That's confusing

Geometry Level 4

Let △ A B C \triangle ABC be a triangle. Let point M be the point on BC that divided BC equally. Let D and E be points on AC that divided AC into three equal part.

F and G are points of intersection between AM and BD and BE, respectively.

If A F : F G : G M = a : b : c AF:FG:GM=a:b:c and a,b,c are co-prime number.

Find a + 2 b + 3 c a+2b+3c


The answer is 17.

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คลุง แจ็ค by คลุง แจ็ค , 26, United Kingdom

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3 solutions

J o i n M E . I n Δ B C D , B D = a n d ∣ ∣ 2 M E . . . . ( 1 ) I n Δ A M E , M E = a n d ∣ ∣ 2 F D . . . . ( 2 ) ∴ f r o m ( 1 ) , ( 2 ) . . B F = B D − F D = 3 F D . . . . ( 3 ) I n Δ s G M E a n d G F B , M E ∣ ∣ B F s o a l l c o r r o s p o n d i n g a n g l e s a r e s a m e . . ∴ Δ s a r e s i m i l a r , ∴ f r o m ( 2 ) , ( 3 ) G M F G = M E B F = 2 3 . . . . ( 4 ) L e t G M = 2 X , ∴ F G = 3 X , F M = 5 X . . . . ( 5 ) I n Δ A M E , M E = a n d ∣ ∣ 2 F D . ∴ A M = 2 A F , ⟹ A F = F M = 5 X . . . f r o m ( 5 ) ∴ a = 5 X , b = 3 X , c = 2 X . ∴ ( a + 2 b + 3 c ) X = ( 5 + 2 ∗ 3 + 3 ∗ 2 ) X a + 2 b + 3 c = 17 Join ~ME.\\ In ~\Delta~ BCD, BD=~and~||~2ME....(1)\\ In ~\Delta~ AME, ME=~and~||~2FD....(2)\\ \therefore~from ~(1),(2)..BF=BD-FD=3FD....(3)\\ In~\Delta s ~GME~and~GFB, ME||BF~~so~all~corrosponding~angles~are~\\ same..\therefore~\Delta s~ are~ similar,\\ \therefore~from~(2),(3)\dfrac {GM}{FG}=\dfrac {ME}{BF}=\dfrac 2 3....(4)\\ Let~GM=2X, \therefore~FG=3X, ~FM=5X....(5)\\ In ~\Delta~ AME, ME=~and~||~2FD.\\ \therefore~AM=2AF,~~\implies~AF=FM=5X...from~(5)\\ \therefore~a=5X,~~ b=3X,~~c=2X. \\ \therefore~(a+2b+3c)X=(5+2*3+3*2)X\\ a+2b+3c=\Large~~~~\color{#D61F06}{17}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

I was waiting for geometrical solution. Now, you post it!

คลุง แจ็ค - 5 years, 10 months ago
Nelson Mandela
Jul 26, 2015

5:3:2 using vector algebra. Take A as origin and take position vectors of B and C and section formula to get the points M D and E.

Now use two point form of line,

t a → + ( 1 − t ) b → t\overrightarrow { a } +(1-t)\overrightarrow { b } and solve them to get F and G. we will find that they are collinear and the ratio is not simple.

So, 5+6+6 = 17.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Use Menelaus' theorm and this problem gonna be 80% easier

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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