## Tuesday, May 27, 2008

### Geometry Problem 110

See complete Problem 110
Contact Triangle, Area, Incircle. Level: High School, SAT Prep, College geometry

1. For trigonometry lovers!
Triangles ADE,BDF,EFC are isoscele,hence we know the angles.
S=(1/2)ED*EF*sin(DEF)
EF=2ECsin(C/2)=2(p-c)sin(C/2)
after substitution
S=(p-a)(p-b)(p-c)/(2R)
because
(p-b)=(a-b+c)/2=R(sinA-sinB+sinC)
(p-b)=4Rsin(A/2)cos(B/2)sin(C/2)
S''=(1/2)CF*DF*sin(DFC)
after substitution and simplification
=(p-a)(p-b)(p-c)/(2R)
because
b=2RsinB and 2sin(B/2)cos(B/2)=sinB
with Heron's formula we can say
S=[ABC]r/(2R)
the transformation of (p-b) is made by
sinp+sinq=2sin((p+q)/2)cos((p-q)/2)...

2. Let A=Area of Tr. ABC. Using standard notation, A^2=s(s-a)(s-b)(s-c). Also A=r*s where r=inradius and s=semiperimeter as usual. Thus,
(rs)^2=s(s-a)(s-b)(s-c) or r^2*s=(s-a)(s-b)(s-c). We now use the result of Problem #82 whereby S/A=r/2R or S=rA/2R=(r^2*s)/2R =(s-a)(s-b)(s-c)/2R or 2/S=4R/[(s-a)(s-b)(s-c)]-----(1).
Now draw a perpendicular,h, from D to BC. We've S1=h*AD/2=h(s-a)/2. Like wise, S2=h(s-c)/2 since Tr. BDF is isosceles and thus 1/S1+1/S2=2[1/(s-a)+1/(s-c)]/h=2b/[sin(B)(s-a)(s-b)(s-c)]. But b/sin(B)=2R hence 2/S =4R/[(s-a)(s-b)(s-c)]-----(2). By equations (1) & (2), we've 2/S = 1/S1 + 1/S2 ; hence etc.
Antonio, if there's still an easier way to do this, would you be kind enough to show us the same?
Ajit

3. Suggestions for another solution:

1. The area of a triangle is equal to one half of the product of its altitude and base.

2. Similar triangles.

3. The two tangents to a circle from an external point are congruent.

4. One more solution as suggested by Antonio: In the given diagram, let FD and CA (both extended) meet in V and let angle DVA=Θ & VA=x. Draw perpendiculars h1, h & h2 to FD extended from A, E & C resply. Now proving that S=2S1/(S1+S2) is tantamount to proving h = 2h1h2/(h1+h2) since the three triangles have a common base FD.
Apply Menelaus's Theorem to Tr. ABC with FDV as the transversal, we've:x/(x+b)*(s-c)/(s-b)*(s-b)/(s-a)=1(ignoring the - sign which is unimportant here). This gives us x/(x+b) =(s-a)/(s-c).
Note that AD=AE=(s-a), CE=CF=(s-c) & BF=BD=(s-b) where 2s=a+b+c. Now, h1=(x)sin(Θ), h2=(x+b)sin(Θ) while h=(x+s-a)sin(Θ). Substitute the values above to eliminate x & s to obtain: h=[b^2 -(a-c)^2]sin(Θ)/2(a-c) & 2h1h2/(h1+h2) also =[b^2-(a-c)^2]sin(Θ)/2(a-c). In other words, h = 2h1h2/(h1+h2) or S = 2S1S2/(S1+S2)
Ajit

5. I propose an alternative solution to problem 110, following the idea showed by Ajit, with slightly different termination.

Let h, h1 and h2 be the altitudes EG, AJ and CK, of triangles DEF, ADF and DFC, respectively, related to the common base DF. Take H on the extension of AJ, with AH = 2*h1. Take I such that HI is parallel to DF and CI is perpendicular to DF. We have CII = h1 + h2.
If we prove that h = 2*h1*h2/(h1 + h2), this implies that S = 2*S1*S2/(S1 + S2).

The extension of KJ meets the extension of CA at V. The extension of IH meets the extension of CA at W. Putting AV = x, we have AW = 2x. Let’s use the Menelaus’ theorem to triangle ABC, with the transversal DF, which meets AB at D, BC at F, and CA at V: (AD/DB)(BF/FC)(CV/VA) = 1, or [(s-a)/(s-b)]*[(s-b)/(s-c)]*[(x+b)/x] = 1, where s is the semi perimeter of ABC. Thus, x/(x+b) = (s-a)/(s-c), and x = b*(s-a)/(a-c) (1).

We have EG/CK = h/h2 = VE/VC = (x+s-a)/(x+b). Substituting (1) and simplifying the result, we get h/h2 = 2*(s-a)/b.

We have also AH/CI = 2*h1/(h1+h2) = AW/CW = 2x/(2x + b). Substituting (1) and simplifying the result, we get 2*h1/(h1+h2) = 2*(s-a)/b.
So, h/h2 = 2*h1/(h1+h2) and finally h = 2*h1*h2/(h1 + h2).

6. Solution without using Menelaus’s theorem.

Using same notations as Joe and Nilton, let h, h1 and h2 be the altitudes EG, AJ and CK, of triangles EDF, ADF and CDF, respectively, related to the common base DF.

Triangles EDF, ADF and CDF having common base DF implies their areas S, S1 and S2 are proportional to their respective altitudes so proving S = 2.S1.S2/(S1+S2) is equivalent to proving h = 2.h1.h2/(h1+h2) or 2.h1.h2 = h.h1 + h.h2.

1. If DF // AC, h=h1=h2, S=S1=S2 and the result is obvious.

2. If DF ∦ AC, h≠h1 , h≠h2 and the extension of DF meets the extension of AC at V.

Triangles AJD and CKF are similar => AJ/CK = h1/h2 = AD/CF = AE/CE.

Triangles VJA, VGE and VKC are similar => VE/h = VA/h1 = VC/h2 = AE/(h-h1) = CE/(h2-h).

So AE/CE = h1/h2 = (h-h1)/(h2-h) => h1.(h2-h) = h2.(h-h1) => 2.h1.h2 = h.h1+h.h2

QED

7. Idea basically as above. Let M,N,P be projections of A,C,E respectively onto DF and we need to prove PE=2AM.CN/(AM+CN) ( * ).
Note that triangles ADM and CFN are similar, wherefrom AM/CN=AD/CF=AE/CE ( 1 ).
Let S, T be reflections of A,C across DF; then ACTS is an isosceles trapezoid and P is the intersection of its diagonals At, CS, thus PE/AS+PE/CT=1 or PE/AM+PE/CN=2, which is equivalent to (*), done.

1. Hello Stan,
In your proof of June 19, is PE/AS+PE/CT=1 a well known property of isoscele trapezoids ? If so, could you tell me what theorem it is ?
Regards.
Greg