Very nice problem.

Simple calculation will yield a very remarkable answer.

It seems that this must also be shown by geometry.

But that is another nice problem.

Using the equation of the radius in each triangle and setting them equal we get

CD = (5 AD - 7 BD) / (BD - AD)

We plug this in Stewart relation and we get an equation in AD & BD. Combined with BD + AD = AB which is known, this gives AD and BD and then CD.

This is a working man's solution which shows that in a right (at least) triangle metric relations can always give a way to solve, but it can be very laborious.

I am sure there must be a smarter solution.

... sqrt(35/2) ... or sqrt(70)/2,

although, the radius of two inscribed circles looks a little unpleasant like this

1/2 sqrt(35/37) (-6 + sqrt(36 + sqrt(1295)))

however, the sine of the angle ADC, or BDC, looks nice like sqrt(70/74)

let's test the area of the triangles AI1C +AI1D + DIC + DI2C + CI2B + DI2B = ABC then if the angle DCB=a Y I1K2=r1 and I2K1=r1 we have AC*r1/2 +AD*r1/2 + CD*r1/2 + BC*r2/2 + BD*r2/2 + CD*r2/2 = AC*BC/2 => CD=(AC*BC-AC*r1-BC*r2) /(r1+r2+AB*(cos(a)/CB + sin(a)/AC)), We can't conclude anything from angle a

Peter Breuer

radius = area/semiperimeter but I used as area altitude times base/2.

The two triangles have the same altitude from C so that cancels out and bases are AD & BD. I tried to avoid trigonometry.

Again, I did not suggest to use any trigonometry. That included the law of cosine. I said Stewart relation.

It may give even an upper order equation, but I am sure it can be solved.

But it will not have any square roots of algebraic expressions.

Continuing with my answer you have to angle ABC=b then AB*cos(a)/CB + AB*sin(a)/AC = (sin(a)*cos(b)+sin(b)*cos(a))/(sin(b)*cos(b)) =>sin(a+b)/(sin(b)*cos(b)) => angle CAB=c, sin(c)/CD = sin(a+b)/AC

BC/AB=sin(c), AC/AB=sin(b), BC/AB = cos(b) => sin(a+b)/(sin(b)*cos(b))=AB/CD

=>CD = (AC*BC - AB)/(r1+r2) => CD=(5*7 - sqrt(74))/(2r1) now we need to find r1

we build a rectangle with vertex E extension of CD then we have the rectangle ACBE because the radii of the circumferences are equal => CD=sqrt(74)/2

In triangles ACD and CDB, since the inscribed circles have equal radii (denoted as "r1"), the sum of the lengths of the segments tangent to each circle from the common vertex D is equal to the length of the hypotenuse AB

r1 + r1 + AK2 + BK1 = sqrt(74)

AK2=(7*r1^2)/(7-10*r1)

AK2=(5*r1^2)/(5-14*r1)

From where r1= 350*sqrt(74)+sqrt(113620-23680*sqrt(74))/296

and with CD=(5*7 – sqrt(74))/2*r1 you have CD

Javier, apparently the number you have under the square root,

113620-23680*sqrt(74), looks negative.

this is a cube equation

The right answer is sqrt(AC.BC/2).

The real problem is to prove geometrically that CD is equal the area of the triangle ABC.

You have to take into account that the radii are equal, so I don't think the solution is that it doesn't take the radii into account. I think the circles touch at a single point and that is where CD is tangent

now having that CD=sqrt(AC*BC/2) prove that the radii of the circles are equal?

In k1k2r/2 you are assuming that the two radii are equal and what you want to achieve is that the 2 radii are equal

Since you already know that CD=sqrt(AC*BC/2) now what needs to be proven is the hypothesis that thus the radii r1=r2

you arrived at the thesis CD=sqrt(AC*BC/2) without having passed because the radii r1=r2 what is missing is that it is proven that r1=r2

I calculate CD2=AC BC /2 using r1=r2. The same way you did.

But my question is: can it be proved without calculation, but on a geometrical way.

a correct answer peter

Let us be a bit general and rotate the axis

instead of C origin, AC=7 horizontal line, BC=5 vertical line

O origin, xy axis, OH=h horizontal line, OV=v vertical line. Eqaution of OP line is y=mx, and P in on VH line (i.e. D. on AB line). Incircle of OHP and OVP triangles are of same radius. One needs to find m, and thus length of OP line.

Note, the equation of VH line is x/h+y/v=1, or vx+hy-hv=0. Origin is at negative side of this line. The 2nd quadrant (x0) is at negative side of mx-y=0, while the other side is at positive side.

Let the incircle radius be r. Since OVP's incircle is tangent to vertical OV line, and OPH's incircle touches vertical line, Let their center coordinates be (r, y1) and (x2,r) respectively.

Now distance between point (x0,y0) and ax+by+c is given as

(+-) (ax0+by0+c)/(a^2+b^2)^0.5, + or - depending on whether (x0, y0) is at positive or negative side of line.

