Keep your distance

Geometry Level 5

Let $\Delta ABC$ have integer side lengths and be right-angled at $B$ . Now suppose $\Delta DEF$ is located within $\Delta ABC$ with the same orientation, such that its sides are parallel to the corresponding sides of $\Delta ABC$ and the distance between the corresponding parallel sides is $3$ .

If the area of $\Delta ABC$ is $4$ times that of $\Delta DEF$ then there are $n$ possible triangles $\Delta ABC$ independent of orientation. If $S$ is the sum of the lengths of the hypotenuses of these $n$ triangles, then find $n + S.$

The answer is 273.

2 solutions

Brian Charlesworth
Jan 26, 2015

Let $DE || AB, EF || BC$ and $DF || AC$ . Also let sides $AB, BC, AC$ have respective lengths $a,b,c$ . Since the area of $\Delta ABC$ is $4$ times that of $\Delta DEF$ we then have that $DE = \frac{a}{2}, EF = \frac{b}{2}$ and $DF = \frac{c}{2}.$

Now break $\Delta ABC$ into trapezoids $ADEB, BEFC$ and $ADFC$ , as well as $\Delta DEF$ . Adding up the areas of these latter four regions and equating the sum to the area of the larger triangle, we have that

$3(\frac{3}{4}a) + 3(\frac{3}{4}b) + 3(\frac{3}{4}c) + \frac{ab}{8} = \frac{ab}{2}$

$\Longrightarrow 6a + 6b + 6c = ab \Longrightarrow ab - 6a - 6b = 6c = 6\sqrt{a^{2} + b^{2}}$

$\Longrightarrow a^{2}b^{2} + 36a^{2} + 36b^{2} - 12ab^{2} - 12a^{2}b + 72ab = 36a^{2} + 36b^{2}$

$\Longrightarrow ab(ab - 12a - 12b + 72) = 0 \Longrightarrow b(a - 12) = 12a - 72$

$\Longrightarrow b = 12 + \dfrac{72}{a - 12}$ .

Since $b$ must be an integer, we must have $a - 12$ as a (positive) factor of $72$ . The possible values of $a$ are then $13,14,15,16,18,20,21,24,30,36,48,84$ .

Upon calculating corresponding values for $b$ and $c$ , we find that there are $6$ possible triangles independent of orientation, namely

$(13,84,85), (14,48,50), (15,36,39), (16,30,34), (18,24,30), (20,21,29)$ .

The desired sum is then $6 + 85 + 50 + 39 + 34 + 30 + 29 = \boxed{273}$ .

Amazing problem. Thanks Sir!

Satvik Golechha - 6 years, 4 months ago

Impressive. I totally failed at attacking this problem.

Ken Hodson - 5 years, 4 months ago

$\color{#EC7300}{ (12,37,35):- a=6*(1+\dfrac{37+12}{35})=\dfrac{72} 5.~~a/x~Not~an~integer.~~No~solution. \\ (9,41,40):- a=6*(1+\dfrac{41+9}{40})=\dfrac{27} 2.~~a/x~Not~an~integer.~~No~solution. \\ (28,53,45):- a=6*(1+\dfrac{53+28}{45})=\dfrac{54} 5.~~a/x~Not~an~integer.~~No~solution. \\ (11,61,60):- a=6*(1+\dfrac{61+11}{60})=\dfrac{66} 5.~~a/x~Not~an~integer.~~No~solution. \\ (16,65,63):- a=6*(1+\dfrac{65+16}{63}=\dfrac{54} 7.~~a/x~Not~an~integer.~~No~solution. \\ (33,65,56):- a=6*(1+\dfrac{65+33}{56})=\dfrac{21} 2.~~a/x~Not~an~integer.~~No~solution. \\ (48,73,55):- a=6*(1+\dfrac{73+48}{55})=\dfrac{108} 5.~~a/x~Not~an~integer.~~No~solution. \\ (36,85,77):- a=6*(1+\dfrac{85+36}{77})=\dfrac{108} 7.~~a/x~Not~an~integer.~~No~solution. \\ (39,89,80):- a=6*(1+\dfrac{89+39}{80})=\dfrac{78} 5~~a/x~Not~an~integer.~~No~solution. \\ (65,97,72):- a=6*(1+\dfrac{97+65}{72})=\dfrac{39}2~~a/x~Not~an~integer.~~No~solution. \\ We~can~see ~that~not~only~a/x~is~not ~an~integer~but~ it~is~also~less~than~1.~so~did~not~check~further.\\ Any~better~explanation~will~be~appreciated. }$ $Better~Solution$ $ANY~PYTHAGOREAN ~RIGHT~TRIANGLE~ WITH ~INRADIUS~ =~ 6 ~IS~ THE~~ SOLUTION$

