What Is The Area? Challenge From Croatia

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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Connect with me

[grahampcharles]

Spoken typo: at 2:28 you say b1 plus b2 when you mean b1 over b2.

[jonk]

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

[jbtechcon]

I just used the phrase "We can math the world a better place" and my parents stopped calling me son.

[blazejecar]

these videos make me feel like the biggest dimwit on the face of the earth

[benjamingeiger]

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.

[matejkovacic]

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!

[CauchyIntegralFormula]

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.

[a.t]

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)

[eddytsang]

A simpler way:
x:(y+2) = 3:4
y:(x+3) = 2:4

[sergiovargas]

The person who drew the triangle put the top area smaller than the triangle of area 4. Amazing

[brandonk]

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.

[Paul_Hanson]

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.

[smchoi]

This could be solved systematically ("trick-free" yet not too quickly) using basic vector algebra:

Let k be the unknown (area) to be found.
Let A, B & C be the vertex points of the triangle forming the outer boundary lines, where A is also the common vertex of triangles with areas 3 & 4, B also that of triangles with areas 4 & 2, and C also that of triangles with areas x & y (as designated in the video, but x & y are not used in the following). Also let P be the point where AC meets the common side of triangles with areas 3 & x, Q be the one where BC meets the common side of triangles with areas 2 & y, and M be where AQ intersects BP.

Call u & v the vector from A to B & that from A to C, respectively.

On one hand, vector from A to M
= [4/(3+4)] (vector from A to P) + [3/(3+4)] u
= (4/7) [((3+4)/(3+4+2+k)] v + (3/7) u
= [4/(9+k)] v + (3/7) u

OTOH, vector from A to M
= [4/(4+2)] (vector from A to Q)
=(2/3) {[(4+2)/(3+k+4+2)] v + [(3+k)/(3+k+4+2)] u}
= [4/(9+k)] v + [(6+2k)/(27+3k)] u

Comparing the u components,
3/7 = (6+2k)/(27+3k)
k = 39/5 = 7.8

[zecrets]

*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_*

[mgc]

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.

[anandanpoornash]

it can also be solved by using ceva's and menalau's theorem

[twwc]

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.

[damiengreenwood]

Agreed I thought 3 straight away.. the diagrams are terrible and misleading. Easy way is to draw it out on graph paper/cad and see what it really looks like

[sklman]

That area sure don't look almost twice as big as the lower one..!

[paull]

All of what you just said is assuming that the 2 line segments are perpendicular to the sides they touch.