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Geometry Difficulty 5.2 AIME, harder Prove it Belarus

Given a right-angled triangle ABCABC with ∠C=90∘\angle C = 90^\circ. Its perimeter is equal to 3030 cm. CHCH is the altitude of this triangle, CKCK and CLCL are the bisectors of the angles ACHACH and BCHBCH respectively.
Find the length of the hypotenuse ABAB if KL=4KL = 4 cm.

Solution

Let a=BCa = BC, b=ACb = AC, c=ABc = AB, α=∠BAC\alpha = \angle BAC, β=∠ABC\beta = \angle ABC, P=a+b+cP = a+b+c. Then ∠ACH=β\angle ACH = \beta, ∠BCH=α\angle BCH = \alpha, thus ∠ACK=∠KCH=β/2\angle ACK = \angle KCH = \beta/2, ∠HCL=∠LCB=α/2\angle HCL = \angle LCB = \alpha/2. Since ∠HKC=∠KAC+∠ACK\angle HKC = \angle KAC + \angle ACK (exterior angle of triangle ACKACK), we have
∠HKC=α+β/2=∠BCH+∠LCH=∠BCK. \angle HKC = \alpha + \beta/2 = \angle BCH + \angle LCH = \angle BCK.
Hence triangle KCBKCB is isosceles and BK=BC=aBK = BC = a. Similarly, AL=AC=bAL = AC = b. Therefore, AK=AB−BK=c−aAK = AB - BK = c - a, so KL=AL−AK=b−(c−a)=a+b−cKL = AL - AK = b - (c-a) = a+b-c. Hence
c=(a+b+c)−(a+b−c)2=P−KL2=30−42=13 (cm). c = \frac{(a+b+c) - (a+b-c)}{2} = \frac{P - KL}{2} = \frac{30-4}{2} = 13 \text{ (cm).}

Figure 1

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