$\color{#3D99F6}{ \dfrac 6 i~must~be~an~integer.~~\therefore~i~must~be~a~factor~of~6\\ (12,37,35):- i=5.~~~\dfrac 6 i~Not~a~factor~of~6.~No~solution. \\ (9,41,40):- i=4.~~~ \dfrac 6 i~Not~a~factor~of~6.~No~solution. \\ (28,53,45):- i=10. ~~~\dfrac 6 i~Not~a~factor~of~6.~No~solution \\ (11,61,60):- i=5.~~~ \dfrac 6 i~Not~a~factor~of~6.~No~solution \\ (16,65,63):- i=3.5. ~~~\dfrac 6 i~Not~a~factor~of~6.~No~solution \\ (33,65,56):- i=12. ~~~\dfrac 6 i~Not~a~factor~of~6.~No~solution \\ (48,73,55):- i=15. ~~~\dfrac 6 i~Not~a~factor~of~6.~No~solution \\ (36,85,77):- i=14. ~~~\dfrac 6 i~Not~a~factor~of~6.~No~solution \\ (39,89,80):- i=15.~~~ \dfrac 6 i~Not~a~factor~of~6.~No~solution \\ (65,97,72):- i=20. ~~~\dfrac 6 i~Not~a~factor~of~6.~No~solution \\ Incidentally~in ~articles~on ~right~triangles~one~can~see~that~there~are~only~SIX~triangles~with~inradius~r=6. } \\$

$\color{#D61F06}{ \\ We~ saw~ that~ the~ required ~triangle~ should~ be~ Pythagorean~ right~ triangle ~with~ inradius~ =~ 6.~~ From ~the ~web~ we~ know ~that~ there~ are~ six~ such~ triangles ~and~their~sides~are~also~given.~~ We~ add~ the~ lengths~ of~ their~ hypotenuses.}$

Niranjan Khanderia - 3 years ago

Mark Hennings
Nov 16, 2016

If we set up a coordinate system so that the right angle of the triangle is at the origin and two vertices are at $(a,0)$ and $(0,b)$ , then the equation of the hypotenuse of the smaller triangle is $bx + ay \; = \; ab - 3\sqrt{a^2+b^2} \; = \; ab - 3c$ and so the coordinates of the vertices of the smaller triangle are $(3,3)$ , $\big(3,\tfrac{ab - 3b - 3c}{a}\big)$ and $\big(\tfrac{ab-3a-3c}{b},3\big)$ , so the non-hypotenuse sides of the smaller triangle are $p \; = \; \tfrac{1}{b}(ab - 3a-3b-3c) \hspace{2cm} q \; = \; \tfrac{1}{a}(ab - 3a - 3b - 3c)$ The area condition that $\tfrac12ab \,=\, 4\times\tfrac12pq$ now becomes $a^2b^2 \,=\, 4(ab - 3a -3b -3c)^2$ , and hence $3(ab - 2a - 2b - 2c)(ab - 6a - 6b - 6c) \; = \; 0$ We cannot have $ab = 2a + 2b + 2c$ , since that would force $p,q$ to be negative. Thus we deduce that $ab \; = \; 6a + 6b + 6c$ which tells us that $p = \tfrac12a$ and $q = \tfrac12b$ . Since $(a,b,c)$ is a Pythagorean triad, and we don't care about its orientation, we can find positive integers $d,u,v$ such that $a = d(u^2-v^2)\,,\,b = 2duv\,,\, c = d(u^2+v^2)$ . The above condition now becomes $dv(u-v) \; = \; 6$ There are $9$ different solutions for $d,u,v$ , and these produce exactly $6$ different Pythagorean triads $(a,b,c)$ , namely $(13,84,85)$ , $(14,48,50)$ , $(15,36,39)$ , $(16,30,34)$ , $(18,24,30)$ , $(20,21,29)$ . This makes the answer to the question $\boxed{273}$ .

Keep your distance

The answer is 273.

2 solutions

0 pending reports