Let \(ABC\) be a triangle and let \(AD\) be the perpendicular from \(A\) on to \(BC\). let \(K,L,M\) be the points on \(AD\) such that \(AK=KL=LM=MD\). If the sum of the areas of the shaded regions is equal to the sum of the unshaded regions, prove that \(BD=DC\).

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TopNewestI solved it using trigonometry – Souryajit Roy · 2 years, 1 month ago

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– Llewellyn Sterling · 2 years, 1 month ago

Same here. Took me less than a minute to solve.Log in to reply

this tym's rmo was the easiest..........and it took me some time to believe that this was a rmo question – Nabarun Dutta · 2 years, 1 month ago

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I have solved it using simple geometry of similar triangles... – Arnab Deb · 2 years, 1 month ago

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its not given in the question paper that the lines are parallel, i think instead we had too think the contrary that if lines arent parallel then what???? thats the real question !!! otherwise the question can be done by a 7 standard student! i am waitin for the official sol. to know whats the correct answer – Aayush Srivastava · 2 years, 1 month ago

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shadedandunshadedregion actually is. – Aneesh Kundu · 2 years, 1 month agoLog in to reply

– Ajeet Gill · 2 years, 1 month ago

is my answer lil higher levelled ?Log in to reply

– Nabarun Dutta · 2 years, 1 month ago

In kolkata, i.e. west bengal region.......professors of ISI came and corrected the problem......and this was what they gave...........u aren't alone bro......evn I was thinking in that lineLog in to reply

Are those three lines parallel to BC? If parallel,proof is direct!If not,what? – Ranjana Kasangeri · 2 years, 1 month ago

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– Raushan Sharma · 1 year ago

Yeah, they are parallel. Actually here, this correction was made. Even there was a small correction in Problem no. 3Log in to reply

– Aneesh Kundu · 2 years, 1 month ago

I assumed them to be parallelLog in to reply

– Aneesh Kundu · 2 years, 1 month ago

The diagram in the question paper looked exactly the same.Log in to reply

– Ranjana Kasangeri · 2 years, 1 month ago

Well,I did for parallel and 'disproved' for non-parallel( Hope it does good)Log in to reply

the question is incomplete as it must be specified whether the horizontal lines are parallel – Prateek Gupta · 2 years ago

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– Aneesh Kundu · 2 years ago

Assume that they are parallelLog in to reply

i got 3 correct, and attempted 5 in delhi rmo what chance do i have to get selected – Sauditya Yo Yo · 2 years ago

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The difference of the area of the circumscribed n the inscribed square of a circle is 35 sq.m.Find the area of the circle!!!!!!!!!!! – Anna Anant · 2 years, 1 month ago

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It must be isoceles or equilateral triangle, AB must be equal to AC – Anna Anant · 2 years, 1 month ago

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There are 4 similar triangles if you observe

let BD = a and DC = b

Observe the left side of the figure

The segment above BD will be 3/4 of a

the segment above will be 2/4 of a

and the topmost segment will be 1/4 of a

Similarly the lengths of segments on the right can be found Let AK = h,

Use the formula for area of trapezium and area of triangle, you will get a = b – Pratik Soni · 2 years, 1 month ago

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assume all lines from the points K,L,M are parallel to the base. Let DC = k*BD ; Name the points on the side AB formed by intersection of parallel lines through K,L,M as - K1 ,L1 ,M1 and for the side AC the points of intersection as K2 , L2 ,M2 . Let the area of triangle AK1K be A1, and the area of the trapezium below it (KK1L1L) be A2, area of trapezium LL1M1M be A3, area of trapezium MM1BD be A4;

"The area of the triangles and trapeziums on the right will be 'k' times the area of the triangles on the left" Since we have assumed that DC = k*BC and the lines are parallel to the base

So area of triangle on the right say A1' = k* A1; area of right trapezium below (KK2L2L) = A2' = k

A2; let area of trapezium on the right LL2M2M = A3' = kA3; area of trapezium MM2CD = (say) A4' = k*A4;Given that area of shaded region = area of unshaded region implies (shaded) A1 + A2' + A3 + A4' = A1' + A2 + A3' + A4 (unshaded) .......equation(1)

"In triangle ABD area of all the trapezium's are proportional to the area of top triangle AK1K " Let area of trapezium KK1L1L = A2 = x

A1 ; (x is the proportionality constant) area of trapezium LL1M1M =A3 = yA1 ; area of trapezium MM1BD = A4 = z*A1;Hence A2' = k

xA1 (since A2' = kA2 and A2 = xA1) Similarly A3' = kyA1 (since A3' = kA3 and A3 = yA1)similarly A4' = k

zA1;substitute A2, A3, A4 , A1' ,A2', A3', A4' in terms of A1 in equation 1....

