#30 of Sept 2016 Grade 10 CSAT (Korean SAT) Mock Test

Geometry Level 5

I've drawn 2 triangles with one inscribed in the other, i.e. with the vertices of the smaller triangle lying on different sides of the large triangle. Given that the large triangle has side lengths 3 2 , 4 , 3\sqrt{2},\, 4, and 10 , \sqrt{10}, as shown in the figure, let P P denote the minimum possible perimeter of the smaller triangle.

If P 2 P^2 can be expressed as M N , \frac MN, where M M and N N are coprime positive integers, find M + N . M+N.


This problem is a part of <Grade 10 CSAT Mock Test> series .


The answer is 149.

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2 solutions

Boi (보이)
Jul 23, 2017

Let B ( 0 , 0 ) B(0,~0) and C ( 4 , 0 ) C(4,~0) and we see that A ( 3 , 3 ) . A(3,~3).

The equation of line A C AC is y = 3 x + 12. y=-3x+12.

Let F ( a , b ) , F(a,~b), and let F 1 F_1 and F 2 F_2 be the points of reflection over A B \overline{AB} and A C . \overline{AC}.

Then it is clear that F 1 ( b , a ) F_1(b,~a) and F 2 ( a , b ) . F_2(a,~-b). Also, b = 3 a + 12. b=-3a+12.

The minimum of the perimeter of triangle D E F DEF is equal to the minimum of F 1 D + D E + E F 2 . \overline{F_1D}+\overline{DE}+\overline{EF_2}.

The minimum would occur when D , E D,~E are on F 1 F 2 . \overline{F_1F_2}.

F 1 F 2 = ( a b ) 2 + ( a + b ) 2 = 2 a 2 + 2 b 2 = 20 a 2 144 a + 288 = 20 ( a 18 5 ) 2 + 144 5 . \begin{aligned} \overline{F_1F_2}&=\sqrt{(a-b)^2+(a+b)^2} \\ &=\sqrt{2a^2+2b^2} \\ &=\sqrt{20a^2-144a+288} \\ &=\sqrt{20\left(a-\dfrac{18}{5}\right)^2+\dfrac{144}{5}}. \end{aligned}

Therefore, the minimum of the perimeter of triangle D E F DEF is 12 5 \dfrac{12}{\sqrt{5}} when F ( 18 5 , 6 5 ) . F\left(\dfrac{18}{5},~\dfrac{6}{5}\right).

P 2 = 144 5 = M N , M + N = 144 + 5 = 149 . \therefore~P^2=\dfrac{144}{5}=\dfrac{M}{N},\\ M+N=144+5=\boxed{149}.

Finding the orthic triangle inside Δ A B C \Delta ABC ... much about the same.

Nice solution!

Michael Huang - 3 years, 10 months ago

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Thank you!

Boi (보이) - 3 years, 10 months ago

they never counted out degenerate triangles

keanu ac - 3 years, 7 months ago

The in inscribed triangle is orthic because orthic triangle have the minimum perimeter.
Let the sides be a + b + c = a C o s A + b C o s B + c C o s C = 4 5 + 5 5 + 3 5 = 12 5 = P . P 2 = 144 5 = M N M + N = 144 + 5 = 149. a' + b' + c' = aCosA + bCosB + cCosC = \dfrac 4 {\sqrt5} + \dfrac 5 {\sqrt5} + \dfrac 3 {\sqrt5} = \dfrac{12}{\sqrt5}=P.\\ P^2=\dfrac{144} 5=\dfrac M N\\ M+N=144+5=149.
To find Cos, Cos Rule Is used.

How do you know it must be an orthic triangle?

Pi Han Goh - 3 years, 10 months ago

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Since the perimeter of the inscribed triangle is minimum. I have added this to my solution. Am I correct ? Thanks for your comment.

Niranjan Khanderia - 3 years, 10 months ago

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Right. What's left to do is to show that large triangle is acute, but that can be easily verified.

Pi Han Goh - 3 years, 10 months ago

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@Pi Han Goh Thank you. I have added that all three Cos are >0, big triangle is Acute.

Niranjan Khanderia - 3 years, 10 months ago

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