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RMO 2014 Delhi Region Q.1

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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Note by Aneesh Kundu
2 years, 7 months ago

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I solved it using trigonometry Souryajit Roy · 2 years, 7 months ago

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@Souryajit Roy Same here. Took me less than a minute to solve. Llewellyn Sterling · 2 years, 7 months ago

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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, 7 months ago

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I have solved it using simple geometry of similar triangles... Arnab Deb · 2 years, 7 months 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, 7 months ago

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@Aayush Srivastava First of all the question was asked on the basis of the diagram provided. If the lines are not parallel then what if they intersect inside the triangle. If yes then they've not told us what the shaded and unshaded region actually is. Aneesh Kundu · 2 years, 7 months ago

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@Aneesh Kundu is my answer lil higher levelled ? Ajeet Gill · 2 years, 7 months ago

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@Aayush Srivastava 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 line Nabarun Dutta · 2 years, 7 months ago

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Are those three lines parallel to BC? If parallel,proof is direct!If not,what? Ranjana Kasangeri · 2 years, 7 months ago

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@Ranjana Kasangeri Yeah, they are parallel. Actually here, this correction was made. Even there was a small correction in Problem no. 3 Raushan Sharma · 1 year, 6 months ago

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@Ranjana Kasangeri I assumed them to be parallel Aneesh Kundu · 2 years, 7 months ago

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@Aneesh Kundu The diagram in the question paper looked exactly the same. Aneesh Kundu · 2 years, 7 months ago

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@Aneesh Kundu Well,I did for parallel and 'disproved' for non-parallel( Hope it does good) Ranjana Kasangeri · 2 years, 7 months ago

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the question is incomplete as it must be specified whether the horizontal lines are parallel Prateek Gupta · 2 years, 6 months ago

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@Prateek Gupta Assume that they are parallel Aneesh Kundu · 2 years, 6 months ago

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i got 3 correct, and attempted 5 in delhi rmo what chance do i have to get selected Sauditya Yo Yo · 2 years, 7 months 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, 7 months ago

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It must be isoceles or equilateral triangle, AB must be equal to AC Anna Anant · 2 years, 7 months 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, 7 months 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' = kA2; 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 = xA1 ; (x is the proportionality constant) area of trapezium LL1M1M =A3 = y A1 ; area of trapezium MM1BD = A4 = z*A1;

Hence A2' = kxA1 (since A2' = kA2 and A2 = xA1) Similarly A3' = kyA1 (since A3' = kA3 and A3 = yA1)
similarly A4' = kzA1;

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

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

    A1 + k*x*A1 + y*A1 +  k*z*A1 = k*A1 + x*A1 + k*y*A1 + z*A1

=> 1 + kx + 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, 7 months 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, 7 months 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, 7 months 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, 7 months 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, 7 months 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, 7 months 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, 7 months 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, 7 months ago

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@Moeid Ibrahim That is a special case and ocurrs only when \(BD=\dfrac{AD}{3}\) Aneesh Kundu · 2 years, 7 months ago

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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, 7 months ago

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@Ajeet Gill 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. Aneesh Kundu · 2 years, 7 months ago

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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, 7 months 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, 7 months ago

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Can be proved by concept of similarity of area of triangle. Navdeep Singh Roheria · 2 years, 7 months ago

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Not equal Yogesh Dawande · 2 years, 7 months ago

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these lines are not parallel Aayush Srivastava · 2 years, 7 months ago

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@Aayush Srivastava Why? Aneesh Kundu · 2 years, 7 months ago

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Ab=ac Ashna Arora · 2 years, 7 months ago

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@Ashna Arora Good observation Aneesh Kundu · 2 years, 7 months ago

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Easily solved, just putting the formulae for areas, it is proved. Nazmul Haque · 2 years, 7 months ago

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it is given that they are parallel in the question paper :) Nihar Mahajan · 2 years, 7 months ago

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@Nihar Mahajan Did the coordinators tell that? Cause we weren't . Aneesh Kundu · 2 years, 7 months ago

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@Aneesh Kundu In Maharashtra and Goa region paper they were given parallel. Pranav Kirsur · 2 years, 7 months ago

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@Pranav Kirsur That's great!!! Aneesh Kundu · 2 years, 7 months ago

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I'm not a math wizard but I presume the figure is an equilateral triangle? Clarence Elmer Quismundo · 2 years, 7 months ago

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@Clarence Elmer Quismundo No, it's actually any general triangle, but with parallel lines cutting it into several pieces. Raj Magesh · 2 years, 7 months ago

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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, 7 months 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, 7 months ago

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i have solved it in 2 min and it was so eassy Ayushadarsh Tiwari · 2 years, 7 months ago

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@Ayushadarsh Tiwari How did u solve Tiwari ji Vrishabh Kumar · 2 years, 7 months ago

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@Ayushadarsh Tiwari HOW DID YOU SOLVE IT? CAN YOU EXPLAIN ME? Ramesh Perumal · 2 years, 7 months ago

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Its parallel :) Muskan :) · 2 years, 7 months ago

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@Muskan :) I hope it is :) Aneesh Kundu · 2 years, 7 months ago

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