = 84/5 sq.units
Thus remaining area =
A-(3+4+2)
= 84/5 -9 = 39/5 =
*_7.8 sq.units_*
Wow!! ...
@Syed Ali Wijdan exactly
your each video open my mind
I constructed a line parallel to one of the lines dividing up the triangle and ended up with a rather more verbose way of solving this... this is so obviously much neater.
Finally i could Solve atleast one of the questions that u post!! Yay!!
Used MPG!!!
what does MPG stand for?
mass point geometry
Felipe Chujoza, thx
I did not know about MPG till this video, but after some study I find the MPG way to solve this problem really elegant.
Thanks for opening that door to me.
Geometry is my favorite math subject. What is yours?
WiseOldDude calculus
Geometry too
Numbers.
fertilization
@@danielroberto3097 Lol. What is "fertilization" in math? Factorization?
I am glad that I'm still able to solve this problem under 20 minutes. Man I should really spend some time to re-pratice my elementary math skills.
It can also be done using side ratios of similar triangles.
Denote the points A, B, C, D and E anticlockwise around the triangle starting from the left. Call the intersection inside the triangle M.
Introduce point F on BD such that EF//AC. Then we have two pairs of similar triangles, DEF~DAC and BCM~BFE.
By applying the base-ratio argument, BM:BE=4:(4+3)=4:7. Similarly, MC:AC=2:(2+4)=1:3. Also, MC:EF=BM:BE=4:7 (BMC~BEF).
The two ratios give AC:EF=12:7. So, AD:ED=AC:EF=12:7 (ADC~EDF). AE:ED=(12-7):7=5:7.
Again, by the base-ratio argument, we have (3+4):(?+2)=5:7, and get ?=7.8.
I figured it out on 5:09
Lol
Well done 👏
It is extremely unclear when you say that the b1 and b2 triangles have the same height that you aren't talking about the midline.
Exactly... it’s unclear
They must have the same height since they share a common vertex.
@@wirebrushproductions1001 I've literally NEVER learnt it this way! It's so eye opening and simple. Thank you!
using the fact that ratio of area preserves under linear transformation, you can transform the triangle in such a way that makes calculation easier
I honestly needed to see this video yesterday....PSATS used this so much
Diagram made me think you said the lines bisected the angles. Dang
No algebra required.
Label points:
A is the lower right.
B is the lower left.
C is the top.
D is the intersection along AC.
E is the intersection along BC.
F is the intersection of BD and AE.
G is the intersection of AB with a perpendicular passing through E.
H is the intersection of AB with a perpendicular passing through F.
We are given area(BFE) = 3, area(ABF) = 4, and area(AFD) = 2.
Find area(CDFE).
Given that the particular dimensions of the triangle were immaterial, consider the particular case where angle CAB is a right angle, and |AB| = 2. The latter simplifies everything because the areas of the various triangles based on AB are numerically equal to their height above AB. Otherwise you have to clutter all the heights up with a factor of 2/|AB|.
By considering the areas of triangles ABD, ABF, and ABE, we determine that their heights |AD| = 6, |FH| = 4, and |EG| = 7.
Triangles ABD and HBF are similar triangles with the latter being |HF|/|AD| = 4/6 of the former. So |HB| = 4/6 * |AB| = 4/3 and |AH| = |AB|-|HB| = 2/3.
Triangles FAH and EAG are similar triangles with the latter being |EG|/|FH| = 7/4 of the former. So |GA| = 7/4 * |AH| = 7/4 * 2/3 = 7/6 and |BG| = |AB|-|GA| = 5/6
Triangles CBA and EBG are similar triangles with the latter being |AB|/|GB| = 2/(5/6) = 12/5 of the former. So |AC| = 12/5 * |EG| = 12/5 * 7 = 84/5
Triangle CBA has base |AB| = 2 and height |AC| = 84/5 so its area is 84/5. Subtracting the given areas 3, 4, and 2 of triangles BFE, ABF, and AFD respectively, we get the area of CDFE is 84/5 - 3 - 4 - 2 = 39/5.
This is a brilliant solution, but I never ever like to use "without loss of generality" in a geometry problem that is generalized to all triangles of a certain nature. In a realistic competition, I would use this method for the sake of making it easier, but in general I would not assign a specific case and genralize from there.
"No algebra required"
Proceeds to label 8 unknowns and uses algebra to solve the question
@@jacobahern6965 I dont see any unknowns in his solution
@Sean Yang:
Well, the proof is easily adjustable to work without any concrete assumptions about angles and side lengths. Just define G and H as intersections of AB and parallels to AC (not perpendiculars to AB) passing through E and F, respectively. The reasoning using similar triangles still works the same. For the length of AB, as Kevin Martin said, just set it to be 2/y (or 2*z) and you will just have to carry a factor of y (or z) around with you until you calculate the Area of ABC when it will cancel out (1/2*2/y*84y/5 = 1/2*2z*84/(5z) = 84/5).