And distance from incenter of circle to three sides are equal; and the incircle radii are equal for both triangles. Thus we have

-(vr+hy1-hv)/(h^2+v^2)^0.5=(y1-mr)/(1+m^2)^0.5=r=(mx2-r)/(1+m^2)^0.5=-(vx2+hr-hv)/h^2+v^2)^0.5

From which we need to solve for m

2nd and 4th side of the equation gives (from left)

y-mx2=(m-1)r

while 1st and 5th side gives

hy1-vx2=(h-v)r

Solving these two for y1 and x2 gives

x2=(2h-mh-v)r/(mh-v); y1=(mh-2vh+v)r/(mh-v)

putting these two values in 2nd and 3rd, or 4th and 3rd side, of first 'big' equation, and dividing with r, multiplying with(1+m^2)^0.5 we have

(hm+v)(1-m)/(mh-v)=(1+m^2)^0.5

With a little side-change and squaring

(1-m)^2/(1+m^2)=(mh-v)^2/(mh+v)^2

Subtracting 1 from both sides and writing as a quadratic equation for m

m^2 (h^2-2hv)+2mvh+(v^2-2vh)=0;

and thus using quadratic formula, and little bit of rearrangement

m=v/2(h-2v)*((2((h^2+v^2)/vh-1))^0.5-1)

only positive value of root is taken, because if both v and h are positive, then, the target line must have positive slope as well

Now, the intersection point between x/h+y/v=1 and y=mx is trivially easy, and (x^2+y^2)^0.5=vh*(1+m^2)^0.5/(v+mh)

Without squaring the m, adding 1 and again square rooting, note from (hm+v)(1-m)/(mh-v)=(1+m^2)^0.5 , we can write

(1+m^2)^0.5/(v+mh)= (1-m)/(mh-v)

and finally

(1-m)/(mh-v)= h (1-vM)/(1+hM)

magnitude of this expression is the length of OP line

here

M=((2((h^2+v^2)/vh-1))^0.5-1)/2(h-2v)

So, what is the value for this case? Here h=7, v=5

M~1.65950672......

OP=~4.048868

Please do check whether I have made any algebraic/arithmetic mistakes there.

Peter Breuer Sumit Bhowmick Javier López Wong Enno Diekema Steftcho P. Dokov

Dear Colleagues! Thank you very much for your interest in this task. I think it's worth it!

Many of you demonstrated the correct course of solving the problem, this is very pleasing.

I invite you to get acquainted with my solution to this problem in a geometric way.

Great job, Liudmyla!

Especially, with ABI similar to I1I2I.

("z = p - c") == because there is a square with diagonal CI, and the other 2 vertices (of the square) are the tangent points of the incircle

Good demonstration Liudmyla-Hetmanenko, you used planimetry, not discrete mathematics or vectors, not even trigonometry, your demonstration is excellent.

Peter Breuer,

"I1I2 / AB = 1 - r/R" !?!

this is true because 'constant of proportionality', I1I2 / AB,

is valid for the ratio of any two corresponding line segments in both triangles.

In this particular case, it is equal to the ratio of the corresponding heights in both triangles from vertex I, a ratio which is (R-r)/R,

... hopefuly, this helps ...

(R-r = is the length of the height in triangle I1I2I from vertex I to the side I1I2)

I would like to summarize and also draw your attention to the fact that in this problem a special case was proposed (I considered a right triangle), but this problem is also valid for the general case (for any triangle).

I chose the lengths of the sides of the triangle according to my wishes; you can create a problem for any lengths of the sides.

Thank you peter for you proof

When angle C is not rectangular then the answer is

4CD^2=(AC+BC)^2-AB^2

Provided that the similarity of the two triangles ABI and I1I2I might be hard to detect,

then such a fact for the inradius can be utilized:

"the inradius r is one-third of the harmonic mean of the three altitudes (heights) in the triangle"

https://en.wikipedia.org/wiki/Incircle_and_excircles#Other_properties

The above fact can be applied to the pair of triangles ADC and BDC, where the common altitude from vertex C cancels out.

Thus, because r1=r2 for both triangles, we get

1/h(A) + 1/h1(D) = 1/h(B) + 1/h2(D), where

h(A) and h1(D) are the heights in ADC from A and D, and

h(B) and h2(D) are the heights in BDC from B and D.

Then the lengths of the altitudes

h(A), h1(D), h(B) and h2(D)

can be replaced by the lengths of AC=7, BC=5, CD=x, and sine or cosine of the angle ACD = denoted by "d".

Hence, the equation:

1/h(A) + 1/h1(D) = 1/h(B) + 1/h2(D)

becomes

1/(7*sin(d)) + 1/(x*sin(d)) = 1/(5*cos(d)) + 1/(x*cos(d)), ((*)).

Another, equation for "x" and "d" comes from

AD+BD=sqrt(74), where AD and BD are expressed from the law of sines for triangles ADC and BDC, that is,

AD/x = sin(d)/sin(alpha) = sin(d)/(5/sqrt(74)), and

BD/x = sin(pi/2 - d)/sin(beta) = cos(d)/(7/sqrt(74)).

So, AD+BD=sqrt(74) becomes

35 = x*(7*sin(d) + 5*cos(d)), ((**)).

And, we can ask “wolframalpha.com” for (x,d) solution of stars equations ((**)) and ((*))

https://www.wolframalpha.com/input?i=solve+7*x*sin%28d%29%2B5*x*cos%28d%29%3D35%2C+5*%28x%2B7%29*cos%28d%29%3D7*%28x%2B5%29*sin%28d%29%2C+for+x%2Cd

which shows x=sqrt(35/2) and angle “d” such that

tan(d/2) = ( sqrt(70) – 5 ) /9, so

d = 2*arctan((sqrt(70) - 5) / 9) = 41.018 degrees

... sorry for this long writing spam ...

(eventually, the only useful thing here is "r being one-third of the harmonic mean of the three altitudes")

"r being one-third of the harmonic mean of the three altitudes"

that is, the equality

r = 1/[ 1/h(A) + 1/h(B) + 1/h(C) ]

can be easily verified by simply dividing both sides by S = area of ABC.

Then, the LHS (left hand side) becomes

r/S = 1/p = 1/ semiperimeter.

For the RHS (right hand side) we use

S = a*h(A)/2 = b*h(B)/2 = c*h(C)/2,

which we plug in the RHS - appropriately choosing S for each term,

and cancel out h(A), h(B), h(C), so that we get "p" in the denominator.