A1 + A2' + A3 + A4' = A1' + A2 + A3' + A4

=> 1 + k

x + y + kz = k + x + ky + z (dividing by A1 which is non zero) => (1-k) + x(k-1) + y(1-k) + z(k-1) = 0 => -(k-1) + x(k-1) -y(k-1) + z*(k-1) = 0 => (k-1)(x - y + z -1) = 0 => k = 1;we assumed DC = k*BD => DC = BD

Hence proved... – Shubham Chandra · 2 years, 1 month ago

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it makes no difference whether the lines are horizontal or not. if the sum of the areas of corresponding sides are equal and the heights are equal, abd and adc are congruent, so their bases bd and dc are equal – Howard Meeks · 2 years, 1 month ago

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to prove that BD=DC, you must think of the drawing as an isosceles triangle, without arguing or just ignore its unsymmetrical form. And if you will reconstruct the drawing to an isosceles triangle, only thus you can prove that BD=CD. – Jacob Olandag · 2 years, 1 month ago

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simply.. equate the area of shaded and unshaded region and then apply the property of similar triangles.. – Rishabh Yadav · 2 years, 1 month ago

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Isn't the prependicular from A is its median,,,,,if it is so,,,,,,,,then BD=DC – Jasvinder Singh · 2 years, 1 month ago

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If the areas of the shaded and unshaded parts are equal, then that means that 1/2 of the triangle is shaded and the other half is not. if we try to arrange the triangle such that the half is shaded and the other other half is not, we will notice that they share the same measures of altitudes.. and so for their areas to be totally equal the measures of the bases of the two triangles must be equal as well.. which are BD and DC.. – Anna Anant · 2 years, 1 month ago

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Make it to a rectangle or square.square ll be formed if it is a equilateral triangle . Isosceles triangle forms a rectangle.( shaded and unshaded regions allign with their respective kind forming rectangles. Two shaded n two unshaded rectangles of equal area ll be fromed. – Anitus Raj · 2 years, 1 month ago

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what if we drop 3 more perpendiculars from the corners of trapeziums and triangles?....we'll get equal squares. – Moeid Ibrahim · 2 years, 1 month ago

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– Aneesh Kundu · 2 years, 1 month ago

That is a special case and ocurrs only when \(BD=\dfrac{AD}{3}\)Log in to reply

if we look at triangles - (area of green triangle) by side AK... then its ratio with the area of triangles ABD and and ACD is 1:16 as AK:AD :: 1:4 ... thus it is possible iff... area of ABD = area of ACD....as they have a common perpendicular... so it also means there bases are equal – Ajeet Gill · 2 years, 1 month ago

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– Aneesh Kundu · 2 years, 1 month ago

Area of the green ∆ is a quarter of ADB. But we cant say the same for ADC, for that we first need to show that \(BD=DC\) which is exactly what we needed to prove.Log in to reply

on the other hand ,the areas of shaded & unshaded region is equal(in given fig.) iff the triangle is Isosceles triangle& equilateral otherwise area must not be same(referring given fig) – Prksh Mirase · 2 years, 1 month ago

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It is an intuitively obvious result. To prove this, - understand that the result is certainly true when BD = DC. - take any point D' on BC and prove that for the regions formed between AD and AD', the shaded regions do not equal the unshaded regions in area. It is simple then. – Anirban Ghoshal · 2 years, 1 month ago

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Can be proved by concept of similarity of area of triangle. – Navdeep Singh Roheria · 2 years, 1 month ago

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Not equal – Yogesh Dawande · 2 years, 1 month ago

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these lines are not parallel – Aayush Srivastava · 2 years, 1 month ago

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– Aneesh Kundu · 2 years, 1 month ago

Why?Log in to reply

Ab=ac – Ashna Arora · 2 years, 1 month ago

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– Aneesh Kundu · 2 years, 1 month ago

Good observationLog in to reply

Easily solved, just putting the formulae for areas, it is proved. – Nazmul Haque · 2 years, 1 month ago

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it is given that they are parallel in the question paper :) – Nihar Mahajan · 2 years, 1 month ago

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– Aneesh Kundu · 2 years, 1 month ago

Did the coordinators tell that? Cause we weren't .Log in to reply

– Pranav Kirsur · 2 years, 1 month ago

In Maharashtra and Goa region paper they were given parallel.Log in to reply

– Aneesh Kundu · 2 years, 1 month ago

That's great!!!Log in to reply

I'm not a math wizard but I presume the figure is an equilateral triangle? – Clarence Elmer Quismundo · 2 years, 1 month ago

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– Raj Magesh · 2 years, 1 month ago

No, it's actually any general triangle, but with parallel lines cutting it into several pieces.Log in to reply

By therom of similar triangles, all horizontal lines are parallel. You have just to use formula Area of triangle is half base x height and Area of trapezium is 1/2(sum of parallel sides x height)and you get the solution. – Ujwala Ranade · 2 years, 1 month ago

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when I was giving the test in Narendrapur Ramkrishna Mission(my test centre) a correction was given from the co-ordinator which said that the lines were parallel. – Souryajit Roy · 2 years, 1 month ago

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i have solved it in 2 min and it was so eassy – Ayushadarsh Tiwari · 2 years, 1 month ago

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– Vrishabh Kumar · 2 years, 1 month ago

How did u solve Tiwari jiLog in to reply

– Ramesh Perumal · 2 years, 1 month ago

HOW DID YOU SOLVE IT? CAN YOU EXPLAIN ME?Log in to reply

Its parallel :) – Muskan :) · 2 years, 1 month ago

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– Aneesh Kundu · 2 years, 1 month ago

I hope it is :)Log in to reply