Edit: Plus, since |AD|, |FH|, and |EG| aren't heights any more, the sine of angle BAC is also a factor that will have to stick around.
I had started modifying my answer to remove the right-angle assumption, but it did not occur to me to put EG and FH as parallel to AC rather than perpendicular to AB, so around the last step I got stuck into solving equations again...
How do you produce the transition effect at 5:20? Thank you.
Also (1/Area of ∆) + 1/4=1/(4+3) + 1/(4+2) simplifying we get Area of ∆=84/5. So area(unknown region)=84/5 - 9=(84-45)/5=39/5 or 7.8
I admit I was unable to solve this problem, and it bugged me a lot. I chased after complex methods such as the system of equations between the angle of the sides after bisecting the unknown triangle. I even almost pursued further shoelace theorem with ambiguous points, but as usual this led to nothing. Then, I did a lot of angle chasing, only to get nothing. Finally, I thought of using Heron's formula. Many of the sides shared common side lengths, so I thought this would be a good idea. In fact, the ratio of area 3 to 2 urged me to write a ratio of b1 to b4 but was belittled - I was completely stuck! Obviously, I considered the base x height at first, yet all I got was the triangle with area 4 for the main base - silly me!
I scrambled for a simpler solution, knowing as this was a 14 year old problem, it would be nothing but a small calculation that utilizes an elegant trick. Needless to say, I did not expect that the only solution to this was such a hard bash!
Nevertheless, I enjoyed the video :)
1:58 OK but how do we KNOW those triangles are the same height? I dont see the logical leap there myself
never mind, I found a comment explaining it. height from the b1/b2 base.
Alternative solution: The given values, as well as the result, speak only of areas, thus they are invariant with respect to any transformation which preserves areas and is affine (roughly: maps parallel lines to parallel lines). You can (i) use skewing (change angles without changing the area as shown with the triangle at the beginning of the video), (ii) multiply the first and divide the second coordinate by the same number. Thus you may - WITHOUT LOSS OF GENERALITY - apply a convenient transformation and solve a specific task, e.g. with the crossing lines perpendicular and for a given proportion of their length. Such a case is completely determined, easy to draw in cartesian coordinates (one possible arrangement of vertices is (0, -2), (-4, 0), (5.6, 3.6)), and it remains to compute the area of a KNOWN quadrangle. This solution is not expected at the age of 14, but there is no need to look for another hint.
It's a basic ssc exam question in indian it's ladder extended form
In triangle ABC two lines meeting at O and extend upto E from C and D from B then Area of traingle 1/ABC +1/BOC = 1/BDC +1/BEC
1/Total Area +1/4 = 1/7 +1/6
x =84/5=7.8
I just looked at it and thought it looked like 3 lolol
Eyyy!!! Im not the only one!! My maannn!!!!!!!!!! I tot the horizontal line which seperates the two triangles and the triangles have same height and base so that means the area is the same.. so i tot 4+2=6 for the first triangle a d 3+3=6 .. like if u get it
Same
@@philbytan284 I thought the same thing
I got 5 by assuming that the triangle is isosceles .I will never assume anything in math!!
proportions
You have not told that lines drawn from vertices are perpendicular to their sides.
This is very important point.
Next, what if lines drawn from vertices are not perpendicular to their sides. Then how to solve it?
Yes ur right??
Those lines are not perpendicular to the base. If they are then the question is simply wrong.
And he did solve your “what if” in the video, because that is literally the question.
Hey i am 16 and from greece and last year i had this exact question in my kangaroo math competition. By the way no one was able to solve this. Thanks for the explanation
Nice video . Keep it up!
We can do with mpg also nd by formula also
What is nd?
@@iamyoda7917 think he means "and"
what is mpg?
@@Not.Your.Business@ mass point geometry................ a beautiful geometry concept created by one of the indian mathematicians.
@@Not.Your.Business czcams.com/video/NBZawFsTrvc/video.html
Thats 7.8
It’s much simpler if you solve it using the fact that area of triangle could also be written as 1/2*A*B* Sin(Angle b/w those sides).
I figured it out after you drew the line. thanks
Good Morning from India... 💐
In the triangle let part having area 3=x , 4= y, 2=z
Bro actually I got a relation in this question that is :
1/Δ + 1/y = 1/(x+y) + 1 (y+z)
Where Δ= area of whole triangle Now after solving thing you will get area of whole triangle after that just subsctract the 3 areas and you have got the answer is just 5 seconds!!
How did you find the equation?
@@Epiciouss
Haha didn't that worked?
@@VaibhavKlashGaming What do you mean?
Troll
Amiya sir ka student h kya bhai...?
which software do you use?
please let us Know!
Coordinate could be used to solve polygon questions like this. This method maybe need much calculations but you don’t need to spend time on thinking how to solve. When coordinate is set, every point and line is fixed, just to find the relations between them. Set (0,0) on the left vertex, assume the central point is (a,b), so the line from the two point is y= (b/a)x. Then right vertex is (0, 8/b) and the top vertex of area 2 is (1.5a, 1.5b). With those two points, you can get this line. Same you can get the another line. The top vertex is the intersection of the two lines. Then do subtraction between triangle and three known areas.
There is a way to get the same answer and no need to solve for x and y.
But unfortunately the comment section is too small for you to give the solution.
Solution subsequently submitted. Typing speed too slow.
In Croatia 7 out of every 10 war criminals solved this problem in less than 5 minutes, the other 3 were busy running the government.
Are you from Croatia?yes ....
I think they are genius I wasn't able to solve that question and I am 19 years old.
Don't worry because the exercise is not so easy : the teacher is wrong !
The answer is 13/3 and not 38/5 which seems too much.
I could prove it.
Sorry : ... not 39/5 which seems too much.
I didn't even realise that those two lines through the triangle are random. At the very beginning I thought that the lines were something like medians. That's why I couldn't solve the problem at first. But honestly, even if I knew that the lines were random, I couldn't have solved it. I've just graduated from school and I've managed to forget many of this simple geometry stuff.
I'm a 7th grader from Romania (that means that i'm 13 years old) and i'm still able to solve the problem! =D
I love this chanel!
Wrong.... Area of x = 1
Area of y = 2
Sum of x+y+2=6
Sum of 4+2=6
But then 1/3 is not equal to 6/6, so x can't be equal to 1 while y = 2. The ratios need to be equal, not the numerator and denominator.
3+x+y=6 and 3+x=2y x=1 y=2
I did it by diving the unknown part with a "horizontal" line. Using ratios, we get the bottom area 1.5. If the top has area x, then x/4.5 = (x + 3.5)/7. Solve x and add 1.5 to find the unknown area.
another way is to match the other diagonal of the quadrilateral. Easy to see that the lower triangle has the square of 1,5, and use the Menelaus's theorem for the triangle of (3+ the quadrilateral) to have the ratio of two sidelines and we get the correct answer.
Good video! Please do some physics problems. I want to practice for my physics competition🙄
My first thought when looking at the problem was to use ratios of triangles. But it seemed a bit inelegant and too complicated for a test for 14 year olds. I was fully expect Presh to then tell me that there was a clever/simpler solution.
So glad I didn't bothered doing the calculation as it's just boring number crunching. Requires very little imagination.
Is it possible to deduct all the main triangle values (base and sides, angles) from this result?
Wait. Does that make the lines medians by any chance or does it have to beentioned specifically for them to be bisecting the angle at the respective vertice?
it can also be solved by mass point theorem and by melenaus theorem
My solution was similar but simpler, I think. Label the vertices of the triangle ABC counterclockwise starting at the top, let D and E be the intersections of the secant lines with side CA and AB respectively, and let P be the intersection of the secants. Now draw DE and calculate the area of PDE (no pun intended). We see area(PDE) : area(PBE) = area(PBC) : area(PCD) = 4 /2, so area(PDE) = 3/2. Continue by setting up an equation for x := area(DAE) by looking at AD : DC. This ratio is equal to both (x + 3/2 + 3) / (4 + 2) and to x : (3/2 + 2). Solving this linear equation yields x = 63/10 and thus 63/10 + 3/2 = 78/10 for the total 'missing area'.
It is interesting to notice that not all triangles with the area of 16.8 can be split as in the drawing. if one vertex coordinate is (0,0) and the other is (t,0) t>0 , then we can select a point on the line y=14/t to serve as the third vertex of the triangle that has an area of 7. Suppose we select (w , 14/t). Now the triangle (16.8) is "defined". It should not be hard to write the coordinates of the other vertex.
I managed to solve it using an alternative, convoluted, inelegant way; the only insight being that there is an infinite number triangles that meet the criteria. You can choose whatever base you fancy and one of the angles of the big triangle, the 2/3/4 areas restrictions establish the rest. The solution is the same for all those triangles, of course.
Mind blowing puzzle
Is the information provided sufficient to uniquely specify the triangle? (ie all 3 edges and angles). (I'm tempted to say yes, but extracting the 3 edge lengths sounds hard)
No. There are infinitely many triangles which can be divided in this manner while retaining the areas of the whole and of subregions.
I ´m hardcore fan of Mindyourdecisions ❤️ . Could you tell how can I get better at math .
Thanks a lot sir, good question .....
I did what I have suggested here below, (0,0) and (t,0) are the "base" vertexes, (w,14/t) is the third vertex of the 7 area triangle... with analitic geometry I found the third vertex of the "big" triangle and saw that the area is not dependent on t,w and it is 16.8
I'm sure this is not the simplest approach, but I set it up as 10 equations and 10 unknowns, which solves for all the information needed to complete the rest of the problem. Granted, most of these equations are trivial, so I think it would quickly simply ...but too lazy to actually grind through once I know I *could* grind through.
3(x+y+2)=7x; 2(x+y+3)=6y. Multiply first equation by 6, second equation by 7, add the two equations together and you can find x+y =7.8.
A brilliant question
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